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
, i2
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:
2k 1 1 2k 1
2k 2 1 2 k 1 2k 1 2 2 k 1 2k 1 2 2 k 2
IH 2k 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
2k 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(XY) xX or xY
x(XY) xX and xY
x(X–Y) xX and xY
xXc xX
(x,y)(XY) xX and yY
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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 BBC AC
( 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={nZ | pZ, n = 2p}
C={mZ | qZ, m = 2q-2}
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Exam 2 Review
15
Does A=D
A={xZ | pZ, x = 2p}
D={yZ | qZ, 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 BA, 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
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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 BA, 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
24
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
25
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
27
Permutations
What if just want 5-letter words from
COMPUTER?
=87654
8!
8 5!
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Exam 2 Review
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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
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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
30
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
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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 nk k
n
a b a b
k 0 k
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Exam 2 Review
32
0
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