CMSC 250
Discrete Structures
Exam 2 Review
Summations

What is next in the series …
9 16 25
4, , , , ______
4 9 16
2

k  1

,

General formula for a series

Identical series

Summation and product notation  2

Properties (splitting/merging, distribution)
k
ak 
, k 1
k 1
ak
k2
i 1
bi 
, i2
i 6
k 1
k
5
 2k
k 1
6
7
1
1 7 1
 

k 0 k  1
j 1 j
k 1 k

Change of variables

Applications (indexing, loops, algorithms)
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k 1
Exam 2 Review
2
Properties

Merging and Splitting
n
n
n
 a   b   a
k m
k
k m
k
k m
k
 bk 
 n
  n
 n
  ak     bk    ak  bk 
 k m   k m  k m

n
i
a  a
k m
k
k m
k

n
a
k i 1
k
 i
  n

ak    ak     ak 

k m
 k  m   k i 1 
n
Distribution
n
n
k m
k m
c   ak   c  ak 
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Exam 2 Review
3
Using the Properties
ak  k  1; bk  k  1
n
a
k m
n
k
 2   bk
k m

 

  ak     bk 
 k m   k m 
n
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n
Exam 2 Review
4
Mathematical Induction

Definition
– Used to verify a property of a sequence
– Formal definition (next slide)


What proofs must have
We proved:
n
i 
n(n  1)
2
– General summation/product
i 1
n 1
n
ar
a
– Inequalities
r  R 1 , a  R, n  Z 0 ,  ar j 
r 1
– Strong induction
j 0

Misc
– Recurrence relations
– Quotient remainder theorem
– Correctness of algorithms (Loop Invariant Theorem)
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Exam 2 Review
5
Inductive Proof
Let P(n) be a property that is defined for
integers n, and let a be a fixed integer.
 Suppose the following two statements are
true.

– P(a) is true.
– For all integers k ≥ a, if P(k) is true then
P(k+1) is true.

Then the statement for all integers n ≥ a,
P(n) is true.
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Exam 2 Review
6
Inductive Proofs Must Have

Base Case (value)
– Prove base case is true

Inductive Hypothesis (value)
– State what will be assumed in this proof

Inductive Step (value)
– Show
 State what will be proven in the next section
– Proof
 Prove what is stated in the show portion
 Must use the Inductive Hypothesis sometime
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Exam 2 Review
7
Prove this statement:
n  Z
3
,2n  1  2
n
Base Case (n=3): LHS : 2(3)  1  6  1  7
3
LHS  RHS
RHS : 2  8
Inductive Hypothesis (n=k):
Inductive Step (n=k+1):
Show:
2(k  1)  1  2
2k  1  2
k
k 1
Proof:
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Exam 2 Review
8
Prove this statement:
n  Z
3
,2n  1  2
n
Inductive Step (n=k+1):
Show:
Proof:
2k  1  1  2k 1
2k  2  1  2 k 1  2k  1  2  2 k 1  2k  1  2  2 k  2
IH  2k  1  2k  Multiply by two   4k  2  2k  2
New goal: 2k  1  2  4k  2
2k 1  4k
1 2k
1
k
2
Which is true since k≥3.
So 2k  1  2  4k  2 
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2k  2
Exam 2 Review
and
2k  1  1  2
9
k 1
Another Example

For all integers n ≥ 1,
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

3 2 1
Exam 2 Review
2n
10
Sets

Set
–
–
–
–
–
–

Proofs
–
–
–
–
–

Notation –  versus 
Definitions – Subset, proper subset, partitions/disjoint sets
Operations (, , –, ’, )
Properties and inference rules
Venn diagrams
Empty set properties
Element argument, set equality
Propositional logic / predicate calculus
Inference rules
Counterexample
Types – generic particular, induction, contra’s, CW
Russell’s Paradox (The Barber’s Puzzle) & Halting Problem
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Exam 2 Review
11
Set Operations
Formal Definitions and Venn Diagrams
Union:
A  B  {x U | x  A  x  B}
Intersection:
A  B  {x U | x  A  x  B}
Complement:
c
A  A'  {x U | x  A}
Difference:
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A  B  {x U | x  A  x  B}
A  B  A  B'
Exam 2 Review
12
Procedural Versions of Set Definitions
Let X and Y be subsets of a universal set U
and suppose x and y are elements of U.
1.
2.
3.
4.
5.
x(XY) xX or xY
x(XY) xX and xY
x(X–Y) xX and xY
xXc xX
(x,y)(XY) xX and yY
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Exam 2 Review
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Properties of Sets (Theorems 5.2.1 & 5.2.2)
A B  A
A  A B
A B  B
B  A B

Inclusion

Transitivity

DeMorgan’s for Complement

Distribution of union and intersection
A BBC  AC
( A  B)'  A' B'
( A  B)'  A' B'
A  ( B  C )  ( A  B)  ( A  C )
A  ( B  C )  ( A  B)  ( A  C )
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Exam 2 Review
14
Prove A=C
A={nZ | pZ, n = 2p}
C={mZ | qZ, m = 2q-2}
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Exam 2 Review
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Does A=D
A={xZ | pZ, x = 2p}
D={yZ | qZ, y = 3q+1}
Easy to disprove universal statements!
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Exam 2 Review
16
HW10, Problem 2

Did yesterday in class …
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Exam 2 Review
17
Counting

P( E ) 
Counting elements in a list
– How many in list are divisible by x



Probability – likelihood of an event
Permutations – with and without repetition
n
Multiplication rule
 c(i)
i 1
– Tournament play
– Rearranging letters in words
– Where it doesn’t work





P ( n, r )  r P n 
n( E )
n( S )
n!
(n  r )!
Difference rule – If A is a finite set and BA, then n(A – B) = n(A) – n(B)
Addition rule – If A1 A2  A3  …  Ak=A and A1, A2 , A3,…,Ak are pairwise
disjoint, then n(A) = n(A1) + n(A2) + n(A3) + … + n(Ak)
 n  P(n, r )
n!
C (n, r )    

r!
(n  r )! r!
r
Inclusion/exclusion rule
Combinations – with and without repetition, categories
 r  n  1
Binomial theorem (Pascal’s Triangle)


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

Exam 2 Review


r
18
Prove: # elements in list = n – m + 1

Base case (List of size 1, x=y)
– y – x + 1 = y – y + 1 (by substitution) = 1

IH (generic x, y=k [where x  k])
– Assume size of list x to k, is k – x + 1

IS
– Show size of list x to k + 1, is (k + 1) – x + 1
– Prove
 Split into two lists …
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Exam 2 Review
19
How many in list divisible by something

How many positive three digit integers are there?
– (this means only the ones that require 3 digits)
– 999 – 100 + 1 = 900

How many three digit integers are divisible by 5?
– think about the definition of divisible by
x | y   k Z, y = kx and then count the k’s that work
100, 101, 102, 103, 104, 105, 106,… 994, 995, 996, 997, 998, 999
20*5
21*5
…
199*5
– count the integers between 20 and 199
– 199 – 20 + 1 = 180
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Exam 2 Review
20
Multiplication Rule

1st step can be performed n1 ways
2nd step can be performed n2 ways
…
Kth step can be performed nk ways
Operation can be performed n1* n2 *…* nk ways

Cartesian product n(A)=3, n(B)=2, n(C)=4




– n(AxBxC) = 24
– n(AxB) = 6, n((AxB)xC) = 24
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Exam 2 Review
21
Multiplication Rule

If there are n steps to a decision, each
step having c(k) choices, the total number
of choices is:
n
c
(
i
)

i 1
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Exam 2 Review
22
Permutation Example

How many ways to take a picture?
–
–
–
–
–

With
With
With
With
With
1
2
3
4
5
person?
people?
people?
people?
people?
Number of ways to arrange n objects
– n!
– 10n < n!
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Exam 2 Review
23
Difference Rule Formally

If A is a finite set and BA, then
n(A – B) = n(A) – n(B)

One application:
probability of the complement of an event
P(E’) = P(Ec) = 1 - P(E)
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Exam 2 Review
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Addition Rule Formally

If A1 A2  A3  …  Ak =A
and A1, A2 , A3,…,Ak are pairwise disjoint
(i.e. if these subsets form a partition of A)

n(A) = n(A1) + n(A2) + n(A3) + … + n(Ak)
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Exam 2 Review
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Inclusion/Exclusion Rule

If there are two sets:
n(A  B) = n(A) + n(B) – n(A  B)

If there are three sets:
n(A  B  C) = n(A) + n(B) + n(C)
– n(A  B) – n(A  C) – n(B  C)
+ n(A  B  C)
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Exam 2 Review
26
r-Permutations

If there are n things in the set, and you want to
line-up only r of them.
n!
P ( n, r )  r P n 
(n  r )!

Example:
– Class = {Alice, Bob, Carol, Dan}
– Select a president and a vice president to represent
the class
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Exam 2 Review
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Permutations

What if just want 5-letter words from
COMPUTER?
=87654
8!

8  5!
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Exam 2 Review
28
Permutation w/ Repeated Elements
What about NEEDLES
 NE1E2DLE3S  7!

– NE1E2DLE3S, NE1E3DLE2S,
– NE2E1DLE3S, NE2E3DLE1S
– NE3E1DLE2S, NE3E2DLE1S
7! 7!
 
6 3!
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Exam 2 Review
29
Combinations

Different ways of selecting objects
– Counting subsets
– Without duplication / all items distinguishable
– Note: order is not taken into account
0
n, r  Z where n  r ,
 n  P(n, r )
n!
C (n, r )    

r!
(n  r )! r!
r

Examples:
– Class = {Alice, Bob, Carol, Dan}
 Select two class representatives
 Select three class representatives
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Exam 2 Review
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Combinations with Repitition

In general
– n – categories
– r – items to choose from
– Repetition allowed and order doesn’t count
xxx |||
r
n-1
 r  n  1


 r 
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Exam 2 Review
31
Binomial Theorem
(x+y)2
 (x+y)3

…
 (x+y)n

 Given
any real numbers a and b
and any nonnegative integer n,
n
 n  nk k
n
a  b     a b
k 0  k 
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Exam 2 Review
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