Further Maths - Discrete

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Mathematics
Topic
Level
Higher
10
-
Option:
Discrete
Mathematics
for the
Paul
Ben
Fannon,
Woolley
IB
Diploma
Vesna
Kadelburg,
and Stephen
Ward
CAMBRIDGE
UNIVERSITY
PRESS
Cambridge,
New
Mexico
Delhi,
Cape Town,
Madrid,
Melbourne,
York,
Singapore, SaoPaulo,
Cambridge
PRESS
UNIVERSITY
CAMBRIDGE
City
Press
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Building, Cambridge CB28RU,
The Edinburgh
UK
www.cambridge.org
Information
on
\302\251
is in
publication
www.cambridge.org/9781107666948
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Cambridge
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title:
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NOTICE TO TEACHERS
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Worksheets
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or used in any way outside the
in the
Press
Contents
this book
How to use
v
viii
Acknowledgements
1
Introduction
Proof
3
of proof
Methods
1
4
contradiction
by
6
principle
Pigeonhole
9
Strong induction
Divisibilityand prime
Factors,
multiples
The
Euclidean
Bl Prime
13
and remainders
J3 Greatestcommondivisor
I
13
numbers
least
and
common
multiple
23
algorithm
26
numbers
Representation of integers in differentbases
How
fingers
many
Arithmetic
Linear
in
need to
do we
between
Changing
different
2J Solutions
5
general
Introduction: working with
Rules of modular arithmetic
and
19 Chinese
linear
remainders
congruences
remainder theorem
little
theorem
solutions
42
44
47
51
arithmetic
Division
42
46
solution
subject to constraints
Modular
Fermat's
integer
solutions are there?
the
33
36
equations
with
33
35
bases
Examplesof equations
Finding
count?
bases
different
Diophantine
How many
19
51
53
56
59
62
Contents
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it*'
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6
67
theory
Graph
to graphs
|#EJ Introduction
E3
Definitions
EW
Some
67
69
important
theorems
79
and
complements
85
a graph
87
E\302\243J Subgraphs
around
[\342\200\24213
Moving
m
Eulerian graphs
89
C2I
Hamiltonian
93
7
graphs
on
Algorithms
98
graphs
98
UJ Weighted graphs
Q3
Minimum
Ed
Finding the shortest path: Dijkstra's
Q3
Travelling
Q3
Visiting
8
Recurrence relations
spanning
First
Kruskal's
102
106
algorithm
algorithm
edges: Chinesepostmanproblem
vertices: Travelling salesman problem
along all the
all the
sequences
|*\302\243J Defining
Q3
tree:
order
linear
134
relations
136
(?9 Second order recurrencerelations
|Jl\302\243JModelling
9
Summary
using
recurrence
118
134
recursively
recurrence
112
139
143
relations
and mixed examination
practice
149
Answers
158
Glossary
172
Index
Contents
177
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This book coversallthe material
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Option) of the HigherLevel
Mathematics
10 (Discrete
Topic
1h
for the International Baccalaureate course.It is largely
of the
independent
syllabus
from the Core topics,soit can be studied
at any point during the course.The only
required
of the core are proof by induction
and
and
series
1.4)
Topic
sequences
(Syllabus
(Syllabus
the
material
that will be examinable.
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Mathematics
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There aremany interesting
highlight some of thesein
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a summary of all the topicsand further
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to simplify
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find
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How
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questions
Calculatoricon
/
to use a graphical calculatorin the final examination
paper
for this Option. Somequestions
can
be done in a particularly cleverway
by
one of the graphical calculator functions.However,
we recommend
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using
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aid of a calculator. These questions
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drill
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differently
are of increasing difficulty.
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levels.
the
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are drill questions.They help you practise the methods described in the book,but
like the questionsin the examination.
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most of these.
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Red questions are at the very top end of difficulty in the examinations.If you
then you are likelyto be on coursefor a grade 7.
Gold questions
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are a type that
are
in order
discussion
and
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to help you
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concept.
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International Baccalaureate\302\256
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a green
if you
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examinations.
guidelines.
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deal with
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section of
colour-coding scheme.
not be surprised if you find
at all areas of the syllabus.
equally
good
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will
guarantee you
get a grade 7; after all, in the
and examination
stress!
are aiming
question you cannot do. Peoplearenever
can do all the red questions
that
does
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examination
typical
follow
for
a grade
6, do
These questions are gradedrelative
to our
of the final examination, so when
first
experience
you
start the course you will find all the questionsrelatively
but
the
end
of
the
course
hard,
by
they
should
seem
more straightforward.
Do not get intimidated!
We
hope
you
find
might also find it
after
lots
of revision
the Discrete
Mathematics Option an interestingand
quite challenging,
but
and practice.
do
not
get intimidated,
Persevere and you
will
enriching
frequently
course.
You
topics only makesense
succeed.
The author
team
\302\273
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1
How
to use
this
book
vii
&$<
Acknowledgements
The authors and publishersaregrateful
for
in either
the original or adapted form. While
/
the
permissions
every
effort
granted to reproduce materials
has been made, it has not always
the sources of all the materialsused,orto trace
all
possible
If any omissions are broughtto our notice,we will be happy to include the
to identify
been
on reprinting.
acknowledgements
IBexam
International
Baccalaureate
Baccalaureate
permission to reproduce
Cover image:Thinkstock
questions
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Diagrams
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Photos:
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Chapter
Acknowledgements
the
book
5 page
were
created
51, Edyta
Organization.
Organization
by Ben Woolley.
Pawlowska/Shutterstock
copyright
holders.
appropriate
We gratefully
intellectual
acknowledge
property.
Introduction
In this
\342\200\242
about
to
how
solutions of
find
integer
find
a remainder
used
in
mathematics
prime numbersare so important
and why
of integers,
divisibility
\342\200\242
of proof
methods
different
\342\200\242
about
will earn:
yo
Option
linear equations in
variables
two
(Diophantine
equations)
\342\200\242
to
how
when one integer is dividedby
calculations and solve equationswith
\342\200\242
about
various
remainders
of remainders,
properties
\342\200\242
about
including the Chinese RemainderTheorem
a graph,
structure called
mathematical
new
a
lines, and can be usedto modelnetworks
how
arithmetic)
(modular
and
Little Theorem
Fermat's
\342\200\242
to perform
how
and
another,
solve
to
problems
optimisation
on graphs,
for
of
consists
which
finding
example
points
the
joined
by
route
shortest
between two points
\342\200\242
how
find
to
by
a
recurrence
relation.
problem
Introductory
Supposeyou
can you
for a sequence given
a formula
have
supply of
a large
20 cent and 30centstamps.In how
many
different
ways
make up $5 postage?
Discrete
several
branches
of mathematics concerned with the study
of
rather than continuous.This meansthat they are made up of discrete
(separate) objects,such as whole numbers or vertices of a cube, ratherthan continuous
quantities
such as distance or time (represented
real
Some aspects of discrete mathematics,
numbers).
by
the
of whole numbers, date back to antiquity.
of the areas started developing
mainly
theory
Many
in the 17thcentury
in
and
found
substantial
the
20th.
Oneofthe main drivers
applications
only
includes
mathematics
are discrete,
which
structures
recent
behind
discrete
fundamentally
Theory is the first
and
divisibility
particularly
around
especially
Oneof
the
main
the 3rd
topics
subject is its
science,
in computer
use
as computers
are
machines.
of Discrete
Five branches
Number
in the
developments
Mathematicsarestudiedwithin
part
the
IB syllabus:
option. It is concerned with properties of wholenumbers,
numbers. Number theory was studiedby Greek
mathematicians,
of this
prime
century AD,
of interest
and
a little
bit later
was Diophantine
by Indian and Islamicmathematicians.
equations -
equations
where
we
are only
interested in integer solutions. We will study these in chapter 4 of this option.Numbertheory
has
held
the interest of mathematicians all over the world for many
centuries.
This is because it is full
of problems
that have remained unsolved for a long time. Some examples of these problemsare:
the
Goldbach
the twin
(that every even integer greater than 2 isa sumof two
primes);
conjecture
there
are
of
consecutive
odd
numbers
which
are
prime conjecture (that
pairs
infinitely
many
Introduction
**-
(*
both prime);
Theorem
Last
Fermats
and
greater than 2), which was first stated
a topic of active research becausethey
n is
are
+
b\342\200\236\302\273Arr~-^
there arestillmany
we
things
don't
'v
,\342\200\236.
xn + yn
(that there are no integerssatisfying
in 1637 but only proved in the 1990s.
Prime
have
know about
found
recently
applications
\342\200\224
zn when
numbers
but
in cryptography,
their distribution and how to find
them.
describe how to get fromoneterm in a sequence to the
next. Given a recurrence relationand a sufficient
number
of starting terms we canuse theserules
for
the
behaviour
of the sequence it is moreuseful
to
to build
the
However,
up
sequence.
analysing
in the sequence.
know
how the value of each term depends
on its position
an
for
equation
Finding
the
nth term of a sequence can be remarkably
difficult
and in many cases even impossible.In the
final
of this option we will learn how to solve sometypes
of recurrence
relations.
We will
chapter
also use recurrence relationsto solve
and
model
financial
situations
problems
counting
involving
are equations that
relations
Recurrence
Logic is the study
and
circuit
and
rules
of
both mathematics
electronic
debt repayment.
and
interest
compound
1h
of reasoning and structureof arguments.It isa part of
as being of philosophicalinterest,it is importantin
science and the study
of artificial
Within the IB
intelligence.
principles
As well
philosophy.
computer
design,
syllabus,logicis covered
in
in Theory
and
Studies
Mathematical
of Knowledge.
can be modelledon a network.
This
is when
a
set of points arejoinedby lines, possibly
with a length assigned to each line.Thesegraphs
could
road
molecular
electronic
circuits
or
for
networks,
structures,
flowcharts,
represent
planning
One class of problems involves
the properties
of graphs, such as various ways
of
example.
studying
from
one
to
another.Another
class
is
where
point
important
optimisation
problems,
getting
graph
for
the shortest possible route between
example,
theory is required to develop an algorithmto find,
two
or the optimal
time.
points,
sequence of operations to completea taskin the shortest
possible
This is very different
from
continuous
optimisation
problems, which are solved usingcalculus.
in
was
first
studied
the
18th
but
of the developments
are more recent
Graph
theory
century,
many
in
and have applications
such diverse fields as computer science,physics,
and
chemistry,
sociology
with problems that
is concerned
Theory
Graph
graph theory in the secondpart of this option.
of the underlying
structure of a collectionof objects,rather
Group Theory is an abstract study
than
themselves.
For example, counting hours on a 12-hourclock,
rotations
and
objects
by 30\302\260,
in that they repeat after
remainders
when numbers
are divided by 12all have the same structure,
12 steps.
can
be applied
Group theory is an exampleof how a singleabstractmathematical
concept
in many concrete situations.It is importantwithin
in the theory of polynomial
mathematics
equations and topology (the study of certain properties of shapes), but it alsohas applicationsin
in
and chemistry
because it can be used to describe symmetries. Groups are covered
the
physics
We will study
linguistics.
Sets,
and Groups
Relations
In the first
of this
chapter
in subsequent
option.
we will introduce
option
theory,
chapter8 Recurrence
relations.
someof
combine
which
lj 2
various methodsof proofwhich
chapters. Chapters 2-5 coverNumber
9 contains
Chapter
from
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different
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theory, and
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have discussedin the introduction,Discrete
Mathematics
both the properties of large systems
as graphs
(such
Can
with
deals
with many
points and lines)and
has
In such systems it is
there are infinitely
many).
impossible to checkall possiblecases,sohow can we be certain
about what is true?In mathematics,
truth
is established
through
proof, which is a sequence of logicalstepsleadingfromknown
(of which
We
used
have
quadratic formula are someexamples.
direct proofs, this meansthat we
we alreadyknewand derived
new
However,there aresome mathematical
be proved in this way. One of the
the proof
is
expansion
An
is an
v2
that
alternative
cannot
it
proved?
the
were
some results
direct
calculation.
by
from
started
results
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that
results
most quoted
examples is
sinceits decimal
that it never
show
is to try
approach
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of these
Most
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infinite,
be true before
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derivative
the
identities,
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this course to
throughout
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results:
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conclusions.
to new
or assumptions
facts
principle
induction.
statement
numbers
of natural
properties
contradiction
pigeonhole
strong
As we
you
chapter
learn:
will
and show
cannot
v2
that
repeats!
as a fraction; but how canyou show by direct
that something cannot be done?In this situation
we
need to use an indirect proof, where
we find
some roundabout
of
that
the
statement
must
be
true.
For example,
way
showing
we couldtry to see what would happen if v2 couldbewritten
as
a fraction
and hope that this leads to an impossibleconclusion;
be written
calculation
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this
is
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by contradiction.
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showingthat
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starting
is true
first
In this
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involves
>
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that
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the
only
statement
directly
If
you
have
shown
than
proving
directly, does
establish
that
something can be proved,
rather
for the
proved for all integers,even if we have
the
proof
statement
proof,
a given
that
starting number,and
for some number we can alsoprove
it for
statement
proved
having
the next
the
indirect
of
by induction, which is usedto show
true for all integers above
a certain
it
that
its truth?
can be
proved
it for
one.
chapter
which
we look in more detail at someindirectmethods
will be needed later in the course.
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rsM
of proof
VS.
^
<Kvv,
.Ok\"
*
^
(a
^<+^
by contradiction
Proof
Sometimesit is
Proof
a
special
general
of
form
is disproved
states
proposition or
its
which
For example,
It is
must
negation
be true.
BCE. One
famous
that
are
been
was his proof
since
then,
validity has been
some,
most
century
in
Oneof
its
the proofthat
disputed by
by the 20th
notably
and
mathematician
Dutch
philosopher
has
it
Although
used tool
a widely
mathematics
many
(see Worked
2.15).
example
can prove that a statement is true by showing
that if
the opposite was true, it would contradictsomeofyour
else we already knowto be true).
(or something
assumptions
You
infinitely
prime numbers
L.E.J. Brouwer.
the
v2
in the
form \342\200\224,
where
using
start
is that
irrational number
of an
impossible)
+a
of proof by contradictionis
examples
is an irrational number. Rememberthat
the
cited
most
definition
form
to express
the
fact
integers. It is difficult
a number
that
Here proof
an equation.
that v2
by assuming
q are
p and
be written
it cannot
is of the
(if
is not
Prove that
v2
be written in
cannot
the form
by contradiction,wherewe
form
is a
\342\200\224,
1
where
\342\200\224,
really
useful tool.
p,qeZ.
q
contradiction: Start#
v2 as a fraction and
Try proof by
by
writing
show that
this
leads
Suppose
that
and that
the
v2 =
\342\200\224
with
p,q
e Z,
to impossible
consequences
The same fraction
can
several ways
so
we
should
be
(e.g. -
in#
written
=
-
=
specify which
that
one
we are using
Topic
10
-
Option: Discretemathematics
p
\342\200\224)
iAm*>AJ**
and
fraction
q have
not
of a certain
orked example 1.1
rsM
this
contradiction
by
most
the
of
examples
there
Proof
*y
4fJjf^
300
around
Euclid
try
1.1
POINT
KEY
Proof by contradiction
could
odd
know whether
don't
we
but
square
must also be an
by contradiction.
proof
was already used by
opposite
an odd
have
not really obvioushow to start.We
root of n2,
is
this
is what
we are
produces an oddnumber(remember,
about
what
would
to
happen
prove!). However,thinking
trying
if n was not an odd numberallows
us to do some calculations:
If n was not oddit would
be even,
and the product of two
even numbers is alsoeven, so n2 would be even. But we were
told that n2 was odd, so this situation is impossible!We can
therefore
conclude
that n must be odd. This way of reasoning
is
in
common
all
branches
of
and
is
known
as
mathematics,
very
a
either
that
the
we
to prove that n
want
we
and
suppose that
must
that something
directly
simpler to show that
much
the square
taking
conclusion.This type of argument
relies on the low of excluded
middle,
impossible.
number.
a proposition
by showing that its
lead to an impossible
would
truth
it is
n2
show
to
difficult
be true, but
number
od
which
in
>
argument,
reductio
absurdum,
1'
more
of a
case
called
is
contradiction
by
v
is in
its simplest
no common factors.
form so
a-
*
(a
*
1 ^ - ^
v
<+/*
\\
it
)
continued...
We
calculations
do some
now
can
\342\200\242
P2
\342\200\224
=
2
Then
(squaring
eo p2
both eldee),
=
2q2-
\302\260l
useful
in
common
for
Looking
(2r)2=2f
=> 4r2
= 2<f
=>2r2
=
eo cf
as
p and
that
of
is even and therefore
Hence p and
reached a contradiction/
assumed
we
also be
eo p must
even,
Then
integers
We have
p2 \\e
even: p = 2r
involving
problems
solving
that
means
This
is#
factors
c\\
have
c\\
even.
is also
factor
common
2,
is a
which
contradiction.
q had no
So V2
common factors
cannot be written
as
\342\200\224.
1
proof by contradiction to prove severalresultsin
this
The exercise below is intended to give
some
option.
you
in
this
of
but
does
not
practice
proof,
represent
writing up
type
We
will use
typical examinationquestions.
One
of
Cantor's
put
all
(infinite)
1. Prove that
if
n2 is
Show that there
X2
-
examples
of
diagonal
proof,
it is impossible
real numbers
into
that
which
to
an
list.
1A
Exercise
2.
most
proof by contradictionis
shows
+G
the
*SHG? fascinating
an even
integer than n is alsoan
are no positive integersx and
y such
even
integer.
that
= 1.
y2
3.
Prove that log2
4.
Prove
5.
The
is no
there
that
mean
age
6. Show that
of five
be at
them must
the
5 is an irrationalnumber.
least
sum
largest even
students is
18 years
of a
integer.
18.Showthat
at
least
one of
old.
rational and an
irrational numberis
irrational.
7.
Prove
converse
the
sidesof
a
triangle
of Pythagoras'
Theorem:
\342\200\224
and
a2 + b2
G =
c2, then
If a,b,c arethe
90\302\260.
1 Methods
'
of proof
t*J
VS.
?3n
.0
'
* ^
(*
v
c+^
8.
calculator to show that
Use your
(a)
has one realroot,and
this
find
the
to 3 significant
correct
root
x +1 = 0
x3 +
equation
figures.
that
Prove
(b)
root is
this
irrational.
principle
Pigeonhole
If you
in a year,
of 367
a group
have
them share
so
a birthday. This
if each
person
people, you can be certain
had a different
KEY
The
principle
pigeonhole
is also
4fJjf^
19th century
German
mathematician
are
example
given
the
real numbers
11 different
of those
two
between 1 and
numbers
100
differ by at
we have 1 1
pigeonhole
two
least
is a
pigeons.
1.2
principle to show that
Since
then there
n pigeonholes,
in
pigeonholewhich containsat
Gustav
Lejeune Dirichlet.
You
principle
If n + 1 pigeonsareplaced
after the
difficult
1.2
Pigeonhole
principle,
orked
POINT
can be
*^f
as
known
Dirichlet's
would
this
birthday
366 people.This simple observation
a surprisingly powerful tool in solving
seemingly
very
in
It
is
stated
the
form:
problems.
traditionally
following
of
366 days
for at most
account
two
that
are at most
there
is because
numbers, to use#
principle we need 10
'pigeonholes'
Divide the
pigeonhole
10.
Interval
10
into
[1,100]
ec\\ua\\
Intervale:
...,91-100
1-10,11-20,
Since there
+a
Use the
inclusive.
most
be in
the
are 11numbers,
same
Interval
two
of them
must
(by the pigeonhole
principle).
What
is the
two
between
largest
numbers
possible
in
the
same
difference\342\200\242
These
two
numbers
differ
by
at
the
most
10.
interval?
Notice that we could alsohave
proved
the
above
claim by
contradiction.
Put the 11numbersin order,
that
x1 < x2 <... < xu, and
suppose
of
consecutive
numbers
differ
more
than
10.
Then,
pair
by
as xx > 1, it follows that x2 > xx +10 > 11, x3 > x2 +10 > 21,and so
with xu > 101,which contradictsthe assumption
that
on, ending
all the numbers
are between 1 and 100.Hencethere
must
be
two of the numbers which differ
at
most
10.
by
each
The pigeonhole principle itself can be proved
A more general form of the principleis:
Topic 10
rsM
3$)
-
Option:
Discrete
mathematics
by
contradiction.
'
^
(0-
-^^c^
1.3
POINT
KEY
General
principle
pigeonhole
If kn + 1 objectsare
be
may
then
empty)
k +
least
v
(some of which
n groups
divided
into
there is
a group which containsat
1 objects.
orked example 1.3
the
Prove
know
don't
We
principle
pigeonhole
general
about
anything
objects are split into
proof seems unlikely.
so
groups,
So
stated above.
Suppose that the statementisfalse,
how the#
a direct
a proof by
try
of the
each
n
at most k
contains
groups
so
objects.
contradiction
Then the
total
most
This
kn.
about the
of
number
of
number
can be
objects
at
the assumption
contradicts
objects.
least one of the groups
contain at least k + 1 objects.
Hence at
must
Not all statementsthat can be proved using the pigeonhole
principle can easily be proved by contradiction. The next
exampleusesthe generalpigeonhole
principle.
orked
1.4
example
Nineteen points are drawn
inside
a square
it is possible to find three pointsthat
of side
that
form
3 cm. Usethe pigeonhole
to show
principle
with an area lessthan 1 cm2.
a triangle
+G
one
want
We
of the
to#
'pigeonholes'
contain a triangle,so we
(since 2x9=
'pigeonholes'
The
Since 19= 2x9
18)
there
at
is the
What
triangle
side 1 cm?
It is not obviousthat
contradiction.
an
area
could
we
can
We
try to show that
large, but sincesomeofthe
anything
their
about
total
powerofthe pigeonhole
principle
would be
to
difficult
prove
using
the
overlap
triangles
area. This
in
into
nine
1, by
three of the
the pigeonhole
squares
Q,
VVj
principle
contains
which
points.
form
a triangle
the area of
the
with
square,
which
is 1 cm2.
statement by
the triangles have
total
it is
area is too
difficult to say
last exampleillustratesthe
proving
any other
statements
which
method.
1 Methods
^
of
squares
the above
that all
assuming
try
than 1 and
more
prove
least
+
of these
area less than
a square
inside
is
is one
Thesethreepoints
area of a#
of
possible
largest
which
be split
can
square
side 1 cm.
9
to define
need
of proof
I ^
*
^
(a
v
^<+^
is an example of a non-constructive
existence
proof. In the last
a triangle with the required
but
not
how to find it. Some
exists,
property
if we
mathematicians
the
of
such
Is it useful
to
that
exists
know
dispute
validity
proofs.
something
don't know how to find it? Can an object be said to exist if there is no known method of finding
The
pigeonhole
we
example
or
principle
found that
>
y
it?
constructing
\302\253^.
^^
the pigeonhole
use
will
We
about graphs,
theorems
principle to
see Worked
example
some
prove
\302\253^.
^^
6.1.
IB
Exercise
1.
A
pick to be certainthat
need to
do you
colour?Proveyour
2. Prove that
if you
(inclusive)
if 9
that
Show
must
be three
5. Showthat
at
friends
Five
the
it is
are
points
whose
8
integers
it is
possible to find 10
a 0.
the
party
and some
of them
are friends.
party.
there are two
of friends.
7.
are two
seated in a row of 12chairs,there
occupied chairs.
(b) Show that
+a
integers there
impossible that there is botha person
with everyone and a person who has no
is friends
who
principle.
it is
why
Explain
same
integers between 1 and
them must add up to 9.
91 positive
in
There are n people at
(a)
of the
five different
people are
consecutive
among
three
have
sweets
by 7.
pick
whosesumends
6.
8 positive
two of
then
you
the pigeonhole
using
any
among
3. Prove that
4.
result
is divisible
difference
different colours.How many
of five
sweets
contains
bag
drawn
on the
people
who
have
the same
number
surface of an orange. Prove that
to cut the orange in half in sucha way that at least
the points areon the samehemisphere
points
(any
along the cut count as beingonboth hemispheres).
possible
four of
lying
8. Each point in the planeis coloured
either
red
or blue. Show that
1 cm apart.
there are two points of the same colour which are exactly
\342\200\242i*
The
of graphs.
colouring
any
in
the
group
It
whom
8
other,
know
-
know
Discrete
Option:
^
cl
\302\243%*..
are selectedinsidea squareof side1m.Show
to find three of thosepointswhich
lie inside
radius
that
it
a
\342\200\224
m.
7
10.
of
rsM
3$)
circle of
In
people, there
or three none
each other.
Topic10
51 points
is possible
is often
form:
are eitherthree who all
each
9.
with
following
of six
^88
Ramsey's
which deals
theorem,
stated
10 is
in question
result
a special case of
mathematics
and diagonal of a regularhexagonis coloured
either
red
or blue. Show that there is a triangle with
all three
sides of
the same colour.
Each side
induction
Strong
|QP
You will have learnt that
about
statements
some
positive
ofmathematical
integers can be proved usingthe Principle
if we know
induction.
The key idea in inductive proofs is that
that the statement is truefora certainpositive
we can
integer,
use this to show that it is true for the next integer.This relies
on
a
connection
between
the
statement
for
n
and
the
establishing
statement for n + 1.
on more
However,sometimesthe statementfor n depends
value.
One
familiar
is
the
Fibonacci
previous
example
=
values
Fn_1 + Fn_2 with
sequence, defined by Fn
starting
\342\200\224
\342\200\224
If
1.
we
want
to
about
need
Fl
F2
Fn we
prove something
than one
about both
information
let us
For example,
formula
by
Fn
\342\200\224 +
Fn_1
Now suppose
true
Fk
=
prove that
terms
for
all
Fk_1
+ Fk_2,
Fn <
2n for
for n
starting values: Fx
up to
we need
all n. First of all, the
> 3, so we needto start
-1
and including Fk_x for
to use the resultfor
As we are assuming that
to and includingFk_x, we
checked
have
we
F2
< 22.
= 1
statement is
some
two
k. Since
terms.
previous
the statement
fact that
use the
can
and
< 21
that the
checked
have
we
that
Fn_2.
applies
only
Fn_2
the two
checking
and
Fn_1
Fk_x
< 2k~l
up
and
Fk_2<2k~2. Then:
=
Fk
Fk-i +
pk-2
<2k~l + 2k-2
=
2*\"2(2
+ l)
<2fc\"2x4 = 2fc
that
shows
This
\342\200\242
The
integers
n,
then
n
for
that the
It follows
as
we
the
for n
is true
2n
n-k.
n
\342\200\224
1 and
for all n < k then we
\342\200\224
2.
can
show
that
it
\342\200\224
k.
statement
can reach
statements
use
statement is also true for
so the
is true
statement
is also true
the
Fn <
statement
\342\200\242
If the
Use
< 2k,
shown that:
we have
Thus
Fk
for
statements
Fn
is true
< 2n
for all positive
any n by repeating the stepsabove:
n = 1 and
iovn
n-2Xo prove it when n -
\342\200\224
2 and
n
\342\200\224
3 to
prove
3,
it
for n - 4, and so on.
This
variant
of proof
by induction,
we needto useseveralprevious
called strong induction.
where in the inductive step
rather
than just one, is
values
* ^
(*
KEY
'
1^
^1
^
V
vc+^
l .4
POINT
Strong induction
Supposethat
a statement
have
we
(or rule)
(a)
the statement
is true for somenumber
(b)
assuming that
the
(this is calledthe inductive
is true
the statement
then
for
base
of
can
that:
show
and
cases,
can
for all n<k,we
is true
statement
about a positive integern. If we
also true
it is
that
show
for n-k
step)
all
n.
integers
positive
number of basecasesneeded
on how
depends
in
the
next
term
many previous terms are requiredto calculate
each
term
was found by
the sequence. In our exampleabove,
two base values.
using two previous terms,sowe needed
Note that the
Sometimes the inductive step requiressomeofthe previous
terms,
don't necessarily knowwhich ones.We then really need to
know that the statement is true for all previous
values of n. This is
but we
wherethe methodofstrong
orked
to show
induction
strong
that every
positive integer n can bewritten
n=
where p e N
It
is
not
each
and
obvious
how1
cases we need,solook
at
the
inductive
obvious
is no
There
2pcv + 1p~xcv_x
0 or 1.
q is either
immediately
many base
step
first
connectionf
connection
k and
between
If
k is
even
we can
+...
Inductive
+ 2q
-
^
q, vv,
\342\200\242V^
+ o
be either
even
2
relate k to
this by
Option:Discrete
can
all
_
-
\302\253
k le
even, then
so we
know that
If
2 to get
km
mathematics
\342\200\224
le an
Integer
leee than
2^cp
+
is also
2?cv_y
k,
k
:
the statement letrue fom = \342\200\224
2
cy-0 or 1.
Then
which
Topic 10
oaeee\\ k
statement for
or odd.
k =
10
+ c0
two separate
for some p and
rsM
form
step:
Conelder
2
Multiply
the
in
Suppose that we have proved the
n< k, and look at n = k.
between n = k and any particular
as the
previous terms. However,
required
expression involves powers
of
2, we
may be able to establish a
+a
useful.
1.5
example
Use
is particularly
induction
+ 2c0
form.
required
+...
of the
+ 22q
a-
*
(a
*
1 ^ - ^
v
<+/*
\\
continued...
k is
If
relate k to
we can
odd,
is odd,
\342\200\242\342\200\242
\342\200\224^If k
\\
k \342\200\224
is an Integer \\eee
2
is true
the statement
that
n =
for
k, so
than
we know
k \342\200\224
1
:
2
= 2'cp+2P-1cH
+ ...
^
for eome p and
To get
need
k we
k =
1
add
and
0 or
=
which
1.
+
2{2Pcp+2P-'cp_,
2P+'cp+2Pcp_,
is also
of
+
... + 2c,+c0)+
that
yet: we
case
base
and
can get
= 3
n
not done
have
we
n
by using
only need one
write
Always
a
... + 22c,+2c0+\\
n = 1,
When
1, so we
=
take
p
\342\200\242
a conclusion
is true for
to assume
of k in
value
each
k. Hence the
strong
get the
n \342\200\224
\\, and
if it
is true for
it is alsotrue for
statement istrue for all integers
Induction.
V^^^^^^^^t^^^iA^
A
,^fc._Ajt -^-n (^
The value used depended on
= 10 and n = 11we used n = 5
we used n = 15. This is why we needed
statement
was true for all previousvalues
of n.
values.
for n
exampleis called binary
ofn. It isa special
a number
= \\to
the above example,
The form from the previous
or base 2 representation
^^form,
^^
the
that
c0
we can show that
k then
n<
^
+a
and
Conclusion:
n > 1 by
we usedoneofthe previous
k; soto prove the statement
and to prove it for n = 31
=0
required
n =
for
k.
base case
a\\\\
the statement
n =
form.
The statement
To prove
for
E3asecaee:
\342\200\242
n = 2
to
/\\
the correct form.
Thereforethe statementis true
Remember
+ c0
Then
double*
to
c =
+ 2q
base, which
will
we
the
case
in more
investigate
in chapter3.
of<^^
detail ^^
Exercise 1C
In
1.
this
that
Given
t*J
a natural
number.
and
a0=2yal=5
=
an+2
5an+1
-
6an, show
that
+ 3n.
a=2n
2.
n is
exercise,
A
sequence
ax
\342\200\224
0,a2
is defined
=
l,a3 = 4.
by
an+3
Prove that
=
3an+2
for
-
+ an
3an+l
all n, an
=
(n-l)
with
.
1
'
^^VNOC,
Methods
of
proof
*
(*
'
^
1^
3.
numbers from
(n
numbers are
in the
17th century
believed
that
and
(He only checked the first
only 100 years later
was
Euler
proved
that
F5
is
The sequence Gn
all prime.
were
they
four!)
It
Gn+3
are
Gn+2 +
Gn+l
that
defined by
Fn
the
Show
and
Xn-fl\\
=
22\"
first
=
by G1
is defined
+ Gn.
= l,x2=2
+1
for
n>\\.
one end in
Prove that
a 7.
1,G2= 2,G3= 3 and
all n>l,Gn< 2n.
that for
that
in fact
Showthat
divisible by 641. It is not known
whether any other Fermat
numbers
=
by x1
numbers exceptfor
all Fermat
4
question
is defined
xn
V
vc+^
Show
l)(*n-l + *n-2) \342\226\240
Fermat
Fermat studied the
^
The sequence
~
=
*\302\253
^1
12
made
be
greater than or equalto
4
using only cent and 5 cent stamps.
of postage
amount
any
can
cents
prime.
that
Show
for
all integers
n>
is
2, the numbercos \342\200\224
irrational.
8.
Letu0 = 1and
thatun
= 2n
9. Showthat
most
+a
12
Topic 10
-
Option:
^
q,vv,
Discrete
mathematics
2n
un
\342\200\224 +
for alln
n straight
regions.
un_x
un_2 +...
+ u2+ul+
2u0 forn>\\. Show
e N.
lines
divide the
plane into
at
^
(a
^<+^
In this chapter
will learn:
you
Divisibility
\342\200\242
about
including
multiples,
and
and
factors
the greatest common
prime
divisorand the least
common
numbers
multiple
for finding
\342\200\242
a method
common
the greatest
divisorof
always be added, subtracted and multiplied
whole number. The only basicoperation
not always result in a whole number is division.For
at factors is an important tool in solving
looking
the
that
does
this
reason,
theory. Some of the materialin this chapter
be familiar to you already,
but
some
of the more abstract
will
results may
expected to
You are
new.
be
be ableto prove many
useful
properties
\342\200\242
why
are
The proofs are also examplesof techniquesyou
to use in later chapters.
need
large
Euclidean
of factors
and multiples
results.
these
will
\342\200\242
various
in number
problems
(the
algorithm)
is a
result
and
of
can
numbers
Whole
numbers
two
prime
numbers
so important.
j
multiples and remainders
Factors,
ES9
number
of
Much
of
multiples
cannotbe
the basic
divided
and
the divisibility
of
properties
this chapter,
Throughout
number
a whole
mean
we
factors or
when numbers
section we review some of
relevant
to division and
numbers.
whole
with finding
at remainders
or looking
In this
exactly.
terminology
'number'
is concerned
theory
numbers,
by
(integer).
Consider two numbers, a and b. If thereisa numberk such
b = kawe can write a \\ b and say that:
+G
a divides
or
by a,
or a
is a factor of b,
or
is
b
Note
in
and
basic properties
a
and
\342\200\242
If a
|b
\342\200\242
If a
| b then
Note
is a
that
1 is
\\
a
c
then
b=
0 is
always
find
any number
for
of every
a factor
>
a\\(b\302\261c).
factor of itself.When
can
that
of divisibility:
\\ (be)
specify whether we want
We
but
zero,
2.1
POINT
Some
we cannot divide by
fact a | 0 for all a.
0 as
a ^
possible
KEY
of a.
a multiple
that
that
b,
divisible
is
b
i
number, and that
about
talking
to
include
numbers
b \342\200\224
ka +
c.
k and
r and
factors,
1 and
every
number
we will always
the number
itself.
r suchthat:
0<r<a
A3
2 Divisibility
\342\200\242
and prime numbers
13
rsM
^
vs.
<K
vv, \342\200\242V^
+ o
'
* ^
(*
1^
^1
^
v
vc+^
*
by a. Note that
write b = ka.
4 we will
In chapter
examine other rules
with
working
l^>for
r = 0
The process of finding
]^>
called
remainders,
the quotient
call k
We
it seems
Although
useful
surprisingly
modulararithmetic.
to perform
called the division algorithm.
very simple, writing b as ka + r or ka is
when proving other results.
is divisible
by 2 if its last digit is even.
\342\200\242
A number
is divisible
by 5 if its last digit is 0 or 5.
\342\200\242
A number
is divisible
by 10 if its last digit is 0.
by 4 if the numberformedby
is
divisible
digits
by 4.
is divisible
two
digits is
three
last
test
Divisibility
using
is divisible
\342\200\242
A number
is divisible
its digits
prove
divisibility
]^> tests for
numbers
in
written
other
bases
theory,
and
and
structure
by 9 if the sumofits digits
is
by 11. (Soifthe numberhas
the sum
\342\200\224
ax
a2 +
a3
of
digits
-...)
the divisibilitytests listedabove.You
reproduce those proofs in an exambut, more
they are examples of types of proofusedin number
similar
you will be asked to produce proofs with
prove
asked to
importantly,
]^>
is
we calculate
al,a2,a3...
now
by 3 if the sumofits digits
by 11 if the alternatingsum
is divisible
\342\200\242
A number
be
digits:
by 9.
divisible
will
all the
by 3.
divisible
could
its
divisible by 8.
is divisible
\342\200\242
A number
We
to
its
by 8 if the numberformedby
is divisible
\342\200\242
A number
will
number
\342\200\242
A number
last
3 you
the
of
1H
last digits:
tests using
\342\200\242
A number
chapter
tests which usethe digits
1-
2.2
POINT
Divisibility
In
r is
k and
there are divisibility
be divided.
+G
can
we
case
long process, it is usefulto have ways of
whether one number divides anotherwithout
having
the division. For divisibilityby many
small
numbers
deciding
KEY
divided
b is
be a
can
division
As
and r the remainderwhen
if and only ifa\\b, and in that
J
some of
using similar
techniques.
10.
than
In orderto
number
using
KEY POINT
'
If AT
then:
is a
those
derive
its
we need
a way of expressinga
digits.
2.3
/c-digit
N = lOkak+10k-lak_1
We use
proofs
number
+ ...
with digits ak,ak_x. ..
a2, a^,
wq
+ 102a2+10a1+a0.
shorthandN = (akak
x...a2a1a0).
1
14
rsM
Topic10
-
Option:
Discrete
mathematics
VS.
^
Q,VVj
^
*
(*
-
*
+
V
*<l+r
* )
orked
2.1
example
number is divisible
Prove that a
and only
4 if
by
if the numberformedby
digits is
last
two
such that
the
number
two
is divisible
its
divisible by 4.
is an
This
we
so
statement\342\200\242
to
Let
(i)
digits
prove
last two
prove.
number
by its
that
Frove
N
\\aet
digits
by
4.
by 4.
is divisible
is divisible
digits
whole
number
As the
statement
of
be a
N
formed
something about the last two
us a way to start, so first
gives
'if the number formed by the
Knowing
the
'if and only iP
have two things
by 4, then
is divisible by 4'
is about
Let
N
=
\\0kak
+... + 102a2+
+ 10^%^
\\0ay
+ a0
sense to write
it makes
N,
digits\342\200\242
N = [akak_}...a2a}a0)
Separate
We are told
that
(0^0),
This
4.
by
4A/I for
it as
two
last
the
proving
a
good
some number A/I
what
a factor
is divisible
We now need to
It is
reverse
look
for
by 4,
\\02a2)
+ (\\0a,+a0)
a good
the previous
an alternative
idea
to
So we write
N
separate
in
\342\200\242
=
both
\342\200\242*
+ a2) + 4M
100(l0^2%+10M%_1+...
other
divisible
\342\200\242
Nbe a number
that the number
(ii) Let
and
try
strategy
of its
the last two
if
are divisible
both parts
by 4.
4 1100,
As
proof (and only
terms
+ 4M
is)
by 4
the
prove
= (/\\0kak+/\\0k-'ak_,+...
+ /\\02a2)
first
so
doesn't
and
+
of 100
divisible
are
terms
direction.
A/I
in the
terms
the
have
bracket
100
\342\226\240\342\226\240(\\0kak+\\0k-'ak_,+...
not
(it's
divisibility, it is usually
idea to take out common
factors. All
But
digits\342\200\242\342\200\242
is lOO] + o0/ is divisible
means that we can write
important
When
2
last
the
that
digits*
which
formed
by 4,
eo
N
is
by 4. Frove
its
last
two digits is
by
is divisible
by 4.
divisible
this
work)
digits\342\200\242
digits
Let
N
=
=
\\0kak
{\\0kak
+
+ 10fc-1flfc_!
= 100(I0fc~2
We can see that the first part is#
divisible by 4. We have also assumed
that N is divisible by 4. If 4 | (b + c)
and 4 I b then it must be that 4 | c
As 4
I
N
\\0k-yak_y
and
+...
+ 102a2
+ \\0ay
+ 102a2)
+...
ak+\\Ok-Z)ak_^+...
+ a0
+ (10^ + a0)
+ a2)
+ {\\0a^+a0)
4 1100, it follows
that4l(10fll+fl0>
2
Divisibility
and
prime
numbers
15
vs.
^^%\302\253.
.0
I
* *~
(a
V
-+*<;
+n
HINT
EXAM
a statement
To prove
to produce
two
'A
form
the
of
proofs:
separate
the
Often (but not always)
and
if
and
A.
B then
if
use similar
directions
different
two
you need
if B'
only
then B,
if A
techniques.
The proofs for
results,
|
I
from the
9 and 11 use the following
Factor Theorem.
2.4
POINT
KEY
I
by 3,
divisibility
follow
which
For
any
integer
For
any
odd power
n,
power
n,
an
\342\200\224
bn
an
+ bn
a factor
has
(a-b).
factor (a + b).
has a
orked example2.2
number
and
to write
sensible
seems
It
then
in
is divisible by 9 thenthe sumofits digits
number
if a
that
Prove
Let
the#
of its
terms
separate
=
N
digits
sum of
the
by 9.
is divisible
/\\Okak
+ ...
+10fc\"1flfc_1
+ 102a2
+ 10^
+a0
Then
the digits
N = (ak
+ak_i +
... + a1
+(lOfc_1
-1)%^
+...
form (10p -1) hae a factor of
the first bracket are divisible
(10-1)
= 9,
+
+a0)
(/\\Ok
-\\)ak
+ (102-1)02+(10-1)0!
All
one
of the
are
form (10
so we can use
from
Key
the
after
brackets
the
point 2.4
+G
But
we
are
divisibleby
bracket
eo
a\\\\
As
N
after
terms
also
first
is divisible
(ak + %_-,
by 9,
+...
+
a-,
+
a0)
use the sametechnique
is
divisible
digits
by 9 then
You can
for divisibility
proofs
We can
to
show
the number
by 3 and 11follow
combine these basicdivisibility
that
if the
KEY
a similar
tests
16
A?
Topic10
-
Option:
Discrete
If a | AT
and
(other
than
of this result
proof
9.
The
argument.
to check
the
for
following
2.5
POINT
the
]^>
sum of the
is divisibleby
divisibility by other numbers.In doingso,we use
result, which we will prove in the next section.
the
also be
must
by 9.
divisible
be
^4
See the end of
]^> next sectionfor
9.
by 9
divisible
^
by
= 1
N is#
so the
must
-1),
with
10,5
told that
9,
Each
result
the
0 =
of the
term
first\342\200\242
b
1),
\\ AT,
then
and
if a
(ab)
and b have no commonfactors
N.
\\
mathematics
I
-
'
* ^
(*
orked
2.3
example
Find the
possible valuesof digitsp and
divisible
by
q such
the five-digit
that
number (386pq)is
18.
18
= 2 x 9, for
sufficient
that the
As
v
-^^c^
is
it
need
We
to be divisible
\"5&6pq
by 9.
2 and
by
for
tests
divisibility
9
and
2
both
We know the
18
by
divisibility
number is divisible by
2 and*
Divisibility
by
2: q
for 9
Divisibility
by
9:
is an even digit
3 + 8+6 +p + q
= 9k
<^\\7 +p+q=9k
As p and q are digits,
they
between 0 and 9, which
are
both
*
+ ci<\\&,
But 0<p
=
p + o[
so
\\
or
\\0.
the
limits
possibilities
now combine this
We can
the
fact*
that q is
even
with
= 1:
Ifp+q
If
= 10:
p+q
q
= 4, p = 6
q
= 6,
=
q &,
As well as showingthat
is divisible
number
a specific
p
= 4
= 2
p
by
that
a whole
class of
prove
something, we can sometimes
numbers is divisible
number. Two main methods
by a given
used
in
such
\342\200\242
in particular
factorising,
the resultsof
+a
\342\200\242
proof
are:
proofs
point
Key
the difference
of two squares and
2.4
by induction.
example illustratesthe useof factorising:
The following
orked example2.4
If
n is
^ 4
We
can
an odd
integer, prove
n2 -1 using
factorise
the
of
= 2x4,
that one of the factors
Since 8
the
odd,
one
other
and
that
hopefully
n2 -1
that
difference\342\200\242
two
we
is divisible
by 8.
H2
- 1=
(n
- 1) (H
+ 1)
squares
can
show*
is divisible
by 2 and
n is
4.
Remember
that
by
other
even number is
every
divisible by 4
(n
\342\200\224
1)
1) are two
so they are both
and
integers,
one of them
Hence their
ae
(n +
is divisible
^
Q,vv,-V
yy\\
+ 0U
divisible
by 2,
and
by 4.
productis divisible
by
2 x
4 = &,
required.
2
it*f\302\273
even
consecutive
Divisibility
and
prime
numbers
'
a-
*
(a
V
^<+^
next
The
shows a
example
such questions in
seen
asked in this
proof by induction. You
the core course,but
they
have
can
already
also be
option.
orked example2.5
by
We
follow
standard
the
start
so
induction,
positive integer n,
for every
that
induction
Prove
format of
proof by#
by checking the case
n= 1
If
is divisible
number
a
useful
be
it
in
a
later
be#
can
some integer a. This
as 7a for
written
by 7,
2n+2
When
h
So it istrue for
it is
Assume
for some
for
rpn+Z
We
need to use the
The
easiest
way
assumption for
is to express one
terms
equation
using
=
n \342\200\224
\\.
true for
= 2
(*)
integer a.
n =
,
k+V.
rz2n+1 _
the
x (7a -
pk+3 .
= 14a+
32M)+
the
are divisible
terms
the Principle of
1. Check
the whole
k +1.
by 7 for
all
n e
Z+ by
Mathematical induction.
whether each of thesenumbers
(a) (i) 333444
(b)
(i)
(d)
(i)
(a)
(i)
(b) (i) (613c&)
Option:
^
q,vv,
Discrete
mathematics
by 3,4
and 11:
8765432
(ii)
(ii) 747472
515152
(32a4fc)
divisible
(ii) 5151515
515151
2. Find the missingdigits
is
33334444
(ii)
(c) (i) 123456
-
7, so
2A
Exercise
Topic 10
by
7.
by
n =
is divisible
expression
18
using (*)
= 1,
As the expressionis divisible
by 7 when n
=
and if it divisible by 7 for \302\253k then it is also
= k +1, it follows that the
divisible
by 7 for n
conclusion1
+a
rsM
32M
9 x
7x3ZM
expression is divisible
Hence it is true for
write
r^Zk+Z)
(*)
Both
Remember to
= k: Then
n
= 2x2k+z+9x3ZM
k*
n = /c. \342\200\242
of
= 5x7
23+33=35
Then
for n
= 1:
2k+2+32k+'=7a
may
calculation
Relate to the expression
by 7.
is divisible
+ 32n+1
so
the given
that
number is divisibleby
(ii)
(llc65&)
(ii)
(2213ab)
36:
*
(a
v
l+r
+ )
Find
by
divisible
pk |
0
not by
for
that
p, q such that
of n the
all values
least two factors (otherthan
Prove by inductionthat
valuesof n.
(a)
Q
4/c +
prove that
digit
values
which
9?
(b)
11?
by 4 for all
[6
marks]
in
one
of
1
+ 2.
takes the
can only
number
have
which
numbers
are
and
which
of n
is the
divisible
form
4/c
or
be of the form 8/c,
[8 marks]
the hundreds
by 15.
number 111-41
divisible
and the
[5 marks]
by
[4 marks]
Number N is given in termsofits digits
+ ... +
N = 10fcaJfc+10fc-1aJfc_1
N is
that
divisible
is divisible
by
as:
102a2+10a1+a0
by 11 if and only
+
%-%-i+%-2-\342\200\242\342\200\242\342\200\242
(-!)
Prove
[3 marks]
4/c \302\261
1, 4/c
number
has at
+ 35
+ 12n2
positive integer canbewritten
a square
1H
that
ndigits
(a)
Show
show
itself).
4.
Q Findallthree
units digits equal
+G
n4
is divisible
-1
forms: 4k,
a square
or 8A; +
8A; +1
For
\\ q,
l.
(c) Prove that
J
p
1-
[5 marks]
number
1 and
5n
Explain why every
the following four
(b) Hence
[6 marks]
integer k.
positive
eveiy
digits) is
44.
integers
positive
for
gfc
1958 (400
19581958
number
22 but
two
Show
[7 marks]
the
by
Given
number (2006ab) is
33.
Showthat
Q
that the
b so
a and
digits
divisible
if
^0
by 11.
induction
[8marks]
that
7n +
4n
is divisible
+1
[8 marks]
by6forallneZ+.
13 Greatest
common
divisor and least
common multiple
the result from Key point
2.5, which
only applies when two numbers do not have
any
common
factors.
For many results in number theory it is
importantto know the common factors of two numbers.
In
the
section
previous
we used
2 Divisibility
and
prime
numbers
19
vs.
?3n
*
^
(a
v
^<+^
)
common divisor (gcd)
The greatest
numberd
the higWco^Jbut
example,
gcd(4,6)
called relatively
prime
the
hcf,
or
factor,
are
notation
IB uses
gcd (a, b).
algorithm
]^>
Section
the
in
]^>
2C.
b. A
more efficient
The greatest
common divisor
35
use to
should
try
theses
in
involved
proofs,
will
the thought
as they can
examination
proofs here. You
processesand strategies
be appliedto proving
results.
other
yo^
to understand
on
required
show
one of the
been
have
papers in the past, so we
greatest
checking
can
you
1H
has someimportantproperties
prove other results.The proofs of these
properties
which is useful
for
are
1-
themselves
divisor,
common
=2
gcd(a,b) is to list all the factorsofa and
slightly
way is to list only the prime
=
factors.
For example, 36 2x2x3x3 and
90 = 2x3x3x5,
so gcd(36,90) = 2x3x3 = 18.Finding
factors
can
prime
be difficult, so in the next section
we
will
see an alternative
numbers.
method, which is of moreuse with larger
calculator
\302\243nfind
d \342\200\224
gcd(a,b).
prime.
relatively
which
^Your
the largest
One way to find
Euclidean
See
common
no
have
b is
and gcd(l2,35)= 1.Two numbers
(or coprime) if their gcdis 1 (sothey
12and
factors other than 1). For example,
For
called
is also
The gcd
a and
b. We write
a and
both
divides
which
of
answers.
orked example 2.6
a b
if gcd
that
Prove
It
is a
a, b and
can
b we
a and
both
gcd
= 1.
d d
idea to write some
good
relating
equation
\342\200\224
d then
(a, b)
sort of
\342\200\242
d. As d divides
a =
write
We can write
to
show
that
a = md
b =
and
=
Qcd(m,n)
nd. Then we need
/i.
md and
b=nd
+G
Give gcd(m, n)
a
about
something
try to
and
name
find
out#
using definitions of
it,
that
a factor. But
a and
both
know
we
common factor
Remember that
f =
wanted
20
rsM
^\\
Topic
10
-
to
of
gcd(m,
know
b have fd
that the
a and
as#
largest
b is d
n), and
that
n
=
oft, eo a
= f.
Then
pf and
m-
= pfd and b = oftd.
n
and
This shows
m
gcd (m,n)
Suppose
we*
this is 1
that
meane
This
the
However,
both a and
Hence f
eof
largest
b is
< 1. E3ut
\342\200\224
\\, ae
fd d'wldee both
number
d, eo fd
f is
that
a and b.
d'wldee
< d.
a positive
Integer,
required.
Option: Discrete mathematics
vs.
^ <k
\302\243%*..
.Ok\"
The above
their greatestcommondivisor,
two numbers
divide
if we
example says that
the
by
are
numbers
resulting
relatively prime.
There are two further
important
properties
happens to the gcdwhen we subtract two
results are summarisedbelow:
|
=1
\342\200\224,\342\200\224
\342\200\242
gcd
b
I
d)
U
= d
\342\200\242
I
All three
= d then:
If gcd(a,b)
gcd(a,a-b)
I
numbers.
us what
2.6
POINT
KEY
which tell
-
\342\200\242
=d
qb)
gcd(b,a
for
Z
q e
any
return to the resultin
2.5: if a and b both
point
Key
=
N and
if gcd(a,b)
divide
N. For
1, then ab alsodivides
x
6
and
5
both
divide
600
so
30
6
(=
5) alsodivides
example,
600.
6 and 4 both divide 36,but 24 (= 6 x 4) does
However,
Let us
Sothe condition
to
Notice
that
that
gcd(a,b)
is divisible
which
not.
for the result
is necessary
12isthe smallest
divide 36, and
by both 4 and 6.
12 does
apply.
number
= 1
important concept. The least common
multiple(1cm) of a and b is the smallestnumber/ such that
both a and b divide /. We write
/ = lcm(a,b).
For example,
=
=
lcm (4,6) 12and 1cm (5,6)
30. The above examplesuggests
the following
extension
of Key point 2.5.
another
us to
leads
This
KEY_PQJNI2.7
If a | N and
b
\\
this result
To prove
then
N
lcm(ab)
\\
N.
we need to lookin
detail
more
properties of and relationshipbetween
and
lcm
find the lcmby listing multiples
a number which is in both lists.A slightly
We can
at the
gcd.
and b until we find
method
is to use
=
120 2x2x2x3x5,
of a
better
and
prime factors.For example18= 2x3x3
so every multiple
of both
18 and 120 must be divisible
23, 32
by
=
=
and 5. Hence lcm(l8,120) 23 x 32 x 5 360. A convenient
way
to see what's going on is to draw
a Venn
diagram
showing prime
factors of the two numbers.
The
two
Any
common
prime
so gcd
sets,
common
appear
in
product
lcm(l8,120)
factors are those in the intersection
ofthe
(18,120) = 2 x 3 = 6.
must
multiple
at
least
one
of the
contain
all the
factorswhich
numbers, so the lcmis the
the prime factors in the unionofthe two sets,
= 23 x32 x 5 = 360.Notice
= 2160 and
that
6x360
of all
18x120 = 2160.This
is
one
, - 3rvV%
example
of a
very important result.
* a-
(a
*
^
1
^
-
v
<+/*
\\
;
2.8
POINT
KEY
x 1cm (a, b) = ab
gcd(a,fr)
Once
The proof
1cm.
papers,
Prove that gcd(a,b)
= md and b=
*
with
nd
may
1H
Let 5 be
IcmlO/b):it'
b, and it is
multiple. How can we
such
express
5 =
md andb
any
ka and
kind
\342\200\224
nd with
b\\
ka
\342\200\224
is a
5,eo
.
is a
,
whole
= 1.
gcd (m,n)
a and b. Then
of both
multiple
.
Then
fact?
first
the
a =
Then
We know two things about
is a
of both a and
multiple
smallest
= $cd(a,b)
Let d
= 1
(m,n)
gcd
the
result has often appearedon examination
example of a more complexnumber
be asked to produce.
1-
x lcm(a,b) \342\200\224
ab.
write a
can
the
2.7
example
We
can be usedto find
also an
you
proof
theory
this result
gcd(a,fo),
of this
it is
and
orked
found
have
we
whole number.
,
number,
eo
km .
\342\200\224
is a
n
nd
whole number.
How
and
can
to get
trying
factors.
no common
n have
that
Remember
/cm?
divide
n
an expression
in
Now we use the
fact
that
of
is the#
lcm(o,b)
As
smallest
S. The
possible value
of
of
=
qna
=
qmnd
= q \342\200\224
smallest possiblevalue
corresponds to q =1.
}
= \342\200\224
d
eo lcm(a,b)
now
can
qn
is the
\\cm(a,b)
S, it
\\cm(a,b)
V
+G
We
k=
Hence
is 1
q
k; eo
divide
q.
5 = ka
and d
smallest possiblevalue
must
Hence
b
of a,
some
for
are
We
terms
As gcd (m,n)= 1,n
mm
prove
the result
xqcd(a,b) = ab.
of Key point 2.7.
orked example2.8
that
Prove
if a
We
\\
common
and
almost
2.7,
example
N
of a
22
3$)
Topic
10
-
N
1cm (a,
then
this
in
have shown
b) | N.
\342\200\242
Worked
that
From the
any
and b can be written
previousproof,
N =
ab
But
\342\200\224
=
lcm{a,b),
eo lcm{a,b)
q\342\200\224.
d
divides
N.
d
as
rsM
\\
proved
as we
multiple
b
S =
for
q\342\200\224
some
q.
Option: Discrete mathematics
^ Q,VVj
VS.
.Ok\"
*~
*
(a
'
K^
V
<+/*
when a and b
special case
In the
result from Key
are
If a andb
\342\200\242
lcm(
\342\200\242
If
a\\
2.5.
point
are relatively
prime then:
ayb)-ab.
N and b \\ N
then
ab
\\
N.
2B
Exercise
the greatestcommondivisor
multiple for the following
pairs
36
and
68
(a) (i)
(ii)
1. Find
(c) (i)
(d)
(e) (i)
(f)
(i)
(ii) 25pq2
Verify that the
pairs
following
(a) (i) 68 and
(b) (i) 120and
(c) (i) 35 and
(p,
Prove
Q
resultsin Key
points
are
q
and
prime
are
q
2.8 are
and
2.6
prime
numbers
numbers
true
that
if a
32
60
16
write
(ii)
56 and
(ii)
100 and
(ii)
42 and
/ =
pa
21
50
25
and / = qb. Prove that
[7 marks]
\\ (be)
that
In
section
we describe
greatestcommon
this
divisor
when
gcd(a
+ b,a-b)
is
[6
marks]
and
Euclidean
numbers
the
for
of numbers:
gcd(a,b)
If d = gcd(a,b) and/is any
a and b, prove that f\\d.
large
105
= 1.
q)
The
for
p and
and 15pq where p
If gcd(a,b)= 1 prove
either 1 or 2.
+G
85
and
(ii) 27 and 18
Let / = lcm(a,b)and
gcd
135
360 and
35 and
9pq3 where
and
42
and
35
(ii)
64 and 72
6p2q
of numbers:
(ii) 49
56
common
least
the
and
(ii)
56 and 81
28 and
(i)
180
and
225
(i)
the
2.9
KEY POINT
(b)
we get
prime,
relatively
other
= 1
then a \\ c.
common
divisor
[5 marks]
of
[5 marks]
algorithm
method for finding the
of two
one which is useful
numbers,
factorisation
is not feasible.
prime
another
2
^ q,
\302\243^%
Divisibility
and
prime
numbers
'
a-
*
(a
^<+^
Euclidean
appears
but
that
believed
It was
solve
is
it
had
been
at least a century
later discovered
both
in
by astronomers
earlier.
India
and
trying
to
needed
equations
The Euclidean
JF/T^
it
independently
China
*y
Euclid's
in
Elements,
known
algorithm
2.10
POINT
KEY
The
V
\342\200\242
Find
the
remainder
when a
\342\200\242
Find
the
remainder
\342\200\242
Find
the
remainder
this
\342\200\242
Continue
to make
for finding gcd(a, b) with
algorithm
this
rx.
when b is
divided by rx.
Call
this
r2.
when
divided
rx
is
at each
divided
by r2.
Call this
r3.
stage finding the
byrn
+
1.
is gcd(a,
remainder
non-zero
last
Call
rn is
when
\342\200\242
The
b.
process,
remainder
accurate calendars.
is divided by
a > b:
b).
I
an
orked
Use the Euclideanalgorithm
the
finding
example.
2.9
example
Start by
to
find
and
quotient
gcd(102,72).
step we
the next
Continue
are dividing
is no
there
until
102 =
remainder*
when 102 is divided by
For
72
1x72 +
gcd
is the
\342\200\242
30.
by
remainder
72 = 2 x 30+
on the#
penultimate
line
says
gcd(a,
b)
But we
+G
gcd (a,
\342\200\224
gcd
(b,a-
divided
b)
statements
This result will
\302\253^.
^^
in
used
to solve
Section
be
4C
Diophantine
equations.
well
W
Topic
10
-
in the
same pattern. We
together so that,
following
important
Worked
in
gcd(30,
can
example
12)
=
these
connect
2.9 we
gcd(12,
have
6) = 6.
common divisorof two
algorithm can be used to prove the
the greatest
Euclidean
the
result.
KEYPOINT2.il
\302\253^.
There exist
^^
The
\342\200\242V^
+ o
have:
we
So
provides:
argument
numbers,
Option: Discrete mathematics
^ q,vv,
a- qb is the remainder
= gcd(r1,r2)
as finding
Euclidean
illustrated
24
that
which we calledrx.
by b,
gcd(102,72)= gcd(72,30)=
As
q so
of
= gcd(b,rx)
continues
this
result from Key point 2.6,
of the
qb).
can choose a value
gcd(b, r1)
And
= 6
because
works
identical
An
0
that:
a is
when
+
.\\gcd(l02,72)
method
which
12
+6
= 2x12
30
remainder
This
30
72
12= 2x6
The
when illustrated with
much clearer
becomes
method
The
integers
algorithm
in the
m
and
n such
that ma
+ nb \342\200\224
gcd(a,b).
provides a way of finding
next example.
m
and
n, as
^
*
(*
orked
-
the previous
rearrange
and 72,
line to express
second
the
*<l+r
then
12
the
in
we
terms
of
want
an
= -2x72
+
= -2x72
+ 5x(102-1x72)
the only
is not
this
that
m = -19, n
= 27
chapter 3 that
line
penultimate
pair of such integers;for
is anotherpossibility.
are
there
In
we will
fact,
many pairs
infinitely
5x30
= 5x102-7x72
.*.
Note
30-2x12
= 30-2x(72-2x30)
and 72,
102
of
terms
we work back from
6 =
rearrange
so on. As
and 30, and
expression for 6 in
72
+ 72n \342\200\224
6.
example, we can#
to express 30 in
102
of
terms
line
first
the
that 102m
n such
and
m
From
V
2.10
example
Find integers
*
+
5, n =
m =
-7
example,
see in
of such numbers.
Exercise 2C
1.
Use
the
Euclidean
a =
35,fc = 12
to find
algorithm
gcd(a,b) in the following
cases:
(a)
(b) a = 122,fc39
=
a =
(c)
(d) a =
a =
(e)
+a
2.
For
each
42
= 80
320,fc
462,fo = 200
of numbers
pair
Use
Euclidean
the
from the
n such that
m and
integers
(a)
63,fc =
algorithm
(b) Find a pair of integersx and
86.x + 45jy = 1.
Let
that 48p
Use
= d.
gcd(48,30)
the
Find two
+ 30q = d.
Euclidean
are always
relatively
algorithm
previous question,find
ma + nb-
gcd(a,b).
to show that
that
y such
integers, p
to show
= 1.
gcd(86,45)
that
and
marks]
[6
marks]
q> such
3/c +1
and
13/c +
prime.
positive coprimeintegersa and b, show that
possible to find two consecutivepositive integerssuch
is a multiple of a and the othera multiple
of b.
Given two
[7
4
[4 marks]
it is
that
one
[4 marks]
2
^ q,vv,
\342\200\242V^
+ o
Divisibility
and
prime
numbers
*
^
(a
^<+^
Prime numbers
9
A
1 is
1 and itself.Remember
number, as it has only
a prime
not
two factors,
has exactly
number
prime
that
one
are not prime are calledcomposite.
Note
all
other
numbers
even
prime
prime number,
only
When
lemma
in
the
appears
\342\231\246y
a
Elements,
collection of
by Euclid of
known
for the
best
a thousandeditions
published
to
theory
most
one of
results.
related
KEY POINT
2.12
believed to
published
the Bible.
book after
lemma,
numbers,
results.
important
date,
must
more
two
many
it is
prime
at least
those numbers. For example,
600 15 x 40 and we know that 31600, so 3 must divide either
15 or 40.Composite
numbers
do not
have this property: For
example, 61600but it does not divide either 15 or 40.Thereare
With over
second
divide
a product of two
divides
if a
=
number
the
thenit
the
Elements also contain
be
that
formal
first
of geometry,
treatment
which states
is Euclid's
ones
important
Alexandria
around 300BCE.Although
the
odd.
are
to division,primenumbershave some
which distinguish them from composite
properties
interesting
results
mathematical
written
2 is
that
it comes
numbers.Oneofthe most
JfjJ^
Numbers
factor.
which
Euclid's
v
(i)
If p is
(ii)
If p
is prime
(iii)
If a
and
and b
be
will
You
proofs,
as
a
are
the
p
\\
and p \\ c2
then
or p | n.
p |m
then
mn
p
\\
c
and
p21
c2.
relatively prime and ab = c2,then botha
b are
numbers.
square
to use
expected
in
and
prime
these resultsto produceother
example.
following
orked example 2.12
that
Show
If we
+a
it is
put all terms
we
equation,
We
relatively prime.
cannot
that
\342\200\224
\342\200\224c2
n.
n2
an \342\200\242
be able to factorise it
may
to check
need
c such
and
side of
one
n on
with
a result
have
positive numbersn
to find
impossible
^> n(n-/l)= c2
about ob
that
Two
two
the
= c2, but
numbers
have any
n -1
n
are
are relatively
numbers
consecutive
and
we*
common factors
are coneecut'we numbers, eo they
Hence both n and n -1 are
prime.
square numbers.
E3utthere arenttwo
numbers,
eo such
n
coneecut'we
and
c do
ec\\uare
not exist.
Prime numbers are 'building
blocks'
of arithmetic,
as all
other numbers can be made
some
by multiplying
together
=
x
x
23
360
32 5; this is called
prime numbers. For example,
primefactorisation.
Every
whole
number
factorisation,and this resultis soimportant
a name.
26
Topic10
-
Option:
Discrete
rsM
^\\
^
<k
\302\243%*..
mathematics
has a
that
unique prime
it is
given
I \\
^
(a
-^^c+^
)
2.13
POINT
KEY
of Arithmetic
Theorem
Fundamental
Thereis exactly
one
a given
to write
way
number N > 1 in
form:
the
1-
N=p1p2...pk
...<pk are primenumbers.
where p1<p2<
so that
< < ... < pk for
prime factors p1 p2
is no danger of thinking, for example,
the
order
We
there
are two
and 3x3x5
3x5x3
1h
convenience,
that
of
Some
factorisations.
different
values can bethe same,and we usually write the prime
factorisation in a shortened form N = pilpp \342\200\242
\342\200\242
\342\200\242
pmm-
the p
It is
things:
positive integer greater
of prime numbers.
product
Every
2. Thereis only
one
than 1 can bewritten
mentions proving
a
as
the
prove this theoremusing a
methods of proof you already
know.
induction.
proved usingstrong
You can
explicitly
syllabus
The
Fundamental
of
Theorem
this.
to do
way
of
Theorem
Fundamental
the
that
really states two
Arithmetic
1.
to note
important
as an
Arithmetic
The first
statement can
be
strong
of
example
and
of facts
combination
induction.
orked example 2.13
that
Prove
every
integer
positive
two or moreprimenumbers.
is a
This
true
be
for
all
statement
positive
possible
it
+G
not
is
greater than
which should be#
integers, so it may
to prove by induction. As
n to
to get from
possible
a multiplication
number, we may need
n +1 using
use
the two cases
n<
related
smaller than
it,
be written
at
look
can apply
inductive
We
should
write a
the
hypothesis
conclusion*
i.e.
k: It
k is
k is
composite,
k
has
a factor
a,b<k,
2 is a
a.
Then
get k written
So the
all
true
a and b can
n
k we
be written
n =
can prove that it
for
all
as
we
together
of primes.
= k.
k; it is therefore true
this
call
k;
them
Multiplying
for
n = k.
for
this means that
a product
as
composite.
k = ab.
write
can
statement istrue for
n <
is true
by definition
so both
statementis
it is
or
prime,
other than 1 and
we
all
for
of primes.
statement
prime, the
of primes.
products
n =
product of
smaller than
integer
every
is either
If
for
2, as
is true
a product
as
If
k to two numbers*
so we
k,
2.
E3ut
have
n =
for
the statement
that
factor
We
is true
1.
^ 4
as a
written
induction.
strong
kfor some
Now
\342\200\242
be
number.
k can
strong
separately
by
prime
induction
Consider
Vroof
The statement
Assume
a prime
by
to
1is eitherprime,or can
Hencethe
2 and
if
it
is true
is alsotrue
n >
2
by
for
strong
induction.
2 Divisibility
^Sr.-v*
+<*..
and
prime
numbers
27
'
^
(\302\243
V
\342\200\242
+
<*ii+r
second
The
*
by contradic-
Proof
Hon was
part
of the
little moredifficult
introduced
from
in Section1A.
point
Key
to
Fundamental Theoremof Arithmetic is a
and we need to use Euclid'slemma
prove,
by contradiction.
with proof
combined
2.12
orked example2.14
that
Show
have
We
proved
already
a product of
written as
what happens if
were possible.
If
two
be#
can
some
common
factors
cancel
we should
don't
they
in
get
in
the
=
N
are
we
If
one
equal, any number that1
must also divide the other.
side
it's important
and
Euclid's lemma
prime,
is where
This
p}
is a
comes
in
that
remaining
reached a
have now
^---%.
conelder
of the
divide one
qe
by them and then none
are
pe
pv As
equal to
are
of the
it divides the
that
rience
E3y
Euclid's
lemma,
all prime,
the
p, must
so p, must be ec^ual
to
qe.
we
have
pe and qe are a\\\\ different,
it is impossible for N to
different
+a
hand
left
of the qe.
This is a contradiction, as
contradiction4
of the
any
equation, it must also divide
hand side.
right
factors on both
qe.
How
E3ut
We
divide
can
side
one
eome common
are
there
eldee we
way
are
sides
two
<q2
Sopip2---Pk =Wz---%
them
remaining
If
prime
q,qz...qm
<...<pkandq^
^p2
withpi
the#
=
p,pz...pkandN
of the
divides
different
two
hae
N
factorleatlone:
using
so
Suppose that
factorisations
This means that
factorisations,
a
as
by contradiction.
Froof
see
Now
primes.
such
two
N
that
proof by contradiction
are
there
is: any given integer N > 2 canbewritten
only one way.
of primes in
product
is unique, that
factorisation
prime
have
assumed
two
prime factorleatlone.
numbers are soimportant,
be nice to have an
the
easy way
finding
Unfortunately,
only way to check
whethera numberis primeisto try dividing
it by all primes
smaller than it. (In fact, you only need to divide by all primes
As prime
< \\fn;
f
we
Although
cannot
number
prime
is an
will be,
approximationto
many
given
28
there
how
primes there are up to a
size.
This result is called the
Topic10
-
Option:
it
limits
of
^
q,vv,>f
also
that
below). However, although numericalexperiments
suggest
there
are infinitely many pairs of consecutiveodd primes,
this
result
as the Twin prime conjecture) has not yet been
(known
proved.
It is
problems like this, with
proofs, which have kept
numbers
for centuries.
by prime
but elusive
Discrete
rsM
3$)
of
The
second
long sequences of consecutivecompositenumbers.
of these claims can be proved relatively
7
(see question
easily
next
of
understanding
and integration.
see why?)
Primes have a very irregular distribution:thereare pairs
odd numbers which are prime,and there
are
Prime Number Theorem,and
requires
functions
you
consecutive
predict
exactly how large the
can
it would
them.
of
yy\\
+ 0U
mathematics
relatively
mathematicians
simple
statements
fascinated
*
^
(a
v
^<+^
+
)
1<*
We
do
that there
know
are infinitely
prime
many
of this is attributed
The first proof
the
to
mathematician
Greek
is anotherexample
ofproofby
Euclid, and
numbers.
contradiction.
orked example 2.15
are infinitely
there
that
Prove
would
What
if
happen
many prime
were
there
numbers.
Supposetherearefinitely
only#
assume
finitely many prime number? We
it leads
that is the case and see whether
to
an impossible conclusion
a prime
How do we characterise
numbers:
\342\200\242
number?
This
means
than
1 is
that
either
of
either
we makea
that
number
divisible by any
of
the
N =
not
is obviously
ec\\ual
by
so
have
we
an incorrect
made
p-s?
J\\
Also,
N
so
pj3
remainder
gives
+G
our
Hence
assumption
many
infinitely
been
larger than all of
1 when divided
it is
^ pj3 as
N
each
somewhere
For many centuries mathematicians
have
it is
not divisible
science,
cryptography
and
distribution is so unpredictableand
is prime
is very
time-consuming,
banking.
whether
checking
primes
are used in
encryption. This encryption is vital to keeping
secure when it is transmittedelectronically,
for
and so
was Incorrect,
many
prime
numbers.
It
is also
there are
a number
numbers
example
over
states
which
length
like to
see some
examples of
use
of prime
4.
result is the
interesting
prime
remainder
Another
recently
proved Green-Taotheorem,
are
would
For example,
which give
arithmetic
you
numbers of
many
infinitely
to prove
infinitely
types.
1 when divided by
data
data
are
many prime
various
their
personal
possible
there
that
the Internet.
If
finitely
there are
in
interested
Because
there are
that
assumption
primes
by
of them.
any
by
them.
is impossible.
prime numbers, mainly becauseof the challenges
presented
their
But recently prime numbers have
properties.
by
proving
in many modern fieldssuchas
become
important
extremely
computer
or
values
p,p2...pk+/l
Clearly
an impossiblesituation/
have reached
p.x
them.
one of
least
at
of the
one
to
E3utthis situation
We
greater
every integer
Considerthe number
which does not satisfy'
the above conditions. How can
a number
to find
prime
P\\>P2\"-->Pk.
it is divisible
Try
many
numbers,
such
can find an
of
sequence
any given
that all of its terms
that
you
prime.
find
out
about
public
key
cryptography.
a2
2 Divisibility
and
prime
numbers
29
VS.
^^VNOC,
.ON\"
*
(a
'
a-
V
1+r
+
Exercise
)
2D
> 1.
(a)
(i)
(b) (i)
(i)
(c)
(a)
(b)
200
120
(ii)
172
(ii) 138
155
(ii)
as a
Write
846
What
is the
to
of the following numbers:
factorisation
the prime
Find
of primes.
product
smallest
1-
235
number
must be multipliedby
that 846
get:
a square
(i)
number?
(ii) a cubenumber?
that a
Prove
divided by
Q
prime number can only
Show
Showthat
if and
51 a
if 3|fcthen9|fc2.
if n
is an
[7 marks]
even number
greater than 2 then 2n
divisible
-1
[4 marks]
no prime
number p
2p +1
which
for
is a
[6marks]
10!+ 2
that
Show
5 when
only if 51a2
that
Showthat there is
square number.
(a)
marks]
[5 marks]
be prime.
cannot
0
1 or
remainder
give
6.
(a) Showthat
(b)
[6
is divisible by 2 and that
10!+
3 is
by 3.
(b) Showthat
10!+ 2,10!+
numbers
the
3,..., 10!+10
are all
composite.
+G
(c)
[8J
This
there
that
Show
question
leads
are 100
you
consecutive composite numbers.
through
one proof
that
v2
is an
irrational number.
(a)
Any
number
rational
gcd(p,q)
= 1.
can be
Show that if yfl
written in the form
= \342\200\224
then
P where
\342\200\224
41 p2.
q
p = 4/c, show
(c) Deduce thatp and q have
(b) By writing
V2
30
Topic10
-
Option:
Discrete
cannot
be written
that
41 q2.
a common
factor, and
as a ratio of two
hence
that
coprime integers.
mathematics
rsM
3$)
?\302\243*\302\273.
\342\200\242V^
+ o
.on\"
'
^
*
(*
1^
^1
^
v
vc+^
*
J
Summary
If a divides b
then
The division
algorithm
such
integer k. We
for some
b-ka
states that
for
a | b.
of numbers
pair
any
write
a and b
find
can
we
k and
0 < r <a
+ r.
that
b-ka
is a
/c-digit
1-
If
AT
number
with digits
N =
In
For any
integer
For
odd
any
If a | AT
The
n, an +
power
b
and
\\ N>
if a
and
common
greatest
-
n, an
power
bn
has
You
-
-
q
is any
N.
\\
a and
both
divides
which both a
number
(ab)
and b
b.
divide.
(a
gcd
b\\
results:
=
1
\342\200\224,\342\200\224
d)
V^
then
gcd(b,
(a, b) \342\200\224
a
\342\200\224
bq)
| N.
lcm(a,b)
be found usingthe Euclideanalgorithm:
\342\200\224
+
a
qb
r
a by b and b
Replace
Stop when
r, and
by
repeat
r =0
gcd(a,b)isthe
done
then
h
prove and usethe following
then gcd
integer
and b \\ N
if a | N
Write
It is
1),
gcd(a,b) = gcd(a,a-b)
gcd(a,b) can
+G
b).
w
if gcd(a,b) = d then
- if
-
- b).
smallest
is the
induction.
ab
how to
to know
need
proof by
is the largestnumberwhich
divisor, gcd(a,b),
1cm (a, b) -
b) x
and
factorising
and b have no commonfactors(otherthan
The least commonmultiple,lcm(a,b),
gcd(a,
use
{a
(a +
a factor
has
bn
we often
a factor
then:
+... + 102a2+10^ + a0
10fc% +10*-^.!
about divisibility
theorems
proving
...a2,a1,a0
ak,ak_x
of r.
value
previous
m and n, such that am + bmalways possibleto find two numbers,
the
Euclidean
by reversing
algorithm.
Ifp is
a
and
prime
and p
lip
is prime
If a
and b are relatively
The Fundamental
p
\\
\\
then
mn
c2 then
p
prime
p
\\
\\
m
c and
and
or p
\\
need
This
be
p2\\c2.
ab =
c2, then
both a and b
to use
strong
every
are
square
integer
induction and
numbers.
has a
proof by
unique prime
contradiction.
2 Divisibility
\342\200\242
can
n.
Theorem of Arithmeticstatesthat
factorisation.To prove it, you
gcd(a,b).
and
prime numbers
31
rsM
^
VS.
Q,
VVj
.on\"
*
(a
'
a-
examination
Mixed
o
Q
of digits
pairs
possible
(4alb) is divisible
by
a and b
Find
Euclidean
the four
digit number
[7 marks]
345m
that
+
that gcd(l
to show
algorithm
show
to
such that
m and n
integers
that
such
12.
(a) Use the Euclideanalgorithm
O Use
68 and
345 are relatively
prime.
68n = 1.
[7 marks]
1/c+ 7,5/c+ 3)= 1for
all
of k.
values
Q
2
practice
all
Find
(b)
[4 marks]
(a)
Write
a4
(b)
Show
that,
- b4
as a product of three factors.
and b are both
if a
odd numbers,
then
a4
- b4
is divisible by 8.
B
that
if d
[5
(a)
Show
(b)
Hence show that 312-1
\\
n then
(3d
-1) | (3n
is divisible
by
Show that if gcd(a,b) = d then
a
If gcd(a,b)
Q
(a) Prove that
the
(b) Prove that
the product of three consecutive
(c) Showthat
if n
and
= 1,prove
b be
that
of two
product
is odd,
two positive
gcd(qa,qb)
n3
- n
gcd(a,
b) x lcm(a,
(b)
gcd(a,
a +
that
26.
[6 marks]
-qd.
[4 marks]
[4 marks]
gcd(a,c).
consecutive integers
is divisible
integers
is always
is always
3$)
Topic
10
-
Option: Discrete mathematics
?\302\243*\302\273.
\342\200\242V^
+ o
divisible
by 24.
by 6.
[9marks]
b) = ab
b) = gcd(a, b)
[13marks]
IB
(\302\251
32
by 2.
divisible
integers.
(a) Showthat
Show
=
gcd(a,bc)
marks]
-1).
O
Q Leta
rsM
V
1+r
Organization
2005)
*
(a
*~
K^
+/\302\273
In
this
will
you
chapter
learn:
Representation
\342\200\242
how
of
\342\200\242
how
\342\200\242
about
that computer
scientists use
of
numbers
way
writing
using only
know
may
is a
This
in
those
bases
tests
divisibility
different
You
other
bases
in
to perform
arithmetic
bases
different
represent
integers
than 10
in
integers
to
in
bases.
binary numbers.
two
0 and
digits,
1.
a
with
16 digits. Throughout
They also use hexadecimal,
system
different
cultures
have
used
different
number systems.
history
In this chapterwe will learn how to write numbers and perform
arithmetic
bases.
different
in
How
many
count?
need to
fingers do we
write numbers: this meansthat
when
we write the number 635, the 5 representsfive
the
units,
3 represents
three tens (30) and the 6 represents
six
hundreds
So 635 means 5 + 30+ 600.This
is an example
of a
(600).
number system which usesplacevalue,
so that
the digits have
different meanings dependingon their positionin the number.
We
use
People
the
have not
different
+G
because
entire
always written numbers in this way.
numerals
in roman
example,
value
system to
decimal
symbols
system
we
50 and
for
V
stands
for
5 units,
For
and there
500. One advantage of the place
arithmetic,
easy to perform written
can work with individual digits ratherthan
is that
are
it is
with
numbers.
number system,as we move from rightto left each
is worth ten times more than the previous
one.
We say
place
that our number system has base 10and it has ten digits, 0 to
9. This system
because
we use our fingers
developed
probably
for
so
onto
the
next
counting,
moving
digit in the number
to using another pair of hands to count. It is
corresponds
In our
reasonableto imagine
write their numbers base
that
in
creatures
with
eight
fingers would
8.
system,we useten digits (0 to 9), in base
n we need n digits, 0 to n - 1.This means
that we can write
numbers 1 to n 1using just one digit. The number n is written
in
as 10,becausethe T on the left is worth n units.Forexample
base
number
8
8, only eight digits are needed:0 to 7. The
Just as in
the decimal
3 Representation
^ Q,VVj
of integers in
different
bases
33
VS.
.Ok\"
*
^
(a
^<+^
so on. Base8 number
63 corresponds
to our number 51, because3+ (6 x 8)= 51.
The highest
is 63 in
two-digit number in base 8 is 77,which
=
the decimal system (because7+ 7x8 63).To write
numbers
in
than
this
base
8
we
need
three
so
100
represents
larger
digits,
number 64. As we move from right to left, each place is worth
so the base eight
one,
eight times more than the previous
number 254 representsournumber 172(4 units, 5 eights and 2
sixty-fours: 4 + 5 X 8 + 2 X 82 = 172).
is then written
cultures
history
have
used
systems
with
Throughout
different
system
present
4fJ\302\245i
different
most
decimal
the
purposes,
duodecimal (base
in
Mf
number
bases, with bases
12, 20
and 60 being most common.
we now use the
Although
for
12) system
time:
measuring
is still
12
24 hours in a
recently, bases 2, 8 and
16 have been used when working
with
Base 2, or binary
computers.
months
day.
in
and
a year
9 is
11, and
So (63)8= (5l) , (77)
=
(63)
When reading out
them digit by
not
three',
they are related to base 2 (for
example, a group of four binary
digits corresponds to one
digit,
the numbersin
so for
other
then 10 we say
would
be readas 'sixexample (63)
bases
'sixty-three'.
numbers in bases greaterthan 10 we use letters to
in the hexadecimal
the
extra
represent
digits.Forexample,
system
the letter B stands
for 'digit'
11, so the hexadecimal number (3B)
the
decimal
(as 11 + 3x16 = 59).
represents
number(59)
we write
When
hexadecimal digit).
orkedexample3.1
Write
It
in base
(1632)7
to work
is easiest
10.
from
to
right
start
The
The second placerepresents
left,
third
+a
(1632)7=2
as we#
2
units digit is
+
3x7
+ 6x72+1x73
= (660\\0
units
with
there
sevens;
3 of
are
The
them
forty-nines ( 72);
are 6 of them
represents 73 = 343
represents
place
there
The
next
Exercise
place
3A
1.
all the
down
Write
(a)
(i)
base
6
(b) (i) base12
2.
Write the
(a)
(i)
W
(i)
(d)
(i)
(e) (i)
(f)
34
3$)
Topic
10
-
Option: Discrete mathematics
^ cl
\302\243%*..
(i)
digits needed
(23),
to write numbersin:
(ii)
base 4
(ii)
base
followingnumbersin
(b) (i) (471)8
rsM
and
(172)10=(254)8.
because
used
number
the subscript.
the base in
number system, uses only two digits
to the
(0 and 1), which
correspond
two states (on and off) of electronic
switches. Bases 8 (octal)and 16
are
it
bases,
10,
confusion when talking about numbersin different
is conventional to write the numberin brackets
with
To avoid
More
(hexadecimal)
v
base
15
10:
(\")
(62),
(ii)
(266),
(286)16
(ii) (591)16
(4A)lfi
(ii) (C7)16
(DA6)16
(ii)
(1B4L
(B25)12
(ii)
(1BB\\2
'
* ^
(*
largest base
is the
What
1^
10 number that
^
as a
be written
can
three-digit number in base5?
many numbers
but 4
digits when written
from
convert
To
where
base
in
9?
base
marks]
12,
[5 marks]
example of
see
^*M0
base
to
conversion
^py.
10 we just needto remember
to base
For an
bases
different
^^
Worked^**
3.1.
example
place values are powersofn.
10 to another base, n, we needto write
form a1+a2xn + a3xn2 +a4xn3...
are digits between
0 and n-\\.
base
from
convert
in the
number
the
base
any
n the
in base
that
in
between
Changing
To
[3
require 3 digits when written
How
v
vC+^
^1
a^, a2 > a^...
The next example
illustrates
how
Worked
1.5
this.
to do
in
showed
We
that
can
example
integer
every
be
written
<^[ base 2. We
can
in
show
integer can
be converted into
<^[
that any
any
base,
using
a
similar proof.
orked example 3.2
Write
in base
(862)
7.
Write out powers of
The
largest
are
There
power of
two 343s
7
in
that
so the
862,
is 343
862
into
goes
7#
of 7:1,
Fowere
\342\200\242
562
7, 49,
343, 2401
=
\342\226\240*\342\226\240
343
2, remainder176
first digit is 2
There is 176 remaining
+G
are
There
three
49s
in
How
2nd digit is 3 \342\200\242
so the
196
many
7's go
into
next
digit
The
There is 1 unit
KEY
a base
To convert
The
the
\342\200\242
Repeat
largest
left
1
is 4
units digits is 1 \342\200\242
down the digits, from
remainder
to right
\342\200\242
1-5-1 = 1
(062)1O=(2541)7
the
the
10 numberinto
base
power of n that
first digit.
n:
goes into the number.
is the
quotient
\342\200\242
Divide
find
29-^7= 4,
remainder29
3.1
POINT
\342\200\242
Find
just write
now
We
so the
left,
\342\200\242
29?
176-^49 = 3,
remainder
next
until
digit.
division
by the
previous power of n to
by 1 has
been done.
3
^ q,vv,
Representation
of
integers
in different
bases
'
^
*
(*
1^
orked
^
from one
To convert
V
vc+^
base to any other base,it is easiestto go
10.
base
via
^1
3.3
example
Convert (9Al)n to base5.
out
Write
with
Start
10
\342\200\242
of 5 up to 1 200
\342\200\242
to base
convert
First
powers
(9Al)fl
Fowere
have
we
remainder
and
0, we
divide
the
-s-
25
= 3,
0-5-5 = 0,
continue \342\200\242
0-5-1 =
by 5 and 1
Read off
5, 25,125,625
575 +125 = 4,
75
Although
of 5:1,
=
= 1,
\342\226\240*\342\226\240
the largest
1200,
x 11+ 9
1200 625
power of 5 below\342\200\242
and do the divisions
x 112
= 1 +10
(1200)1(
remainder
575
remainder
75
remainder
0
0
remainder
0
\342\200\242
digits
.\342\226\240.(9^=04300)5
3B
Exercise
followingbase 10numbers
(a) (i) 62 in base2
(ii)
(b) (i) 183in base7
(ii)
186
in base
3
212
in base
5
(c) (i) 821in
(ii)
601 in base
(ii)
4632 in base
1. Write the
base
(d) (i) 1199in
base
11
16
2. Convertthe following
+a
(a) (i) (I100l)2tobase3
(b) (i) (5Al)l6 to base 2
(c) (i)
(d)
Given
(i)
(31442l)5tobasel5
to base
(A5A)l2
96 =
that
531441,
16
(ii)
15
20
to base
(210012)3
(ii)
base:
given
given base:
(DB3)15
to
base
2
3
(ii) (30132)4to base16
(ii) (F0F)16to base 14
write 531440 in
in
Arithmetic
As we have
to the
numbers
the
in
different
base 9.
[3
marks]
bases
one big advantage of a place
to
arithmetic.
system
easy perform written
However
with
large the numbers are, we are always
working
Think
in
individual
how
add
base
10:
add
pairs
digits.
you
you
of digits, and if the answer exceeds9 you
to the next
'carry'
to add in any other base.
position. We can use the sameprocess
value
already
is that
mentioned,
it is
For example, working in base7
36
Topic10
-
Option:
Discrete
rsM
3$)
^
q,vv,
\342\200\242V^
+ o
mathematics
you
'carry'
when
the answer
'
a-
*
(a
V
1+r
working in a baselargerthan
to 'translate' letters into numberswhen
calculating.
6. When
exceeds
orked
the
may
need
.
+(C19)
\\
^16
Calculate (3A8)
\\
/16
add
we
3.4
example
First
10,
units
this
(Calculate
digits
in
base
10)
\342\200\242
9 =
>+
= 15+ 2
#\302\253
answer
Is the
larger
15?
than
write
pair of digits, plus
the next
Add
represents number
that A
Remember
above.
12
Add the next pair
of
in base
C
is
\342\200\242
15 is
F in
base
2,
10 + 1+
carry
1
1= 12
10
16#
16
C
write
3 +
for number 1 2#
C stands
digits.
from
carried
one
the
17
\342\200\242
12 =
write
15
F
.\342\200\242.(3AS)16+(C19)16=(FC2)K
You can
also set out your
orked
answerin the traditional
format.
column
3.5
example
Calculate (62) x(35)
.
X
2x5
=
10
= 7
+ 3,
so write
6x5
3,
+ l =
31 =
0
are
multiplying
as we
by the
*
'sevens'
Add
(433)7
+ (2460)7
as
in
the
=
5
3
3
2
4
6
0
digit
=6
3x2
6x3 = 18
2
3
1
(43)7
+G
Write
4
1.\342\200\242*
carry
6
(24)7
previous
\342\200\242
3 2 2
1
3
1
example
.(62)7x(35)7=(3223)7
EXAM
HINT
idea to check your answer by converting
to base
in
also
be
able
to
use
calculator
to work
may
your
some common bases, but you must show your full working
and only use the calculator to check your answer.
It
is a
10.
good
You
3 Representation
\342\200\242
rsM
?\302\243*\302\273.
of integers
in
different
bases
'
^
*
(*
v
c+^
which
number
about its factors.
other bases.As
can be askedto prove
orked
if a
the
digits
Let
ends in 0, 5 or A
15 number
base
be o]f
a2, o3...,
with
the
All terms
number can
For example,a base 10
these tests,
as using
well
similar
derive
can
you
them.
3.6
example
Show that
of a
is divisible by 10.We
in 0
ends
tests for
divisibility
at digits
looking
us something
tell
often
saw that
2 we
In chapter
except for the
units
=
\342\200\242*
(...a5a2a^)15
a5 x 152+.
x 15 +
+ a2
ay
digit
divisible
are
first
o1 being
by 5.
divisible
it is
then
= 15(a2+15xfl3+...)+^
by#
15
Look at
first
the
the last
and
part
digit\342\200\242
The
was
9
by
in
the
whole
by 5.
The next exampleshowsa resultwhich
is analogous
to the
base 10 test for divisibility
by 9. The proof follows the same
test
proved
Worked
which makes
by 5,
divisible
number
^y*
^^
part
If ^ is 0,5
is also divisible
separately
The divisibility
is always divisible by 5.
or A (which stands for 10), then it
first
^y*
^^
example
strategy.
2.2.
orked
3.7
example
Show that
if a
7 number
base
is divisibleby
6, then
the sum
digits is alsodivisible
of its
6.
by
+a
Write
If
is divisible
this
take
of
out the number using
a factor
out
its
by 6,
of 6.
also
We
of
-1)
as(7
We
want
the
(...), so
they
+ a5
+...
x72
+an x
=
(#! +
the sum
&2
+
+ %
7\"-1
)+
\342\200\242
\342\200\242
-an
+ a3(72-l)
[a2(7-l)
-1 can be factorised
+ ...+
an(7\"-1-l)]
all
have
a factor
... + an
x6(...)]
\342\200\242
=
of 6
that the sum
+
(fl1+fl2+fl3+...fln)
[a2 x6
of the digits, \342\200\242
first bracket, is divisible by 6
to show
is the
which
form 7k
+a2 x7
the expression into
on and the rest
o}+o2+...
The terms
want
=a,
...a^a^
(anan_^
\342\200\242
we should be able to#
so separate
digits,
digits
As the
divisible
+ a5
terms
by 6,
in
if
x6(...)
the
the
6, then
divisible
by
divisible
by 6.
+
bracket
square
whole
number
the sum
are
all
is also
of the digits
is
i
j
38
W
Topic
10
-
Option: Discrete mathematics
?\302\243*\302\273.
The
number is divisibleby
of its digits is divisible
You are
n -
by
3C
1. Do the following
answersby
(a) (i)
(b)(i)
10.
to base
converting
(324)6+(ll5)6
(ii) (152)6+(214)6
(6A2\\2+(BBl\\2
(ii)
(A61)12+(13B)12
following calculationsin the given
answers by converting to base 10.
(i)
(b)(i)
(a)
(ii)
(2C)16x(4l)l6
(ii) (53)16x(A8)16
ends in 0, what
7 number
(c) In base6,what
divisible
by
(36)
(3A5/c)14
divisible
Find
of
digit
k so that
a base
the number
to be
14 number
values of digits k and m
(3Am/c)14 is divisibleby (91)10.
the possible
(a) Show that
number,
the last two
digits
divisible
9.
by
so
if the last digit of a base 15number
C, thenthe numberis
Show that for a base 15
divisible
t*J
number to be
divisible
13?
number
(b)
for a
by?
is it
by 7.
(b) What is the criterionfor
(c)
numbers
divisible
?
values
(a) Find the possible
is
it be
must
0, which
criterion
is the
your
(5l)7x(35)7
(b) If a base 12numberendsin
divisible by?
by
Check
base.
(22)yx(l6)7
base
If a
Check your
given base.
in the
calculations
Do the
(a)
Q
and only if the sum
test.
Exercise
+a
1 if
n - 1.
often askedto prove particularcasesofthis
divisibility
3.
applies
3.2
POINT
A base n
2.
illustrates a divisibilitytest that
example
bases.
in all
KEY
V
i+s*
Worked
last
*
a-
*
(a
is divisible
by
that
the
[8 marks]
is 0,
if the
^ q,vv,
\342\200\242V^
+ o
9 or
number
formed
by
by 9, then the wholenumberis
[8 marks]
3
'
3, 6,
3.
Representation
of
integers
in different
bases
\\
Summary
\342\200\242
To
an integer
represent
in base n
we use digits 0 to n -
\342\200\242
To
\342\200\242
We
convert
can
from
perform
base 10
to base n, divide
written
addition and
method as in base 10.
\342\200\242
Divisibility
For
)
criteria
the
are similar to
number is divisibleby
number
place
by powers
multiplicationin any
of n look at
by factors
divisibility
A base n
in base n
The
are 1, n,
values
n2,
n3,
etc.
~ai +CL2n+ a3n2 + ... + aknk~l
(akak-i\"M3a2ai)
\\
1.
n -
base,
of n, starting with
using
the same
the largest.
column
those for base 10:
the last digit.
1 if
and only if the sumof its
digits
is divisible
by n
-1.
I
r
1
40
t>
Topic
^
10
-
Option: Discrete mathematics
VS.
> Q,VVj
^%o<,
^
.
I^
.On
* a-
(a
*
^
1
Mixed
examination
o
10.
in base
(136)8
-
to base16.
(b) Convert (l439)10
Q
Convert
(a)
(b)
O
B
8 digits
is divisible by
is also divisible
by
answers to parts (a)and
not divisibleby 7.
(a)
Convert
(b)
Working
(c)
Convert
(a)
Show
if a
that
its
to show
(b)
that 41012
is
[11 marks]
your answerto part (a).
to part (b) to a base 10number.
5, square
answer
your
of
84 from base 10to base5.
the number
in base
7 thenthe sum
number is divisibleby
in its
3 then
In
(c)
Q
Given
that
values
of a
(a)
Convert
(b)
Prove
The positive
base
a base
possible
9 number
8415 to a decimal number.
9 number
is divisible
is expressedin
that
N is
divisible by b
(b)
Show
that
N is
divisible
Q The positiveintegerNis
Show
that when
(b)
Show
that
when
the sum
of its
[8 marks]
Show
(a)
4 if
by
4.
by
integer N
by
if and
b2
expressed
p = 2, N is even
p =
{akak_l..
.axa\302\247)y
if a0 = 0.
only
if and
in
b as
base
only
if a0=a1
base
p as
if and
only
3, N is even if and only
\\anan_x
if its
if the
= 0.
..axaQ)
^
q,vv,
\342\200\242V^
+ o
.
sum
digits is even.
[11marks]
3 Representation
\342\200\242
[10 marks]
last digit is 0.
IB
(\302\251
rsM
5.
[12 marks]
(a)
of its
the
by
and b.
digits is divisible
Q
is divisible by 15find
(2al04b)6
the
that
i
down a criterion for the numberto be divisible
6, write
base
[9 marks]
representation
in base 6 the last digit is 0 or 3.
(b)
marks]
7.
Use your
(c)
[8
8.
41012 in base
number
if a
that
Prove
base
marks]
(l 1000l)2to base7.
the number
Write
(a)
[7
to binary.
[AB2F)l6
(b) Convert
Q
V
i+s*
\\
3
practice
Write
(a)
^
2008)
Organization
of integers
in
different
bases
'
*bL
kh
v
^ ^1 vc+^
1
y>
In
this
you
chapter
learn:
will
\342\200\242
how
to
Linear
a
tell whether
linear equation
has
variables
in
1-
two
Diophantine
integer
solutions
\342\200\242
how
to
one
find
such an
solution of
equations
equation
\342\200\242
that
1H
one
given
solution,
to construct
is
it
possible
Many
infinitely
only
many other solutions.
equations which need
Such equations
are called Diophantine
we will learn how to solvelinear
involve
problems
practical
solutions.
integer
equations.In
this
Diophantine
equations
results
in two
using some of the
variables
2.
in chapter
met
we
chapter
solving
w
h
with
of equations
Examples
integer
solutions
Can
2x + 3y = 5?
the equation
solve
you
and y are real numbers, then for every
satisfiesthe equation.Sothereare infinitely
/ 5-2x^1
form
of the
Xy~
3
V
integers
then
of x. We
can
(1,1),
(4,-1),
pairs,
but
J
5-2*
y =
will
several
find
if we
However,
only
solutions
be
can only usethe values
of x for which
of experimenting shows that x has
+a
of
therefore
can
(3/c +1,1
form
the
- 2k)
for
write
some
for
values
example
solution
many
y an
make
integer;
5 - 2x is divisible
A bit
multiple of 3. We
for some
by inspection,
still infinitely
not all integer valuesof x will
to be
x and y
an integer
There are
(-2,3).
solutions
many
require
which
can find y
x we
If x
we
3.
by
one more than a
to be
of the solutionsare
that all
k.
integer
is an exampleof a linearDiophantineequation;
in several variables where we only
seek
equation
The above
a
linear
solutions.
integer
only linear Diophantineequations
You will meet
but there
are other types.
the solutions,but
solutions
it may
and
exist,
It is
often
be possible
if they
in
do, whether
course,
to find all
impossible
to prove
this
whether or not
there are infinitely
many
of them.
For example, we can show that the equation
x2 - y2 = 122
has no integer solutions as follows.The expression
on the left
-
y){x + y). The two
is an even number, so they must be
as (x
factorises
odd.
42
rsM
Topic 10
-
Option:Discrete
As
their
product
is even,
factors
either
differ
by 2y,
both even
they must both be even.
which
or both
But
mathematics
VS.
^
Q,VVj
*
^
(a
v
^<+^
+
of two even numbers is divisible
the product
is not.
factors
Hence suchnumbersx and;/cannot
and remainders is very useful
when
by
4, which
exist.
Looking
solving
)
122
at
Diophantine
equations.
orked example 4.1
the
that
Show
a
\\5x
factors
is often
of this
terms
\342\200\224
equation
for
Looking
1-
the
= 17
useful:
equation are
then
number,
35y
must
15and
35
Hence
15x
be
\\a\\\\
number
by the same
divisible
two#
divisibleby
term
third
if
integer solutions.
has no
x and
V is
known
x2+
= z2.
y2
Fermat famously
his theorem. Many
was only accomplished
applications so far,
for
y.
not divisible
by 5,
so 15x
- 35y cannot
extremely interesting and
in
a
over
a
is
of a
book, that he
centuries tried to find
margin
the
a proof for
but this
proof,
had
a
by Andrew Wiles, building on new theoriesdeveloped
by
decades. Although
the
result
itself has not found any
over preceding
theories
turned
out to be
developed
during the search for the proof have
in 1995
new
the
claimed,
mathematicians
mathematicians
several
useful.
4A
Exercise
Explainwhy
+G
by 5
example of a Diophantine equationis
infinitely many solutions,known
In 1637,
jfjk^*
5.
by
This has
as Pythagorean triples.On the otherhand, the equation
This
xn + yn - zn with n > 3 has no integersolutions.
famous
result called Fermat s Last Theorem.
+}[
both divisible
- 35y must be divisible
17.
ec^ual
A well
are
1H
the
6x +
equation
solutions.
9y = 137 has no integer
[3
of the equation15.x
-12y
marks]
(a)
Find
(b)
Show
solution
one integer
pair of numbers of the form x =
solution
of the above equation.
that every
= 1 + 5/c
y
is a
=
2 +
18.
4/c,
[4 marks]
(a)
x =
that
Show
1+
equation 1Ix+ 3y
(b) Showthat only one
3/c and
= 14
y
=
for all
of those
l-llk
satisfy the
integers k.
solutions has
both x and
y
positive.
[5 marks]
Darya has
pieces and
15 piecesof paper.She
cuts them into 10 pieces
some of the new
each.
says
she
must
She
that
small
that
pieces
she now
have counted
selects
each.
several
of those
She then
selects
and cuts them into 10 pieces
has 2007 pieces of paper.Show
incorrectly.
[6 marks]
4 Linear
^
Diophantine
equations
43
vs.
Q\342\200\236
\302\243*W
^
*
-+
prime
positive
every pair of relatively
m> n, the numbersx \342\200\224
m2-n2,
m and n with
integers
\342\200\224
2mn
y
x2 + y2
z
and
\342\200\224
m2+n2
(This
so do
kc for
any integer
k.
are there?
How many solutions
We
triple, then
a Pythagorean
c form
and
and
kb
ka,
generating infinitely
triples).
if a, b
(b) Show that
the equation
satisfy
method for
is a
\342\200\224
z2.
Pythagorean
many
V
*<l+r
that for
Show
(a)
[fT]
*
for solving linear Diophantine
in
Before we search for solutions,
we
to find out whether any exist. It turns out that there is a
equations
need
at techniques
two variables.
look
now
this.
very simple criterionfor deciding
4.1
POINT
KEY
\342\200\242
The
\342\200\242
If
gcd(a,b)
I
c, the
c.
has infinitely
equation
no
has
c
by\342\200\224
does not divide
if gcd(a,b)
solutions
ax +
equation
Diophantine
many
solutions.
In Section4C we
\302\253^.
^^
will
the
left
has
to
that
can
be found
starting
really says two things: that
we can find a solution; and that
part
gcd(a,b)
I
c
there are in fact
how to find one solution,and
solution then
\302\253^.
one
just
from
of this
part
second
The
all solutions
see
result is straightforward:If both termson
hand
side are divisible by a number, then that
number
divide the right hand sideaswell.
first
The
^^
particular
solution.
solutions
many
infinitely
To find one solutionwe
They
the 7th
were
also
studied
n such
and
n are
studied
been
is probably
record
known
BCE.
have
equations
Diophantine
earliest
and
in
that am
solutions
by Sun Tzu
in
history
3rd
use
the Euclidean
there
that
century
from
question of
algorithm from
there are numbers m
if c = gcd(a,b), then m
all over the
and
is one
We will first see
c.
world. The
the 3rd
century
AD and Brahmagupta in
*^
J^S^L
*
*?*
+
century.
Ifc^d
but dIc, then
is equivalent
to ax
a (km) + b(kn)
c-kd
+ by
\342\200\224sox
equation ax + by
Topic 10
next section.
saw
Arithmetico
the
many.
to the
+ bn \342\200\224
So
gcd(a,b).
of the equation ax+ by-
throughout
Diophantus'
return
if there
in the
can
the previoussection.We
infinitely
will
whenever
-
Option:
Discrete
mathematics
kd,
- c.
for some
\342\200\224
kd.
Since
= km
integer k. So the equation
am + bn
and y
\342\200\224
it
\342\200\224
kn are
d,
follows
solutions
that
of the
orked
'
^
*
(*
1^
whether
Check
27.x +
that
such
find one
Hence
V
vc+^
4.2
example
(a) Find x andy
(b)
^
^1
15;/
\342\200\224
3.
integer solution of the equation
gcd (27,
21 x
Euclidean
algorithm
\342\200\224
21.
[a) gcd(27,15)= 3, sowe
15) divides the#
RHS
Apply
+ 15;/
by
\342\200\242
27 =
Euclidean
can
find
a solution
algorithm.
1x15 + 12
15= 1x12 3
+
+0
12 = 4x3
the
Reverse
steps
to write 3
= 27x
+
3 = 15-1x12
15ym
= 15-1x(27-1x15)
=
+ 2x15
-1x27
So x = -1,y
21 =
As
from
the
3 x 7, we can use the answer'
the
previous part and multiply
whole
equation
by 7
next section
In the
linear
we will
see
how
1. Which of these equationshave
7x +
+ 15(2)
=3
by 7: 27(-7)
Multiply
So x = -7, y
+15(14) =
21
= 14
solutionsof
all the
4B
Exercise
(a)
27(-1)
equations.
Diophantine
+a
to find
(b)
= 2
8;/ = 29
(c) 15*+ 27;/= 3
(e) 15*+ 27;/= 35
2. Use the Euclidean
algorithm
solutions?
integer
+ 6;/
(b)
3x
(d)
15*+
(f)
2bc
to find
= 17
27;/=
+ 37;/
30
= l
one solution
of the
followingDiophantineequations:
45* + 20y = 5
(a)
(i)
(ii) 66*+ 42y = 6
(b) (i) 102*+ 72;/= 6
(ii) 165*+105;/=15
3.
Q
Find
one
solution
(a)
(i)
66*+
(b)
(i)
81*
20;/=
+ 36y
of the
following Diophantine
12
= 27
(ii) 45*+
equations:
20;/= 25
(ii) 100*+ 35y
= 15
(a) Findgcd(45,129).
(b) Henceexplainwhy the equation
has no integersolutions.
45.x +129y
= 28
[5
marks]
4
l
^
^SV^
+ c,
Linear
Diophantine
equations
'
* ^
(*
v
c+^
Find
(a)
gcd(132,78).
(b) For which valuesof c doesthe
have
one solutionfor
g (a) Forwhich
\342\200\224
c
solutions?
integer
(c) Find
+ 78y
132.x
equation
c =
of X
values
6.
[10 marks]
does the
equation
Ajc
+ 5y
= 7
have
integer solutions?
(b)
for X =
solution
one
Find
14.
[8 marks]
the general solution
Finding
how to find one solutionofthe
divides c. For example,
equation
gcd(a,b)
we found that the equation
27.x + \\5y - 3 has a solution
x = -l,y = 2. This is called a particular solution. But there are
other solutions, for example x = 4yy = \342\200\2247.
How
can we find all
such solutions?
In
the
+ by
ax
In the
above
needs
to be
we saw
section
previous
\342\200\224
c when
if x
equation,
for the expressionon the left
So let x = -1+P,y = 2 - Q. Then:
decreased
same value.
to
keep
the
27(-l + P) + 15(2-Q)= 3
-27 + 27P + 30-15Q = 3
<=>
<=>
= 0
27P-15Q
27P = 15Q
<=>
= 5Q
9P
<=>
The last
integer
P = 5/c and Q-9k
for
equation is satisfied
by
any
/c. Hence the original equation has solutionsofthe form
+G
x
of a
solutions
+ 5k, y
= 2-
2k for k e Z.
line of the abovederivationwe divided
This gives us the generalform
for
all
linear Diophantine equation.
4.2
POINT
is one
If (x1,y1)
d
-1
is gcd(27,15).
3 which
KEY
=
the last
on
that
Note
by
by a certain amount,y
is increased
\342\200\224
gcd(a,b)
kb
x-xx-\\
yy
=
d
This is
solution of the equation
all the
then
called the
ax
other solutions
+ by-c
and
are given by:
ka
tor
yx
ke\302\243
d
general solutionofthe Diophantine
equation.
We
now
equations.
46
rsM
Topic10
-
Option:
^
q,vv,
Discrete
mathematics
have a
complete method for
solving
linear
Diophantine
orked
'
^
*
(*
1^
result from Worked
To
find
solution already
one
found
have
to find
4.2
example
27*+15;/=21.
We
^
V
vc+^
4.3
example
Use the
^1
gcd(o,
now
can
We
\342\200\242
need#
solution we
the general
the general
solution of the equation
One
solution
qcc\\
(27,15)
\\e x
= -7,y=
14
= 3
b)
use the formula
x=
\342\200\242
= -7 +
-7 +
5/c
3
*y = 14
27k
3
= 14- 9k
Exercise 4C
1.
Find
the
of the
solution
general
following Diophantine
equations:
(a)
(i)
45*+
(b)
(i)
102*+
(c)
(i)
66*+
(d)
(i)
8bc
(a)
Use the
6
72;/=
12
20;/=
+ 36;/
(ii) 66*+ 42;/= 6
(ii) 165*+ 105;/= 15
= 5
20;/
(ii)
(ii) 100x+ 35;/=
= 27
of
solution
162*-78;/=
Show
a general
equation
[11 marks]
38 are
relatively prime.
solution
equation55.x- 38;/ 1.
that 55m
n such
m,
integers
Find
the
12.
that
55 and
(a)
(b) Finda pairof
(c)
15
Euclidean algorithm to find gcd(162,78).
(b) Findthe general
+G
45*+ 20;/= 25
of the
- 38n= 1.
Diophantine
=
Find all
pairs of points with
[11
plane which lieon theline 3x+ 5y
2\302\273J
Solutions
subject
Sometimesa Diophantineequation
there are somerestrictions
on the
One of the mostcommon
constraints
solutions to be positive.
t*J
'
in the
coordinates
integer
to
= 13.
x-y
[8 marks]
constraints
is set
types
marks]
in a
context where
of solutions
is requiring
allowed.
all the
4
?3n
Linear
Diophantine
equations
orked
'
^
*
(*
1^
the
Notice
by
15 to
has only
the whole
divide
can
we
that
45;/ = 300
75.x +
equation
equation
^
V
vc+^
4.4
example
Show that
^1
both x
with
solution
one
5x
\342\200\242
+
and y positive.
3y=20
make the numbers
smaller
of
the general
then look for
solution
find
can
We
k and
the
in
which x, y
In
this
we can
case,
inspection.
(If
you
find
don't
solution
one
see
it, use
solution:
Particular
terms\342\200\242
of k
values
x = 1,y=5
for
>0
solution:
General
by
the
x = 1+ 3/c,y=3-5/c
Euclidean algorithm)
Use
the
condition
x >
=1
god (27,15)
x>0=>3/c>-1=>/c>0
0, y > 0*
y>0^>5>5/c^>/c<1
= 1, so
.\\k
is
there
only
one
solution
positive
(* = 1,y=5).
Exercise
4D
(a)
one
Find
pair
(b) Find all the solutions
with
x, y
x, y such that
of integers
of
the
equation
integers.
positive
5x
5x +
= 42.
+ 8y
8y = 42
[11 marks]
(a) Find gcd(62,74).
+a
2.
equation 62.x- 74y \342\200\224
show that the above equation has infinitely
many
with both x and y positive.
[13marks]
the
(b)
Find
(c)
Hence
solutions
(a)
Find
the
general
solution
of the
general
solution
of the
132*+78;/=6.
(b)
[4~J (a)
Find
You
all the
a pair
have
with | x
solutions
of scales with
12 g and 9 g weights.What
measure using these?
(b)
Could
you
are
weights
(c) If you
the
48
Topic
10
-
Option: Discrete mathematics
^ q,vv,
are
scales,
measure
how
allowed
does
< 20.
weights
by 13 g
[14 marks]
an unlimited supply
more different
replaced
only
\\
Diophantine equation
of objects
of
can you
weights if all the 12g
ones?
to put the weights on one sideof
that affect your answerto part (b)?
'
^
*
(*
1^
^
^1
V
vc+^
*
J
Summary
\342\200\242
The
\342\200\242
One
m,n
equation
Diophantine
solution
by-c has integersolutions
solution)
(particular
that
such
ax +
am + bn
can be
= d; then one solution
found by
is x1
=
if and
reversing
the
c
c
solution
general
\342\200\242
Sometimes
there
is x
= x1H
are constraints
, y
on the
=
y1
if d
Euclidean
- gcd(a,b)divides
algorithm
c.
to find
1-
\342\200\224m>y1=\342\200\224n.
d
d
\342\200\242
The
only
for k
solution so that
1u
e Z.
only
some
values
of k
are allowed.
+G
4 Linear Diophantineequations
\342\200\242
rsM
49
vs.
^^N*.
.on\"
*
(*
'
^
1^
examination
Mixed
Find
(b)
integers
Use the
such that
m and n
+
= 41
that 27
and 59 are relatively
Diophantine
[13 marks]
to the
of solutions
and justify
common divisor of 7854and
the greatest
find
to
diophantine equation
your answer.
[7marks]
IB
(\302\251
(a) Find
the
why
(a) For
of X
(b) Find the generalsolution
find
50
Topic10
-
Option:
that
^
q,vv,
there
102p +
Discrete
rsM
3$)
a general
\342\200\22417
has
no
integer
for
solution
solutions.
equation
equation Xx + 3y
X =
such
= 5
have
4.
that
of the
integer
solutions?
[9 marks]
5m +
3n = 17
equation 5x+ 3y
= 17
x, y integers.
(c) Find the two
that
2008)
[13 marks]
does the
(a) Find a pair of integersm, n
(b) Hence
the
\\y
x, y e Z.
\342\200\224
15 with
values
which
45.x +11
equation
of
solution
45.x +111 y
Show
Organization
gcd(45,lll).
(b) Henceexplain
(c) Find the general
with
prime.
= 1.
+ 27n
20.
number
the
59m
the
of
EuclideanAlgorithm
state
to show
algorithm
(c) Find the generalsolution
59.x + 27y =
equation
3315.Hence
7854.x 3315y
V
vc+^
4
practice
(a) Usethe Euclidean
^
^1
\342\200\242V^
+ o
72<j
solutions
are no
= 1800.
mathematics
which
minimise
positive integersp and
|x
|+1 y
\\.
[13 marks]
q such
[11 marks]
*~
*
(a
K^
+/\302\273
In
this
Modular
\342\200\242
rules
for
with
working
called
remainders,
arithmetic
you
chapter
learn:
will
modular arithmetic
\342\200\242
to
how
solve
equations
involving remainders,
called linear congruences
\342\200\242
to solve
how
simultaneous
congruences
Can
find
the
number? little
last digit
you
A
thought
\342\200\242
a shortcut
of 333333without evaluating the
leads us to realise that
this
is really
powers
for calculating
modular
in
arithmetic.
about divisibilityand remainders:What
is the
remainder
when 333333 is divided by 10? In this section we will
with
remainders
that enable us to
investigate rules for working
a question
answer questionslikethis
one.
Introduction:
Look
after
about
how
After
clock
the
7 hours?
When
you
numbers
time will it show
after 12 hours?What
After 19 hours?
hours
many
answers:
many
infinitely
what
clock:
the
at
working
remainders
with
is concerned,
clock show 5 o'clock?
Thereare
14, 26,... We could say that, as far as
will the
2,
all these numbersarethe
is 2.What other
by 12 the remainder
2 when divided by 12? Again, there
remainder
give
many answers:2, 14,26,50,...As
is concerned, all thesenumbersarethe
are
congruent modulo 12.
they
by 12
If
If
divide 38
are infinitely
+G
same.
far
same.
as division
We
say that
are
you
VSB? Maths,
you
as an
example
in the
classes
equivalence
meet
also
will
congruence
of
Further
studying
Sets,
Relations and Groups option.
divided by 12,what
give? Trying some examples
that
the
remainder
is
suggests
always 4. Similarly,threetimes
the
number
remainder
6. In the next sectionwe will
gives
in general,but first let us look at
that
similar
rules
hold
prove
a few more resultsconcerning
division
by 12. We will need the
a
number
remainder
remainder
gives
does
twice
2 when
that number
>
following result:
KEY
POINT
5.1
number n gives remainder r when
n-kd + r for some number
k.
If a
divided
by d
then
is
This
<^1
division
met in
just
the
algorithm we <^]
Section
2A.
A3
5 Modular
\342\200\242
arithmetic
51
rsM
VS.
^
Q,vv,
orked
'
^
*
(*
1^
(a) If x givesremainder6 and 7 gives
when the following are divided by
(ii) x
+y
-y
(iii)
5.1
point
and
then
do
the remainders
find
12,
12.
xy
\342\200\224
does
2.
give remainder
not
y
form
in the
numbers
the
Write
divided by
3 when
the remainder
x
an exampleto show that
(b) Find
^
v
vc+^
5.1
example
(i) x
^1
from
x = /\\2k
(a) Write
Key#
= 12m +
+ 6,y
3
calculations
the
Write the resulting expression
the
12q + r and identify
the
in
form#
(j)
y =
x +
(12k + 6) +
= 12(/c+ m)
remainder
Hence x
+ 3)
(12m
+ 9
+ y gives remainder
9.
= (12/c+ 6)-(12m +
+3
12(k-m)
(ii) x-y
3)
=
Hencex \342\200\224
y gives
remainder
=
6)(l2m + 3)
(iii)
We want
the remainder to
be
even
and
when
always a
it
is,
we
can
+ 36k
+ 3k
= 12(12km
+ 3k
Hencexy gives
12
whole number
(l2k +
= 144km
= 12(12km
smaller\342\200\242
than
The result is not
xy
3.
+15
6m) + 13
+ 72m
+
+ 6m
remainder
6
\342\200\242
find several
(b)
different remainders
If
x =
which
436
and y =
gives
remainder
27, then
examplesuggeststhat we can add,
multiply remainders, but we needto be careful
them. In the next sectionwe will prove
these
circumstances
general, and investigate under what
perform
Exercise
5A
In
1. (a)
example
4 try
5.1.
to produce
and
when
dividing
in
assertions
we
can
gives
remainder
2 when
If x
12, what
divided
by
divided
by 12, what
remainder does3xgive?
(i)
subtract
proofs similarto thosein
(i) If x givesremainder5 when
remainder does 2x give?
(ii) If x
(b)
16
division with remainders.
1 to
questions
Worked
- =
6.
The above
+a
+1) + 6
gives remainder 5 when divided by
does 4x give?
13,
what
reminder
7 when
(ii) If x gives remainder
remainder does4x give?
52
Topic
10
-
Option: Discrete mathematics
rsM
3$)
^ q,vv,
divided
by 13, what
*
^
o
(c)
(i)
(ii)
2.
(a)
*Vt>.
If y gives
remainder
remainder 4 when divided by
if y gives
remainder
remainder 2 when divided by
does
(c) If a number givesremainder4 when
what remainder does it give when
divided
(d) If a number givesremainder4 when
what remainder does it give when
divided
divided
(a)
(b)
for
by
12,
by 3?
by
12,
by 4?
divided
by 3,
what
by 5,
what
by 9?
divided
by 15?
by
15 and b
gives
remainder
what
does
divided by 13and n gives
by 13, what remainder does
4 when
remainder
divided
when
gives remainder 2 when divided by 3 and y gives
remainder 1 when divided by 3, what can you say
about x + y\\
If x
Describe
which
2
both
and
3.
by
all numbers
start
some new
by introducing
give
1 when
remainder
[4marks]
arithmetic
Rules of modular
We
15,
give?
divided
+a
what
by 5?
divided
(b) If a number givesremainder3 when
remainders can it give when divided
4. (a) If a givesremainder4 when
divided
remainder 5 when divided by 15,
a + b give?
by
divided
3. (a) If a number givesremainder1when
remainders can it give when divided
mn
what
by 3?
divided
remainder8
12,
by 9,
divided
(b) If a number givesremainder3 when
what remainder does it give when
gives
what
give?
yA
If a number givesremainder1when
remainder does it give when
divided
(b) If m
12,
give?
y3
does
V
+r
+<*<
notation and terminology
arithmetic.
modular
We say that
a is congruentto
This is written
as a = b
b
(modm).
modulo
m if
m
The definition
\\ (a
-
b).
has several
implications.
KEY
5.2
POINT
If a
= b (modm)
\342\200\242
a and
\342\200\242
we
b give
write
can
then:
remainder when divided by m
the same
a
\342\200\224
km +
b
5 Modular
It*'
^
<K
vv, \342\200\242V^
+ o
arithmetic
*
(0-
^
'
vC^r^
V
+r*
-+\302\253<;
+ )
is usually used to checkwhether
two
numbers
are congruent
modulo m, and the secondfact is often used in
if a < b, then k needs to be
Note
that
proofs.
negative,but this is
first fact
The
not a problem.
1-
orkedexample5.2
Show
if a
that
= b(mod m)
and c = d(modm)
We need to write equations
a and b, and c and
d. Use
We
have
+ (b +
Nm
a +
written
d), so
use
the
a+c=b+
then
Write a
connecting*
the second
Then
point above
a +
c
in
\342\200\242
form
the
second
point
above
again
= km
+ b,
c =
b) +
(\\m
c = (km +
= (k + l)m
+ (b
a +c=b
+ d(mod
d)
+ d)
w
h
POINT
\342\200\242
ka =
other
rules
of modular
5.3
= b(modm)
\342\200\242
a +
+
+ t
m)
proofs can be producedfor
arithmetic, summarised here.
If a
\\m
Hence
Similar
KEY
m).
rf(mod
kb(mod
c = b
c =
and
m) for
m) then:
d(mod
all k e Z
+ d(modm)
\342\200\242
a \342\200\224
c = b \342\200\224
d(modm)
+G
\342\200\242
ac =
bd(mod
\342\200\242
a\" =
b\"(modm)
With
practice,
m)
for all n e
doingnormal
and
arithmetic,
becausewe areworking
x = 5(mod6)then 3x
In
with
calculations
'reduce'
to the
with
= 15
multiple
in some
smaller
is just as quickas
sense even easier,
numbers.
For example,
if
= 3(mod6) so
(3.x) =9 =
like
this, we should always
steps
3(mod6).
possible remainder
smallest
are never
arithmetic
modular
doing
N.
after
than
each
step,
so we
m.
calculating
larger
It is sometimes convenientto use negative
remainders,
if this produces
smaller numbers. For example:
98
=
i
U
54
Topic 10
-
Option:Discrete
-
Q,vv^^ VVl
+
-2(mod
with
lOO),
so 982
numbers
= (-if
= 4(mod100).
mathematics
vs.
01.
I ^
.on\"
^
o
3
'
v,h
+4,
orked
5.3
example
Find the
remainder when 230
by 13.
is divided
\342\200\242
Start by evaluating small
powers,
then
look at the smallest possible
and
V
-f^
23=3(mod13)
remainder
Square
both
To get
power
30, raise
to
the
26=64
sides*
power
5
\342\200\242
= 12 =
(26)5= (-1)5= -1=
.\". 2?\302\260
gives
If you
41
= 4
42 =
43
look at
find
will
-1(mod13)
remainder
12(mod13)
12.
a number is raised to a power you
start
to repeat. For example:
they eventually
when
remainders
that
10)
(mod
6(modl0)
= 4(modl0)
44 =
6(modl0)
So all even powers of four give a remainderofsix when divided by
ten and all odd powersoffour
a remainder
of four. If we were
give
askedto find the last digit of 4312(which is equivalentto finding
the
remainder
when dividing by ten) we couldthen
write
immediately
down the answer: six.
+G
5B
Exercise
1.
Find
the
(a) (i)
(b)
(i)
(c) (i)
2.
Find
following
513(mod7)
(ii) 419(mod7)
615(modl2)
(ii)
35131(modll)
(ii) 29223(modll)
the
(a) (i)
remainders:
following
522(mod25)
remainders:
316+ 7(mod5)
(ii) 212+3(mod7)
47-22(mod7)
(ii) 66-37(mod5)
(b)
(i)
(c)
(i) 532+323(mod6)
(d) (i) 6x518(mod7)
Findthelast
Find the
digit
of
(ii) 753-653(mod4)
(ii)
8x321(mod5)
22223333.
remainder when 5544
[4 marks]
+ 4455
is divided
by 7.
[7 marks]
5 Modular
it*'
^
Q, VVj
arithmetic
*
(a
'
a-
V
+/\302\273
it
number suchthat
4, find the last digit of 23x +1.
x is
that
Given
last digit
3 Given that
a whole
+ b)5 =
(a
2 (mod
has
2X
[4 marks]
5) show that
b) = 2(mod5).
{a +
>>
[4 marks]
1-
Division and linear congruences
So
we have
far
used addition,
powers
to be ableto divide.
wheredividing
both
subtraction, multiplication
To be ableto solve
in congruences.
and
we also
equations
need
we have already seen examples
the
same number doesnot work.For
by
but 5 ^ l(mod 12). In this
section
we will
example,15= 3(modl2)
see that there are some situations
where
the division
is possible.
However,
sides
= b(modm)
d. To start
to be divisible
with, both sides have
by
by d, as we
can only work with whole numbers. For example, starting from
48 = 3(modl5)we cannot
divide
both sides by 2. However,
3 = 18(modl5),
sowe can replace 3 by 18 on the righthand
side
to get 48 = 18(modl5). This can be divided
by 2 to give
24 = 9(modl5), which is correct.
Suppose
to divide
want
between congruencesand
POINT
KEY
both sides of a
equations.
5.4
If a = b(modm)
a =
then
m)
b\302\261m(mod
words, we can add or subtract
one
sideofthe
just
congruence.
In other
+G
can
We
use
this
to make
already.It turns out that
coprime
POINT
both sides divisible
can
we
d
This rule
10
-
they are
not
number which is
both
a and
b and
gcd(d,m) = 1,
=
\342\200\224
\342\200\224(modm).
then
Topic
by any
d if
to
1
If a = b(modm),d divides
56
divide
by
of m
5.5
Division rule
if 5x
a multiple
m.
with
KEY
difference
an important
illustrates
calculation
above
The
we
that
d
says that,
= 15(mod24)
for
example:
then x = 3(mod24)as
When d and m have
some
common
factors
change the modulus
when
dividing.
For
5 is
coprime
with 24.
to
example, we saw that
we have
Option: Discrete mathematics
t*J
W
^ Q,VVj
vs.
^
o
11* >b.
+4\302\253
l(mod 12),so
by 3. However,5 = l(mod4),so
15 =
but 5 ^
3(modl2)
sides
if m is also divided by
seems
division
the
both
divide
just
to work
3.
5.6
POINT
KEY
cannot
we
-f^
Division rule 2
If a =
i
a
b,
.m.
d
d
d
a, b and m,
divides
d
and
b(modm)
then:
= \342\200\224
\342\200\224
(mod\342\200\224)
'
I
a =
If
3
rule
Division
I
m) and
b(mod
d divides
b, then:
a and
both
I
a
_b
m
mod
d
d
gcd(d,m)
For example,if 6x=
But
gcd(6,15)
we
6(modl5)
so we
= 3,
also have
can divide
to
both sides by 6.
by 3.
the modulus
divide
Hence x = l(mod5).
see
can
We
this works
why
if we rewritethe congruence
as
equations:
this
At
we can
stage
= 5k
2x
then 6x = 15k+ 6.
= 6(modl5)
If 6x
only divide
both sidesby
3:
+2
As 2x and 2 areboth even
k
must
also be
even, so write
k-2n to get:
2x
=
+ 2
10n
can divide by
Now we
x
2:
- 5n +1
and hencex =
We can
x for
linear
KEY
now solve linear
ax =
which
equations,
congruences;
that
is find
the values
The method is similarto solving
b(modm).
some of the rules are different.
although
5.7
POINT
If ax
5).
l(mod
= b(modm) then:
\342\200\242
we
divide
can
both sides
by d if we alsodivide
gcd(m,d)
\342\200\242
we
add
can
^
Q,vv,
a multiple
\342\200\242V^
+ o
of m to
only
one
side.
m
by
of
'
\"~
*
(*
V
c+^
orked
5.4
example
Find the valuesof x such
2x =
(a)
that:
7(mod5)
divisible by 2.
can
We
It is
both sides by 2,
add
5 to the right
= 1, so
gcd(2,5)
= 8(mod9)
not*
7 is
but
divide
<=>
2
by
2x =
(a)
side
hand
we can now
7(mod 5)
(mod 5)
= 12
2x
^> x = 6
\342\200\242
smallest positive*
to reduce to the
customary
+ 5
(d) 25x = 10(mod30)
to divide
need
We
6x
(b)
(c) llx = 5(mod8)
x =
<^
\\
(mod5)
(mod5)
remainder
be any number which gives
= l, 6,1 1,...
divided
by 5, i.e. x
x can
that
means
This
remainder1
when
or-4,-9,...
Start
We
to divide
need
by 6,
by
so add 9 to
so
side
sides
both
Divide
by 6
x on
isolating
gcd(6,9)
the
one side*
right
keep adding to
by 1 1. Alternatively,
RHS
the
5 = &(mod9)
6x +
<^>
6x
= Z)(mod9)
<^=>
6x
= \\2(mod3)
divisible by 6
it is
= 3, so divide
until
it
by 3
is divisible*
as 8x is always a multiple
we can subtract 8x from
= 2(modZ))
<^ x
the*
modulus
We could
\342\200\242
hand
(b)
of
the
8,
left
(c)
\\\\x =
<^
Z>x = 5(mod&)
<=>
3x
^>
b{mod&)
= 13
x =
= 21
(mod
&)
7(mod&)
+G
We
don't have to
are divisible by
divide by 25
one
in
5, and we must
go.
can
by 5
divide
now
5x
to divide
remember
the modulus by
We
(d) 25x
Both sides*
= 10(mod30)
=
2(mod6)
5
again
\342\200\242
5x
x
= &=
=
]4 = 20
(mode)
4(mod6)
numbers
x
examples there are infinitely
many
the
all
the
same
However,
satisfy
congruence.
they
give
In the first
remainder when divided by the final modulus.
modulo
5, and in the second
example the solutionis unique
In all
of these
which
exampleit is uniquemodulo3.Note
the solution is not unique
modulo
modulus): x couldgive remainder
Linear
congruences
the next
58
Topic10
-
Option:
fsM
3$)
^
q,vv,
Discrete
mathematics
example.
do not
that
in the
9 (which
1, 4
always have
second
was the
or 7 when
a solution
example,
original
divided by
9.
as shown
in
*
^
o
orked
*V*>.
5.5
example
Solve 3x = 5(mod
12).
keep
to the
12
adding
divisible
Solve the following
linear
15x = 7(mod11)
(c) (i) 5x = 30(modl5)
=
(d) (i) 12x + 2 4x+ 7(mod9)
(e) (i) 2hc= l8(mod9)
= 9(mod
(ii)
4x
(ii)
1 lx
(a)
be a
multiple
are
no
of
of 3
3.
solutions.
(ii)
3x
(ii)
lbc-2
(ii)
15x
15)
9)
= 18(mod9)
= 3x
+ 5(mod9)
= 60(mod24)
[4 marks]
13).
linear congruence 3x = 12(mod21)
giving
in the form x = a(modm).
Solve the
Q
never
because 5 is not a multiple
So there
= 7(mod
linear congruence 5x = 7(mod
Solve the
answer
5 + 12/ccan
\342\200\242
congruences:
(i) 3x = 7(mod 11)
(b) (i)
3. This
by
...(mod12)
5C
Exercise
(a)
it
adding 12, which is a multiple
of 3, to 5, which is not
we are
because
is
never make
can
we
seems
41 =
by 3
divisible
It
3x = 5 = 17= 29=
sides by 3 we needto*
right hand side until it is
to divide both
In order
v
+^
+<*<
the linear
why
Explain
your
[4marks]
6x = 4(mod9) has
congruence
no
solutions.
+G
(b)
Solve 6x
(c)
List
all
| x | <
= 3(mod9).
the
equation in (b) which satisfy
of the
solutions
10.
[8 marks]
6x = 3(modl5),
(a) Solvethe linearcongruence
answers in the form x = a(modl5).
(b)
How
many
2
Q
Given
that
modulo
solutions
have?
= 147(mod63)
be
3x + y
= 21
(mod
smallest positivevalue
of
the linearcongruence
[7 marks]
your answer.
63 does
Justify
5) and x
a such
giving your
\342\200\224
=
that x
y
7(mod5)
= a(mod5).
find the
[8marks]
9 Chinese remainder theorem
Consider
remainder
the
following
2 when
find all numbers which give
3
and
remainder3 when divided
by
problem:
divided
the language of congruences from the previous
to solve simultaneous congruences:
by
5. In
we
are trying
section,
5 Modular
it*'
^
Q, VVj
arithmetic
'
a-
*
(a
V
^<+^
fx = 2(mod3)
|x
can list
We
=
3(mod5)
each congruenceseparately
both lists.To keepthings
see
and
simple,
let
us
the positive solutions.
just list
x
solutions for
ones appear in
which
=
x =
2(mod3):
x = 3(mod5): x =
2,5,8,11,14,17,20,23,...
3,8,13,18,23,28,...
found two solutions:8 and 23.We can see that
in the first list we need to add a multiple
number
any
of 3 and for any number in the second list a multiple of 5.
Henceto get another number which is in both lists we needto
adda multiple
of 15. This means
that every solution is of the
form 8+ 15/c,in other words x = 8(modl5). Although there
are infinitely
numbers
that satisfy this congruence, the
many
we have
So far
to get
The
remainder
Chinese
*jt
theorem was first
by the
down
written
dpra^
Chinese mathematician Sun
in the 3rd century CE and
solutionis
Tzu
republishedby
1247.
were
Its
in
first
Qin
This is
applications
Diophantine
also
solving
but it has
equations,
found uses in public
in
cryptography
20th
in
Jiushao
an example of the followinggeneralresult:
5.8
POINT
KEY
the
If
century
and
ml
theorem
remainder
Chinese
key
15.
modulo
unique
then the
are coprime
m2
simultaneous
congruences
x
=
x =
a(modm1),
b(modm2)
have a unique solution(modm1m2).
result
The
moduli
the
orked
pairwise
5.6
example
of linear
the system
Given
are
of congruences,as longas
coprime (so any two are relatively
prime)
to any number
generalises
congruences:
x = 2(mod5)
+G
x
=
7)
3(mod
x =
5(mod 13)
(a)
Find
(b)
Hence
We
down
write
can
solve
idea
good
x which
number
one
the solution
the first two
to start
satisfies all three congruences.
with
the
of the system
equations
two
first.
smallest
It is
in
a
the
x =
form
\342\200\242
x =
{a)
moduli
x =
60
fsM
W
Topic
10
-
modulo
Option: Discrete mathematics
^ q,vv,
\342\200\242V^
+ o
35 x 13
17(mod35)
5,13,31,44,57,70,33,96,109,122
17(mod 35)
is unique
3,10,17,...
x = 1(mod13):
We now make a list for the third equation*
and compare to all the numbers which are
answer
2,7,12,17,...
2(mod5):
x = Z){mod7):
So
This
a(modm).
x
*
=
17(mod35):
(b) .\\x =
122 (mod
17,52,37,122
455)
^
o
'
11* >b.
+4\302\253
becomes
finding a solution by inspection
An alternative is to rewrite the congruence
more time-consuming.
as
a Diophantine
We will illustrate this method by an
equation.
numbers
As
get larger,
method
The
<^[ of
was given
32 are
and
modulo
unique
relatively prime,
31x32 = 992. We
each
is*
just
to
need
solution
equation for x*
as an
congruence
32), x
12 (mod
solution
the
find one
Write
4B.
= 21(mod 31).
remainder theorem,
I3y Chinese
x = c(mod992)
\\Ne
can
x =
write
32a + 12
21
+
Hence 32a
<^=>
is a
so we know
equation,
Diophantine
to
find
one
how*
solution
+ 12= 31b+
-311?
32a
W
(*). Both
x from
should give us the
solution
(we
This is the
solution
same
equations for xshould check!)
god (32,31) = \\,and
So 9 = 32 x 9 - 31
From
(*),
x =
21
= 9
1
= 32
- 31
x 9
:.a
We can now find
the
is
solution
x = 31b
This
<^[
in
5.7
example
Solve the simultaneouscongruences
x =
31
equ-
Diophantine
ations
Section
As
for
finding a solution
example.
orked
V
+~
= 9,b
=9
300.
/. x = 300(mod992)
mod(992)
+G
Exercise 5D
1.
the
Solve
following
(i) x = 3(mod5),x
(a)
=
x =
(ii)
(i)
(b)
X =
By
first
the
(a)
2(modll),
0(mod2),x
x =
(ii)
2.
simultaneous
0(mod3),
solving
4(mod7)
x = 4(mod5)
= 2(mod5),x
x =
= 4(mod7)
l(mod4), x = l(mod5)
an appropriate
simultaneous
following
congruences:
Diophantine
equation, solve
congruences:
(i) x = 6(modl9), x = 6(modll)
(ii)
x =
(b) (i) x
(ii) *
=
6(modl3),
15(mod31),x
x =
3(mod22)
= 2(mod30)
= 4(modl7),;c = 9(mod24)
5
^
q, vv,
Modular
arithmetic
*
(*
'
\"~
V
C+r*
)
of congruences:
this system
Solve
(x = 2(mod3)
[x
=
Solve this
x
=
(a)
[6 marks]
5(mod7)
system of linear congruences:
3),
2(mod
Find the smallest
5x = l(mod7).
1-
5), x = 6(mod
x = 0(mod
[8
7)
positivevalue
the
of
(b) Hence
system
5x = l(mod7), x 3(mod5)
solve
of
x such
that
linear congruences
=
Solve these
5)
(mod
[8 marks]
2(mod 3)
Solvethe system
x =
0
[7 marks]
simultaneous linear congruences:
3.x= 2
5x =
marks]
of
linear
congruences:
31).
3(mod 17), x = 21(mod
[10
marks]
(a) Prove that the system of linear congruences:
x = 2(mod6),x = 4(mod9)
has no solutions.
Find the generalsolution
6m-9n = 3.
(b) (i)
(ii) Hence
the
solve
x = 2(mod6),
Diophantine
equation:
of congruences:
system
x = 5(mod9).
[10
marks]
theorem
little
Fermat's
the
of
We have already answeredquestions
such
as: find the remainder
when 517 is divided by 7. We start by evaluatingsmallpowers
if we can spot a pattern:
and
see
+G
52=25= 4(mod7)
53=5x4
=
We can
6 =
get closeto the power17by
515
Hence
517
gives
-l(mod7)
raising
this
to the
power of 5:
515=(-l)5= -l(mod7)
= -4 = 3(mod7)
= -1x4
x52
remainder
3 when
divided by 7.
in these
Our strategy
is to keep evaluating powers
problems
until we get remainder1or-1 and then 'jump to a multiple
close to the requiredpower.But how do we know that this will
ever happen? For example,
lookingat powers of 3 modulo 12,
we find that:
31= 3(mod 12),32
34=3x3
=
^
W
62
Topic
10
-
=
9(mod
12), 33
= 27 = 3(mod 12),
9(modl2)
Option: Discrete mathematics
^ Q,ri
vs.
%**
+ \342\200\236
.Ok\"
^
o
'
11* >b.
+4\302\253
V
-f^
*
the remainders alternatebetween3 and
It seems that
will never
get remainder 1.In this
powers give
In allthe examples
to
need
we
theorem
little
I
Fermat's
|
lip is a prime and a is any
the congruence
littletheorem.
by
not a
a is
following special
multiple
both sides of
can divide
caseof Fermats
The proof
from
of p
then
now
a good
have
Group
strategy for finding
of
Fermat's
uses ideas
Theory,
a
branch
of Discrete Mathematics
studied
in
Option
Relations and
ap-l=\\{modp).
We
theorem
*SEG? little
is a prime and a is not a multiple
If p
a).
when
we
prime
get the
a to
-
p\\(ap
a(modp).
5.10
POINT
KEY
that
extract even moreinformation
of p: as a andp arerelatively
ap =
then
integer
This is equivalentto saying
We can
1-
5.9
POINT
KEY
When
pattern.
the following
a prime,
modulo
the remainders of powers
we are looking at remainders
result tellsus how many powers
find before they start to repeat.
a periodic
follow
odd
powers remainder 9.
so far
seen
have
we
so we
9,
that all
it seems
case,
all even
3 and
remainder
/
8, Sets,
Groups.
remainders.
orked example 5.8
Find
when
remainder
the
little
Fermat's
theorem
jump
2828 is
divided by 13.
is useful
because we
to
a high
quite
We can squarethis
to
get
can*
24*
=>2<324=12=1(mod13)
+a
We now
just
by 284#
to multiply
need
28 = 2(mod 13)
When
we are
little
Fermats
same
theorem
little
=>28>12=1(mod13)
power
power
Fermat's
looking at divisibility by a compositenumber,
theorem
may need to be usedseveraltimesin
=> 2&2&=
.\\
2&z&Q\\vee
by
13.
2d4 = 24
remainder
= 16
= 3(mod13)
3 when divided
the
question.
orked
Prove that
We first
example
for
5.9
n,n7
all integers
factorise 42
see
to
- n is divisible
which
primes
by
42.
we need
to
look
\342\200\242
42 =
2x3x7
at
5 Modular
arithmetic
63
vs.
^SV^
+
c,
.on\"
*
(a
-V
'
a-
1 ^ - ^
\\
v
+^
continued...
there is
Because
n7
Fermat'slittle
in
the
theorem
we can
question,
for
divisibility
by 7:
Divisibility
use*
little theorem,
I3y Format's
by 7
=
n7
n(mod7)
-n).
5o7\\(n7
little*
Now lookat divisibility
by 3: Fermat's
theorem can only be used to get to n3, but we
can then apply other rules of congruences
by 3:
Divisibility
little theorem,
I3y Fermat's
=
ri5
=^>
n6
^
n7
n(modZ))
= n2
(mod3)
(square both
sides)
= ri5
=
=^> n7
Similar
strategy
works for
\342\200\242
2
by
divisibility
-
=
=
=^> n7
have
We
proved
divisibility
(mod2)
n4 = (n2 )2 =
n7
-
m
and hence by
= n
\302\253
)
(mod 2)
is divisible
by 2,
3 and 7
42.
\\
worth noting alternative
divisibility by 2 could have
an even number to any
any
is odd;
power
odd,so
their
n7
+a
=
factorising:
n(n3-l)(n3+l)
l)(n2
+n
+ l)(n
1 are threeconsecutive
of them is even and at least one of
Exercise
product
Find
the
is divisible
41)
n +1)
integers,
them is
so at
least one
divisible by 3. Hence
by 6.
(b)
11)
(i)
411 (mod
(ii)
Find
Option:Discrete
S^-V*1
the
(mod
37)
(ii) 518(mod7)
724(modl3)
remainder
539
(ii) 517(modl7)
(d) (i) 525(mod3)
?
\342\200\224
\\)(n2
remainders:
following
(a) (i) 342(mod
(c) (i)
-
+
5E
1.
Topic 10
and an
either both
and n +
their
noting that
-l)
= n(n n,n-l
the
odd number to
even or both
proved
Secondly, we could have
n are
3 together by
\342\200\224
\342\200\224n
n(n6
First,
by simply
proved
is even
and
n7
is even.
difference
here.
approaches
been
power
hence
by 2 and
divisibility
64
(multiply
n2
sides)
by
-n).
2\\(n7
It is
U
(cube both
L*
I
i
little theorem,
n4 (mod2)
Therefore
3 and 7*
by 2,
line)
n(mod2)
=>n6 =n5
But
first
m).
I3y Fermat's
So
(from the
by 2:
Divisibility
n2
by n )
(multiply
n(mod3)
31 (n7
So
(modi)
when
(ii) 1231(modll)
3532 is
divided by 11.
[5
marks]
mathematics
f
-
*
o
Kx^,
h*
+(3.
(a) Prove that
x25
= x
'
+ ^
+
^
\"
, ,
(mod 13).
(b) Hence solvex25 = 5 (mod 13).
x12
(c) Explain why the linearcongruence
has no
= 5(modl3)
solution.
Q Leta
[8 marks]
17. Showthat
a17
+ b17 is
number show that
a prime
p is
that
Given
numbers such that a
also divisible by 17.
two whole
b be
and
p
that p
| (x* + j^).
\\ {abp
-
that
all
apb) for
[6
Show
by
[3marks]
number greater than 2 and let x and
a prime
be
divisible
+ b is
positive integersa, b.
Letp
v
-f^
marks]
such
y be
[7 marks]
p2\\xp+yK
Summary
\342\200\242
a =
that:
means
b(modm)
m\\(a-b)
a and b
-
we
can
add,
\342\200\242
We
can
add
are
-
subtract
\342\200\242
The
congruences=
x
\342\200\242
Fermat's
and
remainder
Chinese
a{modmx),
ifp is
theorem
same
power.
congruence.
division:
a, b
d divides
d divides
theorem
x =
and m, then
both a
a
b
m
d
d
d
=
\342\200\224
\342\200\224(mod\342\200\224)
f
a _b
=
and b, then ~j
mod
-
~j
m
gcd(d,m)
says that if m1 and m2 are coprime then the
b(modm2)has a unique solution(modmlm2).
either by inspectionor by
little
and raiseboth sidesto the
a multiple of m from just one sideofa
if a = b(modm) and
be found
that
and multiply congruences,
rules for
if a = b(modm)
by m
+ km.
or subtract
special
when divided
remainder
same
a-b
write
can
\342\200\242
We
\342\200\242
There
the
give
solving
says that ifp
prime which doesnot divide
a Diophantine
is a prime then for
a then
ap~l =
system of linear
This
,t>
^
^ q,vv,
\342\200\242V^
+ o
can
equation.
all
a, ap
= a(modp).
A
consequence
l(modp).
5 Modular
l,sM
solution
arithmetic
is
*
(*
\"~
'
Kx*
V
C+^
examination
Mixed
5
practice
Use Fermats littletheoremto find
14112is
when
remainder
the
divided by 11.
[5
[3
of 432.
last digit
the
Find
marks]
Solvethe linearcongruence
Show that
(a)
(b)
Find
[6
13.
by 7.
is divisible
[6 marks]
[6 marks]
[4 marks]
=
congruence 12.x 36(mod32).
the linear
Considerthe
1=
x12 +
equation
Fermat s little
7y, where
theorem, show that
x, y
this
D+.
e
has no
equation
IB
(\302\251
the remainder when 45
(a) Find
show that
Find the last two
410w+2
=
(c)
digits
in the
Solve
the
the simultaneous
Solve
x = 0(mod2),x
UseFermats
Show that 536 gives
(b)
Find
the remainder
divided
x =
when
that 7120-1
536
is divided
is divisible
by
143.
[5marks]
17.
by 13.
simultaneous congruences,find
the
when
remainder
536 is
[8 marks]
be a primeand gcd(a,p)
show that x = ap~2b(modp)
(b)
[4 marks]
= 3(mod5).
221.
by
=
[7 marks]
[7 marks]
remainder 13when divided by
= 1.
(a) Letp
ax
expansion of 442.
14(modl7).
to show
theorem
(a)
(c) By solving
base 5
congruences:
= 2(mod5),
little
2007)
Organization
16(mod25).
congruences x = 2(modl3),x
of linear
system
solution. [9 marks]
by 25.
is divided
(b) Hence
marks]
[6 marks]
gcd(32,12).
Solve
Using
1).
in base
written
when
+ 55552222
22225555
= 5(modl
3x
last digit of 560
Find the
marks]
is a
Use Fermat
s little theorem to
solution of the linearcongruence
b(modp).
Hence solve the systemoflinearcongruences
2x=
1 (mod
31J,
6x =
5(mod 11).
[7marks]
Find
your
(a)
the
solutions
answers
in the
all
State
Fermat
linear congruence 120.x= 110(mod65),
giving
=
form x /c(mod65).
[5marks]
of the
s little
theorem.
(b) Prove that, if p > 3isa primethen
p-
i^Jcp
k=\\
66
fsM
W
Topic
10
-
Option: Discrete mathematics
?\302\243*\302\273.
\342\200\242V^
+ o
=
0(modp).
[4
marks]
^
(a
+r
\342\200\242+<*<;
In this chapter
theory
Graph
you
will learn:
\342\200\242
that
different
many
of real-life
types
problems can be
solved
graphs
using
social
do
What
of a
the
structures, traffic flow and
common?
One answer is that
molecular
networks,
\342\200\242
about
In this
chapter we will
at basic
look
graphs, and in the nextwe
routes arounda network.
will
graph
\342\200\242
how
to
\342\200\242
how
some
recognise
types of graphs
special
properties of
to finding optimal
them
and different
ways of representinga
mathematical
apply
of
features
main
a graph
syntax
language
mathematical
structures
they can all be studiedusing
graphs,
which are normally represented as a set of pointsjoinedby
lines.
seem an abstract mathematical concept but they
Graphs
in a wide variety
have
of unexpected
applications
places.
many
in
have
to
prove
important
some
theorems
about the numberof
verticesand edges
in
to
Introduction
In this sectionwe
precise
for
and
the
the problems
started
different
of moving
ways
around a
graph.
are intended
the development
the need for
are introduced in this
we develop in the next.
understand
and theorems that
definitions
chapter
\342\200\242
about
typical problems that can
to find a solution,
attempt
some
be solvedusinggraphs.You should
if you
but don'tbe disappointed
can't;
to illustrate the sortofdifficulties
that
of graph theory. They should
help you
the
graph
graphs
introduce
will
a
algorithms
Thetable belowshows
of direct
cost
the
flights between
ten cities.
Beijing Cairo \\Dubai
cost($)
213
+G
Frankfurt
96
52
412
450
526
186
250
212
250
315
234
320
52
Frankfurt
450
186
250
36
526
250
315
325
212
234
Moscow
350
312
Sydney
225
524
1.
Can
from
you find
Athens
2. Canyou
each
To answer
428
216
combination
cheapest
once and
the first question,
approaches.
in this caseis Athens-Dubai-Sydney
whichthe initial flight is
We
387
the
at least
city
215
combinationof flights
can
either
it
to
of flights to
return to the starting
312
350
225
312
524
387
216
428
243
243
get
visit
city?
one of two
fewest legs,which
or we can try the one in
expensive, which would be
is tempting
to try
try the route with
the
least
6
^^VNc*.
Sydney
386
215
312
the cheapest
to Sydney?
find
325
36
386
486
Moscow
486
320
65
York
York
65
412
Johannesburg
New
New
Johannesburg
Graph
theory
67
*
^
(a
TTzere ts
^<+^
this
solving
called
problem,
Dijkstras
it
in
section 7C.
Travelling
in
w///]^>
its solution
investigate
7E.
Section
testing
possible
It is
not even
obvious
sometimes
exactly
the directroutebetweenthe two cities is not the cheapest one.
Even
all the routes in which each city appearsjust once
checking
doesnot seem to be an easy task. It turns out that usually the only
routes,
way to solve this type of problemis to checkallpossible
and
this is not feasible. However,therearesomeclever
generally
to
estimate
the cost of the cheapestroute.
ways
Salesman
]^> problemand we
best
is in fact
The second problemis even moredifficult.
that the bestroute shouldvisit each city
the
called
is
This
more
Athens-London-Johannesburg-Sydney. It is not
clearthat this is the cheapest without checking lots of possibilities.
You
can
that if we had 100citiesinstead
of 10, the
imagine
in a reasonable
time.
problem would be impossibleto solve
route
e>
Algorithm.
meet
will
We
Athens-Frankfurt-New York-Sydney.
However,
reveals that neither of theseis the cheapest.
The
an efficient
method for
fe>
v
The
the table
from
information
once, as
represented as a
can be
cities) are joined by
diagram which points (representing
lines(representingdirect
Such
a diagram
is called a
of problems
are used in practical
graph. Where these
the
can
consist
o
f
of points
or
applications,
graphs
lines. An example similar to our problemis a satellite
which
tries to find the shortest routebetween
places
system,
in
routes).
types
hundreds
navigation
two
and
choice of many
has a
impossible
for
time,so
we
found the best
Is this
as proving
to program
roads
and junctions.
It is
such a problem in a reasonable
a computer
to do this.
\\
it
mathematically?
introduces
a computer
the computer
to program
possible solution?
the same
different
to solve
a new set of problems. We need
to solveall problemsof this type,
not just one particular one. This means that
we need
to have
a method, calledan algorithm,which
finds
the
best
always
possible solution. Suchan algorithmhas to have a well-defined
based
on guessing'
sequence of steps, and cannottake shortcuts
or experience'. It must also be efficient
so that
it can perform
the task in a reasonabletime.Checking
all possible
routes is
Using
Would
you say that if we
a computer to try
program
all possible
routes, we can
be certain
that we have
need
a human
not practical, and the timeit takes
size of the graph.A typical
computer
+G
increases
rapidly
can perform
with
the
around 3
billion operationspersecond;checking
all possible
routes for
if
our second problem would take arounda millisecond,
but
we increase
the number of citiesto 15it would
take
around
5
minutes and with 20 citiesthis goesup to 500 years - not very
practical if you are waiting at the travel agent!
Here is a different
This
studied
in
by
result
first
\342\231\246Jf
Euler
Leonhard
and
1 836,
regarded
68
was
problem
4fjjfk
is usually
as the first
of graph
any lines
published
theory.
Topic10
-
Option:
Discrete
fsM
3$)
without
^
cl
\302\243%*..
mathematics
lifting
problem:
your
more than
Is it
pen off the
once?
possible
to draw
paper and without
this shape
going
over
'
^
(*
After
I^C+/-
for a while you
trying
is equivalent to saying, for
may decidethat
prove
the
that
that
example,
out ofa singlepieceofwire. Can
the shapewould
need
to change
Later we will
V
of
out
cannot be made
why? What feature of
become
possible?
This
problem is the
one of the
point.
we will study in the rest of
for this
This is
each
that
features
to
feature
key
numberoflinescoming
of graphs
important
this chapter.
it
This
impossible.
the shape
explain
you
for
it is
is
problem
to Eulerian
related
]^> graphs,
meet
]^>
6F of
Section
in
will
we
which
this chapter.
sUfl Definitions
In
solve
to
order
some
new
In this
terminology.
conceptsassociated
with
are called
graphs.
of vertices
consists
A graph
with graphs it is usefulto introduce
section we will define various
problems
called adjacent if they
So
adjacent.
The
the
relevant;
the
may
edges
only
So the
AC,
BC
of the
vertices
graph are
graphs above have
and
vertices
edges are
the shape
that matters
we draw
of the
edges
is which verticesare
diagrams below represent the samegraph.
intersect
at points other than vertices, soit is
the
The vertices
thing
edge.
Two
Two
two
importantto label
+G
and
vertices
edges.
an
vertex. When
a common
have
a graph,the positionofthe
arenot
connected by
adjacent if they arejoinedby
clearly.
usually
vertices
A, B,
C and
by capital
letters.
Sometimes we need to showa situation
where
we can only
move in one directionbetweenvertices,for example
when
In
a
road
network
with
roads.
such
cases we
modelling
one-way
can put arrows on the edges to indicate the alloweddirection.
In
The
is called a directed graph, or digraph.
resulting
graph
to get from D to A but not from A to D,
1, it is possible
Graph
while both directions between A and B are allowed.
4fjjfk
beginnings
and many
many
century. Because
were
its findings
of
century
developments
important
early 20th
in the
1 8th
the
in
its
with
mathematics,
motivated by applications it has
had contributions
from many
different communities:
mathematicians, computer
For these
linguists.
the
terminology
different
slightly
books.
other
reasons some
is not
so you
standardised,
even
and
chemists
scientists,
of
Graph 1
relatively new branch of
D and edgesAB,
BD.
is a
theory
Graph
\342\231\246Jf
labelled
yet
may
definitions
For example,
and edges are alsocalled
and
firmly
find
in
vertices
nodes
The terminology we use
will be used in your
what
arcs.
here
is
IB examination.
6 Graph
^
Q\342\200\236
t**W
theory
69
'
* *-
(*
V
-^^<+/-
*%**
a simple
In
is joinedto itself(no
no vertex
graph
and
loops)
SoGraph 2 shown
multiple edges betweenvertices.
Some
theoremsof
simple.
alongside,
graph
theory
only
when
apply to simple graphs, so it is importantto note this
them. A graph in which there are somemultiple
learning
edges
are no
there
is not
called a
is sometimes
a graph
When
2
Graph
to
complicated
multigraph.
canbe
convenient
has many vertices and edges it
In such cases it is more
draw.
to representit an adjacency
numberof edgesjoining
You
pair
A
B
C
D
thai a
assume
graph
A
0
1
0
0
this
unless
is simple
B
1
0
1
0
explicitly
C
0
0
0
0
0
is
stated.
for Graph
table
the
0
1 1
D
^
and
2 is:
v
B
c
2
1
2
0
0
1
0
0
A
6.1
POINT
KEY
the
1 is:
Graph
not
should
shows the
For example,
of vertices.
each
adjacencytable for
This table
table.
by
graph, all entries on the leadingdiagonal
to bottom right) are zeros, and all other entries
0 or 1.
a simple
\342\200\242
For
left
(top
either
are
\342\200\242
The
adjacency
table
of an
undirected graph is
symmetricalabout the leadingdiagonal.
+G
I
of edges
number
The
of that
coming out of a vertex is calledthe
For example, in
vertex.
degree
Graph 2:
deg(C) = 1.
Notethat when calculating
the degree, we count the edge
A to A twice,
as
each
joining
edge has to have two ends.
in the adjacency table the numberin the (A, A) cell
However,
is 1, because thereis only one edge joining A to itself. The list
deg(A)
= 5,
deg(B)
= 2,
of degreesofalltheverticesofthe graph is called its
sequence. It is usuallygiven in descendingorder;for
2 has degree sequence 5, 2, 1.
Graph
example,
6.2
POINT
KEY
degree
of a vertex is equalto the sumofthe numbers
of the adjacencytable corresponding
to that vertex. If there is an edge connecting a vertexto
itselfthat edge needs to be added twice.
The degree
in
X?
70
Topic
10
-
Option: Discrete mathematics
fsM
vs.
w
^ cl
\302\243%*..
the
row
or column
*
^
(a
^<+^
we will prove severaltheoremsabout
the
number
of edges and vertices of different
of graphs.
types
will
show
that
there
are
restrictions
on
what
the
They
degree
of a graph can be. Herewe show just one interesting
sequence
result about the degreesequence
ofa simple graph.
In
section
next
the
v
You
to
want
may
review the pigeonhole
principlefrom Section
<&
<& IB before
reading
Worked
next
the
example.
orked
6.1
example
Showthat
This sounds like the
type
of
principle.
and
be 'pigeons',
can
vertices
The
we*
problem
using the pigeonhole
solved
are always two vertices with
graph there
a simple
in
the possible values
the
of
the
same
degree.
Suppose the graph
has
graph is simple, the
of a vertex is 0 and
smallestpossible
the
n
As the
vertices.
degree
one is
largest
n
-1.
degree
'pigeonholes'
There seem to
be
n
pigeonholes!
they all appear
in
the
same
is a
graph?
there
are
in
only n
fact
the
verticesmust
group of
each
4fjfi
in
a
football
same number
+G
be expressed
can
in
people there are
league,
different
various
who
two
at each point
in
the
have
ways in
the same
season
there
graphs
So there
n
different
1
n
to
principle, two
same degree.
pigeonhole
have
2, or
the
the
for
- 1.
of the
For example,
contexts.
number of friends
the
in
have
that
path betweenevery
cannot be splitinto
two
two
vertices.
parts.
that the graph
results in this course
This means
Most
of the
the
at
look
will
We
various
of
types
paths in moredetail
in
6F
Section
For
]^> chapter.
moment
of this
we
just
the ]^>
use
mean a
route between two
graphs.
vertices.
Not
are some
or
group;
played
the term to
Connected
There
in
situations
is modelling
need to move between vertices.It is therefore
to know
whether
it is possible to get from onevertex
important
to another.A graph is called connected if every two vertices are
in other words, there is a
connected(directly
or indirectly);
connected
to
-
are two teams
where we
with
- 2.
of games.
Oneofthe main applicationsof
deal
is n
the
so
-1 possiblevalues
either 0
Hence,by
result
degree
possible degree
degrees:
This
with
no
0, then
If
other vertex is connected
to it,
largest
\342\231\246w
vertex
But can*
special types of graphs which you
connected
need
to know.
6 Graph
^^%\302\253.
theory
*
\"~
*
(\302\243
v
+r
)
graph is a graphwhere
A complete
\\
are joined by an edge.A
vertices is K\342\200\236.
for
symbol
pair
every
a complete
of vertices
graph with n
1-
1h
KEY
You
may
6.3
POINT
be asked
fn\\
to give explanation8
and prooMor
mod
results
of edges in K\342\200\236
is
vh
this
m
0Hhe
The number
chapter.
This is
because there is an edgecorresponding
to every
vertices, and
selectingtwo
n can
from
vertices
pair
of
fn\\
be done in
v2y
ways.
vertices can be splitinto two
X and Y, such
that edges only join vertices in X to those
groups,
in
there are no edges joining verticesin the same group.
Y, and
If every vertex in X is joinedto every
in Y the graph is
vertex
calleda completebipartitegraph
and
denoted
Kr s,
where r and s arethe numberofverticesin groups
X and Y,
A
graph
bipartite
is a
graph whose
respectively.
+G
orked example 6.2
*
\342\226\240^
Show
that
the
How many
of edges in
number
edges
How
from
start
many
each
vertices
Kr
s is
vertex
are there
rs.
are e
in
X?*
There
in
X?*
There are r
So there
edges
vertices
are re
each
from
in
vertex
in X.
X.
edqee in
the
graph.
J.
72
Topic10
-
Option:
Discrete
mathematics
fsM
3$)
v Brn**1
+ 01.
*
^
(a
^<+^
connected graph in which there areno closed
paths
i.e. it is not possible to find a sequence of distinctedges
returns
to the starting
vertex.
is a
tree
A
(cycles),
which
Not a tree
Tree
KEY
POINT
6.4
vertices has n
A tree with n
this
that
Note
to
is the
with
a graph
make
a tree
- 1 edges.
smallest possible number of edgesneeded
n vertices connected, so if any one edge of
is removed the graph
You need
is no
longer
to be ableto prove the above
connected.
induction
is common
theory becausewe can think
of'building
adding verticesand edgesoneat a time.
orked
example
that
the
base
up'
the number
and
is given
in graph
any graph
by
6.3
Use strong inductionto prove
Prove
about
result
of edges of a tree. The proofusesstrong
in the next example.Proofby induction
+a
v
case.
n = 1 is the only
We
that
will
a tree
check
base case
with n vertices
later*
needed
has n - 1 edges.
When
n = 1:
The tree
with
one
has no
vertex
the statementistrue for
Assume
How let
the statement
n
= k
and look
edqee, eo
n = 1.
is true
for
n<
all
at a tree with
k.
k
vertices.
6 Graph
^SV^
+ c,
theory
*
^
(a
v
^<+^
)
continued...
It is
to
tempting
and one edge to get
k - 1 verticesand k - 2 edges.
tree, and
of the
tree
the
need
We
We can
However,
'end'
the
from
to
easy
to
explain
not
have
apply the
why
this is the
removed
any
inductive
edge.
any
This splits
trees-. If the
the
connected
there
but
of the
a cycle,
is
which
numbers of verticesof
trees be a and b.
case)
vertices*
=k
Then
a + b
Both
a and b
Hence the
.
are less than
of
number
edges
+ (b-1)
+1=
(fl +
So the statementistrue
that
Note
we only
need
case, as when
graph
into
two
n
=
n
=
2 we
graphs
a conclusion.\342\200\242
1 as
the
for
also
can split the
true
true for
vertex
of the
n <
all
for
all
k we
n =
n by
b)-1
for
istrue for
The statement
true
the base
with one
k, so
trees
two
original
was
(a-l)
should always write
1h
new
two
the
inductive hypothesis applies: The
have a -1 and b -1 edges.
hypothesis*
1edge;
impossible.
Let the
tree
We
removed
edge back would
removed
the
form
separate
graph was still
would be another path
remaining
putting
then
two
into
graph
betweenthe endpoints
identify
we can use strong
can take out any edge to
into two smaller trees (we
we
induction,
split
not
is
it
with
a tree
out
Take
But since
a vertex.
such
to be
need
would
vertex
this
vertex*
out one
take
and
try
n =
= k-1
n =
k.
], and
if
it
is
is
statement is
can prove that it
k. Hence the
strong
Induction.
each
Another
We
will
graph
the
in
]^>result
this
prove
type of graph is a planar graph.This is any
be drawn in the planesothat the edges do not
important
that can
cross. Not every graph is planar;for example
itis impossible
K3 3 without any edges crossing.
next\"^^>
section.
to
draw
+G
A
This
]^>
is
called a Hamiltonian
cycle, and we w///]^>
study it
in more
in Section
74
of path
type
Topic10
-
by
detail
6G.
Option:
Discrete
planar
graph
its adjacency
may
edges crossing,or given
redraw the graphin its planar form,
be drawn with
table. To
strategy is to lookfora closedpath which includes
of the graph, draw
this
first, and then fit other
path
around
it. Note, however, that such a closed path does not
edges
exist.
always
one good
all vertices
mathematics
fsM
vs.
3$)
^
Q,VVj
*
(a
'
a-
V
^<+^
+
example 6.4
orked
Graph G is given by
the
adjacency
following
B
o
1
0
1
1
1
in planar
G
G with
alphabetical order and
path
1
1
1
c
1
1
0
1
1
D
E
1
1
1
1
1
1
0
0
0
0
1-
form.
draw
First
table.
A
1
Draw
)
vertices
a
find
which includes all the
in*
closed
vertices
Dm
Hamiltonian
Redraw G so that
this
cycle: A5DCEA
a*
forms
path
C
pentagon
Then fit
many
and
as you
then
in
the
other
can inside
edges.
the
Put
as
pentagon,
the others round the
outside
+G
theory
Graph
^SV^
+
c,
/
-
75
.ON
* ^
(a
'
K^
V
+/^
)
Exercise
6B
1. For eachofthe
shown
graphs
below
state the
number of vertices
andthe numberof edges.
1-
1h
Gl
2. For
3.
Copy
ticks
and
to
the
of
each
+a
graphs
complete
mark
G2
G3
G5
G6
above list
the table
correct
the degrees of the vertices.
to classify the graphsabove.Use
options.
G2
G3
G4
G5
G6
wrTTnTtfSfZril
Simple
Complete
Tree
Bipartite
4.
76
Topic10
-
Option:
Discrete
Represent
each
of the
graphs
above in an
adjacencytable.
mathematics
fsM
vs.
3$)
^SV^
+ c,
a-
*
(a
5.
represented by the following
the graphs
Draw
(a)
1 ^
(i)
A
A
(b)
(i)
A
(c)
(i)
1
a
0
0
0
1
0
0
1 1
0
1
1
1
1
1 0
0 2
1 0
the
drawing
vertexforthe
1
1
0
1
0
0
0
1
0
1
0
1
0
0
1
1
1
\\B
1
\\D
\\C
Bin
0
2
0
1>
0
1
1
0
9
o
1
2
0
1
0
0
2
E3 o
1
0
1
2
2
0
0
1
2
0
2
write down
graphs,
the
0
0
0
1
1
1
0
0
the degree of each
B
A
C
\\D
i
i
1
1>
0
0
1
a
0
0
1
a
i
1
0
0
\302\243\342\226\240
1
0
0
tables:
adjacency
following
1
2
E
+G
(b)
(i)
A
\\B
\\C
1
1
0
1
2
0
0
0
1
0
0
2
0
2
0
3
o
\\2
lfsM
^
\\E
\\D
o
1
Q,
\302\243^%
\\E
0
1
1
0
1 1
1 1
\\D
\\C
o
A
(a) (i) \342\226\240mraramn<\302\273>
0
1
1 1 0
1 0 2
0 2 1
1
0
with
graphs
(ii)
D
\\C
0
0
D
\\E
tables
0 2
\\B
o
BB
\\D
\\C
1
Without
1
1
0
6.
9
1
\\B
A
0
1
A
(ii)
1
1
0
0
1
D
\\D
\302\243
\\C
\\B
0
0
2
1 2
0 1 0 0
1 2 0 0
2
i+s*
\\
adjacency
1
0
2
0
1
(ii)
D
\\C
\\B
- ^
2
2
0
3
0
(ii)
A
\\B
2
\\C
D
1 1 3
1 2 0
1 2 0 0
D 3 0 0 0
1
*
(a
^
'
K^
V
+/^
Draw each
7.
graph in planarform:
8. Showthat
(b)
Z^t
A
(a)
of the
each
graphs
by listingthe two
is bipartite
sets
of
vertices:
B
A
(a)
F
tu\\
(b)
V
C
9. Draw an exampleof eachofthe
(a)
A simple,
connected graph with 6 vertices,eachof degree2.
(b)
A
non-connected
simple,
graph with 6 vertices,
degree2.
+G
M
following:
A
simple,
(d)
A
multigraph
(e)
A
simple,
(a)
Draw
with
bipartite
the graph
(b) Draw the graph
(c)
Prove
graph with 6 vertices,
connected
(c)
each of degree3.
degrees 2, 2,4,4,4, 4.
graph with 7 vertices and 11edges.
6 vertices with
K5.
K2 3.
f\342\200\236\\
number of edgesin the graph Kn
that the
each of
is
K2J
[7 marks]
The
graph
(a)
Write
Kn has
in
(b) Showthat
vertices
the
down
a vertex
36 edges. Find the value
smallest
a simple
in
with
and the
connected
a simple
the same
connected
of
largest possible degreeof
graph with n vertices.
graph there
78
Topic
10
-
Option: Discrete mathematics
3$)
^ q,vv,
\342\200\242V^
+ o
must be two
degree.
(c) Draw an exampleofa connected
multigraph
vertices
with degrees 1, 2 and 3.
fsM
[3 marks]
n.
with
three
[8
marks]
'
^
V
I^C+z-
Some
In this sectionwe
will
graphs.
You
so they
are shown
orked
be
will
as Worked
coming
the
exam,
6.5
of the
degrees of all the verticesin
number of
vertex is the
of a
out of
proofs for
examples.
of edges.
degree
important
to learn these
expected
sum
the
properties of
some
prove
example
Prove that
The
theorems
important
it;
of a vertex we are actually
Every edge
edges'
the
This result is known as TheHandshaking
lemma,
can
be
of
as
handshakes
edges
thought
representing
as people. It is used in proving many
other
theorems,
the one in the next example.
two
so each
edge
vertlcee.
two
vertlcee,
the
because
and
ends,
the number
when adding up the degrees
of all
each edge is counted twice.
Hence
edges
counting
has
corresponds to
the degree
in determining
so
to twice
is equal
a graph
vertices
as
such
orkedexample6.6
are
{Below
graph, the
in any
that
Prove
two proofs
number of verticesof odd degreeis even.
of this result; the first oneis shorter,
one may
second
the
but
be
more obvious.)
>
is the
What
or to
y
main
better
give
of
function
intuitive
proof:
To establish
understanding
the
the
of
truth
of the result? Which
of
result, to be elegant,
the two proofs do
you prefer?
vertices?The
that
the
can
to
try
we
do
What
know
about
degrees oftells us
lemma
Handshaking
of all degrees
an expression
write
sum
is even.
So we
of
for the sum
all
degrees
First
Froof
Let a
be the
number
of vertices
degree, and b the number
odd degree.
The sum of all degrees is the sum of
even numbers and b odd numbers:
^degrees
= (a evens)
odd
+ odd
= even
but
odd
+ even
+ odd
+ odd
= even \342\226\240
= odd
I3y the
handsha
the
I3ut
the
even
If
king
lemma,
LHS is
even.
always
even,
be even.
of odd numbers can only be
is an even number of them.
bracket must
second
sum
there
a
+ \\b odds)
is
of even numbers
so the first bracketis even.
The sum
Hence
with
even
even
even
of even
of vertices
Henceb is even,
as
required.
6 Graph theory
^^%0c.
f
-
*
^
(a
v
^<+^
+ )
continued...
Froof
Second
can
We
think
edges,
up a
building
vertex by adding
a single
from
about
how
about
think
and
graph
\342\200\242
single vertex and
of vertlcee of
with a
a graph
In
vertices and
edgee,
the number
the degrees
degree
is zero,
change as we do this
build
can
\\Ne
is even.
which
up any graph
a sequence
by
types of operation:
an edge between two existing
of two
add
1.
no
odd
vertices
2.
1 only
Operation
the
which
vertices
two
need
We
of*
are being connected.
the degrees
changes
different
to consider
cases,
depending on whether thosedegreeswere
or odd before the extra edge was
even
added
an
add
1: The
With
extra
of the two
degrees
vertices
\\ncreaeed
are
connected
being
vertex
by
1, and
remain unchanged.
were even, they are
two degrees
odd
so the number of odd verticesis
the rest
If
the
now
increasedby
2.
two degrees
even so the number
If the
wereodd, they
vertices
of odd
now
are
is
decreased by 2.
If one of the degrees was even
the
and
other one odd, the first one becomes
the eecond becomee even. So
odd
and
the
number
of odd
degreee
remains
unchanged.
Operation2 does
not
existing vertices.
of*
degrees
change
The new vertex has no
edges
that
Remember
we started
vertices of
the
with
yet
of*
number
odd degreeequal to
zero
With
2: The
degree
of the
zero,
which
is even.
The other
don't change.
odd degreeremains
number
So the
Hence
number
as we
build
of vertices
new
vertex
is
degrees
of vertices
of
unchanged.
up
graph, the
the
of odd
Increaseor decreaseby
degree can
2, or
remain
as an even
always remains even.
unchanged. As it started
number,
The
rest
of the
results in this
it
section concern planargraphs.
the
edges do not intersect,
is divided
into a certain number of regions.Counting
the
plane
outside'
as one region, the graph shown overleafdivides
the
into 6 regions. In what follows,we denoteby/the number
plane
of regions,
v is the number of verticesand e isthe numberof
edgesofthe graph. So for the graph overleaf, v = 5, e - 9,f= 6.
The resultin the next example is called Euler's relation.
When
80
Topic10
-
Option:
Discrete
a graph
is drawn
so that the
mathematics
fsM
3$)
^
vs.
cl
\302\243%*..
.Ok\"
^
*
(a
'
K^
V
+/^
)
1-
Euler's relation
\342\200\242
\342\200\236
in
the
in me
is
\"
Formula
\342\200\242
given
booklet.
orked example 6.7
Prove
(You
sure that
As
we
are
can
try
to
apply
with
working
of
the number
edges and verticesof
v
a
graph,
we
from the
a strategy
previousexample:build up the graph
see
by adding vertices and edges, and
how
the numbers
+a
e,
To keep the
graph connected,
whenever we adda vertex
join
we
it
to
e +
f
= 2.
a connected
In
have
graph.)
with a
a graph
v=\\,e=0,f=\\,eo
\\Ne
we will
proof
up any connected
build
can
single vertex and
v-e + f = Z.
of operatlone of two
sequence
add
an
edge betweentwo
(i)
no
graph
edges,
by a
types:
existing
vertices
f change
and
v
\342\200\224
is not true if the graph is not connected.
In the
always
is about
result
this
planar graph,
this result
that
check
can
to make
a connected
for
that
have
a vertex and
(ii) add
of the existing
to
an
edge
joining
it to
one
vertices.
something
(i), v remains
With
and
flncreaeee
one of
So
now
the
we
v -(e
by
regions
unchanged, e
1
the
because
increases by
new edge
splits
into two.
have
+
\\)
+
(f +
\\)
= v
- e+f
theory
Graph
^
Q,VVj
1,
/
-
81
*
^
(a
v
^<+^
continued...
a new vertexand an \342\200\242
edge 'sticks out', so it
any of the regions into
we add
When
new
the
edge,
not
does
split
With
elncreaee
(ii), v and
by
1
f remains
but
unchanged, because none of the existing
regions
into two.
are split
two
So
now
we
have
(v + /l)-(e
In
case
either
change, so
a
stretching
Euler'sformula
in
the
the value
is always
+f
v-e
v
\342\200\224
+ f
e
doee
not
ec\\ual to 2.
first
was
In that
faces).
imagine
=
for polyhedra (three dimensionalshapeswith
discovered
polygons
for the number of vertices, edges and faces.You
can
v, e and f stand
polyhedron
being turned into a planar
graph
by making a hole in one face and
it out
flat. The face with
the
hole becomes
the 'outside' region.
Euler's relation
for
it
+ i) + f
case,
was
study
objects
EXAM
HINT
Both
of these
branch
in the
holes,
There are two
results are
first
The
^
booklet.
+G
other
one
simple,
connected,
example
of vertices
for planar
graphs,
and the numberof edges.
example 6.8) appliesto any
planar
graph. The second one (provedin
6.9) applies only to certain planargraphs.
(proved
Worked
results
important
number
the
only
involving
Formula
the
in
given
with
platonic solids. It is now an important tool
of mathematics called topology.
to classify
used
originally
of solid
\342\231\246Jf
4fJ\302\245i
in Worked
orked example6.8
for a
that
Show
(a)
show that
(b) Hence
There is
no
simple, connected,
the
complete
connection
of verticesand edges of
except
that
e>n-l
if the
graph
is connected.
So there is something special about planar
graphs.We know one equation that could be
relevant: the Euler'srelation
fsM
W
82
Topic
10
-
Option: Discrete mathematics
^ q,vv,
three
or more
vertices,
not planar.
in general,
graph
graph
K5 is
the number*
between
a
planar graphwith
(a)
Euler's
relation:
v-e+f=2
e < 3v
- 6.
*
^
(a
v
^<+^
continued...
Each region has at
and each edgebelongs
Can we find athe number of regions
need to eliminate f.
We
between
connection
and the
of vertices
number
or edges?
means
to
belongs
regions,
twice. So
counted
it is
that
two
to
two
exactly
\342\200\224
,l.e.^
5oe>\342\200\224
Then
which
in
three edges,
regions.
Each region has at least three edges. Sothe
total
number
of edges is at least 3f, except
each edge
least
e=v+f-2
fact
e<v +
=>
e <
^>
3
+ 2e
3<5 < 3i/
^>
2e
o2
-6
- 6
3i/
ae required
the result to
Apply
k5
(b)
k5 has
So
3i/
v
= 5
and
= 9 < e, hence
- 6
e
-
= 10
[\\)
K5 is
not
planar.
We saw in the
previous sectionthat
the
complete
graph
bipartite
= 6 and e = 9, so it
K3
planar. However, this graph has v
satisfies
the
e < 3v
6 for planar graphs. To show that
inequality
is
not
we
need
a
second
result,which applies only to
K3 3
planar
three
planar graphs with no 'triangles'(cycleswith
edges).
3 is not
orked
(a) Showthat
+a
if a
then
triangles
(b)
6.9
example
e < 2v -
that
show
Hence
simple,
K3 3
planar graph with
connected,
4.
is not
more than threeverticeshas
no
planar.
than
the one in the previous Worked example, because
(Note that this is a strongerinequality
- 4 < 3v - 6
\342\200\224
2v
v > 3. For
has
v = 4 and
e \342\200\224
6 so e>2v
A, but the graph is still
example, K4
for
3.)
planar The result does not apply to this graph because it containsclosedpaths
of length
This looks
example,
cycles of
very similar to
region
previous*
for the condition
except
length
the
3.
This means
has at
that
least 4
(a)
relation:
Euler's
of no
v-e+f=2
each
edges.
Each region has at least four edges
(as the
smallest closed path is of length 4), and each
edge belongsto
exactly
So e>
f <
i.e.
\342\200\224,
two
regions.
\342\200\224
2
2
Hence,
e<v+--2
2
<^2e<2v
<=>
+
e-4
e<2v-A
ae required.
6
^SV^
+ c,
Graph
theory
*
(*
'
\"~
V
C+^
+ )
continued...
need
We
contains
Apply result from
to make sure that
no cycles
of length
(a) to
this
\342\226\240
k3 3
graph
the
for
three
(b) K55 has v =
It has no triangles,
3x3
because
group
So 2v
- 4
include two
=
<3
< e,
= 9.
=
\302\243
vertices
to
each
other.
joined
to
have
result to apply
6 and
a triangle
from
hence k3i3
the
would
same
1-
is not planar.
1h
Exercise 6C
1 to
Questions
3 concern the following
A
four
graphs:
_D
A^
(a)
(b)
(c)
(d)
+G
1.
Verify
2.
Write
3.
Q
that
down
the
the
Handshaking
number
lemma
of vertices
holds for
of odd degreein
Draw each graph in planar form and verify
holds for each graph.
A
planar
graph
has 5
vertices and
each graph.
that
each
Euler
divides the planeinto
graph.
s relation
4
regions.
84
Topic10
-
Option:
Discrete
have?
edges does the graph
an example of such a graph.
(a)
How many
(b)
Draw
[4marks]
mathematics
fsM
3$)
?\302\243*\302\273.
.ON\"
'
\"~
*
(*
V
C+r*
*
Draw a simpleconnectedgraph
in
or
following degreesequences,
which
the vertices
have the
does
not exist.
one
why
explain
)
(a) 3,3,2,2
(b) 3,3,2,2,1
A
10 vertices
has
graph
(a)
How
(b)
Show
[9 marks]
2,1,1
2,
2,
(c)
many
that
and each vertex has degree5.
does the
graph have?
the graph cannot be planar.
edges
and
\302\247J Subgraphs
Sometimes we want
to
in
coloured
G.
and
blue
In
the
on the
concentrate
of both
When
we
graph.
Hence if the graphshave
G'
marks]
vertices which
The
is the graph with the same
each pair of verticeswhich are
the diagram shown here, the edgesofG are
edges of its complement G' are red.
the edges
take
[7
complements
are not joinedto each other, rather
G
complement of a simplegraph
verticesas G, and an edge joining
not adjacent
1-
than
those that are.
G and G' we
v
vertices
a complete
get
and
G has
e edges,
\342\200\224
e
has
edges.
K2J
Sometimes
subgraph
A
we are only interestedin certainedgesin a graph.
of G is a graphwhich
contains
some
of
the
only
edges
of G. In the diagramalongside,
a
subgraph
red. Note
that
every
complete
graph
with
graph
n vertices
is highlighted
in
is a subgraph of the
Kn.
+G
6D
Exercise
1. Draw the complements
ofthe
(a)
.
(i)
following
graphs:
(ii)
Graph
theory
85
fsM
^
q, vv,
\342\200\242V^
+ o
/ -
.ON
* ^
'
K^
V
i+s*
(b)
2.
(i)
Find the
adjacencytables for
the
complements
of the
following
graphs:
(a)
B
o
1
1
H
|o
3.
For
of the
each
exactly
(b)
D
1
1
0
o
0
0
0
0
0
1
0
1
0
0
following
C
B
e
1
0 1
1 0
0 0
0
1
1
graphs, draw all subgraphs with
4 edges:
(a)
4. Show that
each
of the
following
graphs contains
subgraph:
(a)
Topic
10
-
Option: Discretemathematics
m
(b)
K4
as
a
D
1
0
0
0
a-
*
(a
*
v
-^<+\342\200\236
+
Draw a graph G with
that G
such
5 vertices
and its
complementareboth
connected.
Show that a
(a)
marks]
[3
bipartite graph cannotcontain
K3 as
a
subgraph.
show that
(b) Hence
1-
bipartite graph with more than
is not bipartite.
complement
is a
if G
4 vertices,then
its
)
[7 marks]
Moving around a
In
on
we
problem,
we
whether
which
you
to return to the starting
All
names
must learn.
of adjacent edges.A path
and
a trail isa walk with no
vertices,
is any
walk
be required
ways of moving around a grapharegiven
different
these
may
every vertex, or there may be restrictionson
are allowed
to use each edge morethan
once.
visit
vertex,
A
of getting
ways
the
theory we are interestedin
from one vertexto another.Depending
of graph
applications
many
different
graph
no repeated
sequence
is a
walk with
repeated
edges.
B
c
ABCDE: a path
ABCBD:
not a path,
but
ABACD:
a trail
not a
trail
+G
ends at the same vertex and has
it
is a
(so is a closed path).A circuit
repeated
walk that starts and ends at the same vertex and has no repeated
edges(soitis a closed trail). Note that if there are no repeated
vertices
then
there
are also no repeated edges;henceevery
cycle
is a
A cycle
no
walk that
is also
starts
and
vertices
other
a circuit,
but not every
circuit
cycle.
JO
B_
D
ABCDE
is a
A:
cycle and circuit
A
ABCDBEA: circuit
but
not
cycle
Graph theory
^SV^
+ c,
/
-
87
.ON
'
* \"~
(*
V
C+^
+
orked
6.10
example
For the graphshown
(a) List
(b)
in
the
diagram:
all paths of length3 fromA
List all
to
D.
the cyclesof length4 starting
(c) Find a trail
that
uses
This can only
be
done
A
)
each
of length 4
A cycle
Some
vertices
will
1h
Think
where
can
go from
different
vertices
has four
different
vertices*
appear
more
than
{a) A3CD, AE3D,
we*
four
3 has
1-
at B.
ending
exactly once.
edge
by inspection.
of length
path
and
AECD
A
(b)
3CEA3,
3ECD3
(c) EA3EC3DC
once*
^^ In the nextsectionwe will see that
^^
can only be solvedfor somegraphs.
questions
like part
(c) ^^^
^^
Exercise 6E
+G
1.
down
Write
all walks
of length
3 from
A
to
D in
the following
graphs:
(a)
88
Topic 10
-
Option:
Discrete
A
b
(b)
mathematics
vs.
^
q,vv,
**+\302\253.
.on\"
a-
*
(a
*
1 ^ - ^
v
<+/*
\\
+ )
1\302\253
(d)
(c)
1-
2. Find all cyclesof length4 in the
from
graphs
the previous
question.
Find the numberof different
given verticesin K4
if n
Find the
(c) 4
number of different
given verticesin K4
if n
[5 marks]
of length
walks
two
n between
is:
(c) 4
(b) 3
(a) 2
J\\
two
n between
is:
(b) 3
(a) 2
of length
paths
[5marks]
Eulerian graphs
The
+G
islands
bridges
is considered
problem
to be the first
in graph theory. Two banks of the river
are represented by verticesof a graphand the
and
problem
published
two
bridges
Konigsberg
by its
seven
edges. The resulting graphis:
Euler
Leonhard
>
1783) was a
S
who
contributions to
is responsible
problem
is to
find a
closed walk that
each
uses
once. This corresponds to crossingeachbridge
and returning
to the starting point.It turns
out
impossible, and
the searchfor
developinggraph
verticesofthe graph.
theory.
The
an
explanation
reason
lies in
edge
that
exactly
once
exactly
this
is
led Euler to start
the degrees of the
today:
mathematics.
of
for
much
still
notation
mathematical
The
the
base
of natural
in
use
a
f[x) to represent
function, X to represent
as
made
mathematician
immense
many branches
He
(1707Swiss
sums, e
logarithms
and i as the simplest imaginary
number. Is new notation
new
mathematical
knowledge?
A3
6 Graph
theory
89
VS.
^ Q,
\302\243%*..
'
\"~
*
(*
V
C+^
each edge exactly
walk that uses
A closed
circuit.
an Eulerian
A
circuit is called
an Eulerian
with
graph
is called
once
an Euleriangraph.
6.5
POINT
KEY
Eulerian graph every vertexhas even degreeand it is
to find a closed walk
which
uses each edge exactly
possible
In an
once.
This result canbe explained
vertexwe must go in
out
and
to the
also applies
This
follows.
as
which uses
of it,
starting vertex,as
end. So the numberof edgesat
time we
Every
edges.
to it
at the
be even.
must
vertex
each
up two
return
we
visit a
orked example 6.11
In
the
that
Show
an
in the
graph
Determine
the
Eulerian
graph
degrees
Konigsberg bridges problemisnot Eulerian.
of all vertices*
all vertices
have even*
number of
+G
odd
with
^u,
^^
must be
(Key
odd
each bridge
the
and
walk
graph
at
the
is
there
odd degree, and the walk
degree.
to
the
starting
point,
it
exactlyonce,becauseofthe
each edgeexactly
once
is
called
semistarting point)
called an Eulerian trail. In a
which uses
to the
semi-Eulerian
end
is
graph
Eulerian.
a walk
returning
Eulerian,
^u,
^^
eo the
6.6
with
graph
(without
degree
even
point 6.4)so we cant
have just one vertex
with
A
odd decree,
to cross
KEY POINT
vertices
have
result:
following
that the
vertices
are not requiredto return
is impossible
Remember
3, 5,
not
if we
Even
vertices:
All
degree
3, 3
Degreesof
are
must
exactly
start
two vertices with
at one
of them
and
other.
trail is equivalentto drawing the graph
without
up your pen and without going over any edge
picking
more
than once. This is a good practicalway
to check
that you
have found an Eulerian trail.
Finding
The
the end
that can
90
3$)
Topic
10
-
Option: Discrete mathematics
^ cl
\302\243%*..
is similar
explanation
difference
fsM
an Eulerian
that
the starting
vertex only an 'in
have
odd
degree.
an Eulerian graph,with the
vertex only needsan but' edge and
to that for
edge,so
they
are
the
only
two vertices
a-
*
(a
*
1 ^ - ^
v
<+/*
\\
* )
orked
6.12
example
Showthat
the
graph
following
is semi-Eulerian
and find an Eulerian trail.
C
B
the
need
We
Degrees of the
of all the vertices*
degrees
A
are
eothe
vertices,
starts
trail
and
can start at
so we
odd
ends at the
A
and
end
Route
algorithm
inspection
exactly
graph
vertices:
= 4JP=1IE=2
two vertlcee
of odd degree,
is semi-Eulerian.
trail: A5CAECD
Eulerian
at D
graphs
<^^ Eulerian and semi-Eulerian
^^
\342\200\242
= 2,C
= 3I0
There
The Eulerian
1-
in Section
will
be
used
7D.
in the
<^^
^^
Exercise 6F
1.
For
each
graph shown
(a)
yJK
below say whetherit is Eulerian,
semi-Eulerian
or neither.
(b)
+G
theory
Graph
^
Q,
vv,
/
-
91
.ON
* *~
(a
*
v
\342\200\242+<*<;+'-
+
)
(d)
(c)
1-
1h
2.
the
of
Which
graphs
given by
the following adjacencytables are
Eulerian?
LA
(a)
0
1
o
\342\226\240a
11
H
1
J* C\\D |
1
E
1 1
0 0 1
0 0 1
0 0 1
D
(b)
El
0
0
0
0
1 1
1
o
o
\342\226\240H
K3
D
o B
0 i
0 i
i
i
1
i
D
0
0
0
0
Bt
m
A
B
lo
1
ll
1
0
Two graphs, G and H, areshown
(c)
1
1
i
0
1
10
1
lo
0
lo
1
C
D
E
F
1
0
0
0
0
1
0
1
0
1
1
1
1
0
1
0
0
0
0
0
1 0
0
1
*
below:
C
D
S\\
'E
if
i
+G
(a) Explain why
(b)
i
H
G
Find
Explain
Eulerian
an
G is not Eulerian.
Eulerian
why
the graph
circuit
in H.
shown below
[5 marks]
and
is semi-Eulerian,
find
an
trail.
^ 4
^
[5 marks]
i
U
92
Topic 10
-
Option:Discrete
mathematics
VS.
I -
.ON'
*
A Hamiltonian
path in a graph visits each vertex exactly
once
and
cycle visits eachvertexexactly
starting vertex.Not every graph has
if it does it is calleda Hamiltonian
graph.
Show that the following
a
a Hamiltonian
the
cycle;
Hamiltonian
graph is Hamiltonian:
cycle
by inspection*
is a
AC5DEA
easy way to check that a graph is Hamiltonian,
except by finding a Hamiltonian cycle.Thereis onecriterion
that
is sometimes
useful: If in a connected, simplegraph
Hamiltonian
is Hamiltonian.
graph
There
cycle, so
once.
returns
6.13
example
Find
\\
graphs
to the
orked
V
^ ^1 c+^
1^
Hamiltonian
A Hamiltonian
'
\"~
is no
n vertices
each vertex
has degreeat
least
graph is Hamiltonian. This implies that
is Hamiltonian (as each vertexhas degree
there are many
degrees.Two
Hamiltonian
examples
graphs
are
shown
with
n
\342\200\224
edges,
complete
every
where
here: a
then
the
graph
n -
1). However,
vertices have smaller
cycle and a cube.
/I
/
/
6
^S^Nn,.
Graph
theory
*
(a
v
i+s*
+
Problems
is no good
or not. will
Hamiltonian
about
Exercise
is Hamiltonian
salesman
Travelling
7E.
in Section
problem
difficult as there
a graph
tackling the
this when
see
can be
whether
decide
to
criterion
We
graphs
1-
6G
1. Find a Hamiltoniancyclein
each
of the
following
graphs:
B
(a)
(b)
F
K
E
G7\\
i
\\H
V
N
C
D
(c)
(
f
\\A
\\d
\\g
B
T
E
I
c
(d)
(^
9
+G
2.
(i)
(b) (i)
Find
cycles,
Topic10
-
Option:
Discrete
3$)
?\302\243*\302\273.
\342\200\242V^
+ o
mathematics
following
graphs and find a Hamiltonian cycle:
K4
(ii) K5
k44
(ii)
the
number
one of
identical.)
fsM
of the
each
Draw
(a)
94
)
of distinct
K3;3
Hamiltonian cyclesin K3
which is the reverseofthe
other,
are
3. (Two
considered
[4 marks]
1h
'
\"~
*
(*
V
C+^
*
)
Summary
A
of all the verticesis equalto twice
of edges.
number
the
verticesof odd degreeis even.
A
In a simple
no loops
are
there
graph
In a connected
it is
graph
In a
vertex is
The number of
1-
by its adjacency table.
be represented
can
graph
of each
direction.
edges have
a digraph
In
out
vertex.
of degrees
sum
The
joined by edges. The numberof edgescoming
of vertices
consists
graph
the degree of that
or repeated
to find
possible
edges.
a path betweenany
complete graph every pair of verticesis joinedby
an
two
vertices.
The complete
edge.
graph with
fn\\
n
has
K\342\200\236,
vertices,
edges.
V27
is a graph with
A tree
A
A tree with n
cycles.
vertices has n
groups of edges suchthat
has two
graph
bipartite
no
-1 edges.
from
edges
only
different
groups are
adjacent.
A
The
\342\200\224
e +
drawn without
\342\200\224
2
planar graphs:
relation)
(Euler's
f
edges crossing.
for connected
hold
relations
following
v
can be
graph
planar
e<3v-6
e < 2v
- 4
are no cyclesof length3 (e.g.for
if there
of a simple graphconsists
ofall
The complement
the
a bipartite
edges
which
graph).
are not
present in the
originalgraph.
A
contains
subgraph
Different
+G
only
of walk
types
some of
the edges of the originalgraph.
around
a graph
Path:
A
walk
with
no repeated
vertices.
Trail:
A
walk
with
no repeated
edges.
Cycle: A
closed
path.
trail.
A closed
Circuit:
In an Euleriangraph eachvertex
which
each
uses
edge
A semi-Eulerian
trail
between
A Hamiltonian
include:
exactly once
graph
those two
has
even
\342\200\242
possible to find an Eulerian circuit
and returns to the starting
point.
has exactly two verticesof odd degree.
Itis possible
tofind
verticeswhich useseachedge exactly once.
graph has a Hamiltonian
returns to the starting
It is
degree.
cycle,
which
visits
each
vertex
an
exactly once
^
Q,vv,
\342\200\242V^
+ o
and
point.
Graph
fsM
Eulerian
theory
/ -
95
.ON
*
(*
'
\"~
V
C+^
*
examination
Mixed
practice
(a)
Draw
the
(b)
Show
that
the complete
(Two cycles,onewhich
(c)
List
all the
graph
is the
Hamiltonian
which of the two
(a) Decide
6
graph with 4
complete
vertices, K4.
has
Kn
reverse
\342\200\224
of the
cycles in your
graphs
shown
K4
Hamiltonian
different
(n-l)!
(a)
Construct
(b)
List
(a)
(b)
[8 marks]
graph.
below
is Eulerian,
each
-
Option:
walks
in the
this graph.
of length 3 from B to D.
Discrete
ten matches.
team
plays
and justify
Eulerian graph from part (a).
A
teacher
exactly five
impossible.
[5
marks]
[7
marks]
table for
Prove
that the sum of degrees of all theverticesin a graph
the numberof edges.
Eleven
teams
take part in a school football league,so each
this is
Topic10
circuit
an adjacency
all the
to play
96
Eulerian
an
Find
cycles.
other, are consideredidentical.)
your answer.
(b)
)
wants
to produce
matches in the first
is equal
team
a schedule
term.
Show
to twice
has
such that
that
[7 marks]
mathematics
fsM
vs.
3$)
^
Q,VVj
'
\"~
*
(*
V
C+r*
given in the
G is
Graph
diagram
A
show that
(b) Hence
Q
the
Find
(a)
a planar
that
Prove
(a)
G
graph cannot have
be drawn
cannot
in planar
form.
[7 marks]
of a complement,G', of graph
of edges
number
3.
of length
cycles
G
with
n vertices
and e edges.
(b)
Write
(c)
Hence show that
of edges in
number
the
down
G and G'canboth be trees
(d) Draw a graph G suchthat
A
graph
(a)
e edges
G has
the
that
Show
a tree with
sum
G
and
G' are
n vertices.
if n-
only
4.
both trees.
[8 marks]
and n vertices.
of the
degrees of the verticesis twice
the
of
number
edges.
(b) Deduce
(c)
(i)
an even
G has
that
number of verticesof odd degree.
connected, planar and divides
If
three
regions. (n -1) verticeshave degree
G is
Graph
degreed, determine
(ii) For eachpossible d),
the
(n,
described
in
draw
a graph
exactly four
one
vertex has
exactly
plane
and
of (n,
values
possible
the
into
d).
which satisfies the conditions
[9 marks]
(c)(i).
IB
(\302\251
(a) Prove that
a bipartite graphcannotcontain
(b) Show that
the
(c) Is K4 4
Eulerian?
bipartite
complete
Justify
your
graph
K3
4
cycles
of odd
is not
Hamiltonian.
two
connected
length.
answer.
(a) State Eulers relationfora simple,connected,
planar
graph.
resultfora simpleplanar
(b) State and prove the corresponding
with
2007)
Organization
components.
[8 marks]
graph
[12 marks]
6
^SV^
+ c,
Graph
theory
In this
you
chapter
will learn:
\342\200\242
how
of
Algorithms
a tree
find
to
minimum
1-
on
weight
all the
connects
that
graphs
vertices of the graph
algorithm)
(Kruskal's
\342\200\242
how
to
the shortest
find
path between two
vertices (Dijkstra's
in question are very large, it
routes.
It is thereforeimportant
optimal
to have algorithms
which
can be implemented on a computer
and are guaranteed
to find
the best solution. In this chapterwe
will
meet
several such algorithms. We will also see an example
ofa problemfor which such an algorithm does not existand we
can only estimate
the length
of the shortest possible route.
algorithm)
\342\200\242
how
In
to
the shortest
find
route around the graph
each
uses
which
at least
edge
once (Chinese
postman problem)
\342\200\242
how
to
the shortest
find
to find
difficult
be
the graphs
where
applications
can
route around the graph
each
which
visits
vertex
at least
graphs
Weighted
once
(Travelling salesman
In
problem).
it.
a
that
is called
number
This
time
distance,
edge has
each
graph
weighted
or cost
a number associatedwith
the weight, and can represent
the
of travel
between the two
Note
vertices.
not the actual lengthof the edge, and that there is no
and
draw
the graph to scale. Also rememberthat
not
intersections
of edges are verticesof the graph.
this is
need to try
all
For
the
example,
A
network the
to
between
and
travel
^7
B
17
\302\243L
+c<
travel
between
go either
alongside
could
graph
numbers
different
a road
(in minutes) taken
So it takes 17 minutes to
junctions.
B and E. To get from B to
junctions
via A (which
could represent
be times
takes 19 minutes)or via
D
C, you
(which
can
takes
12 minutes).
When a graph is large,it may
]y^
D^
J
V
5
to draw it
be possible
not
it using
instead, we can represent
in
the
numbers
weighted graph
of the edges.The adjacency
table
above graph.Notethat all graphs
12
an
table.
adjacency
table
the
represent
the
showing
here represents the
we consider in this chapter
the
the cost
adjacency
It is
matrix.
possible
to perform
algebraic operations on matrices
to
r^
*v
various
deduce
of a
98
-
4fJ\302\245i
called
is also
edges
\342\231\246Jf
of
weights
properties
graph.
Topic
D
B
10
-
Option: Discrete mathematics
:- 4,
rnV
+ <*..
7
7
-
12
-
-
7
-
17
For a
the weights
shown
simple.
The adjacency table
and,
12 - 7
- 5
5
-
-
17
-
-
-
-
-
are
**\"
^
orked
b..ir*i
V
'
v.
/
7.1
example
Graph G has the adjacencytable:
B
C
D\\
E
F
-
-
9
6
-
6
-
-
-
6
9
\302\273
-
-
\302\253
6
-
4
6
-
6
-
4
-
9
5
A
I\302\253
(a)
Write
down the
(b)
Draw
G in
4
-
4
7
5
7
-
degree of vertex C.
planar
The degree of
non-zeroentries in
form.
a
is the
vertex
the
(a) deg (C)=3
number of
row or
corresponding
column
We can start
in order
by
To draw the
Hamiltonian
cycle
edges.
the graph
drawing
A,B,C,D,E,F,
Finally,
graph
in
planar
first, and the
write
the
vertices*
with
and then redraw
fit
(b)
in
planar
form
form,
find a
in the
weights
it
*
Hamiltonian
cycle:
ACDE3FA
rest of the
on the edges
7
?
^k^
+ 0[._
i-
Algorithms
on
graphs
99
7A
Exercise
1. Construct the adjacencytablefor eachweighted
below:
shown
graph
A
2.
(a) (i)
(ii)
(b) (i)
(ii)
Draw a weighted graphwith
in the following
(a)
the
weights
of the
1-
edges given
tables:
(i)
E
10 3
4 10
4 10 12 9 6
-
-
6
+\302\243i
C D
ii
A
10
3
-
(ii)
-
9
-
-
C
D
-
7
9
-
5
4
-
5
-
-
4
4
-
9
-
7
9
-
-
A\\
B
C
D\\
E
F
-
-
7
-
-
9
-
-
5
8
8
8
?
5
-
-
8
-
-
8
-
-
7
-
-
8
8
7
-
3
1'
8
-
-
3
-
1-
12
-
B
A
1'
^
l\302\273
(b)
(i)
A
\342\226\240-JL
\302\273
<-
u
H
\342\226\240
H
\342\226\240
\342\226\240
H
H
\342\226\240
I
I
H
N4V
-
\342\226\240
\342\226\240
D
H
\342\226\240
B
D
10
m
\342\226\240
\342\226\240
\342\226\240
\342\226\240
\342\226\240
Topic
(ii)
H
H
\342\226\240
\342\226\240
\342\226\240
100
*
D
D
D
\342\226\240
\342\226\240
B
\342\226\240
\342\226\240
7
Option: Discrete mathematics
t ^k^
+
Q[._
.ON
**\"
^
3.
Draw
in
planar
b..ir*i
given by the costadjacency
the graph
form
matrix:
(a)
R
C
A
B
5
5
-
7
5
-
-
10
9
12
E
D
(b)
C
-
9
7
10 12
5
4
4
-
9
-
9
-
-
8
8
10 8 are
G
E
10 8
- - 8 15
8 15 8
8 10
- 8
10
of edges in a simplegraph
The weights
D\\
shown
1-
8
-
in the
following table:
c
A
B
B
D
B
B
\342\226\240
D
B
B
B
EI
B
BB B
B
BB B
Rb
(b)
Draw
Hence
the
and list
graph
find
*
B
B
B\302\253
\342\226\240
(a)
\302\273
\342\200\242>
all Hamiltonian cyclesstarting at A.
cycle of
the Hamiltonian
least weight.
Draw
find
the
the
[7
weighted
graph corresponding to the given table and
of an Eulerian path starting at vertex
A.
length
H
A
B
I
-
I
14
9
-
-
-
12
-
11.
Consider weightedgraph
G
A
B
I
E
I
F
given
by the
-
-
10
-
[7 marks]
following table:
B
c
D
*
4
6
~
-
~
-
6
5
Q
4
-
Q
6
5
5
-
B
-
6
-
5
B-
-
5
5
-
the
its
D
12 14
9 - - - 10
9
-
9
1-
drawing
I
\302\261u
-
-
cycleand find
C
1Z
1\"
+<i
Without
marks]
graph,
length,
explain
explaining
why there
is an Eulerian
your calculation
[5 marks]
clearly.
7 Algorithms
? ^Ui)^
+
Oi..
on graphs
101
graph G is:
the weighted
for
table
The
B
c
D
~
7
-
7
-
-
8
-
-
-
-
8
5
-
A
\342\226\240
D
Q
D
8
Q
H
10
12 9
10
\342\226\241
Without
(a)
8
-
b
-
-
5
10
-
12
10
-
9
-
-
-
explain why G is not
the graph,
drawing
E
Eulerian.
(b)
H is
Graph
that
H is
obtained
from G by deletingoneedge.Given
Eulerian:
(i) State which edgeshould
(ii)
Find the
deleted.
be
length of an Eulerian cyclein H.
Minimum
tree:
spanning
[6
marks]
Kruskal's
algorithm
Six
The
shows
graph
where
a direct
length
of the
internet networkby
distances (in kilometres) betweenvillages
to be
are
villages
to an
connected
is possible.
connection
What
is the
cables.
minimum
cable required?
minimum connector problem.We only
edges to createa connectedsubgraph.
In particular,
if A is connected
there is no need for any cycles:
to B via C, then the edge AB is not required.This means
that
we are trying to find a subgraph that
is a tree and has minimum
Such
a
is
called
the minimum
possibleweight.
subgraph
This
need
of a
example
to include enough
is an
spanning tree.
Thereisa simple
for
algorithm
finding
the minimum
spanning
one edge
tree, called Kruskal'salgorithm.It works
by adding
at a time, starting with
the
and checking that we never
shortest,
create cycles.This is an example of a greedy algorithm, where
at each
we pick the best possible option(in this case, the
stage
shortest
not
possible edge).
always
work
for
In our example,
the
We
other
will
see
types of
later
that this
strategy does
problems.
is AC, so we add it to the
BCand
CD.Therearethen
graph
three
of length 6, but AB and CFarenotrequired,
as they
edges
would create cycles, so we just add CE.All the vertices
are now
our spanning
tree.
connected, so we have completed
first.
Topic 10
-
Option:
Discrete
mathematics
Next
shortest
we add
edge
BF, then
**\"
^
v.
/
7.1
POINT
algorithm
\342\200\242
Start
by marking
each
stage
a time.
vertices.
shortest remaining edgeas long
a cycle.
create
vertices
until all
edges
adding
Keep
all the
add the
asit doesnot
\342\200\242
the graphoneat
adds edges to
Kruskals
\342\200\242
At
'
V
is summarised below:
This algorithm
KEY
b..^
been
have
connected.
minimum
The
The next
example shows how to set out
is clear.You
your method
your
so that
working
which
ones
you
edges,
you skipped
would
create
a
and
draw
(because
they
cycle),
your completed
tree.It helps to list the edges in order of length first. Remember
that a tree with n vertices has n-\\ edges,soyou
know
when
have
the
tree.
It
is
alsoa
idea
to
draw
the
completed
you
good
tree as
note which
you go along.
orked
List the
there
more
may be
than
one
tree,**
the same
v/e-g^Vou
are
required
only
to find
one answer
onlessexpliatlytold
otherwise.
7.2
example
minimum spanning tree for
Find the
not
unique;
necessarily
the order in
to show
need
the
considered
have
is
*ee
spanning
edges in the
order they
were
edges
in order*
this
graph:
added
to the
tree, and
state the weight
of
your
tree.
+c<
We start by
of
first.
length
the
listing
If
two
have
edges
same length, we can list
CD 2
EF2
the
FG2
in any
them
order
AG
2
3CA
each edge
Consider
add
We
it
skip
to
the
graph
after
if
and
order
it does
\342\200\242
not
create a cycle
AE because we already
have
As there
in
EF,
EG,
GA
AE
4 skip
ED4etop
AC
5
E3 6
AD 7
are 7 vertices,we can stop
the 6th edge (ED) has been
added
7
v
^k^
+ G(._
i-
Algorithms
on
graphs
continued...
We
drew the tree
so we
edges,
tree.
The
weight
as we added the\"
have the complete
is the sum of the
weights
of
the
all
edges
Kruskal's algorithm is quick and
it is difficult
to check if you
graph
implemented on
cycles, and
called
1.
Use
this
easy to
are
a computer,we would
can
be
very
on
implement
creating
need
time-consuming.
a long
an
small
cycle. Also, for
additional
So for
graphs.
algorithm
large graphsthere
However on
an
algorithm
a large
to be
to check for
is an
alternative,
Prim's algorithm.
Exercise
7B
Kruskals
algorithm
and state its
(a)
(i)
Topic
10
to find
spanning tree for
the
graphs
below
weight.
1
-
a minimum
Option: Discretemathematics
(ii)
A
Draw each
tree
**\"
^
26
B
13 j>
io
(b) (i)
(c)
Use Kruskals
by
(a)
the
(D
following
12
Jv
/
D
5
8
5
-
7
4
7
7 10
4 9
7 10
- 8
10
9
10 8
10
-
(i)
E
- 30 - 35 40
50 30 70
-
70
-
35
40
50 45 A
-
4
-
6
-
10
10 -
50
-
-
15
15 -
-
20
B\\C
D E
4 - 5
5 - 6
6
6
-
5 10 - 4
-
7
-
-
(ii)
F
50
-
-
20
-
15
-
15
-
10
12
of the
14
15
-
16
13
13
-
11 14
15
-
16
19
18
18
A
B
C
D
50
45
40
40
-
-
\\A\\B\\C
5
O
5
-
Q
7
6
Q
O 8
F 8
-
5
-
^^
+
0[._
i-
18
18
-
-
45
35
D
19
50 35 40 45 45 - 20 35
-
1-
16
35 40
25
25
35
40
-
10
-
11
7
6
-
20
35
30
D
E\\F
-
8
5
4
5
-
4
4 4 5
3 2
7
?
graphs represented
16
12
-
45
-
*
10
5
4
7 - 2
2 -
each
m
-
D
C
(c) (i)
\"M
E
C
\342\226\240nnrann
10
8
(b)
v.
R
6
A_
tree
for
algorithm to find the minimumspanning
tables. Draw each tree and state its weight.
-
+\302\243i
'
V
(i)
Q~
2.
b..ir*i
-
15
15 25 -
40
30
25
-
-
8
3
2
-
Algorithms
on
graphs
1
B
8
The diagram shows graphK.
L>
^20
17
tree for K.
spanning
a spanning
(b) It is requiredto
which includesthe edgeCF.
Kruskals algorithm to do this.
Explain
C
8
12
14
10
12
10
12
12
14
of a
\342\200\242e
D
11
14
13
15
11
10
can adapt
you
not
need
to find
[6marks]
is meant
what
briefly
Explain
tree
10
do
You
weight
'15
(a)
cr
how
the tree.
16^
l'n
of least
tree
find
15
10
minimum
the
Draw
(a)
10/
12y
graph.
the minimumspanning
the
table alongside.Draw
by
(b) Find
E
9
10
11
15
14
11
13
10
F
by the minimumspanning
of the
tree
tree
your
graph given
state its
and
[7 marks]
weight.
(a)
(b) In a
12
A
can't
this
connected
simple
and 7 edgesof
20 and 22. Find the:
has 5 vertices
graph
13, 15,18,
10,11,
lengths
edge has weight 8. Sharon
spanning tree has weight 139.
be right.
[6 marks]
the minimum
why
Explain
have different
the shortest
and
that
finds
a tree with n vertices.
all edges
12 vertices
with
graph
weights
12
of edges in
number
the
State
(a)
minimum
possible
weight
(b)
maximum
possible
weight
of the
minimum spanning tree
of the minimum
spanning tree.
[5marks]
minimum
The
tree of
spanning
diagram has weight
Find
51.
the
the graph shownin the
range of possible values of x.
[5marks]
Finding the shortest
path: Dijkstra's
algorithm
We
often
two
between
a
represents
or
need
the
There can be many
so calculating
the length of each oneisnot
section
we will meet an algorithm which can be
routes,
practical.In
this
on
implemented
two places.
between
route
quickest
possible
jj
graph theory to find the shortestpath
An example might be when the graph
vertices.
road network and we want to find the shortest
to use
the
computer.
is that the shortest path doesnot necessarily
start
the
shortest
the graph alongside.The
edge. Consider
A is AB, but the shortest path from
shortestedge
to D
is via C. So we needto look different
of getting through
ways
The difficulty
with
A
from
at
the network, but avoid
to look
having
Dijkstra'salgorithm
all
considers
but only
keeps
of the
track
the currentvertex.
recover the shortest
When
path
Topic
10
-
Option: Discrete mathematics
the
at all
possible paths.
vertices
shortest path from
the
from
algorithm
the
start
in the
is completed,
to any
graph,
the start to
we can
vertex. In order
^
3u.
V
I
this, we needto keepsome
we go along.This
each
vertex as
a box nextto each
by drawing
distance
previous
from start vertex
of
order
for
information
done
is usually
/
follows:
as
vertex
v.
^C^e
h-v**-!
to do
'
labelling
temporary
labels
A
far.
in the
labels
The
top row
are called permanentlabels.They
we know that
to
this
vertex.
possible path
in once
written
only
Orderof
labels.
permanent
we
The starting
certain
end
vertex
end
that we
Distancefrom startis shortest
vertex to the currentvertex.
last
the
This
path.
box
allows us
recover
the
the previous vertex in
we write
shortest
the
the final
shortest
to work backfromendto startvertexto
path.
on
The algorithm
is much clearerwhen illustrated
show in red the information
stage.
an
example.
addedat each
We
orked
+c<
be
only
once
the starting
from
distance
the
In
the vertices are given
vertex will always be labelled1,
will not necessarily be last.We can
have found the shortestpossible
path
vertex has a permanent label.
the
but
are
the shortest
found
have
in which
order
is the
labelling
distance from start found
whenever we find a shorter
is updated
information
This
the shortest
shows
label
temporary
so
path.
(a)
example
7.3
Find the shortestpath
from
S to
T in this
graph:
A
C4
down
(b)
Write
(c)
If we are not
the length
allowed
WD
of the shortestpath
to use
the edge
from
S to
D.
BD, what is now the shortestpath
7
v
^k^
+ G(._
i-
from
Algorithms
S to
T?
on
graphs
continued.
Start
S and
by labelling
vertices\"
the
to it.
adjacent
least distance from
Vertex A has the
a permanent label,
in the top row
it gets
so
S,
write
we
which
B
adjacent
For
8,
to
lis
A.
4 =
7, but
label 6, so this
3 +
The vertex
labelled
with
3 + 11
= 14.
has
8 already
is
not
updated.
the
with
label is C,
temporary
which are*
at vertices
look
now
We
so
this
smallest
is
made
permanent
Vertices adjacent to C
8: 4 + 1 = 5 < 6, so this
are 8 and
Vertex
label,
now
that
noting
the
in
\342\200\242
is updated.
D: 4
8
D.
+4
=8
gets a permanent
the previous
vertex
of length
path
5 was
C
3
Vertices
D:
T: 5
to 8
adjacent
5 +
+ 8
are D and
2 = 7 < 8, so it
= 13 < 14, so it
D now
Vertex
label,
noting
is
updated
is
updated
T.
4
\342\200\242
gets a permanent
that
it was
reached
from 8
3I4|S
r^
,v
108
Topic
10
-
Option: Discrete mathematics
S
^ *~
'
V
k^h
^
/
continued...
12< 13,so
T:7 + 5 =
is
T
\342\226\240
updated
again
It is
remaining vertex, so it
the last
gets
labels,
permanent
can stop
so we
The
of the
length
shortest path is the*
middle number
We can
label
a permanent
have
vertices
All
work back from
find
Tto
actual
the*
back
Working
use the
back
to go
(b)
diagram*
completed
D to
from
Working
If
back
can't
this
not changed
stay at 13.
8
from
notice
got
in
to 7,
T:
path is
from
back
5-C-3-D-T.
D:
shortest
path
from
5 to
D is
5~C~3~D,of
7.
without
3D\\
5-C-D-Tboth
5-C-3-T'and
have length
13.
so T would
13
that
can be
different
two
(c)
then D is
be done,
But
added
we
when
to
12.
D-3-C-S
S
length
We look
from
shortest
So the
BD:
path is
shortest
T-P-3-C-S
path
So the
We can
of the
length
T's box
the
in
The
ways.
+c<
EXAM HINT
for each
list
to
the
a
graph
shortest
is given
Dijkstra'salgorithm
vertex
a diagram
contain
vertex, like the
final
its
a
in part
diagram
state
path and
with
box
completed
(a). Don't forget
length.
HINT
EXAM
If
must
solution
Your
in
as
cost
a
you
and all the edges
labels before moving
on
adjacency
draw the
matrix you can
graph. Begin
from
starting
to more
edges
it. Then
with
follow
the
start
add temporary
and vertices.
7 Algorithms
v
^k^
+ G(._
1-
on
graphs
7C
Exercise
1. For
each of the following
- fill
in the
- find
-
boxes to carry out Dijkstrasalgorithm
the shortest
down
write
networks:
the
path (or paths) from S to F
of the shortest paths from Sto
lengths
D,\302\243andF.
For
versions
larger
filled in,
of the
see the endof this
diagrams
chapter
that
(pages
can
be photocopied
and
131-133).
XT
(ii)
(a) (i)
-D
10
10y
r H
12
Hsj
5
-\302\243
VF
TT
r3
c-
(b)(i)
(ii)
(c) (i)
(ii)
+c<
-6
E
I\342\200\224Fpx
2.
Use
Dijkstras
(a) (i)
A
x
110
Topic
10
algorithm
-
to find
2\342\200\224D.
the shortestpath
12
y
Option: Discrete mathematics
from
S to
T:
^ **\"
b..ir*i
'
V
v.
/
(ii)
*-
3.
Use
algorithm
Dijkstras
to find
shortest path from Sto F in
the
weighted
given by
graphs
the
following tables:
(a)
(i)
IB
s
6
D
O
O
Q
6
-
8
4
8
-
D
7
Q
-
D
D
B C
8 8
4 - 6
6 -
A
7
7
-
-
6
(ii)
H
-
-
-
-
12
9
5
11
-
G
6
-
-
\342\226\241
\342\226\241
F
5
-
8
8 7
5 - 12
*
8
-
12
11
12
6
9
6
-
3
5
-
8
5
3
5
-
S\\A\\B\\C\\D
6
4
5
-
+\302\243i
4
5
5
-
6
6
2
-
B
(b) (i)
A
5
6
6
4
B
2
5
6
4
3
3
6
5
5
2
C
6
5
4
5
2
D
4
3
5
-
1
-
-
-
2
3
2
1
-
G
H
I
J
E
7
4
\302\273
8
Note an unusal
3
format
15
12
18
18
of the
v/hich has
7
8
22 21
exam
\342\200\242in
appeared
in
questions
the past.
30
16
56 31
table
24
40
50
42
26
39
33
23
7 Algorithms
? ^Ui)^
+
<*..
on graphs
111
(ii)
17
31
H
41
32
D
38
25
28
7
H
D
X
17
A
V
yP 8 H.
8/
'
Q*.
6
\\
2
V3
5/
7\\
3/
^0
/
The
C
B
6,
J^
6s
29
31
23
29
16
46
41 22
42
M
/
the cost
represents
graph
21
of train
journeys between
towns.
different
(a)
Use Dijkstras
algorithm
S to M.
State your route
(b)
Write
/5
\\l
16
of the
the cost
down
to find the cheapestroutefrom
and its cost.
cheapest routefromS
to
/.
[8 marks]
L
F
<R>16X^\\J
AK
\\
13
^7
9
22
31 19 12
\\3
yw
4\\
24
H
2/
10 V/
Bt
D
B
showsa networkof roadsand their lengths.
A to G.
from
(a) Find the length of the shortestpath
Find
the new shortest
(b) The road BEis closedfor repairs.
The diagram
\\2
8
^^Cr
y^i?.
D
C
path from
A
to
G, making
your method
clear.
[8 marks]
The diagramshows
a weighted
graph. The unknown
shortest
length x is a positive integer.Ifthe unique
A
from
to
G
is
show
that
there
is only one
ABCFG,
path
[6 marks]
possiblevalue of x.
+\302\243i
|\302\273j
Chinese
The
This problem was originally
posed and studied by the
Chinesemathematician
jTlki
**++*
which uses eachedgeat
vertex.
about
Think
travel
every
along
even
which useseachedge
they
sum
met
We
<^[ graphs
in
Section
<^[
Eulerian,
your
postman
112
*\\?
Topic
10
-
the
weights
If there aresome
Eulerian
there
6E
of
Option: Discrete mathematics
so
some
find
least
a route
once
all
of all
vertices
edges
the
length
to the starting
letters: He needs to
returns
Eulerian circuit
If there is more than one
once.
same length, which is equalto the
degree,
exactly
have
the Route
of minimum
and
a postman
delivering
street at least once.
If all the verticeshave
suchcircuit
also called
problem,
postman
edges:
problem
postman
inspectionproblem,isto
+^
1962.
Mei-KuKuanin
Chinese
the
all
along
Travelling
there
is an
the edges.
of odd
need to
degree the graph is not
be repeated. We know that
be an even number of verticesof odddegree.
For
need
to
know
how
to
solve
the
Chinese
exam, you only
for graphs
with two or four verticesof odd
problem
can only
**\342\226\240*
*
of odd
vertices
degree you
v
/
can use the
algorithm:
following
KEY
are two
If there
degree.
'
V
POINT
7.2
Chinese postman
\342\200\242
of odd
vertices
the
Identify
algorithm
degree.
path between those two vertices,
eitherby inspection
or using Dijkstras
algorithm.
\342\200\242
The
in
this
shortest
need
to
be
usedtwice,
path
edges
and
all the other
edges are used only once.
\342\200\242
Find
shortest
the
\342\200\242
The
edges
the
plus
orked
route is the total weight of all the
of the above shortest path.
length
of the
length
7.4
example
(a) Find
the length of the shortestChinese
postman
(b) State which edgesneedtobe used
(c)
Find
2
A_
Check
+c<
the
of all
degrees
Find the
In
this
shortest path
case,
8 to*
from
do
we can
\342\200\242
A.
B
3
(a)There
are
C
two odd
vertices: 5
and
F
vertices
the
F.
graph.
twice.
route starting and finishing at
one such
this
for
route
this
&toF:&-l-F{\\er\\Qth=7)
by
inspection
The length
is the weight
edges
of all
the*
Total length
=42+7=49
plus 7
(b)Theedqee31and
Find
one
inspection, remembering
IF can be used twice.
way
to achieve
include
route
possible
this,
if
8/and
The
easiest
possible,
BIFIBat
by*
that
(c) A
possible
route:
IF are
weed twice.
A3IFI3CPIPEFGHIHA
is to
some point
7 Algorithms
v
3*1,**
+ Q[._
on graphs
113
If there arefour
of the
lengths
combination.
B, C
of odd
vertices
degree we
AB and
in three
up
best
A,
ways:
possible
CD
BD
and
AC
the
paths between all possiblepairsand pick the
For example, if the four oddverticesarecalled
then they canbe paired
and D
need to look at
AD and
BC
To calculatethe length
of the repeated
edges for the first pairing
A to B and
need to add the length of the shortest
from
path
the length of the shortestpath from C to D. When we have done
we
lengths of
for
one
that
gives
the
length of repeatededges.
the Chinese
To solve
obvious
seems
7.3
POINT
KEY
even
combinations,
one is the
which
we then choosethe
roirtes
all possible
tfit
three pairings
shortest total
the
Wr\\te down
for all
this
to
need
You
shortest.
postman problemfor
four vertices
of odd degree:
\342\200\242
consider
all three
possible
a graph
with
ways of pairing up the four
vertices
\342\200\242
find
the
up those
\342\200\242
pick
the
shortest
lengths
length of the optimalChinese
for the graph shown in the diagramand
edges need to be used twice.
Identify
the
Write down all
and find
Remember
odd
vertices*
three pairings*
shortest
that the shortest
do
not
need
to be
paths.
paths
Topic
10
-
with the
repeated.
route
postman
state
which
Odd
vertices:
3, C, D,
E
3Cand DE:10 +
b
3D and
CE:
10 +
9 = 19
3E and
CD:
18 +
17 = 35
= lb
Option: Discrete mathematics
add
lowest total; those edges
(using 3AC)
(using
direct
Select the lowest total;
those*
need to be repeated
edges
114
combination
7.5
example
Find the
+c<
each pair of verticesand
for each
combination
needtobe
orked
path for
Repeat edges 3A, AC and DE.
The length of the route is:
+ lb = 73 + lb = 91
(all
edges)
3DE
and
CED)
^ **\"
b..ir*i
V
v.
/
7D
Exercise
1. For eachgraph
G
below:
(ii)
(i)
(a) Showthat G is Eulerian.
(b) Write down the lengthofa Chinese
postman
(c)
2.
'
a Chinese
Find
For each
route.
postman route starting and finishing
C.
at
graph:
(ii)
(i)
+c<
(a)
it is
why
Explain
and finishing at
(b)
Use the
impossible to
Chinese postman
finishing
your route.
at
A,
walk alongeachedge
of
the
graph
exactly once,
and
using
algorithm to find the route of
every edge of the graphat least
minimum
once.
State
length, starting
the weight of
7
v
^k^
starting
A.
+ Gl._
Algorithms
on
and
graphs
1
3.
For each
of the
^F
(i)
graphs below:
(ii)
^*
4
(a) List
Find
(b)
and
(a)
(b)
Write
State
vertices
the verticesof odd degree.
the
length
of the
state
which
edges
the
down
the
optimal Chinese postmanroute
need to be repeated.
two vertices
length
of the
of odd
degree.
of odd
degree.
shortest path between the two
the lengthofthe shortestroutewhich
starts
at C and uses every
of
the
edge
graph at
least once.
[6 marks]
(c) Hence find
and
finishes
The graphrepresents
the
in a small town,
the
weights
edges representing
lengthsofthe
roads.A salesman
wants
to visit all the houses in the town,
so he must
walk along each road once.Hewants
to start
and
finish at his shop, which is locatedat junction
C.
with
Topic 10
-
Option:Discrete
mathematics
the
of the
network
of roads
*v
^
V
h^A*1-
B
0.2
A
\342\226\240^5-iJc
+^
^
D
0.3
/
\302\260-5
o.<N
0.2
' W
/%
E
l
06 \\0.6
0.5
X
0.4
J7|'
>J
0.5
F
C
0.4
0.8/^
\\^
0.4
A
A
0.4
Usethe Chinese
route for
the shortest
to find
algorithm
postman
G
0.6
if
salesman.
the
possible
[8 marks]
(a) The diagram shows the town
of
The
islands and sevenbridges.
with
Konigsberg
bridge
bank (NB) and the secondisland
(I2)
corresponding graph, the weights
time, in minutes, to walk between
between
the North
is closed.
In the
the
of
its two
represent
edges
different
the
places.
NB
around the town
A tour
every bridge
+u
(b)
returnsto the
North
taken by
a tour.
The
bank, crosses
closed
one), and
(except
bank.
Find
the shortest possible time
You must make your methodclear.
such
minutes
between
for
time
at
least
once,
for the
NB and I2 reopens, and it takes
the two places. Find the
between
bridge
16
to walk
shortestpossible
North
at the
starts
once
least
at
a route
starting
crosses every bridge
at the North bank.
which
and finishing
[9
(a)
For
the graph
starts and
which
^1
shown in the
graph
at
of
length
least
finishesat
once.
Make
diagram, find
shortest
the
marks]
route,
uses each edge of the
method
clear and state the
your
A
and
route.
your
F
D
/n
12\\
/U
12s
15
E
16
21
21
G\\
X14
12y
i >\342\200\224\342\200\224
18
A
B
16
c
7
? 3,ry\\k
Algorithms
on
graphs
117
Kn
+ Oi.
.OS
(b) The edge GHof weight
once,
and finishing at A.
starting
uses eachedgeat
[9 marks]
in
the
table.
following
B
A
21
D
C
18
21
14 42
8
18 8
14
42
K is
why
Explain
16
6 4
21
4
21
11
6
4
4
16
(a)
graph. Find the
of the edgesshown
the weights
K has
graph
Weighted
route which
shortest
new
the
of
length
least
to the
is added
16
11
not Eulerian.
vertices
C and F in K.
(b) Findthe shortestpath between
and
at D, which
(c) Findthe shortestroute,starting
finishing
uses each edgeof K at least once. Make your method clear
and state the length
of your
route.
uses
each
(d) It is required to find the shortestroutewhich
edge
of K at least once,but does not need to return to the starting
vertex.
Find
(ii)
What
this
least
the
(i)
possible
and
start
length of such a route.
should be usedto achieve
finish vertices
route?
shortest
[10 marks]
all the vertices:
Visiting
salesman problem
find the shortestroute
around
a graph
which visits each vertex at least
once
and returns
to the starting point.Think
about
a salesman
who wants to visit
versions
of this problem,
every town in a region. Thereareseveral
on
whether
the
is
and
whether
we are
complete
depending
graph
allowed to repeat vertices.If the graph
is not
it may not
complete
have
a Hamiltonian
so
each
vertex
once
cycle,
visiting
exactly
may
The
Hamiltonian
<^[
were
introduced
cycles
in
Section6G.
<^[
Travelling
salesman
travelling
problem
is to
be impossible.
The version
in
weight
we will
find a Hamiltonian cycleof least
In
other words, we want to find the
graph.
solve
a complete
is to
route around a completegraphwhich
once
and returns to the starting
point.
exactly
shortest
One
and
find
Hamiltonian.)
inefficient,
Topic 10
-
Option:Discrete
mathematics
solve this
way to
their
weights.
This is
or even
visits
each
vertex
problem is to list all Hamiltoniancycles
(We know that a complete graphis always
to find the shortestcycle,but is
guaranteed
not feasible, for
large
graphs.
V
il^x
h-v**-
a complete
is because
This
Hamiltonian
+^
graph with n verticeshas -{n -1)!
becomes very
number
this
and
cycles,
' W
large,very
quickly.
1-
difficult than it seems. In fact, there is no
known algorithm (other than listing all possiblecycles)
which
the shortest Hamiltonian cycle.The bestwe
guarantees
finding
can
do is find upper and lower boundsfor the problem.
These
are two numbers such that we can guarantee
that the shortest
This
is more
problem
possible Hamiltoniancyclehas
between
somewhere
length
them.
For example, in the graphshown
is ABCDA with length 19.We
bound
the
for
salesman
travelling
bestpossible bound.
length16so know that
For
upper
cannot
possible
say that 19
is an upper
This is not the
problem.
the cycle
L < 16.It turns
ABDCA has
out that
16
is
bound),
possible length (so the bestpossible
upper
know this without checkingall possiblecycles.
the shortest
but
we cannot
On
the
other
we can
hand,
be certain
cycle shorter than 12. This
Hamiltonian
every
cyclein a K4 graph
Hamiltonian
the shortestedgein
our
has
graph
that
is no
there
we know
is because
that
edges, and
weight 3. Therefore L > 12;
consists
of 4
bound for the travelling
salesman
say
problem. So in conclusion, we can say that the length of the
12 < L < 16.
shortest possible Hamiltoniancycle,L, satisfies
if they are quite
are
Upper and lowerbounds
only useful
closeto eachother. Knowing
for example, 3 < L < 97is
that,
not very useful. However if, in a large graph, we can say that
21 < L < 24, then we have
a lot of information
about the
quite
In
this
level
of
problems
graph. many practical
accuracy is
a lower
12 is
that
we
+<l
shortest
19. We
cycle
there is a
one
shortest
the
example,
in fact
we
that
know
Hamiltonian cycleof length19,sothe
be longer than this. In otherwords,
Hamiltonian
cycle has length L <
Hamiltonian
one
here,
therefore
sufficient.
The
of L. We
to give
now
will
but
complicated,
KEY
above for
used
method
is unlikely
7.4
POINT
Nearest
neighbour
\342\200\242
a starting
pick
\342\200\242
go
to the
\342\200\242
one
algorithm for finding an
upper bound:
vertex
closest vertex
until
repeat
\342\200\242
add
finding upper and lowerbounds
us a sufficiently
narrow
for the value
range
look at some methods which are more
in much better upper and lowerbounds.
result
all the
more
not yet visited
vertices have been used
edge to
return to the starting vertex.
7
:- * r*k
hn
+ ft
Algorithms
on graphs
119
clearly gives a Hamiltonian cycle,soit provides
the shortest possible Hamiltonian cyclewill
bound;
This
In the graphbelow,
this value.
givesthe cycleADBCA
of
choosing
be
upper
vertex
the starting
A as
than
less
17.
length
B
4
A_
7.5
POINT
KEY
an
can be found
upper bound
An improved
starting
by
a
from
different vertex.
In
the
want
We
this
16 (ABDCA), so this illustrates
cycle
length
neighbour
algorithm will not always find the
has
Hamiltonian
nearest
the
that
we get the same
the shortest possible
We know that
of 17.
bound
as smallas possible.
we pick
vertex
whichever
example,
upper
to be
bound
upper
bestpossiblesolution.
at finding
look
now
We
bound.
7.6
POINT
KEY
a lower
bound for the travellingsalesmanproblem
can
by using the following algorithm:
A lower
found
\342\200\242
remove
the
\342\200\242
find
(and all
vertex
one
be
the associatededges)from
graph
the
length
minimum spanning treeforthe
of the
remaining
graph
\342\200\242
add
shortest
two
the
edges connecting
vertex.
We
For
want
shown
subgraph
A
are
BD
and
AD
and
we
may
not
the
conclude
can
be
immediately
as
10
-
Option: Discrete mathematics
A from
Its minimum
travelling
this
that
with
salesman
the upper
15 < L < 17.We
problem
bound we found
saw
earlier
that
the
Hamiltonian cyclehas length16.
bound,especially the
Topic
as largeas possible.
our graph leavesthe
spanning tree consistsof
CD and has weight 8. The two shortest edgesfrom
either AB or AC, with
the
total
7. Hence
length
actual minimal
It
to be
vertex
in red.
the lower bound for
is 8 + 7 = 15.Combining
earlier,
bound
removing
example,
edges
lower
the
the removed
clear why this method givesa lower
selected
above do not even form a
edges
^ **\"
b..ir*i
'
V
v.
/
have selectedthe correctnumber
of edges
(if there are
n vertices, any Hamiltonian cyclewill have n edges), including
the two shortest edgesfromA and two shortest edges
that
connecting the remaining vertices.Soit seemslikely
any
Hamiltonian
cannot
be
shorter
than
this.Ifthe
cycle
edges
in this
selected
found
way happen to form a cycle,then we have
the solution to the travelling salesman problem.
cycle. We
If
1-*
the graph we get a
different
lower
bound.
vertices B, C and D fromthe
Removing
above
lower
bounds
of
14, 13and 14,respectively.
graph
gives
Remember
that we want to make the lowerbound
as large
as
because
that
it
closerto
the
actual
solution.
So
possible,
brings
in this case the best lowerbound
we can find is 15.
vertex from
a different
remove
we
orked example 7.6
For
the
in the
shown
graph
diagram:
at A
(a) Usethe nearestneighbour
algorithms
starting
and going to D first, to find an upper bound for the
travellingsalesmanproblem.
vertex A find a lower bound for the
(b)
By removing
salesman
travelling
Write down an
(c)
problem.
inequality satisfied
by
the shortest Hamiltoniancyclein
By removing
(d)
vertex C find
Explain why we have
(e)
this
an
improved
found
actually
the
the
L, of
length,
graph.
lower
bound.
of the travelling
the solution
salesman problemfor
graph.
from A
Start
closest
and always go to the#
vertex not yet visited
(a)
AD
(2)
D&{2)
+c<
3C(Z>)
CE(4)
We
visited
have
all the
vertices, so#
to
return
EA(4)
upperbound
A
- = 15
A
Find a
the
graph
minimum spanning tree for#
with A removed, then add
two
the
shortest
(b)
remove
A:
from A
edges
3D {2)
3C{3)
CE(4)
Add
edges
lower bound
AD and
=
(2 +
A3:
3 + 4)
+ (2+
7
v
^k^
+ G(._
i-
2)
Algorithms
= 13
on
graphs
121
continued...
L
the
and
minimum
a
Find
the
between
is somewhere
spanning
the graph
tree for#
(d)
C:
remove
C
vertex
without
(c) 13<L<15
upper\342\200\242
bound
lower
3D {2)
AD {2)
AE(4)
\\
Add the two
shortest edges from
Add
C*
CE and
edges
lower bound
The
upper
and lower
have found
are
=
CI3:
(2 +
2 + 4)
+ (3 +
4)
= 15
bounds we0
the
(e) Upper and
same
are
bounds
lower
the eame:
15<L<15
so
+<i
Note that
in
a cycle.
creates a
-^/
15. The improved lowerbound in
This
is the bestpossible
because
cycle.
cycle
POINT
one
we have
The lower bound and
We
have
found
lowerbound.
r^
*V
122
Topic
10
-
Option: Discrete mathematics
exact
the
found
travelling salesman problemin
(2)
(d)
L >
the
solution
either
of these
upper
bound
a cycle with
the same
also
15, so
15.
than
shorter
part
77
We know that
(1)
(a), we created
one, becauseit
L<
we cannot find
KEY
solution.
bound in part
necessarily the best
is not
cycle
be that
could
is the
an upper
finding
This
= 15
L
to the
two cases:
are equal.
length as the
* *-
tv.
1
fc
'
\\J
^
/
7E
Exercise
1 to 2 refer to the travelling
Questions
the following graphs:
salesman
*A
(a)(i)
on
problem
A
(ii)
\342\200\242B
(b)(i)
(c) (i)
a
m
D
O
+<i
1.
2.
Use
By
E
32
28
26
19
14
21
26
22 .
18
18
21
22
22
17
23
22
31 31
14
Q
26
21
18
26
18
deleting
22
starting at
vertex
A, find a lower bound for each
of edges of graph G are given
in the
neighbour algorithm
I
B
16
16
12
\\\302\273
List
(b)
Hence
all
Hamiltonian
solve the
cycles
I
A
A
(a)
A
D
28
The weights
(ii)
C
Q
the nearest
22
b
32
O
(ii)
starting
I
C
to find
C
22
21
. 22
an upper
E
D
17 22
,
23
.
31
18 31
18
26
26
bound.
graph.
table:
\\d
12
8
18
8
9
18
8
A
B
9
and finishing
travelling salesman problemfor
at
A.
G.
[5 marks]
7
?
^k^
+ 0[._
i-
Algorithms
on
graphs
1
The
edges of a simple graphG aregiven
of the
weights
table:
this
DIE
A\\B\\C
4 3
3
3
9 8
7 5
4
3
7
6
(a)
the
why
Explain
vertex in turn, to find an
A
|
B
, 5
G.
marks]
[11
| F
10
11
8
7
6
7
7
8
7
10
6
11
7
7
1\"
8
10
1'
7
12 10
an upper bound for
vertex
deleting
minimum
the
| E
|
10
9
lower
Explain
6
6
spanning
how
from
starting
salesman
travelling
B and
10
find
A to
on G.
problem
using Kruskals algorithm to find
tree for the resulting graph,find
a
travelling salesman problem.
know that you have found a solutionto the
and write down the weight of
problem,
for the
bound
you
salesman
travelling
the
12
9
the nearest neighbouralgorithm
(a) Use
(c)
for
bound
a lower
D
C
|
5
A
By
the graph find
the graph given by this table:
Consider
(b)
each
with
starting
salesman problemfor
the travelling
salesman
improved upperbound.
D from
vertex
deleting
By
for the travelling
algorithm,
neighbour
(c)
7 6
9 7
8 5
9
9
upper bound
is at most 29.
problem G
(b) Use the nearest
for
in
the optimal
route.
[9
marks]
Summary
\342\200\242
In
a weighted
\342\200\242
A minimum
graph
edges have
spanning
tree of
which is also a tree.
\342\200\242
KruskaPs
for finding
algorithm
weight and skipping edgeswhich
\342\200\242
To
-
Dijkstra's
perform
algorithm
associatednumbers,calledweights.
a weighted graphis a subgraph
weight
possible
adding
edges
of
in order
a cycle.
form
for finding
Start by labellingeachvertex
connected
from the start vertex.
smallest
spanning tree involves
a minimum
would
of
the shortestpath
to the
between
start vertex with
two
vertices:
a temporary label which
its distance
-
At
each
then give
and
Topic
10
stage
-
find
the vertex with
to it.
temporary labelsto all verticesconnected
Option: Discrete mathematics
ji^
the smallesttemporary label,makeit a
permanent
label,
is
^ *~
-
-
If a vertex already
The labelsfor
has a
it
is updated
in a
if the
only
permanent label,we canwork
are recorded
vertex
each
label,
a temporary
has
vertex
end
the
Once
V '
k^h
^
/
new one would be smaller.
the shortest
to find
backwards
box, with permanent labelsin
the
path.
row.
top
distance
from previous
vertex
start
order of
labelling
1-
temporary
labels
The
Chinese
edge
at
- If
- If
problem
postman
and
once
least
returns
the
graph
is Eulerian,
the
graph
is not
\342\200\242
find
\342\200\242
the
\342\200\242
the
of the
weight
shortestpath.
The
salesman
travelling
path need to
-
The
can only
solution
exact
be usedtwice,allthe otheredges
is to
problem
uses each
the Routeinspectionalgorithm
route is the sumofallthe edgesin
graph (i.e. the shortest routewhich
which
pairs of verticesof odd degree
path between
in this
edges
we use
a graph
is any Eulerian path.
the solution
Eulerian,
shortest
the
is to find the shortestroutearound
to the starting vertex.
visits
the
graph
used
are
once
the weight
plus
of the
in a complete
find the Hamiltoniancycleof leastweight
vertex
and
returns
to
the
every
starting point).
be found
by checkingall Hamiltonian
cycles,
which
is
inefficient.
-
-
An
can be
bound
upper
An improved
found by using the nearestneighbour
upper bound canbe found
by
starting
from
algorithm.
vertex.
a different
can be found by removing onevertex,finding
the
of the minimum
weight
tree
for
the
and
the
of
the
two
shortest edges
spanning
remaininggraph,
adding
weights
A
lower
bound
fromthe removedvertex.
+c<
-
The
required
and
the
least
upper
In orderto
possible
and
a more
get
the
of a Hamiltonian
LB<L<UB.
weight
bound,
upper
accurate
bound
estimate
as small
cycleis somewherebetweenthelowerbound
for L we aim to makethelowerboundas largeas
as possible.
7
v
^k^
+ G(._
1-
Algorithms
on graphs
125
examination
Mixed
7
practice
(a) Explainwhat
by a
is meant
(b) Use
Kruskals
to find
algorithm
the order
List edges in
spanning tree.
minimum
spanning tree for
the minimum
you addedthemto the tree,draw
your
this
graph.
(a)
Outline
(b)
Use
[7 marks]
Kruskals
briefly
Kruskals
algorithm
by the
represented
state
and
tree
its weight.
to find
the minimum spanningtree.
for finding
algorithm
the minimum spanningtreefor
following table.Draw the treeand
state
14
15
15.5
8.5
8.5
12
10
22.5
22.5
13
16
8
20.5
8
20.5
16
25
16.5
25
21
21
15.5
graph
its weight.
D
B
14
the
19
16.5
12 13
18
19
11.5
18
11.5
15
[9
marks]
The
weights
the following
of the
edges of a graph with
vertices
A
B
H
10 15
10
12
15 12
Use any
for this
126
Topic
10
-
C, D
and E are given in
table.
\342\204\242\"
(a)
A, B,
method to find an
graph.
Option: Discrete mathematics
11
19
18
16
13
14
D E
11 16
19
13
18
14
17
17
upper bound for
the
travelling
salesman
problem
^ **\"
b..ir*i
Use Kruskals algorithmto find
(b) (i)
the
'
V
obtained
subgraph
by removing
the vertex E
the
from
the total weight of this minimumspanning
lower bound for the travelling
salesman
problem
(ii) State
/
spanning tree
a minimum
draw
and
v.
graph.
find a
hence
and
tree
for this
for
graph.
[11marks]
IB
(\302\251
to solvethe travelling
We want
(a)
nearest
the
Use
bound for
neighbour
the
(b) By removing
vertexB find
Hamiltoniancycle
has edges with
34
34
-
22
-
20
1-
11
1
H
-
1-
-
1-
1-
1-
1(a)
Use Dijkstras
(b)
It is
by
L, the
upper
shortest
of the
length
marks]
[10
given
in this
table:
J
24
-
-
26
-
-
-
34 21 - 16 - 26 - 10
- - 26 - 25
- 26
10
-
-
12
25 - 18
-
12
-
18
-
algorithm to find the shortestpath
the lengthofyour
find an
11
-
16
21 - - -
B to
bound.
I
-
24
34
-
vertex
problem.
a lower
weights
C D
22
- 20
-
at
graph K shown here.
K.
in
1
algorithm starting
salesman
travelling
for the
problem
down an inequality satisfied
(c) Write
A graph
salesman
2006)
Organization
from
A to
/. Write
down
path.
required that the
path includesedgeFI.Find
the
length
of the
new
shortest path.
(c)
The
path
does
not need
to include
from the graph.Findthe new
shortest
FI any
path
more,
and
but
the edge
its length.
7
?
^k^
+ 0[._
i-
GJ is removed
Algorithms
[12 marks]
on
graphs
By
salesman problem for
the travelling
Consider
each
deleting
in turn,
vertex
this graph:
find a lower bound for
the
salesman
travelling
[10 marks]
problem.
The diagram showsa plan of a house,with
the plan on a graph,
representing doors.
(a) Represent
(b)
why
Explain
the starting
it is
with
doors
vertices
connecting
adjacent
representing
rooms
walk througheachdooronce
not possible to
rooms.
and edges
and
to
return
room.
The table shows distancesbetweenthe centresofadjacent
2
3
J_ 4 6
4
6
5
rooms.
7 8
5
5
3
m
4
]
4
2
I
7
4
2
3
S
5
m
\\3
to find
the shortestroutebetweenrooms1and
(c)
Use
(d)
Hence
find the
at least
once and returnsto the starting
Dijkstras
algorithm
shortest tour of thehousewhich
passes
through
each
Topic 10
-
Option:
Discrete
mathematics
door
point.
[13
128
8.
marks]
*
*-
tv.
1
fc
'
\\J
^
Q
For the
(a)
List
of odd
vertices
the
the
list
clearly,
the optimal
The
/
/>
graph shown in the diagram:
degree.
(b) Solvethe Chinesepostman
Q
^
below
graph
edges
for
problem
this
graph.
Show your
to be usedtwice,and
which need
state
the
method
represents
of
length
route.
[9marks]
a network
of paths connectingseven
in
fountains
a park:
A
(a) Explain why it is not possible to start from
path exactly once and returnto A.
Find
the length of the shortest routewhich
(b)
fountain
uses
walk
A,
each
path
along
at least
each
once
and returnsto A.
(c)
A new
path
path is
exactly
fountains
to be built
once
should
and
that
so
return
this new
it becomes
to the
possible to
starting point.Between
path be built?
walk alongeach
two
which
[11
marks]
7 Algorithms on
- * v>,Ar
Vh
+ ft
graphs
ft
The table
showsthe lengths
(in
km)
10 12
7
12
8
8
8
11
10
10
9
12
11 10
9
12
8 12
8
5
12
8
A
(i)
Explain
why
to
exactly twice, and
each
road
Find
the distance
G has
can start from village A, travel
the snowplough
the following
\302\273
l>
(a) By deleting eachvertexin
for
Option: Discrete mathematics
needs to cover.
turn,
JEM
E
6
10
9
7
-
8
8
9
9
-
8
6
8
10
find
6
-
travel
along
along
[9
marks]
5
a lower
10
5
-
bound for
the travelling
G.
why this lower bound is in fact
salesman problem.
Explain
to the
cost adjacencytable:
I.
salesmanproblem
return
return backto A.
the snowplough
\302\253
ji^
have
using
answer.
your
-
-
without
dothis,it
(ii)
10
does not
if he
it.
A snowplough needs to clearall the roads.To
must
each road twice (not necessarilyin oppositedirections).
(i)
Topic
once
exactly
starting village? Justify
(b)
8
6
needs to
inspector
road
Graph
8
11
drive along eachroadto inspect
Explain
why he cannot return to the starting
village
some roads more than once.
road
(ii) Canhe use each
(b)
5
7
11 6
7
(a)
eight villages.
9
9
7
roads between
of main
the
solution
to the
travelling
^
Fill-in
-
(a)
(i)
shortest
the
down
write
D,
boxes
the
the
find
for
and,
in
fill
'
v.
/
E and
for Exercise 7C
diagrams
Photocopy
-
V
b..ir*i
A
Appendix
1.
**\"
of the
each
following networks:
out Dijkstras algorithm
(or paths) from S to F
to carry
path
of the
lengths
shortest paths from Sto
F.
)
J_
A\342\200\224
\342\200\224D
y
X
y
rrrir/
\\
4
\342\200\224rf
^\\
V
\\
E3N
\\
V
\\\342\200\224
~
v
yA
V
+\302\243i
7 Algorithms
?
^Ui)^
+ <*..
on
graphs
1
(b)(i)
1-
(ii)
+\302\243i
132
Topic10
-
Option:
Discrete
mathematics
*
(c)
\"-
K \\\\
^k^a^t-
--i^n,
(i)
1H*
\302\243
+\302\243\302\253
oL
7
?
5^>C^
+ 0i.
Algorithms
on
graphs
133
In this
you
chapter
will learn:
\342\200\242
about
different
define a
\342\200\242
how
to
for the
some
ways
sequence
find
term
nth
relations
the formula
for
by recurrence
relations
how
to
sequences
defined
\342\200\242
Recurrence
to
relations
use
recurrence
to model
various situations.
often
Sequences
or
more
for example,
terms;
previous
which link
rules,
satisfy
ux
each term to one
\342\200\224
2, un+1
=
2un. By
we can sometimes spot
looking at the numbersin the sequence
2n. But
a pattern; in this casethe sequence
is2,4,8,...soun \342\200\224
sometimes
the pattern in the numbers is not so easy to spot. In
this chapter we will learn a systematic
to find a formula for
way
some specialtypes
of
sequences
recursively
from the core course that there
(\302\24323Defining
main ways
not
of rule
define
fully
un+l
-
for example
sequence,
2un
could generate
=
un+l
is alsocalleda
a sequence.
- n2.
un
new terms
link
definitions
This type
two
value of the termto its position
link the
for example
sequence,
\342\200\242
Recursive
the
that
rules
the
are
we can definea sequence:
\342\200\242
Deductive
in
seen
have
should
You
sequences.
to previous termsin
2un.
recurrence
relation,
For example,
and
does
the recurrence relation
the sequences:
1,2,4,8,... or 3,6,12,24,...or
0.7,1.4,2.8,5.6,...
define
the sequence
fully, we need a starting value.For
To
example,
+c<
if we
also know
relation generatesthe third
Recurrence
relations
that
ux
- 0.7
then the
recurrence
sequence.
are classified
by how many
previous
terms
to find the newterm.In a first order recurrence
of each term dependsonly on one previous term;
=
If a new term
order
recurrence.
un+1
2un is an example of a first
this is called a second order
terms,
depends on two previous
recurrenceand an example of this is the Fibonaccisequence,
\342\200\224 +
values
to fully
un+2
un+1
un. In this case we needtwo
starting
are needed
the value
in
are
defined
sequence. Higher orderrecurrences
but in this coursewe will only consider
first and
second order recurrence relations.
define the
the same way,
r^
*V
134
Topic
10
-
Option: Discrete mathematics
^ **\"
orked
b..ir*i
=
un+2
3un+1 -2un
with
u1
relation
the recurrence
Apply
third*
from
u1=lyu2=2
=1
u2
=2
u5
-
u4
onwards
term
first five
terms suggestthat
sequence which lookslikeun
guarantee
that
continues
pattern
relations
recurrence
about
Results
this
induction.
orked
A
Prove
by the
is defined
by induction
As each
There
=4
=
3u5
-2u2
=e
\342\200\224
=16
2u*
pattern in the
is, of course, no
See
<^[
can prove it.
are often best proved using
unless we
1C
Section
more
for
of <^[
examples
proof
by
induction.
8.2
example
sequence
is a
there
= 2n_1.
-2uy
\"
3^2
= 3l/a
uR
These
/
relation:
by the recurrence
sequence, defined
of the
terms
five
v.
8.1
example
Find the first
'
V
that
term
on
terms,
two
previous
two base
need
we
un+2
=
-
3un+1
with
2un
2.
\342\200\224\\,u2\342\200\224
ux
- 2n~l.
un
depends
recurrence relation
\342\200\242
cases
n = 1 : u,
When
=
Whenn
= 1=
21~1
= 22-1
2:u2=2
So the statementis true
for
n = 1 and
n
= 2.
+c<
As each
one previous
term
on
depends
need
we
term
The expressionun
n
than*
to use
strong
induction
= 2n_1
= /c-1
more
and
Suppose the statement
Then for n = k:
\"k
2uk-z
= 2(V2)
=
conclusion*
all n<k.
= 3(2k-2)-2k-2
k-2
So the
Always write a
-
3tVi
for
= 3(2k-z)-2(2k-z)
for
is valid for*
n =
=
is true
The
n=
2k~'
is true
statement
statement
2, and
un
if
it
for
= 2\"~1 is
is true
for
all
n \342\226\240
true
n <
for
n
= 1 and
k then we
can prove that it is alsotrue for n = k.
Hence it is true for all n by strong Induction.
8 Recurrencerelations
-
* v>,Ar
Vh
+
Gl._
8A
Exercise
the
1. Find
(a)
un+1=4un+l,
(b)
= 4-un_ly
un
(c)
un+2
(d)
u\342\200\236=nun_19 \302\253!=2
that
Given
2.
=
A
un
a2
Q3
In
a
first
the
is fully
order
order
recurrence
if we
Although we can find
is an
recurrence
relation
recurrence
example
a
of
logistic
can be used to
model
growth.
population
first
do
In
defined by
Sequences
on the starting
(depending
value). These range from
a few
where
periodic,
values
repeat again and again, to
chaotic,
where
repeats.
In
a
chaotic
a small change
value can have
sequence,
the
in
starting
a large effect
To analyse
terms.
subsequent
never
the pattern
these sequences we need a
range
of
mathematical
from quadratic
techniques,
equations
136
r^
*V
to
differentiation.
Topic10
-
Option:
Discrete
by induction
-1.
that
un
term
any
in the
times,
a
2an
\342\200\224
2n + 3.
that
an
- n2
for all n
-
an_x
relation
recurrence
the
Show
-
> 1.
relations
the sequence dependsonly
The sequence
sequence by applyingthe
solution for un in terms of n
using your calculator, you will find
to
follow
appear
any recognisablepattern.
this
course
the
we
that
they
only linear recurrence relations,
involves only multiples of un and no
will study
function
f(un)
powers, roots, or any other functions.We will only
higher
learn a general method for solving
recurrences
with constant
coefficients and so we will not consider
recurrence
relations
=
such as un+1
where the coefficient of un depends
on
(n +1)un,
behaviours
of
recurrence relation an+l
we can write un+l - f(un).
know the first term.
terms
twenty
not
where
the logistic map exhibita
range
by
can only be found in somespecial
cases.For example,
even
for a
=
\342\200\224
0.6
4un(l
un) with ux
simple-looking recurrence like un+1
there
is no known expression for un in terms of n. If you plot the
4fJf^
mop, which
ux
many
sufficiently
^f
by the
each term in
term; so
determined
2un prove
linear recurrence
First
previous
3un+1
+1.
\342\200\224
7. Prove
with
-un+2n-\\
un+\\
This
- 2n
-
\342\200\224
un+2
The terms of a sequence satisfy
4.
on
=2
= l,u2
3, u2
\342\200\224
5 and
ax
ux
is defined
sequence
with
Ui=3
\342\200\224 \342\200\224
5 and
induction that
3.
\302\253!=1
un+1un,
ux
by the
relations:
recurrence
following
sequences defined
of the
terms
five
first
on
n.
form of a first order recurrence
relation
with constant
coefficients is un+l - aun + g(n) (so although the coefficient
of
an
does);
un does not depend on n, another part of the equation
will
is
We
look
at
cases
where
un+l
-un+n.
example
mainly
g(n)
a constant,
to deal with
although the method can be extended
The general
many
other
functions
g(n).
So the most generalfirst order recurrence
relation we will learn
how to solve in this sectionis un+l
aun + b. We have alreadymet
two
cases
of
this
recurrence:
When
a \342\200\224
1, un+l \342\200\224un+b
special
mathematics
is
' W
V/
*\302\253~
^l*^*
^
sequence, so ww
an arithmetic
produces
we get
ux +
(n
-
l)b; if b
= 0
a geometric
uYan~l.
sequence,un \342\200\224
solution for other valuesof a and b, let us
the
To find
a particular
' +1'
was no
If there
with
start
example:
-
un+l
3un +1 with
the
in
whose
a constant
be
would
value
1-
- 2
ux
the sequence
of the form
equation,
geometric and the solution
c is
=
/>
term. When
need to
equation
way we can modify the
and this works if we selectthe
a constant,
adding
by
were
the first
on
depends
we checkthis, it doesnotsatisfy
our
The
simplest
try something different.
solutionis
would be
c x 3n
so we
constant:
right
Ifun=cx3n+dthen:
=
cx3n+1+d
+l
3(cx3n+d)
application
+l
=^cx3n+1+d= cx3n+l+3d
\\
d =
3d +
=>
dA =
\342\200\224
a philosophical
razor,
that
principle that states
the simplest solution is often the
correct
one.
role can this
What
l
=^>
an
of
example
of Occam's
is an
This
*'
1
2
of principle
sort
in
play
mathematics?
We
can check that
- c
un
x
3n
\342\200\224
relation for all valuesof c. We
ux
You
+<l
which makes
of c
value
the
first
X*f9%
can
1
= 2^>c = 5
sequence is
for our
with
1
\\3n
complicated
relations
recurrence
To solve
\342\200\242
set
the recurrence
- cx
un
model
l^>to
8.1
POINT
an
relation un+1
aun
+ b:
but
relation to find
the recurrence
into
\342\200\242
substitute
in the
counting,
you
need to
+ d
\342\200\242
substitute
value of
the
first
situations ]^>
involving
=
term
to find
8D
Section
we will use more
linear
+ b.
aun
the
on
point
the
In
=
the
using
\342\200\224.
into
order
gives
'+c which
curve.
relation.
first
by
of one
coordinates
to
equation of a
constant
arbitrary
can be found
(5)
-
-
un
We can use this methodto solve
any
recurrence relation of the form un+1
KEY
analogies
integration
using
find the
an
works by substituting
that this
check
recurrence
some
has
This
\342\231\246^f
curve.Integration
formula
the
the
\342\200\224
2:
cx3l--
So
the general solution of
formula
for our particular
this
call
the recurrencerelation.Tofind
sequence we needto choosethe
term
recurrence
the
satisfies
2
know
not
will
how
to
solve them.
d
c.
,We can extendthis
This method
works whenevera ^
know how to find
given
by
un
-
the
uY + [n
solution,
\\)b
= bn
1. When
as it
is an
+ {ux
- b).
a =
method
1 we already
arithmetic sequence
form
functions
to deal
linear
g.
8 Recurrence
- *
v>,Ar
with
recurrence relations of the
=
un+]
oun + g(n) for various
relations
137
w
Vh
+ d..
.o
orkedexample
8.3
Find
Un+\\
the
for the
formula
\342\200\224~Un~^
Follow the
With
Ul
general
the
into
of the
term
defined by the recurrencerelation
sequence
- 6-
down the#
Write
above:
method
Substitute
nth
form
of
\342\200\224
I + d
solution
the
to find
recurrence
=
c I
u\342\200\236
Let
Then:
dm
\\\"
+<<=!L1
cii
+ d
-3
\\/1+1
+ --3
2
2
d =
-6
\\\"
=c
u\342\200\236
Use
the
first term to
find
-6
u, =6
c\302\253#
6 =c
c
=
-
U.
-6
24
-6
= 24
u\342\200\236
/.
-i*\342\200\224>_+~
,_!*->_f
Exercise
8B
+<l
1. Find
(a)
the formula for
un+1
(i)
(b) (i)
(c)
f2J
Apply
=
-3un
-1,
the
by
=
\302\253! 3
+ 2,
ux
method
ww+1
is
sequence
with ux
8.
(a) Write
=
rww and
first term
down the first
10
-
Option: Discrete mathematics
wM+1
=
=
2un
sequencesbelow:
+39u1=l
+ 5, mx
-3w\342\200\236
= -2
(ii) un+1=-un-2,1^=8
this section
given by
for the
wM+1
(ii)
from
to a geometric sequence
uv
arithmetic sequenceun+l
the recurrence relationun+1
three
(b) Find an expressionfor
Topic
(ii)
= 1
the method to an
Try applying
where d ^ 0. Where does it fail?
A
138
of n
terms
un+1=-un +1,^=3
(i)
defined
[3.J
un+1
= 2un
in
un
terms
un in
terms
of the
of n.
=
-un+d,
5un
+ 8
sequence.
[9 marks]
/
(a) Find the generalsolution
relation
recurrence
the
of
u\342\200\236+i=2un-l.
with first term
the sequence
Find
(b)
ux
=
the above
1 satisfying
recurrence relation.
[8marks]
order recurrence
Second
relations
look
at second
depends
on two
now
will
We
where
un+2
This means that
order recurrence relations,
previous terms: un+2 - f(un+1,un).
we need to know the first
to
terms
two
fully define the sequence. As before, a solutionis only
easy
to find for linear recurrences with constant coefficients,
=
un+2
aun+1
+ bun
case
Moreover,we will
+ g(n).
when g(n) = 0;
are called
relations
recurrence
such
at the
look
only
homogeneous.
\\
To develop a methodfor solving
us look at a particular
example:
un+2
-
-
5un+1
with
6un
by the method for
if
to
see
thereisa solution
try
Motivated
constant k. Substituting
-1 and
ux
of
form
the
u2
- -1
un-cxkn
the recurrence
into
let
relations,
recurrence
order
first
the
this
recurrence
these
we can
for some
gives:
cxkn+2 =5cxkn+l -6cxkn
-6xkn
=5xkn+1
^>kn+2
=>A;2=5A;-6
+ 6=0
^k2-5k
^>
like there aretwo
It looks
un-c2x
it
is not
-1.
3n.
can
You
possible
However,
cx
substituting
3
2 or
solutions:
possible
check
to find
k =
2n and
un-cxx
solution by itself works,as
or c2 to make the first two terms 1 and
into the recurrenceshows
that
the sum
that neither
two possiblesolutions,un-cxx2n +c2x3n satisfies
recurrence
as well. It is now possible
to usethe values of
to find the constants:
of the
-1
ux
u2
-
=> 2q
-1 =>
the
ux and
u2
+ 3c2 = 1
4q + 9c2
i
= -1
i
of the
We
two
these
Solving
sequence
can apply
form
un+2
-
-
2,c2 =
\342\200\224
2 x
2n
\342\200\224
method to any
recurrence
equations
is given by un
the same
aun+l +
bun.
Trying
gives
cx
a solution
-1, so the nth
term
3n.
of the
relation
of the
form un-cxkn
8 Recurrence
-iaVNoi.
relations
139
in a
will result
quadratic equation k2
auxiliaryequation.
KEY
+ b.
-ak
This is calledthe
below.
is summarised
method
full
The
8.2
POINT
order recurrencerelation:
a second
To solve
un+2=aun+l+bun
the
\342\200\242
Find
\342\200\242
The
the
of
values
above example
real roots. We will see
In the
rootsor
complex
formula
for the
is
solution
general
\342\200\242
Use
kx and k2, of
= ak + b
k2
solutions,
-
cYk\" +
u2 to
and
ux
the auxiliary
c2k2\342\200\242
values of cx
find the
had
equation
two
and
c2.
distinct
deal with the caseof repeated
us apply the method to find the
how to
later
let
First
roots.
nth
un
the auxiliaryequation:
term
of the
-
-1.
Fibonacci
sequence.
orked example 8.4
the
Solve
relation
recurrence
down the
Write
un+2
=
un+l +
auxiliary equation
solve
and
un
with
\342\200\242
un
ux
u2
= cxkn
where
it
kz=k + \\
,k2-k-/\\=0
>k =
Y
'\\+S
+ c9
u\342\200\236
=d
\\
Use
u}
and
u2*
u1
= 1:
cy
+
-
\\-S v
S 1-V5
+ c9
= 1
[1]
+c<
u9 =
1: d
'\\
+
S
3
If
we
subtract
the two
can
equations
we\302\253#
\"'\"IS
Option: Discrete mathematics
\342\226\240-c.
= 1
1_
1
a
s\302\260
-
1-Vs
\\+S
^5=1
1
10
A
lZ]-ir]=>Ci+c2=0=>c2=-Ci
<=>a
Topic
3-V5
V5
=1
eliminate
D] =>q
140
+
'\\-S
+ c9
Y
/,
/^V^
1--s/5
[2]
^ **\"
V
'
v.
/
also workswhen the rootsofthe auxiliary
complex. The constants cx and c2 will be such that
This method
are
equation
b..ir*i
the imaginaryparts canceland
real.
un are
all
orked example 8.5
relation
recurrence
the
Solve
= 0 with
-Qyu2-1.
ux
equation and*
down the auxiliary
Write
+ 4un
un+2
+ 4
k2
solve it
=> k
= 0
= \302\2612i
un=c,(2\\)n+c2(-2\\)n
of
values
the
Use
and
u}
U)
u2*
= 0
=0=>2cj-2c2\\
u2 =1=>-4d
[1]
D]
-4c2 = 1
[2]
=^=^2
1
[2]
-((20\"+(-2i)\
a complex number
easier if it is written
Raising
to
a
in
polar
-i(
form
2\"
5
=
cis(-#)
= 2cos#
2\"
^
H7I
^
2
= -2n
If the
last answer
seems complicated,
to see that
few terms
confirms
all
However,
won't
that
cases
you
un
the
+
look at what
a repeated root.
We now
Consider
ux
9. The
\342\200\2246,u2\342\200\224
the
2 J
H7I
\342\200\224
2
z
coe
\342\200\224
first
The final answer
are
real
numbers.
sequences
in most
have to simplify fully,
when
happens
relation
the auxiliary
un+2 =
equation
auxiliary
an acceptable
be
I would
(\342\200\2242i)\"
recurrence
the
out
-nn^
is correct.
sequence
always
\342\200\224((2i)\"
you can write
)1
of the
terms
=
^
x2cos
&
+c<
2c\\e
2j
l
\"
cis#+
+
2cis-
is#
power
is
6un+1
k2
- 6k
answer.
equation has
9un with
+ 9=
0 and has
- c x 3W,
- 6
If we try to set un
then
ux
the
solution
2, but then
u2 ^ 9. Sowe needto modify
implies c \342\200\224
slightly. If we try the next simplestform,un=(c+ dn) x 3W, we
can use the values of ux and u2 to find constants c and d:
only
one
root,
k = 3.
8
-
* v>,Ar
Vh
+
Gl._
Recurrence
relations
+ d)x3
ux =6^>(c
=6
u2=9^>(c + 2d)x9 = 9
this
Solving
un-(3-
n)3n
orked
example
down
set
has
a
homogeneous recurrence
equation has a repeated root k,
the auxiliary
when
+
un \342\200\224(c
dn)kn.
8.6
Solve the recurrencerelationun+2
Write
relation
I
equation
of
form
order linear
a second
solve
I
This
whenever the auxiliary
8.3
POINT
To
You
root.
repeated
I
-1.
can
check
that
satisfies the recurrencerelation.
will work
the solution
KEY
3,d =
c =
gives
the auxiliary
\342\200\224
2un+1
-
un
that
given
equation and#
\342\200\224 \342\200\224
2.
ux
\\,u2
0=>k = 1
-2k + 1 =
solve it
Let
Use
the
values
of
u}
and
=
un
(c +
dn) x 1\"
Then
u2*
u1
= 1 =>
c +d =1
u2
=2^>c
+ 2d
=2
:.c = 0,d = \\
il =
5o
n
+c<
Exercise
8C
1. Solvethe following
(a)
un+2
=
5un+l
(b) un+2=4un,
(c)
un+2
=
-6un,
142
Topic
10
-
-
(b)
un+2
Option: Discrete mathematics
5,u2
= 13
for
u2=13
ux=9,
?>un+l -2un,
the
nth terms
of the
relation:
recurrence
un+2
=
uY
=
4un+l +
2un+1
relations:
Ul=4,u2=0
2. Find the formula
(a)
recurrence
order
second
4un
-un,ux
= 0,
uY
=
4,u2
= 5,u2=7
= 12
sequencegiven by
the
^
3u.
*\302\253~
V
I
^
h^A*1
3.
the
Solve
recurrence
following
relations,
the complex
'
W
-^5*\302\273c+^
answerin
leaving your
form.
(a)
un+2
(b)
un+2
=
2(un+1 -un\\
-
=
ux
= -6,
ux
= -10,
13un,
4un+1
4. Solvethe following
u2
= 0
= -40
u2
and
relations
recurrence
solutions
write the
explicitlyin realform.
(a)
un+2
(b)
un+2
=
=
-un\\
2(un+1
ux
=
-6,\302\2532
= -12
is defined by the recurrence
relation
=
3. By
1, u2 \342\200\224
2un for n>ly and uY
A sequence
un+2
-1
un+1 -un9u1=l9u2=
\342\200\224
4un+1
recurrence relation find an
expression
for
the
solving
of n.
un in terms
[11 marks]
Two
an and
sequences,
bn, satisfy
the recurrence relations
an+l=3an+K
with
(a)
ax
Find the
value of a2.
(b) Showthat
(c) Solve
(d)
an+2
\342\200\224
2an+1
Hence find
an expressionbn
we will
section
terms
an.
of n.
[18 marks]
examplesof the application
look at some
of recurrence
be used to model
involve
that
situations
increases, for examplesavings
relations
recurrence
order
First
relations.
+c<
in
for
using recurrence
relations
this
+ 8an.
the second orderrecurrence
relation
[SJD Modelling
In
K+l=5<*n-K
\342\200\224
\342\200\2246,bx -6.
repeated
or debt
accounts
can
percentage
repayment
schemes.
orked example 8.7
0.3% monthly interest added to the accountat the endof eachmonth.
month $100 is paid into the account.
Let
of money in the
un be the amount
accountat the end of month n (after the interest has beenpaid).Find an expression
for un in
in the account after two years.
terms
of n and hence calculatethe amount
of money
A
At
account
savings
can
We
to
pays
of each
start
the
find
a recurrence
n +
un. In month
then
relating
1, add
increase
*
un+]
:1.003(u\342\200\236+100)
100 and
by
0.3%
8 Recurrence
- *
v>,Ar
Vh
+
Gl._
relations
continued...
a linear
This is
recurrence relation, so we*
of solution
form
what
know
to
un =cx/\\.OOZ)n
Set
1.003\"+1 +
c x
try
+d\\
d = 1.003cx 1.003\"
+ 100.3
1.003^
+
= 100.3
=>-0.003d
=>d = -33433-
3
and use
first term
the
Find
it
to
c#
find
of first month
= 100.3
end
the
At
=100x1.003
Ul
33433- = 100.3
cx1.0031-
=>
c =
3
33433-
1
3
.\\u\342\200\236=33433-(1.003\"-1)
3
'After 2
years' is the
of month
end
\342\200\242
24
2 years:
After
u24
= 2492
There
will
be
in the
$2492
account.
k ^^ ,T^^^^'*^^^^^*^^l&^**^^*M**.^****tM*-\302\243^\342\200\224l^r
use recurrence relationsto solvesomecounting
In fact the Fibonacci sequence was first
written
down
problems.
to model the size of a rabbit population.We look
at this in the
We
can also
next example.
+c<
orked example 8.8
with
starts
farmer
A
the second
a pair
onwards
year
of rabbits. Assume
each
(a)
Find a secondorderrecurrence
(b)
Hence find
how many
produces
pair
relation
pairs
of rabbits
do
rabbits
one pair
for
not breed
per year,
a male
year, but from
a female.
and
of pairs of rabbits, un
the number
the farmer will
in their first
have after
ten
years
in
n.
year
(assuming
all
survive).
In
the
year
n, there
previous
year
are all the
rabbits
plus all the new
born
that
from#
(a)
The number
ones
all
year
PLUS
the
rabbits
144
Topic 10
-
Option:Discrete
mathematics
year
from the
n is:
previous year (tv-i)
(un_z, because they only born
which are two or more yeare old)
new rabbits
pairs
of pairs in
to
**\"
^
b..ir*i
V
'
v.
/
'(:
continued.
We need
the
starting
(b) Year 1:1^=1
values\342\200\242
Year2:u2=1
1We
for
sequence
Fibonacci
this as the
recognise
\342\200\242
Fibonacci
we already know
which
the formula
u\"
sequence:
] + y/E
=
s
u,0 =
Hence
are
There
other recurrence
relations that
such as death rate and competition
many
factors
various
interacting species give particularly
model.
equations for the predator-prey
interest
model
population
resources.
for
Find out
results.
interesting
is invested
at the
taking into account
that involve two
about the Lotka-Volterra
bank account, which pays compound
rate of 4% per annum.By first writing
a
in a
amount of money in year n + 1 to
the amount of money in year n, find how much the
investment will be worth at the end of the 15th year (after
account
savings
5% interest
pays
year $2000 is paid into
in
year
per year. At
the
start
of each
and the interest is added
amount of money in the
the money for that year is paid in but
the
at the endofthe year.
account
the
[8 marks]
is added).
+c<
A
growth,
Models
relating the
recurrence
interest
55
8D
Exercise
\302\2431000
v\302\253\\
1-S
n, after
account
un be the
Let
before the interestis added.By solving
an expressionfor un in terms of n.
a recurrence
relation
find
[10 marks]
paid off in monthly instalmentsof
\342\202\254350.
The
interest
on the loan is 1%permonth,
and
is
charged
added before the payment for that month
is made. Let un be the
A
loan
amount
(a)
(b)
of
to be
\342\202\25416000is
owed at the
Explain why un+l
Solve the above
(c) Hence
paid
find
how
start of month n (so ux
=
1.0\\un
=
>
16000).
\342\200\224
350.
recurrence relation.
many
off.
months
it will take
for the debt to be
[11
marks]
8 Recurrence
- *
v*,>r
Un
+ Oi..
relations
145
/',
climbs
Chung
un
(a)
number
the
be
up
Explain why
Write down
either one or two steps at
in
of ways
which she can reachthe
the stairs
-
un
(c)
un_Y
each
that
that
spent
previous two days.
Omar spent on his
day n.
on
homework
(a) Write down a recurrence relationfor
oftn
spent 15 minuteson his
that Omar
27 minutes on day
1 and
day
terms
tn + 2in
1<mdtn.
+
Given
(b)
the average
he spends
day
on the
in minutes,
of time,
amount
2,
on
homework
an expression
find
time he spendson his homework
(c) Describe the long-termbehaviour
for the
day n.
on
the
of
\\
sequence
tn.
[11
the planesothat
are drawn in
n lines
no three passthrough
intersection points.
(a) Explainwhy
Pn+1
down the
(c)
Prove
by induction
be
not
(a)
contain
of
n,
length
be
Pn
\342\200\224
Pn+n.
that
the
considering
By
point.
explain
Pn
n2 \342\200\224n
\342\200\224
for
all
for the
possibilities
why
un
\342\200\224
un_x
Write down
(c)
Hence find
n >
2.
[10 marks]
of Osand Is,
last digit of a string
+ un_2.
the valuesof ux and u2.
the number of 16-digitstrings
and Is which do not contain
two
consecutive
(b)
marks]
are parallel and
the number
of
two
Let
of sequences consisting
two consecutive Os.
number
the
do
no
%
value of P2.
Write
Let un
same
the
(b)
which
a
on his
he spends
of time
amount
the
mathematicshomeworkso
of the amountof time
he
be
can climb
Chung
ways
[10 marks]
Omar decidesto vary
Let tn
step.
u2.
12 stairs.
of
flight
nth
Let
+ un_2.
the valuesof ux and
Hence find the number of different
(b)
a time.
consisting
Os.
of Os
[10 marks]
of Towers of Hanoi there aren rings
of different
in
three pins. The ringsstartonthe first pin, arranged
order of size so that the largest
one is at the bottom. The object
of the gameisto end up with all rings on a different pin, stillin
In
the
game
sizes and
decreasingorderofsize.
At
ontopof
a
Let
the
Hn
be
the game
smaller
with
(a)
Topic10
-
Option:
Discrete
mathematics
Find
no
point
larger ring
be placed
,
/
ring.
minimum
number
of moves
n rings.
the values
can a
of H1 and H2.
needed to complete
**
^
The
-
for the
strategy
all
Move
rings,
secondpin.
/>
game is as follows:
other than the largest one, to the
third pin.
Move
all the other
rings from the secondto the third pin.
=
(b)
2Hn_x +1.
Explain
why, if this strategy is followed, Hn
(c) Find an expressionfor Hn in terms of n, and hence find the
the
Move
to the
ring
largest
minimumnumberof
the game
to complete
needed
moves
with 10 rings.
Q
A
bank
[13marks]
interest on a loan. At
5% annual
charges
year, the interest is addedand
(a)
(b)
If the
amount
owed
at the
(c) For
start of year n is (1000-
the
end
of each
R, is paid off.
amount
+ 207?.
20#)(l.05)w_1
to the nearest dollar,
after
10 years.
repaid
what is the total amount repaid(to the
R, correct
of
value
of R,
value
this
the
amount,
is completely
loan
the
that
a fixed
borrowed is $1000,show that
minimum
Find the
so
then
nearest
dollar)?
(d)
loan
the
minimum
is the
What
is eventually
annual repayment requiredsothat
[12 marks]
paid off?
Summary
In
this
\342\200\242
A first
\342\200\242
To
we
chapter
order
solve
linear recurrence
a second
solvethe auxiliary
\342\200\242
The
un+1
order linear
equation
k2
for solvinglinearrecurrencerelations.
=
aun
distinct roots /q,
If the
equation has a repeatedrootthen
need
financial
can be
has solution of the form
=
aun+1
un
\342\200\224
can +
+ bun
d.
we first need to
+ b.
\342\200\224ak
equation has two
constants
+ b
homogeneous recurrenceun+2
If the
values of
\342\200\242
You
at methods
looked
k2 then
the
the solution
solution
is un=(c
is
un
\342\200\224
cYk\" +
c2^2-
+ dn)kn.
found by substituting into the recurrence
relation
and
using
starting
un.
to be
able to
write down recurrencerelationsto modelcounting
problems
and
situations.
8 Recurrencerelations
examination
Mixed
Find an expressionfor
=
- 0,un+1
ux
8
practice
The value
of
5un
a car
n>
1.
[8 marks]
by 8% each year.
decreases
an equation for the value
the previous year.
Given
that the value of the car after
will
take
for it to fall below $4000.
down
Write
(a)
the sequence defined by
of n for
terms
un in
for
+1
of
the
car after n
year
was $8500
years,
un,
of
terms
in
its value in
(b)
relation
recurrence
the
Solve
-
un+2
one
find how
long it
[9marks]
5un+1
-
that
given
6un
-1
ux
and
u2
- 3.
[10 marks]
relation
Q Solvethe recurrence
+ 4un+1
un+2
+
= 0
4un
with
ux
-
and
-4
u2
- 4.
[10 marks]
Given that
un
in
-36
and
ux \342\200\2246,u2\342\200\224
for n >
=0
+ 9un
un+2
1, find
an
of n.
terms
for
expression
[10 marks]
Solvethe recurrence
relation
- -(an_1
an+l
+ an) given that
4.
ax -\\,a2-
[10 marks]
A
several new
produces
crystal
magic
crystals every day. The crystalsthat
were
the previous day produce only one new
but the older ones
crystal,
9
new
each.
produce
crystals
If
is
the
number of crystals on day n:
(a)
cn
Write down the numbers of crystals on days
n - 1 and n - 2.
(i)
one
(ii) Write down the number of crystals which are exactly
day old on
day n.
produced
(iii) Find an expressionfor
Hence
(iv)
(b)
explain
(i) How many
A
=
0\302\253+l
crystals
an expression
of term.
has
sequence
the first
expressionfor
Topic10
-
through
(a)
Explain
(b)
Prove
Option:
8un_2.
of term,
first day
has
on the
nth
day
[16 marks]
term
an in
n straight linesaredrawn
148
n.
ax
-
\342\200\224
and
the recurrence
satisfies
relation
~ b.
3an
Find an
pass
2un_x +
on the
magic crystals
does
he have on the second day?
for the number of crystals
Harry
Find
(ii)
un
crystals createdon day
two newly-formed
was given
Harry
why
-
of new
number
total
the
Discrete
in
the same
why
un+1
by induction
mathematics
of n and
terms
the
un+n
that
+
[6 marks]
so that no
plane
point. They
\342\200\224
b.
divide
the
two are paralleland
plane
into
un
no
three
regions,
\\.
un--n2
+ !.
+\342\200\224\302\253
[8 marks]
mixed
and
Summary
examination
a large supply
of 20
can you make up $5 postage?
)
Suppose
I
ways
Letx
revisited
problem
Introductory
you have
the number
y be
and
equation 20.x+ 30y = 500
gcd(20,30)
one
Hence
The
cent stamps. In how
x and
different
many
We
the
to solve
want
integers. This is a linearDiophantine
y non-negative
method from chapter 4.
= 10, we write 10 =
20(-l) + 30(1).
50.
solution
of the equation is x \342\200\224
-50, y \342\200\224
=
\342\200\224
solution is x -50 + 3/c, y
50
2A.
so we
equation,
and 30
cent
of 20 cent and 30cent stamps,respectively
with
As
practice
general
use the
We need x,y > 0 and, rememberingthat
are 9 pairs (x,y) satisfying
these
conditions,
x, y
are integers,
so there are
this gives k > 17 and
9 different
k <
25. There
up the
to make
ways
required postage.
number
Although
theory
and
graph theory
have a surprising numberof realworld
applications,
of the proofs
which are associated with these
significance is due to the beauty
areas.In this option we focussed on three new types of proofwhich allowed us to justify the
methods we use:proofby contradiction,
the pigeonhole
principle and strong induction.
their
mathematical
The mainthrust
of the number
To do this, the Euclideanalgorithm
theory
was
section was working towards
Diophantine
solving
needed
which arose based upon the study
of
equations.
greatest
divisors.
common
remainders
when a number is divided, provides another
Easier modular arithmeticproblems
can
be solved
using linear
is in the solution
of hard problems in
congruences. One application of Diophantineequations
modulararithmetic,
the
Chinese
remainder
theorem. Also, Fermats little
theorem
particularly
a
tool
for
the
remainder
when
we
are
with
provides
powers
very powerful
determining
working
Modular arithmetic,
to look
tool
useful
the study
of
at divisibility.
of
numbers.
Graphs are
a way of representing informationabout
connections
between
several
can
be
asked
about
such
as:
it,
questions
objects.
has been drawn
\342\200\242
Can
a walk
around
the graph
go throughevery
\342\200\242
Can
a walk
around
the graph
go alongevery
There are alsooptimisationproblems
which
\342\200\242
the
minimum
\342\200\242
the
shortest
Finding
Finding
spanning
be
once
exactly
solved
^3^%.
_.
(Hamiltonian
path)7.
trail)7
(Eulerian
such as:
two vertices (solvedusing
Dijkstras
9 Summary
\342\200\242
^
edge
once
exactly
graph
tree (solved using Kruskalsalgorithm)
path between
n*'
v?
can
vertex
Once a
algorithm)
and mixed examination practice
-
.
149
.O
the
\342\200\242
Finding
problem,solvedusingthe Route
the
\342\200\242
Finding
travelling
find
shortest
salesman
upper
and
a graph usingeachedge
route around
shortest
inspection
at
least
once
(the Chinese
postman
algorithm)
visits every vertex and returnsto the starting
(the
point
in
This
cannot
be
but
we can
solved,
problem
problem).
general
efficiently
lower
route which
bounds.
based upon previous terms in the
sequences
Finally, recurrence relationsarea way of defining
can
be solved
the
constants.
sequence.
They
by applying standard trial functions and fixing
in applied fields
defined
recurrence
relations
ariseboth
within
mathematics
and
Sequences
using
such as financeand biology.
r
1
150
t>
^
Topic
10
-
Option: Discrete mathematics
- Q,
VVj
VS.
^%o<,
^
.
I^
.On
* ^
(*
Mixed
'
1^
examination
1. @)
S3
242
Show
that
if d
A
\\
n then
connected
simple,
-1) | (2n
(2d
-1).
is divisible
63.
by
values of
possible
d7.
[8 marks]
each with
has 5 vertices,
graph
the
are
What
the
d >
degree,
0.
answer.
your
Justify
same
possible value of d:
For each
(i)
divisible by 43.
show that 242 -1
Hence
2.
-1 is
that
^
V
vc+^
9
practice
Show
^1
an
Draw
of such
example
a graph,
(ii) Statewhetherornotyour
and if it
is Eulerian,
graph
is find an Eulerian
circuit.
For
the
A
ak +
N has
State
+ ax +
+...
ak_x
+... + \\Qax
+ a0.
a0 is also divisible
result if AT
a similar
so that:
digits {akak_x.. .a^),
Prove
by
divisible
that a
digits
is divisible
Use
^H
number is divisible
Show
show
the
that
6.
so
the number
that
and only
is not
algorithm
divisible by
to find
(199 lab)
is
if the sumof its
base
6
366.x
5.
[21marks]
gcd(610,366).
= 732(mod610).
Define the complementofa graph.
If H is a simple graph with 20 edgesand
16 edges, how many vertices doesH have?
Define the
3 then
by
by 5.
(223412)6
Eulidean
5 if
by
RSI Solvethe linearcongruence
1DJ
is divisible
33.
by
Hence
that if N
3.
by 11.
is divisible
Find all possiblevaluesof digits a and b
4.
a
[12 marks]
+ ak_x\\Qk~l
10*
ak
possible value of d find
cycle.
number
N=
the largest
with
graph
Hamiltonian
its
[13 marks]
has
complement
[6
marks]
following terms:
(i) bipartite
graph
(ii)
degree
of a
vertex.
9 Summary
^SV^
+ c,
and
mixed
examination
practice
* *~
(a
The
'
K^
two graphs,
shows
diagram
V
+/\302\273
K
K and L.
L
D
3J
trail
Eulerian
an
Find
UseFermats
the linear
in L.
ax
congruence
[6 marks]
that x
to prove
theorem
little
=
= ap~2b(modp)is a solutionof
is a prime which doesnot
where p
b(modp),
a.
divide
3]
K is bipartite.
or not
whether
Determine
or otherwise,
Hence,
the
Solve
solve 2x = ll(modl7).
of linear
system
congruences:
5x
2x = ll(modl7),
Show that if a base 12numberis divisible
its last two digits is divisible by 9.
=
by
[10 marks]
l(modll)
the number
9 then
formed by
when
350 is divided
by 13.
2J (i) Find the remainder
(ii) Find thelast digit in the base 13 expansion of 350.
Show
that if a base 13 numberis divisible
the sum of its
(i)
by 12 then
digits is divisible
(ii)
by
Hence show that
Show that the
12.
(3114CC)l3 is not divisible
number of distinctHamiltonian
with n vertices
by
cycles
12.
in a
[16 marks]
graph
complete
is \342\200\224In
-1)! }
2V
List all
the Hamiltonian cyclesfor
the
shown
graph
in the
diagram.
A
Hence
152
Topic10
-
Option:
Discrete
solve
the travelling
mathematics
salesman problem
for
this
graph.
[9 marks]
(\302\243
**~
'
*>***
V
+^
*
10.
J
Graph K is shown in the diagram
P
K is
that
Show
Which
that
is planar.
of K
subgraph
every
from K so that
be removed
should
edge
but
not planar,
the
is
graph
resulting
Eulerian?
Find a spanning tree for K.
Show
that K has a Hamiltonian cycle.
11.
Find
two
the
Find
(i)
m and
integers
of the
solution
general
90m + 48n = gcd(90,48).
n such that
Diophantine equation
90.x+ 48;/=624.
that
Show
(ii)
the linear
Solve
solutions
there
are no
solutions with both x and
List
congruence 48.x= 54(mod90).
y positive.
all
the
non-congruent
(mod 90).
12.
The weightedgraph
vertex
A from
G is
shown
Use Kruskals algorithmto find
and state its weight.
2J
the
find
Hence
G is producedby
above. Graph
deleting
G.
weight
of a
the
minimum
lower bound for
tree of
spanning
Hamiltonian
the
graph G
cycle
in G
beginning at vertex A.
a complete
for
Prove
(n-\\)\\
Hamiltonian
graph with n vertices (with n
cycles have
to
be
examined
> 3), that
to find
no
than
more
the Hamiltonian
cycleof leastweight.
gg
How
many
least
weight?
cycles in G would
have
to
be
examined
to find
the one with
the
[12 marks]
IB
(\302\251
9
Summary
and
mixed
Organization
examination
2007)
practice
153
vs.
^f*!*!**
+ 01.
I ^
.Ok\"
*
(a
13.
a-
*
1 ^ - ^
For the
V
i+s*
\\
graph shown below:
C
6
4
E
Explain why it is not possibleto find
which uses eachedgeexactly
once.
Use
to find
algorithm
Dijkstras
G
a route,
starting
and finishing at A,
the lengthofthe shortestpath
from
S to
T.
starts
at A, goes over each
length of theshortestroutewhich
edge at least once, and returns to A. Give an exampleof sucha route.
Show
that
the number
of vertices of odd degreein a graph must
be even.
[17 marks]
Hence find the
14. ^H
fig)
the
Draw
show
that
Usethe resultof part
not
K5.
graph cannot containcyclesof oddlength.
K5 is not bipartite.
why a bipartite
Explain
Hence
with 5 vertices,
graph
complete
the complete
that
(b)
to
n is
the graph
prove
bipartite graph K4
3
is
Hamiltonian.
For what
values of m,
Kmn
Hamiltonian?
Justify
your
answer.
[12 marks]
15. Forthe graph
shown
in the
diagram:
4
B
Explainwhy this graph
2) Find the minimumlength
16. (gj
and
returns
Use
the
Euclidean
Ba) Explainwhy
For
to the
which
the
values
Eulerian.
is not
of
a route
starting point.
which uses
each edge exactly
once
[6marks]
to find
gcd(66,42).
66.x + 42y = 9 has no integersolutions.
equation
= m have
of m does the equation 66.x+ 42y
integer
algorithm
solutions?
Solve the equationfor
154
Topic
10
-
Option: Discrete mathematics
^ q,vv,
\342\200\242V^
+ o
m
- 18.
[12 marks]
(*
*
17.
'
^
1^
^
to
BS) UseEuclid'salgorithm
Use the resultof part (b)
to
find
b,
x gcd(a,b)
= ab.
are always
coprime.
lcm(a,b)
2 and 4n
In +
that
show
+1
[12 marks]
lcm(702,401).
18. (Qj Prove that in a planar graph, e < 3v - 6.
n > 3 vertices
and e edges suchthat
fi51 G is a simple graphwith
its complement, G', are connectedand planar,
(i)
Write
down the
(ii)
Show
that
V
vc+^
two positive integersa and
Prove that for
^H
^1
G and
both
number of edges in G'.
n<10.
[10 marks]
19. Consider this graph:
.14 9.
/
B
v
13
k12
y
5.5/
12
11
n.2
15
V
21
irs.
G
^S.
r
16.5
*H
Find a minimum spanningtreefor
this
Draw
graph.
your tree
and state
its weight.
How
found in part
20.
Solve
^H
da)
x
= 7(modl
Explain
21.
A
sequence
ux
--\\yu2
^
1). How
your
the
as
3x = 6(mod5), 4x = 6(mod2).
congruences
3x =
congruences
6(mod5), 4x = 6(mod2),
solutions (mod550)arethere?
many non-congruent
[7 marks]
reasoning.
is given
by the
recurrence relation un+2
=
un+1 +
6un with
=17.
Write down
the first
terms
five
of the
sequence.
relation.
S) Solvethe recurrence
22.
The weighted
Use Kruskal's
added, to find
[10
4
c
Algorithm, indicating
and draw
the
minimum
the orderin
which
the
tree for
spanning
9 Summary
+ c,
marks]
graph H is shown below.
B
^SV^
tree you
[8marks]
simultaneous
the
Consider
of the samelength
(a)?
simultaneous
the
are there
trees
spanning
many
and
mixed
edges
are
H.
examination
practice
* a-
(a
*
^
1
ai (i)
-
v
<+/*
\\
number of edgesin
State the
v vertices.
has
tree
A
^
the
tree,
justifying
your answer.
(ii)
call
will
We
of which
each
components
a 'forest'if it consistsofc
is a tree.
with v vertices
a graph
Here is an exampleofa
4 components.
with
forest
r
yYy
How
A
graph
one of
forest with
will a
edges
many
and
vertices
v
has an odd number of vertices.Prove that
the verticesmust be even.
the
OJ
The
distances
shows
table
vertices has n
a tree with n
that
Prove
(in metres)
3
I
I
(i)
Find
(ii)
State the
the
6
3
3
8
5
5
minimum
betweenfive
6
[17marks]
2008)
Organization
tree for
in an
workstations
3
I
4
6
spanning
least
- 1 edges.
3
3
of at
degree
IB
(\302\251
23.
have?
c components
office.
8
I
5
5
4
6
3
3
the graph represented
by
minimum length of cablerequiredto connect
all
table.
this
the
workstations.
you cantellthat the graph is Eulerian.
(ii) Find an Eulerian cycleand state its length.
(i)
24.
Explain how
Consider
the weighted
following
table:
graph with the
weights of the edgesshown
MSM
Use
Forthe
Kruskal's
156
rsM
Topic10
^
Option:
Q,vv,
36
36
-
41
38
44
42
35
-
50
48
41
44
-
52
40
52
50
48
travelling
salesman
Remove vertex A
(ii)
Remove vertexB to find
(iii)
State which of the two
Discrete
\342\200\242V^
+ o
mathematics
to
find
41
52
44
50
38
-
on this
problem
(i)
40
35
44
graph:
a lower
bound.
another
lower
lower
bounds
52
48
-
spanning tree.
the minimum
to find
algorithm
42
50
48
41
bound.
is better.
[18
marks]
in
the
(a
*
'
a-
V
-^<+^
+
(q)
By considering
an inequality
the cycleABCDEFA
by the
satisfied
find
solution
an
upper
bound,
of the travelling
and hence
)
write
salesman problem.
[15
marks]
25.
borrowed
Julie
5%.
At
Let un
Pj
end
the
be
the
Explain
Find an
which
\302\2431050
of each
amount
why
un+1
be paid
back with the annual interestrateof
year, Julie pays back\302\243100before
n.
Julie owes at the start of year
=
1.05un
expressionfor un
BHj Find how longit will take
Sal
is to
the
interest
is added.
-105.
in terms
for
of n.
Julie to
pay off the debt.
9 Summary
[12
marks]
and mixed examination practice
157
vs.
^^VNc*.
.0
+1.
9.
the
Divide
Answers
10.
HINT
ANSWER
prove
answers
supplied;
been provided.
in
any
hints have
Express
Fn+l in
6.
You can
go from
7.
1
Assume
terms of
Assume
the
9.
that
n2 is
even
that
there
are such
odd, and
n is
but
In the
L is
that
considerL + 2.
5.
If
6.
was not
this
can
what
you
inductive
step,
2.
+b=c
that C ^
Assume
7.
the largest even integer and
be under
c are
8. (a) -0.682
(b)
x = \342\200\224
and show
Write
that p and q must
4,11
(ii)
3
(c) (i) 3,4
(d) (i) 4,11
of regions
(ii)
(a)
(8,6)
(3,2),
(i) (0,8),(4,4),(8,0)
(ii) (2,8),(6,4)
=
(4,0), (7,3)
(0,7),
(\302\253,&)
5.
Write
6.
Factorise the expression
9.
525,
10.
(a)
=
q
585
555,
n
of 9
a multiple
(b)
(a) (i) 4,612
(ii) 7,210
(b)
(i)
(ii)
45,1080
(c)
(i) 1,4536
(ii)
1,4165
45,900
i. n
(e) (i) 8,576
(ii)
2.
(f)
about
6.
What
when a number
and 8), (2 and
7),
when
happens
last
have
the same
(b)
How many
you add
10 numbers
then
numbers
Cut
a great
along
of friends
circle passing through
are
triangle
equilateral
Answers
-
*,v>,Ar
i>i
+
d\\a,
Gl._
b
2C
i. (a) 1
two
of
the points.
8. Consideran
q-nd
of side 1 cm.
2.
and show
a + b = md, a-b-nd
Exercise
different
= md,
which
digit?
9,54
3pqy18p2q
Write p
4. Write
possible?
7.
(i)
(3
(4 and 5).
even
(ii) 75pq2
3.
be (1
'pigeonholes'
and 6) and
5.
remainders
possible
by 7.
is divided
n
2B
Exercise
IB
kp
(ii) 35,105
3. Let the
increases
4,11
(d) (i) 28,56
Think
contradiction
(i) (fl,b) = (0,0),(9,0),(5,4),(l,8)
(ii)
both be even.
Exercise
(ii)
(ii) 4
3.
triangle.
resulting
(i) 3
rational
two possible
C > 90\302\260.
In each case draw a
from
C and apply Pythagoras
perpendicularline
(i)
(b)
(b)
\342\200\224
a.
3,4
(a)
18;
consider
and
90\302\260,
cases: C < 90\302\260
and
to the
and
a
where
irrational, and consider c
b is
and
use proof by
2
Chapter
1.
this
true, all of them would
say about the average?
a
that
Assume
digit.
base cases
Exercise 2A
that log2 5 = \342\200\224
and rearrange
Suppose
n. How many
about how the number
another line is added.
Think
numbers and factorise
expression to get even = odd.
4.
at the last
needed?
when
consider
expression.
Assume
3.
look
and
Fn
4 to
n \342\200\224
nxn.
2.
Then
of those lines and the
colour.
and double angle identity.
Exercise 1A
1.
from
five lines
same
three
4.
are
Chapter
squares and find the
one of the squares.
1C
Exercise
be
cases
some
the
the
at
must be the
endpoints
lines connecting them.
three
asked
Option, when you are
or show, there will
not
For this
to
consider
covers
and look
of them
Three
it:
circlewhich
vertex
one
Pick
into 25 equal
square
of the
radius
(b)
1
(c)
21
(d)
80
(e)
2
(a)
m =
3,n = -l
(b)
m =
-25, n
=
and
that
d < 1
show
that \\fd>2
h
(c) ra =
1
-l,n=l
2.
m
= 1,
(e)
ra
=
4. p
2,
= -3
g
4.
2D
Exercise
1. (a) (i) 23X3X5
2,3,4,6,12
(c)
ends in 00
(a) 0,7
5X47
(ii)
(b) (i) 94
1.
26508
(ii)
94
(a)
1. (2,2),(5,2),
2. (b)
m
(8,2),
= -27,
practice
(4,6), (7,6)
(1,6),
is one possibility
n-137
4. (a) (a-b)(a +
(b)
(ii) 0, 1, 2, 3
(i) 0,1,2,3,4,5,6,7,
2. (a) (i) 19
(b) (i) 313
(c) (i) 646
(i)
74
3494
1613
(ii) 50
(ii) 182
(ii) 1425
(ii) 199
(ii) 436
(ii) 287
(b)
The sum
(c)
(\302\253,&)
=
111110
(i)
(b)
(i) 351
(c) (i) 687
(d) (i)
B, C,
D, E
4
Exercise 4A
1.
6
9 are
and
both divisible
by
137 is not
3, but
2. (a) x = 2,7= 1
3.
(b)
4.
Equation
4AF
20220
(ii)
(ii)
(ii)
(ii)
1321
2A1
BBC
(i) 221
k =
0
15 +
9x + 9y
= 2007 has
no
solution
4B
(b)
3.
(a)
(b) (i) 1000111100
)
(ii) 1102012C
(i)
(d) (i)
5E6
4. (a)
(b)
5. (a)
3. 888888
(b)
3C
Exercise
(i)
(b)
(i) 1693
410
(ii)
(ii)
443
(a)
(ii)
(i)
(ii)
x =
x =
(i)
x = 2,y = \342\200\2246
(ii)
x =
(ii)
31E
170B
(ii)
(ii)
3226
BAO
= -2
x = 2,y = -3
5,y = -7
29y = \342\200\2243
l,y = -l
(b) (i) x = 3,y
(ii) 1000111100
1.
by 5
is divisible
(0,3), (5,3)
6170
2. (a) (i) x = l,y
3B
(a)
(c)
of the digits
(3,0),
1. a,c,d,f
Exercise
(a)
7056
Exercise
124
4. 999
2.
(c)
Chapter
8, 9, A, B
1, 2, 3,4, 5,6, 7, 8,9, A,
(f)
(a) 314
5.
(ii) 0,
(e) (i)
4.
120064
0,1,2,3,4,5
(i)
(i)
(a)
6. (a)
(a)
(d)
3.
(b) 211211
3
Chapter
59F
(b)
(b) 100
2
b)(a2+b2)
Exercise 3A
1.
practice
1010101100101111
2. (a)
3.
divisible by 13
(0,D), (7,6)
examination
Mixed
examination
Mixed
1.
(0,0),
(k,m)
2x3x3x47
(a)
0
of digits
=
sum
(ii) 2x3x23
(c) (i) 5X31
2.
(b)
(b)
(c)
23x52
(ii)
(i) 22x43
(b)
3678
(ii)
3. (a) 7
3. (b) x = ll,y = -21
=
(ii) 2505
(i) 1065
(a)
(b) (i) B2C
n=0
67, n = -29
(d)
+1.
= -6
x = -3,y
3
3
28
6
Multiples
of 6
(c)
x = 3,y = -5
6. (a)
gcd(?t,5)= l
(b)
=9
x =
3,y =
(other
answers
are possible)
-7 (other answersare
possible)
Answers
-
*,v>,Ar
i>i
+
Gl._
3
o
h
4^,7 =
1. (a) (i) jc = 1 +
(ii)x = 2+
jc = 3
(i)
=
7^,7
+
(u)jc= 2
+
(b)
-2-9fc
5.
\342\200\2243
\342\200\224llfc
12k,y = -7
-17fc
= 3
jc
(i)
+ 4k,y
= -6-9k
7k,y
= 9-20k
(ii)x = -3 +
2.
2.
(there are
= 84
4.
(a)
Yes,
All
ll,n
(c) x = 220
51; no
3.
(a) 3
4. (a)
(a)
(
5.
5
gcd(^,3)= l
2 + 3k,y
x =
(a)
m =
(b)
jc = 4
= -480
4
(ii)
divide
41
4.
(b)
= -10-15k
= -l-4k
(a)
5.
x =
0(mod3)
x =
4(mod8)
6x = 9k + 4 is impossiblebecause3
and 9 but not 4.
x = 2(mod3)
(a)
* =
3,8,orl3(modl5)
(b) 7
6. a = 2
Exercise 5D
5
(a)
(i)
x =
18 (mod 35)
(ii)
(mod 55)
(b) (i) x = 32 (mod70)
(ii) x = 21 (mod60)
(ii) 6
(a)
(i)
(b)
(i) 7
(ii) 2
(c) (i) 4
(ii) 4
2. (a) (i) x = 215(mod209)
(ii)
x =
201(mod286)
(b) (i) x = 542(mod630)
(ii)
x =
225(mod408)
(b)
3
3. x = 5(mod21)
(c)
1
4. x = 20(modl05)
3. (a)
(b)
^
(mod 9)
x = 24
(d) 0
^
x = 8
(c) -7,-4,-1,2,5,8
2. (a) 1
\\
5C
(e) (i) x = 0(mod3)
Exercise 5A
10
3
(ii) 4
3. x = 4(mod7)
+ 59fc
4, n = -1 (there are others)
+ 3^,7 = -l-5fc
4 and
x = 4,jy = -1
Chapter
4
(ii)
2. x = 4(modl3)
y
+ 37k,y
(ii)
(d) (i) x = 4(mod9)
x = 2 (mod 9)
(ii)
1.
160
(ii) 4
(c) (i) x = 0(mod3)
practice
(c) x = 1,7=
1.
2
3
59 others (investigate!)
solutions as 51 doesnot
(b)
B
(ii)
(ii)
= -24
+27k,
(c) x = 25
5.
weights
> 117 and
weights
ra =
2.
(
(ii)
examination
(b)
(c)
4.
all integer
Mixed
1.
0
of 3
Multiples
(b)
(c)
(ii)
2
1. (a) (i) x = 6 (mod 11)
(ii) x = 6 (mod15)
x = 10 (mod 11)
(b)
(i)
< 0
(a) jc = 3 + 13^,7 = -5-22fc
(b) (-10,17),(3,-5),(16,-27)
3.
0
Exercise
(b) x = 6-37fc,;/ = 5-31fc
x and 7 are positive for all k
(c)
+<l
others)
2
(a)
( )
(c) (
(d) (
3. 2
4D
1. (a) x- -126,y
(b) x = 2,;/= 4
1(ii) 4
(b) (
+ 5^,-13-3^)
Exercise
3
5
(b)
(c) x = 9-3Sk,y= U-55k
(26
by
5B
i. (a) ( )
6
= 4-27fc
(b) x = 2-13fc,;/
(b) m = 9,n = 13
4.
(a) Divisible
(b) 6fc+l
Exercise
2. (a)
3.
6
= -3-llfc
7fc,^
(c) (i) x = 2 + 10k,y = -6-33k
(ii)x = l + 4k,y = -l-9k
(d)
+<J-
4. (a) 9
4C
Exercise
(b)
1
5.
1,4,7
3
6. x = 4(modl5)
3,8,13
Answers
C ******
(a)
+ <*..
(b)
x =
3(mod35)
divides
6x
'
V
+<J
x =
7.
3.
207(mod527)
8. (a) Show that x = 6a + 2 = 9a
(b) (i) ra = 5 + 3a, n = 3 + 2a
(ii) x = 4(modl8)
+ 4
Simple
9
(i)
(b)
(i) 4
(ii) 5
(c) (i) 1
(ii) 1
(ii) 1
(ii)
0 (mod 13)by
FLT
examination
5
practice
2. 6
0
0
1
1
IR o
i
0
Qo
i
2
2
0
G2
4.
1
6.
(a) 4
7.
Apply
8.
(a) 24
= 9(modll)
H
little
28
to
theorem
find
x12(mod
0
0
0
0
1
0
1
0
0
1
1
0
1
0
0
0
1
0
0
o
1
o
o
o
o
7)
65)
(mod
10.
x = 82 (mod170)
12.
(b)
1
(c)
183
(b) x = 16(mod341)
14.
fc
0
1
0
1
0
El
o
1
G6
Q
\302\260
0
|R
o
0
iga
a
B
c
'
1
2
1
1
PR
G2
G3
4
5
7
7
3
5
5
5
8
6
6
6
D
1
B B
3
3
3
3
1
2
1
1
4
2
3
2
3
1
1
3
1
1
3
1
2
5
2
5
2
3
2
3
2
D
'
o
6B
G4 G5
1
\342\226\240filo
6
Chapter
o
0
1
0
0
0
0
0
^9
D
Gl
1
HT>9B
\342\226\240uH
= 2,15,28,41,54
Exercise
0
\302\253
13.
2.
1
D
D
3
C
G6
E9
D
'
0
2
1
A
B
0
0
1
0
1
0
0
0
1
0
0
0
1
0
1
1
0
0
0
0
1
0
1
0
0
1
0
1
1
0
0
0
1
0
0
0
0
0
D D
0 0
0 1
0 0
0 1
1 0
0 0
0 1
D
G
0
0
0
0
1
1
0
o
1
0
'
0
1
Rl
o
1
0
D
'
0
1
o
1
0
|R
D
1
0
1
0
1
0
0
0
0
0
0
1
E
0
1
0
1
0
Answers
-
* v>,Ar
i>i
+
Gl._
/
/
A
\302\253
(c) 31
1.
/
(b) x = 3(mod8)
Fermats
9. x =
o
1
1
o
o
B
x
/
/
/
0
1. 9
3.
/
/
B
= 5(modl3)
Mixed
/
/
i
ER
(c) x12 = 1 or
/
4.
4
JC
G5 G6
S
Bipartite
H
(a)
3. (b)
G4
^^^^^H
(d) (i) 2
2.
G3
Complete
Exercise 5E
1.
/
Connected
is impossible
G2
Gl
o
+<t.
7. There are many
5. (a) (i)
possible
answers,
(a)
8. (a)
(b)
9. (a)
+c<
oL
6.
162
A-2, B-3, C-2, D-2, E-2
(ii) A-3, B-2, C-2, D-3
(b) (i) A-4,B-7,C-3,D-7,E-7
(ii) A-9, B-5, C-3, D-3
(a)
(i)
Answers
C
^i-V^
+ 0i.
A,D,E,G
and
A,B,D,H and
B,C,F
C,E,F,G
for example:
o
(e)
(There are
10.
other possibleanswers)
(a)
%
n=
11.
9
12. (a) l,n-l
(c)
(b)
Doesn't
exist (odd number
of odd
vertices)
(c)
+c<
6. (a) 25
(b) e>3v-6
oL
Exercise 6D
1.
(b)
(a)
(i)
A.
C
Answers
C ^i-V**
+ Oi.
163
*
a
b\302\273*V^T7
^r,
1(i)
(b)
2.
(a)
I
A
HI
I
|
|
C
|
D
0
n
n
0
0
l
0 1
1 0
0 1
l
0
1 0
(b)
9
A
BCD
o
i
0
0
0
0
0
\302\2601
1
]
3
o
3
+\302\243i
B
0
1
1
0
3. (a)
164
A?
Answers
t
******
+ 0[._
i-
a
h
1
+1.
3.
2
2
(c) 0
(a) 2
(b) 4
(c) 20
(a)
(b)
4.
6F
Exercise
i.
(a)
(b)
(c)
(d)
Eulerian
Semi-Eulerian
Semi-Eulerian
Neither
2. (b) and (c)
3. (a)
have
vertices
Four
odd degree
(b) SMNPQNRQMRS
4.
are
There
two vertices
of odd
degree;
YWXVYZWUVZ
6G
Exercise
4. (a)
1. (a)
(c)
2.
ADBFCEA
AFGHEDCBA
(b)
ABCFEHGDA
(d) ABCDEFGHA
(a) (i)
ABCDA
A
(ii)
ABCDEA
+1,
5.
are
There
6. (b)
At
one group
least
(b)
answers
possible
many
(i) AEBFCGDHA
ABCD
of vertices must
least 3 vertices, so in the complement
will
be cycles of length 3.
have
at
there
E
6E
Exercise
1.
(a)
(b) AECD, ADAD,
ACAD
(c) ADAD,
(d)
2. (a)
(b)
H
ADCD, ADAD
ADCD,
ABCD,
ABAD, AEAD
AABD,AAAD
ABCDA
ABCDA,
ABCEA, ADCEA
(c) None
(d)
G
(ii) ADBECFA
ADBC,
ABCD,
F
3. 6
None
Answers
A?
- *
v*,>r
n-,
+ Oi..
a
h
examination
Mixed
1.
(a)
practice
1
6
A,
+1'i'
8.
(c)
Yes;
have degree 4, which
all vertices
is even.
9.
(b) v-e
+f =3
7
Chapter
Exercise
7A
1. (a) (i)
(c)
ABCDA, ABDCA,
2. (a) G2;degreesare
ACBDA
all even:
2,4,2,4,2
-
7
8
-
12
7
-
-
-
13
8
-
-
-
8
-
-
-
-
4
12
13
8
4
-
(b) ABCDEBDA
3. (a)
(b)
A
a
c
0
A
\302\253
B
o
0
0
C
o
l
0
D
1
l
1
E
1
l
0
BCBD,
BCBC,
1
D
a
1
1
1
1
0
0
0
1
0
0
-5-75
5-5-7
-5-57
7-5-5
BEBD, BEAD
5
7
7
5-
-
3
3
4
-
4
-
-
-
6. (a)
(b) (i)
(n
\342\200\224
= \342\200\224
l
n
l)
1
-
-
6
6
7
4
-
6
2
7
4
2
-
-
4
5
4
-
-
4
-
-
-
-
-
5
-
-
-
4
3
4
-
-
-
-
3
-
-
4
-
-
-
-
-
3
3
-
-
-
1
-
6
-
(ii)
+1,
7.
(c)
(i) (n, d) = (1,6),(2,5),(3,4),(5,2) or (6,1)
Note:
(ii)
(n9d)
n = l9d
= (4,
3) not possible
=6
n = 2,d
= 5
^P
n
= 3,d
= 4
n
= 5,d
= 2
2.
n
(a) (i)
= 6,d=l
CN
10
166
^
^
^
Answers
C ******
+ Oi..
WC
4.
(a)
5.
ABCDEA
= 76
Length
are even;31
6. All vertices
7.
(a)
odd degree,
F have
B and
(b) (i) BF
59
(ii)
7B
Exercise
1. (a) (i)
(b) (i)
AD, BE
BC,
AB,
= 41
(or CE); weight
AC, BC,AE;
(ii) AD,
weight
= 28
BG,EH,CD,ED,BH,AB,
FG;
= 141
weight
(c)
C
4
B
(b)
AEBCDA,AEDCBA
ABCDEA, ADCBEA,
= 30
(ii) AG, BC, ED, BG, GD, AF; weight
(i) RS,RP (or PS),SK, SL, LM, RQ, PN; weight
= 47
PU, PY,
(ii) PW,
weight
2. (a) (i) BD,BC,AD,
(ii) AD, AB, BC,
(b) (i) ED,FG,
weight
3. (a) BD,
(b)
4. (a)
Add
CD,
ED, AB,
ED;
weight
= 24
FC, CE,AB,
CE, AB,
least
BE
AF (or EF)
then
weight
start
Kruskals
algorithm.
which includes
every
vertex,
(b) BD,AC,
AF, BE;
AE,
weight = 45
5. (a) n-1
(b)
6. (a)
(b)
7.
It has
YZ;
weight = 155
= 21
BE), AD; weight
=
19
(or DE); weight
AC, BD;
EG,
CF, BC (or
CF first,
Treeof
(or WS), TX,
AE; weight = 52
(or EF), CG, AC, BD;
DG
EF,AB,
(ii) DF,
PV, PS
= 120
(ii) EF,
(c) (i)
PT,
= 91
to be
at
least
8 +
9+-
+
18 =
143
49
58
x>18
Answers
a
1.
(c) (0
7C
Exercise
2|4|
5
6 115 |
S
4
(a)
^
(i)
15
|K)7
-D
A-
X5
3/
MB
mrj H
V
*E
5
-E
c-
3 |7 |
/7
\\
X3
4\\
/
4 8 |A/S V
8
3|4|5
S
5 12
7
C
12
6\\8\\B\\
or
SBDF
7, 8,
SCEF;
SC\302\243F;15,12,19
11
(ii)
(ii)
2|4|g
V
4X
m
c
V^L2
|5|15|5/C|
>615
7
un
8
V
7
JD
uss
\\
4\\7\\S
7
\302\243/
15
3|6|S
6ll9|D
2419
6
S5DFF;14,19,22
(b)
3.
(b)
(i) SVQT
(a)
(i) SCF
(b) (i)
4. (a)
7
|12|D|
(b)
5.
SBGJF
(ii)
u
1.
SCQRHKM
cost = 21
17
(a) 21
(A5FFG)
(b) ACDEFG
(b)
(i)
(ii)
50
(ii) 32
e.g. CDFFAFCA5C
2. (a) (i) Vertices D and
F
IP
3. (a) (i)
(ii)
(b)
SBDEF; 4, 8, 10
^
C ******
+ <*..
odd degrees
odd degrees
e.g. ABCEHGFDFGEBDA;
A,B,E,G
B, D,F,G
(i) 78, repeat AB and FG
(ii) 107, repeat BE, EF and
4. (a) 0,C
Answers
F have
(ii) Vertices D and G have
(b) (i) e.g.ABDEFBEGCDEFGA;70
(ii)
^
or SBPRHKM;
or SBQRHKM
(c) (i) e.g.CDEFABDAEBC
\\5\\6\\S
6
A;
(ii) SQ5FF
or SBCDGJF
Exercise 7D
SC5DF; 8, 8, 12
168
S5DFT
(ii) SBDTor SABDT
(ii) SCF
6. jc=1
2|4|S
+1,
SCDF; 15,19,22
2. (a) (i) SCFT
(i)
/4
\\/
c-
SBDF or
/
A
V
<^9
\\
-F
(b)
9
(c)
87
DG
40
7|22|D
22
h
5. e.g.
6.
CBDFDBAJIHGFHJCFEDC(length
1
(b) 84 (ACFIJ)
9.1)
6.
7. (a) e.g.AFGAEGBHEDHDCBA,
8. (a) C and F have
(b) CEF(15)
7. (a)
odd degrees
(c) e.g.
CEFECDEBACFDBC,
(d) (i) 156
(ii)
F
(ii) 85
36
(a)
(i)
(b)
(i) 63
(c) (i) 103
2. (a) (i) 32
(ii)
99
(ii)
110
(b) 1 and 8 have
(c) 1-3-5-8 (length
(d) 66
8. (a) A, B, C, D
(ii) 77
56
(c) (i) 95
3. (a)
= 180
length
C and
7E
Exercise
(i)
(ii)
82
(ii)
104
9.
ACDBA, ADBCA,
ACBDA,
ABDCA,
ABCDA,
ADCBA
(b) L = 45
4.
(a)
(b)
The cycle AC&EDA has
28
40
(b)
40
2.
examination
of minimum length
every vertex
BH,NF, HN, HA, BE, CN
equal.
7
which
(weight
=
one edge
add
We
vertices,
1.
than
degree.
of
two vertices
between each pair of
so now all vertices have
A
At
km
(a)
1,5,21,85,341
(b)
3,1,3,1,3
bound
form a
8
weight of the
10
_
edgesgives
95.
1.
tB
(a)
(i)
(b)
(i)
2\"+l
(ii)
2\"
~(-3T+\\
12
2
2l3 J
(ii)
= 33;
2.
(b)
89
(c)
89<L<114
33|
-
|
-3
LB = 60
4. (a) 114
ACGJ,
lower
Exercise 8B
Ad
weight
the
(d) 2,4,12,48,240
maximum
X the
306
(c) 1,2,2,4,8
=117.5)
11
5. (a)
of odd
Exercise 8A
BJ, AB, EG, CH,DE
tree gives 92;
(ii)
degree.
ED)
AE,
Chapter
answers:
the length
(a) There are severalpossible
of any cycle, e.g. ABCDEA
73;
gives
of the minimum
doubling the weight
spanning
(i)
of odd
vertices
11.(a) 33
(b) The edges includedin
cycle.
48)
order
CG, I],
(weight
(b)
EB, CG,GD
contains
of increasing length,
would
create a cycle.Stop
any
the vertices are connected,
all
BC,
EF,
(ii)
that
skipping
5
are
odd degree,
practice
A tree
when
3.
(b)
890 (repeat
repeat AE,
even degrees,
(a) Add edges in
(b)
There
(b) (i)
Mixed
(b)
(a)
adjacent
(a)
(a)
length
=12)
10. (a) (i) There are vertices
(ii) No; there are more
29.
length
(c) Upper and lower bounds are
1.
= 92,
(b)
odd degrees
(c) AandD
(c) 26
5.
276
= 208
length
(b) 230
(b)
ACFJ/(84)
(c)
(a) 99
(b) 131
1.
+<J-
3.
length
Un
-I
It gives
4. (a)
= 55
HL
\\*.n-urn-l
d=0
8,48,248
Answers
-
*, v>,Ar
i>i
+
Gl._
a
h
5.
(b)
w\342\200\236=2x5\"-2
(a)
w\342\200\236=cx2\"+
(b)
W\342\200\236=l
1
(b)
3.
4.
1
8C
Exercise
1. (a)
w\342\200\236=2\"+3\"
(b)
+1'i'
3.
un=(3-n)(-2)\"
w\342\200\236=(2-i)(3i)\"+(2
6.
+
,\342\200\236=3
(c)
+ 2\"+1
m\342\200\236=5
(a)
l)2\"
u\342\200\236=(\302\253
(b)
= 3
2\302\253
u\342\200\236
(a)
h,
(b)
+
=(l
u\342\200\236
+ i)(-3i)\"
4(-lJ
\302\253\342\200\236=2\"-(-2)\"
(a) (i)
un_lyun_2
(ii)
un_x-un_2
(iii)
9un_2+(un_1-un_2)
+
+
4. (a)
+ i)\"-3(l-i)\"
=-3(l
=
(b) (i) 4
2i)(l + 3i)\"
(ii)
+(l-2i)(l-3i)
8.
cos
2
u\342\200\236
(b)
=-6,
\302\253!
h2
I(4--(-2)-)
an=-(l-3\"\"2)
=\342\200\22412
9
Chapter
w\342\200\236=-6^)
2.
u\342\200\236--
6.
(a)
12
(c)
fl,=4--(-2)-
(d)
6B=4-+(-2)-
examination
Mixed
Sin(^j
5.
\342\226\240$+tf+$-*r
(a)
9
practice
4
2 and
(b) (i)
8D
Exercise
i-
un=3\"~l
5.
7.
2.
11 years
=
\302\253\342\200\236+i
1.04^,^=1000,
= 1800
\302\25315
2.
=40
000(1.05\"-1)
\302\253\342\200\236
3.
(b)
un
(c)
62 months
=35000-18812(1.01\
4. (b) ux = 1, u2
(c) 233
5. (a)
+1,
=2
\302\243n+2=-(tn+1+0
+ 16
\342\200\224
(b)
tn =23
(c)
tn approaches
below that
23, being alternately
above
and
value.
(c)
(b)
ul=2yu2=3
(c)
2584
8. (a)
Hx
= 1,
3.
H2 = 3
(b)
130
(c)
1288
(d)
\302\243>50
1.
un
=
2. (a)
\342\200\224
+
(b)
ak
(c)
22,55,88
ak_x
(a)
It is
a graph
the
not in
(b)
^
^
^
is divisible
\342\200\224
+...
ak_3
11
by
practice
8
all the
containing
edges
which
graph,
original
6. (a) (i)
A
graph
that
with
two sets
the vertices
of verticessuch
from the same set are
adjacent,
(ii) The number
vertex,
\\&~l-\\)
4
nn=0.92ivi
Answers
C ******
+ <*..
are
9
(b) K is bipartite
170
ak_2
122
(b) x = 2(mod5)
5.
examination
Mixed
ABCDE
4. (a)
(c) Hn=2\"-1, 1023
9.
ABCDEA,
ABCDEACEBDA
6. (b) 1
7.
are Eulerian;
Both
(ii)
of edgesstarting
from
that
not
o
h
1
+ o
(c) EDCABC
(b) Vertices from
7. (b) x = 14(modl7)
(c)
x =
9.
(b)
A5CDA,A5DCA,AC5DA
(c) L = 37
11.
(a)
(b) (i)
15.
(c) e.g.PQflSl/V
m=
between
or AC5DA)
(A5DCA
(b) RS
10.
9
(ii)
= 2
n
\342\200\2241,
jc = -104
+ 8^,7
= 208-15k
16.
(c) x = 3,18,33,48,63,78
(mod90)
12.
(a)
Choice
1
Total
Edge
b)
a)
6
b)
66 and 42 are
Weight
17.
c)
18.
b) (i)
19.
a)
1
=2
KQ
2
= 2
QF
2
= 4
FE
3
b)
=4
PB
3
20. a)
6
ER
4
=7
=7
9
PQ
5
BC
CD
5
6
281502
BC,
AB,
b)
48
(a)
S and
= -9 + llk
AI, DE,
=100)
DG, DF, (or EFor EG),BD,
= 84
x = 2(mod5)
is a unique
10; there
solution
(mod
55)
22.
a)
CF,EF,BC,CD,AB
b)
(i) v-1
b)
(use v+/=e
+ 2with/=l)
v \342\200\224
c
(i) AB,AD,DE,BC
(ii) 12m
T have odd
(b) 13(SBEGT)
(c)
not
6
-1,17,11,113,179
b) un=3n+2(-2)n
23.
Vertices
9 is
degree
21. a)
(d) 1814400
13.
even,
odd
3
(ii)
(b)
by
(taken
IH;weight
31
=
of vertices.
sets).
vertices have
All four
34
a)
HP
weight
two
the
c) m divisible
d) x = 6-7k,y
so 9 choices)
(10 vertices
even number
3 and 5.
c) It has cycles of length
d) A Hamiltonian cycle would have 7 vertices,
which is odd.
e) m = n (because the vertices in a cycle alternate
31 (modi87)
8. (b) (i) 9
alternate, so any
two sets
the
have an
must
cycle
c)
degree.
24.
61; ASBSCEBEGTG
Each vertex has
(i)
degree 4, which
46
is even,
(ii) ABCDEACEBDA,
EDTFDBA
a) CD,AB,BC,BF,CE
b) (i) 213
14.
c)
+c<
25.
(ii)
239
(iii)
239
239<L<253
b)
un
c)
16 years
=2100-1000x1.05\"
Answers
A?
K
^V****
+ Ql..
\302\253-
*
^
(a
v
+r*
\342\200\242+<*<;
)
Glossary
that
Words
appear
in bold
in the
definitions of other termsarealsodefined
abstract nature of this optionmeansthat
terms
more simple
of other,
Term
terms
defined
some
adjacency
1-
concepts.
Example
vertices in a
A table showing which
are
graph
by edges.
joined
(ofverticesor edgesin
adjacent
The
glossary
only be explainedin
can realistically
Definition
table
this
in
to each
joined
a graph)
other.
The following
edges of the
are AB
graph
A
B
A
0
1
B
1
C
0
0
1
c
0
1
0
if they
are adjacent
vertices
Two
that the
and BC:
shows
table
are joined by an edge; two edges
are adjacent if they are joinedby
a
vertex.
algorithm
A specified
sequence of
guaranteed
to answer
Kruskals
steps
a certain
equation
auxiliary
A
used in
equation
quadratic
solving a secondorderlinear
recurrence
relation.
base 10
for writing numbers using
system
ten digits, 0 to 9, where moving
a digit a place value to the left
A
increasesits value
base cases
A
graph
two
into
un+l +2un has
A:2 = k + 2.
equation
un+2
the
a graph.
relation
recurrence
The
sequence
find
=
auxiliary
'3' in 438has value
In base ten, the
= 30.
3X10
times.
need to
be tested in
a proof by induction.
whose
vertices can be split
such that only
groups
to start
order
bipartitegraph
that
cases
Initial
ten
to
spanning tree of
minimum
question.
is a
algorithm
of steps guaranteed
vertices from different groups are
joined.
Chinese
postman
problem
To find the route of shortest length
which includes every edgeof a
and returns to the starting
graph
In an
Eulerian
graph, every
Euleriancircuitisa solution
Chinese
of the
postman problem.
point.
Chinese remainder
A
theorem
solutions of
result
the existence
about
a system
of
of linear
complement
(of a
graph)
A
closed
trail.
A
graph
were
consisting
not in
bipartite
graph
that
complete
bipartite graph
A
graph
A
graph
vertices
172
in which
is joined
remainder
Chinese
of congruence
1 (mod
9) has
a
unique solutionmodulo 63.
of all the edges
the original graph.
where
every vertex
in one group is joinedto every
vertex
from the other group.
complete
the
theorem, the system
3x = 5 (mod7),2x=
congruences.
circuit
Accordingto
every pair of
by an edge.
If a graph
its complement
The
not
A
has
4 vertices
bipartite
complete
and 2 edges,
4 edges.
has
graph
K3 3
is
planar.
complete
graph
with 4
vertices
has 6 edges.
Glossary
rsM
^ Q,VVj
VS.
.Ok\"
| Definition
Term
composite
An
| Example
is not
that
integer
15 is a composite
prime.
=
i 15
\342\226\240
modulo 12
congruent
When
the same
give
5 and
congruent modulo 12.
29 are
remainder when divided by 12.
A
in which it is possible to
graph
connectedgraph
in which coefficients
An equation
do not
with)
degree (of a vertex)
un+l =nun
path.
coming out of
of edges
number
The
relation
recurrence
The
does not have constant coefficients.
on n.
depend
A closed
cycle
any
vertex.
other
constantcoefficients
from any vertex to
a path
find
(equation
numbers
two
because
number
3x5.
the vertex.
degree
of all the vertices
of degrees
list
The
sequence
in a graph.
The
is important
sequence
degree
in determining whether
is
a graph
Eulerian.
algorithm
Dijkstra's
An
the
for finding
algorithm
shortest path between two
in a
Diophantine equation
An
vertices
graph.
Some Diophantine equations
can
by looking at factors
we are only
where
equation
looking for integer solutions.
be solved
and divisibility.
For
equation
solutions
because
even and the
direct proof
A
by
directed
graph
(or
digraph)
fact is derived
a new
where
proof
or
calculation
test
A
test
b2
edge has a
that
is odd.
- b)
(a
(a+b) =
by expanding
directly
A directed
side
simplifying.
graph
could be used to
of one-way
a network
represent
roads.
whether
for determining
one integer is divisible
We
becausethe
another
by
divisible by 5
475 is
that
know
units
digit
is 5.
carrying out the division.
without
division algorithm
-
brackets and
facts.
specifieddirection.
divisibility
directly
reasoning
from previously known
A graph in which each
a2
has no integer
the left side is
right
We can prove
the
example,
= 5
+ 4n
2n2
The processof
the
finding
and remainder when
dividing
quotient
We can write
23
We often use
Euclid'slemma
= 4
X
5 +
3.
two
integers.
edges
'Lines'
Euclid'slemma
A
connecting
vertices in a
graph.
numbers,
divides
a
prime
it must
divide both
about
statement
saying
prime
if a
that
product then
proofs
involving
in
factorising.
factors.
Euclidean
algorithm
A
method
without
prime
Euler's
relation
A
rule
the greatest
for finding
common divisorof
explicitly
factors.
giving
two
finding
integers
all their
a relationship
the number of edges,vertices
number
of regions of a planar
between
We can use Euclidean
to find a particular
Diophantine
in
algorithm
solutionof
a
equation.
A complete
and
vertices
graph.
satisfies
graph with three
has v = 3, e = 3,/ = 2,and
v \342\200\224
e + f = 2 .
Glossary
| Definition
Term
circuit
Eulerian
A
trail
closed
|
a graph which
around
uses each edgeexactly
Eulerian
that has an
A graph
graph
A
to
Eulerian
An
once,but
Fermat'slittle theorem
uses each
which
trail
does
edge exactly
to the
return
not
starting
point.
allowing us to find the
power of a number which
givesremainder 1 when
Chinese
the
Eulerian
divided
circuit is a solution
problem.
postman
can be drawn
graph
pen and
up the
without
picking
without
repeating
edges.
any
A graph which has an
Eulerian trail
semi-Eulerian.
is called
A result
smallest
Every Eulerian
once.
circuit.
Eulerian trail
Example
1512=
l(modl3)
by
a prime.
first
recurrence
order
A
in which
relation
recurrence
term in a sequencedepends
Arithmetic
generalsolution
of
an
The
largest
(gcd)
two
given
greedy
An algorithm
way
3 X
The general
x+y
equation
=
5-t
= 5 is
(teZ).
The greatest common divisor
and 42 is 6.
integers.
of a
solution of the
Diophantine
x = t,y
integer which divides
Kruskals algorithm
of a greedy algorithm,
in which the best
every
includes every
A closed path which
there is no other
30 as a productof
numbers.
The travelling
is to
graph.
is an
find
12
example
is not.
salesman
the
of
whereas
algorithm
Dijkstras
vertex
5 and
to write
prime
stage.
cycle
recurrence
order
lines.
possibleoption is chosenat
Hamiltonian
can be
sequence
geometric
30 = 2 X
equation.
of points connectedby
greatestcommon
algorithm
described by a first
The result stating
that
every positive
can
be
written
as a product
integer
of prime factorsin exactly
one way.
An expression describing all
A set
divisor
A
one
relation.
possible solutions of
graph
each
term.
previous
Theorem
Fundamental
on
problem
Hamiltonian
shortest
cycle in a graph.
Hamiltonian graph
A graph which has
Hamiltonian
A
a Hamiltonian
path
complete
graph
is
Hamiltonian.
cycle.
path
Every
which
every vertex
includes
of a graph.
equation
homogeneous
A
The recurrence relation
is not homogeneous.
with
relation
recurrence
linear
no constant term.
indirect
proof
A
where
proof
a new
using previously
not
inductive step
in
A part
a direct
how,
facts,
Kruskal's
least
algorithm
common
multiple
linear congruences
n,
we
can
next value
for
the
An
algorithm
Proof by contradictionis
example
of indirect
have already
a certain
use this to
prove
it
of n.
for finding
the
Theleast common multiple
divisibleby two
9 is 18.
interested
the
actual
given
integers.
in which we are only
in remainders,
rather
numbers.
Thelinear
than
of 6
and
congruence
6 (mod 7) is satisfiedby all
which give remainder 2 when
integers
3x =
dividedby 7.
Glossary
one
proof.
minimum spanning tree of a graph.
The smallest integer which is
Equations
un+2
where
induction
if we
proved the statement up to
value of
but
=
way.
of proof by
we show
fact is derived
known
un+1
Term
|
linear recurrence
relations
A
| Example
terms
of the
linear
terms.
bound
A
sequence appear as
be
is known to
smaller than (or equal
desired solution.
u
n+l
, = u2n
is
linear.
not
a
connecting
edge
which
number
The recurrence relation
in which the
relation
recurrence
(in a graph) An
vertex to itself.
loop
lower
Definition
A
for the
bound
lower
travelling
salesman problemcanbe found
to) the
removing
by
one vertex
from the
connector
is the
graph.
minimum connector
problem
minimumspanningtree
all the vertices of
The tree
contains
arithmetic
modular
remainders.
A
multigraph
solution
One possible
x = 2,y
of
equation.
A walk
permanent labels
(in Dijkstra'salgorithm)A
assigned
prime
proof by contradiction
by
induction
can be drawn
exactly
of proving
A method
that
assuming
its opposite
leads to impossibleconsequences.
A
'step
by step' method of proving
A
is not
50 = 2x52
5 is a prime number,
a statement by
is true for all
a statement
numbers.
showing
relationship
It can
but
1 is
not.
be provedby contradiction
is an irrational number.
that
V2
You
used
by induction
proof
in the
Core syllabus to prove De Moivres
Theorem.
(integer part of the) result
two integers.
dividing
relation
its
two
The
recurrence
graph fc5
factors.
natural
quotient
contains
graph
same degree.
the
with
planar.
An integer that has
factors - 1 and itself.
that
pigeonhole principle
every
The complete
without
as a productof
a number
Writing
showing
proof
that
use the
can
to show that
two vertices
intersecting.
prime
prime number
We
two objects.
at least
graph
edges
factorisation
the
from
objects are divided into n groups,
then there must be a group
A
graph
number
that if n + 1
stating
principle
containing
planar
particular solution
is one
= 5.
y
vertex.
starting
principle
pigeonhole
x +
= 3
vertex showing the
shortest possibledistance
A
divided
gives remainder 3, then
remainder 5.
repeated vertices.
path
to a
algorithm.
vertex.
one
solution of an
with no
KruskaPs
using
are
some edges
connected by more than
particular
found
same
tree.
spanning
spanning tree can be
Minimum
by 6 and y
x + y gives
where
graph
minimum
If x gives remainder 2 when
with
calculating
minimum
The
as the
a graph.
of smallestlength
which
all the vertices of a graph.
for
Rules
tree containing
shortest
the
find
To
in a sequencedepends
how a
on
The
of
quotient
5 is 4.
term
un+2
-
when 23 is divided by
un+\\un is
recurrencerelation.
previous
an example
of a
terms.
relatively prime
Two integers
whose greatest
(or coprime)
common
remainder
The amount left
divisor
divide
Route
inspection
12 and
35 are relatively
prime.
is 1.
over when we try
another.
to
one integer by
The remainder when
by 5 is 3.
23
is divided
Sameas Chinesepostmanproblem.
Glossary
Term
|
Route inspection
algorithm
Definition
An
| Example
algorithm
Chinese postman
secondorder
recurrence
A
the
for solving
problem.
in which
relation
recurrence
term in a sequencedepends
graph
A
each
The Fibonacci
two
describedby
terms.
previous
semi-Eulerian
on
Eulerian trail.
order
relation.
recurrence
has an
that
graph
sequence can be
a second
In a semi-Euleriangraph
we can
solve the variant of the Chinese
postmanproblemwhere
simple
graph
A
graph
multiple
strong induction
which
has no
edges
between
A variant on
only to simple
graphs.
In this
we use
more
to use
value of
previous
Many theorems
vertices.
proof by induction
need
we
where
loops or
one
inductive
the
in
n
than
Option
in
do not
we
need to return to the starting
the solution
is any Eulerian
vertex;
trail.
course
this
apply
strong
induction to prove Euler
for planar graphs.
s relation
step.
subgraph
A
is a
which
graph
part of another
temporary
algorithm)
(in Dijkstra's
shortest distance from
lemma
number
the
starting
found so far.
vertex
Handshaking
A
vertex showing the
to a
assigned
The
is a
tree
spanning
subgraph.
graph.
label
The minimum
The result stating that the sum of
degrees of all the vertices in a graph
is equal to twice the number
of
edges.
The Handshakinglemma
implies
that the number of odd vertices
in a graph has to be even. This
is
in the Route inspection
important
algorithm.
trail
A walk
travelling salesman
To find the route of shortest length
which includes every vertex
of the
graph and returns to the starting
problem
no
with
repeated
edges.
problem.
point.
tree
upper
A graph
bound
A
town in a region
every
in the shortest possibletime is an
salesman
example of a travelling
Visiting
with no
larger than
solution.
(or
be
the desired
is known to
which
number
A tree
cycles.
equal
to)
An
has 9
10 vertices
with
bound
upper
for the
edges.
travelling
salesman problemcanbe found
using
nearest
the
neighbour
algorithm.
vertices
(vertex)
walk
weight
'Points'in a graph.
Any
A
sequence
number
of adjacent
associated
edges.
with an
edge
in a graph.
The weight
between
weighted graph
A
graph
associated
Glossary
in which
weight.
of an
edge could
represent the cost of
each edge has an
A
travelling
two vertices.
weighted
graph
represent a
weights
being the
could
road network
with
lengths of the
roads.
-
-+<*<;+\"
Index
95, 98-99,172
70, 74-75,
tables,
adjacency
adjacent (ofvertices),
69,172
decimal (base10)system,
degree
defined, 172
algorithm,
postman
problem,
89-91
graphs,
Eulerian
bases, 36-39
in different
see also modular
degree sequenceof
arithmetic
fill-in
base
general solution,
bases,33-35
36-39
in different,
changing between
directedgraph,
35-36
different,
34-35
mixed exam practice,41
40
summary,
numbers,
33, 34
bipartite
graph,
72, 95,172
remainder
59-61, 65,
theorem,
172
complement (ofa graph),
exercises, 85-87, 97
85, 95,172
completebipartite
72,172
and
graph,
proof,
subgraph of, 85
complexroots,
relations,
141
composite numbers, 26, 28,173
by, 63-64
linear, 57-59
modulo
congruent
modulo
95, 173
81, 83-84
trees as,73-74
recurrences, 136,139,173
to, 47-48
contradiction,
proof
by, 3,4-6, 7, 28, 29,175
coprime
prime) numbers, 20, 21, 23,26,175
(relatively
Chinese
remainder
theorem,
60, 61, 63
constant
linear
coefficients,
division
solutions
rule,
cost adjacency
subject
56
matrix,
98,
cycle, 73, 95, 173
Hamiltonian,
whole
13-19
numbers,
graphs, 69, 70, 95,173
79-85
involving,
with
integer
42-43
solutions,
95,174
89-92,
graphs,
101, 109,130
173
13-18
Fermat's
Last
Theorem,
Fermat's
Little
Theorem,
example
of a
formula
for
for
2,43
proof by
first order
recurrence, 134
140
term,
second order
the nth
growth, 144-45
population
induction,
strong
recurrence
linear
65, 174
62-64,
modelling
9
relations,
accounts,
modelling a savings
Fundamental Theorem of Arithmetic,
general solution,
recurrence
graph
31, 174
137,140
algorithm,
Dijkstra's algorithm,
106-12
Kruskal's
102-6
algorithm,
exam
27-28,
98
Chinese postman
minimum
136-39,174
143-44
48, 49,174
46-47,
relation,
algorithms,
mixed
93,118-21
34
Fermat numbers, 12
12, 51,173
53-54
proofs,
graph
56-59
Eulerian trail, 90-91, 95,174
Euler's relation, 80-82, 83,95,
m rule,
connected graphs,
71,
exercises, 78, 85
constraints,
edges,
59-62
simultaneous,
planar
of
Fibonaccisequence
congruences
congruent
number, 63-64
congruences,
duodecimal (base 12)system,
factors,
recurrence
divisibility
linear
Eulerian
cycle, 93,118-19
82-83
Hamiltonian
173
Euclidean algorithm,
23-25,
31, 173
Euclid's lemma, 26,28,173
Eulerian
circuit,
90, 95, 112,174
72, 95,172
graph
composite
and
equations
Eulerian,90,112
planar
by a
theorems
172
complete graph,
38-39, 40,
14, 31, 173
14-17,
algorithm,
113
112-14,125, 172
exercises,61-62
87, 95,
tests,
and remainders, 51-52
exercises,115-18
circuit,
divisibility
division/divisibility
Chinese postman algorithm,
Chinese postman problem,
Chinese
69,173
principle, 6
5
proof,
diagonal
47-48
constraints,
173
Dirichlet's
division
binary
Cantor's
proof, 3,
direct
46-47
subject to
solutions
arithmetic
exercises,
equations, 1,42-43,49,173
of solutions for, 44-46
number
finding
73-74
strong induction,
131-33
diagrams,
Diophantine
10,11,172
cases,
106-12,173
Dijkstra's algorithm,
base 10,33-35,40,172
70-71, 173
a graph,
69, 95,173
digraph,
140-42,172
equation,
auxiliary
79-80
theorems,
of, 27-28
Theorem
Fundamental
112-14
Chinese
and
arithmetic
33-34
70, 95, 173
(of a vertex),
112-18
spanning tree, 102-6
practice,
126-30
Route inspection problem, 112-18
Ind
, ?^%
^
'
+
106-12
shortest path,
minimum spanning
tree,
102-6,175
mixed exam practice, 151-57
124-25
summary,
salesman problem, 118-24
vertices, 118-24
travelling
graphs,
weighted
Diophantine
98-102
Eulerian
mixed exam
subgraphs
85-87
complements,
summary, 95
divisor (gcd), 19-23,31,174
common
greatest
23, 25
exercises,
method
for
70,175
multiples, 13-18
23-25
102,174
Green-Taotheorem,
62-65
65
summary,
multigraphs,
finding,
greedy algorithm,
56-59
congruences,
Fermat s little theorem,
rules of, 53-56
69-78
terminology,
theorem, 59-62
linear
and
division
41
51-52,149,175
arithmetic,
Chinese remainder
87-89
a graph,
and
modular
148
integers in different bases,
representing
practice, 96-97
around
moving
recurrencerelations,
79-85
theorems,
important
arithmetic, 66
modular
graphs, 93-94
Hamiltonian
32
96-97
theory,
graph
89-92
graphs,
50
equations,
and prime numbers,
divisibility
graph theory, 2, 67-69
126-30
on graphs,
algorithms
all the
visiting
V
+,
-+<*<
nearest neighbour
29
119,120,121
algorithm,
exercises, 123-24,127
cycle, 93, 95,174
Hamiltonian
exercises,94,
problem, 118-21
travelling salesman
number
graphs,
path, 93,174
Euclidean
79, 176
lemma,
homogeneous (recurrence relations),
solving second order, 142
inductive proofs, 3, 9-12,
18,73-74,135
26-29
numbers,
Occam's razor,
10,174
step,
integers
63-64
of, 13-18,
divisibility
different
representation
42-44
solutions of equations,
see also prime numbers
in
33-39
1, 149
problem,
introductory
bases,
irrational numbers, proof by
Kruskals algorithm,
3,4
tree,
spanning
a graph,
shortest
all the
route
law of excludedmiddle,
least common multiple
linear
(1cm),21-23,31,174
57-59,174
congruences,
linear Diophantine
42-44,49,173
equations,
exist, 44-46
if solutions
determining
the
finding
4
solution, 46-47
general
mixed exam
practice,
50
136
maps,
loop (in a graph),
lower
bound,
planar
population
pigeonhole
proof by
prime
26-30,175
relatively prime,
by contradiction
connector
102,175
175
18
example,
relations, 135
induction,
Pythagorean
9-12
problem,
20, 21,23
, 3,4-6, 7, 28,29,
by induction, 3
strong
4-6
Theorem,28
numbers,
divisibility
6-8
contradiction,
strong induction,
minimum
21-22
factors,
Prime Number
3
principle,
6-8,175
graph proof, 71
74-75, 80-85, 95, 175
graphs,
144-45
growth,
modelling,
recurrence
methods of proof,
75
93
path,
powers
factors of integer,
16, 31
modular arithmetic, 55, 62-63,64
factorisation,
26-27, 28, 175
prime
proof
119-22,125,175
vertices, 118-24
simple
proof
70,175
106-12
112-18
label, 107,108,109,124-25,176
permanent
logic,2
logistic
73, 74,
(cycle),
Hamiltonian
prime
solutions subject to constraints,
47-48
linear recurrence relations, 136-39,143-44,175
vertices,
46,48,49,175
solution,
closedpath
107-9
algorithm,
two
visiting
pigeonholeprinciple,
Dijkstra's
102-6
path, 87, 88, 95,175
102-4, 124,174
exercises,104-6,126-27
2, 67-68
graphs,
shortest path between
shortest route around
particular
contradiction,
137
problems,
optimisation
minimum
labelling,
13-18
3,174
proof,
inductive
23-25
algorithm,
prime
indirect
theory
1-2
factors, multiples and remainders,
19-21
greatest common divisor,
least common multiple, 21-23
139,174
graph
33-34
systems,
number theory,
93-94,174
Hamiltonian
see graph algorithms;
existence proof, 8
modelling
non-constructive
96, 99,101
Hamiltonian
Handshaking
network
9-12
triples, 43
quadratic
equation
quotient,
14, 175
see auxiliary
equation
Index
I
-
8
theorem,
Ramsey's
175
relations, 2,134,147,
recurrence
of, 143-47
applications
defining sequences, 134-36
order
linear, 136-39
first
mixed exam practice,149
secondorder
139-43
quadratic,
148
summary,
recursive definitions
reductio ad absurdum,
4
relatively prime (coprime)
division rule, 56
s little
Fermat
and
134-36
of sequences,
numbers, 20, 21,23,
26,175
63
theorem,
remainders, 13-18,175
little
Fermat's
62-64
theorem,
see also modular
arithmetic
roots, recurrence relations,
repeated
Route
141-42
algorithm, 125,176
inspection
Chinesepostman
112-18
problem,
see optimisation
route optimisation
problems,
second order
recurrence, 134,176
secondorder
recurrence
modelling
139-43,147
relations,
144-45
growth,
population
graphs
graph, 90-91, 95, 176
semi-Eulerian
sequences
134-36
recursively,
defining
degree sequenceof a graph,
proofs by induction, 9-12
seealso
problems,graphs,
70, 95, 176
graphs,
simple
relations
recurrence
shortest route
70-71
106-24
of, 85
complement
degree sequence of, 71
simultaneous
59-62
congruences,
9-12,176
strong induction,
numbers
prime
relations
tree proof,
73-74
spanning tree, 102-6
minimum
124-25,176
107,108,109,
label,
temporary
135
proof,
95,176
85-87,
subgraphs,
27
proof,
recurrence
trail, 87, 95,176
90-91
Eulerian,
travelling
along the
edges, 112-24
travelling
salesman
problem,
118-24,176
tree, 73-74, 95, 176
minimum
upper
spanning
vertex/vertices,
95,176
labelling
of, Dijkstra's
theorems involving,
algorithm,
all: travelling salesman
vertices, 118-24
a graph),
(around
87, 95,176
closed,89-90
weight
(of an
edge), 98,124,176
minimum
weighted
spanning
trees,
102-4
graphs, 98-102,124,176
, -
107-9
79-85
all the
visiting
walks
102-6
176
119-22,125,
bound,
visiting
tree,
3rvV%
problem, 118-24
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