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Discrete Mathematics Lecture 4 Harper Langston New York University

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Discrete Mathematics Lecture 4. Harper Langston New York University. Sequences. Sequence is a set of (usually infinite number of) ordered elements: a 1 , a 2 , …, a n , … Each individual element a k is called a term, where k is called an index - PowerPoint PPT Presentation

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Page 1: Discrete Mathematics Lecture 4

Discrete MathematicsLecture 4

Harper Langston

New York University

Page 2: Discrete Mathematics Lecture 4

Sequences

• Sequence is a set of (usually infinite number of) ordered elements: a1, a2, …, an, …

• Each individual element ak is called a term, where k is called an index

• Sequences can be computed using an explicit formula: ak = k * (k + 1) for k > 1

• Alternate sign sequences• Finding an explicit formula given initial terms of the

sequence: 1, -1/4, 1/9, -1/16, 1/25, -1/36, …• Sequence is (most often) represented in a computer

program as a single-dimensional array

Page 3: Discrete Mathematics Lecture 4

Sequence Operations

• Summation: , expanded form, limits (lower, upper) of summation, dummy index

• Change of index inside summation

• Product: , expanded form, limits (lower, upper) of product, dummy index

• Factorial: n!, n! = n * (n – 1)!

Page 4: Discrete Mathematics Lecture 4

Review Mathematical Induction

• Principle of Mathematical Induction:

Let P(n) be a predicate that is defined for integers n and let a be some integer. If the following two premises are true:P(a) is a true

k a, P(k) P(k + 1)

then the following conclusion is true as wellP(n) is true for all n a

Page 5: Discrete Mathematics Lecture 4

Applications of Mathematical Induction

• Show that 1 + 2 + … + n = n * (n + 1) / 2(Prove on board)

• Sum of geometric series:

r0 + r1 + … + rn = (rn+1 – 1) / (r – 1)(Prove on board)

Page 6: Discrete Mathematics Lecture 4

Examples that Can be Proved with Mathematical Induction

• Show that 22n – 1 is divisible by 3 (in book)• Show (on board) that for n > 2: 2n + 1 < 2n

• Show that xn – yn is divisible by x – y• Show that n3 – n is divisible by 6 (similar to book

problem)

Page 7: Discrete Mathematics Lecture 4

Strong Mathematical Induction

• Utilization of predicates P(a), P(a + 1), …, P(n) to show P(n + 1).

• Variation of normal M.I., but basis may contain several proofs and in assumption, truth assumed for all values through from base to k.

• Examples:• Any integer greater than 1 is divisible by a prime• Existence and Uniqueness of binary integer

representation (Read in book)

Page 8: Discrete Mathematics Lecture 4

Well-Ordering Principle

• Well-ordering principle for integers: a set of integers that are bounded from below (all elements are greater than a fixed integer) contains a least element

• Example:

• Existence of quotient-remainder representation of an integer n against integer d

Page 9: Discrete Mathematics Lecture 4

Stepping back: Algorithms

• Last lecture we talked about some elementary number theory such as the Quotient Remainder Theorem (pg 157)

• Algorithm is step-by-step method for performing some action (such as finding remainder in Q.R.)

• Cost of statements execution– Simple statements– Conditional statements– Iterative statements

Page 10: Discrete Mathematics Lecture 4

Example: Division Algorithm

• Input: integers a and d• Output: quotient q and remainder r• Body:

r = a; q = 0;while (r >= d)

r = r – d;q = q + 1;

end while

Page 11: Discrete Mathematics Lecture 4

Greatest Common Divisor

• The greatest common divisor of two integers a and b is another integer d with the following two properties:– d | a and d | b– if c | a and c | b, then c d

• Lemma 1: gcd(r, 0) = r• Lemma 2: if a = b * q + r, then gcd(a, b) =

gcd(b, r)

Page 12: Discrete Mathematics Lecture 4

Euclidean Algorithm

• Input: integers a and b (a>b>=0)• Output: greatest common divisor gcd• Body:

r = b;while (b > 0)

r = a mod b;a = b;b = r;

end whilegcd = a;

Page 13: Discrete Mathematics Lecture 4

Exercise

• Least common multiple: lcm

• Prove that for all positive integers a and b, gcd(a, b) = lcm(a, b) iff a = b

Page 14: Discrete Mathematics Lecture 4

Correctness of Algorithms

• Assertions– Pre-condition is a predicate describing initial state

before an algorithm is executed– Post-condition is a predicate describing final state

after an algorithm is executed

• Loop guard• Loop is defined as correct with respect to its pre-

and post- conditions, if whenever the algorithm variables satisfy the pre-conditions and the loop is executed, then the algorithm satisfies the post-conditions as well

Page 15: Discrete Mathematics Lecture 4

Loop Invariant Theorem

• Let a while loop with guard G be given together with its pre- and post- conditions. Let predicate I(n) describing loop invariant be given. If the following 4 properties hold, then the loop is correct:– Basis Property: I(0) is true before the first iteration of the loop– Inductive Property: If G and I(k) is true, then I(k + 1) is true– Eventual Falsity of the Guard: After finite number of iterations, G

becomes false– Correctness of the Post-condition: If N is the least number of

iterations after which G becomes false and I(N) is true, then post-conditions are true as well

Page 16: Discrete Mathematics Lecture 4

Correctness of Some Algorithms

• Product Algorithm:

pre-conditions: m 0, i = 0, product = 0

while (i < m) {

product += x;

i++;

}

post-condition: product = m * x

Page 17: Discrete Mathematics Lecture 4

Correctness of Some Algorithms

• Division Algorithm

pre-conditions: a 0, d > 0, r = a, q = 0

while (r d) {

r -= d;

q++;

}

post-conditions: a = q * d + r, 0 r < d

Page 18: Discrete Mathematics Lecture 4

Correctness of Some Algorithms

• Euclidean Algorithm

pre-conditions: a > b 0, r = b

while (b > 0) {

r = a mod b;

a = b;

b = r;

}

post-condition: a = gcd(a, b)

Page 19: Discrete Mathematics Lecture 4

Basics of Set Theory

• Set and element are undefined notions in the set theory and are taken for granted

• Set notation: {1, 2, 3}, {{1, 2}, {3}, {1, 2, 3}}, {1, 2, 3, …}, , {x R | -3 < x < 6}

• Set A is called a subset of set B, written as A B, when x, x A x B. What is negation?

• A is a proper subset of B, when A is a subset of B and x B and x A

• Visual representation of the sets• Distinction between and

Page 20: Discrete Mathematics Lecture 4

Set Operations

• Set a equals set B, iff every element of set A is in set B and vice versa. (A = B A B /\ B A)

• Proof technique for showing sets equality (example)• Union of two sets is a set of all elements that belong to

at least one of the sets (notation on board)• Intersection of two sets is a set of all elements that

belong to both sets (notation on board)• Difference of two sets is a set of elements in one set, but

not the other (notation on board)• Complement of a set is a difference between universal

set and a given set (notation on board)• Examples

Page 21: Discrete Mathematics Lecture 4

Empty Set

• S = {x R, x2 = -1}• X = {1, 3}, Y = {2, 4}, C = X Y

(X and Y are disjoint)• Empty set has no elements • Empty set is a subset of any set• There is exactly one empty set• Properties of empty set:

– A = A, A = – A Ac = , A Ac = U– Uc = , c = U

Page 22: Discrete Mathematics Lecture 4

Set Partitioning

• Two sets are called disjoint if they have no elements in common

• Theorem: A – B and B are disjoint• A collection of sets A1, A2, …, An is called

mutually disjoint when any pair of sets from this collection is disjoint

• A collection of non-empty sets {A1, A2, …, An} is called a partition of a set A when the union of these sets is A and this collection consists of mutually disjoint sets

Page 23: Discrete Mathematics Lecture 4

Power Set

• Power set of A is the set of all subsets of A

• Example on board

• Theorem: if A B, then P(A) P(B)

• Theorem: If set X has n elements, then P(X) has 2n elements (proof in Section 5.3 – will show if have time)

Page 24: Discrete Mathematics Lecture 4

Cartesian Products

• Ordered n-tuple is a set of ordered n elements. Equality of n-tuples

• Cartesian product of n sets is a set of n-tuples, where each element in the n-tuple belongs to the respective set participating in the product

Page 25: Discrete Mathematics Lecture 4

Set Properties

• Inclusion of Intersection:A B A and A B B

• Inclusion in Union:A A B and B A B

• Transitivity of Inclusion:(A B B C) A C

• Set Definitions:x X Y x X y Yx X Y x X y Yx X – Y x X y Yx Xc x X(x, y) X Y x X y Y

Page 26: Discrete Mathematics Lecture 4

Set Identities

• Commutative Laws: A B = A B and A B = B A

• Associative Laws: (A B) C = A (B C) and (A B) C = A (B C)

• Distributive Laws:

A (B C) = (A B) (A C) and A (B C) = (A B) (A C)

• Intersection and Union with universal set: A U = A and A U = U

• Double Complement Law: (Ac)c = A

• Idempotent Laws: A A = A and A A = A

• De Morgan’s Laws: (A B)c = Ac Bc and (A B)c = Ac Bc

• Absorption Laws: A (A B) = A and A (A B) = A

• Alternate Representation for Difference: A – B = A Bc

• Intersection and Union with a subset: if A B, then A B = A and A B = B

Page 27: Discrete Mathematics Lecture 4

Proving Equality

• First show that one set is a subset of another (what we did with examples before)

• To show this, choose an arbitrary particular element as with direct proofs (call it x), and show that if x is in A then x is in B to show that A is a subset of B

• Example (step through all cases)

Page 28: Discrete Mathematics Lecture 4

Disproofs, Counterexamples and Algebraic Proofs

• Is is true that (A – B) (B – C) = A – C?(No via counterexample)

• Show that (A B) – C = (A – C) (B – C)(Can do with an algebraic proof, slightly different)

Page 29: Discrete Mathematics Lecture 4

Boolean Algebra

• A Boolean Algebra is a set of elements together with two operations denoted as + and * and satisfying the following properties:Commutative: a + b = b + a, a * b = b * a

Associative: (a + b) + c = a + (b + c), (a * b) *c = a * (b * c)

Distributive: a + (b * c) = (a + b) * (a + c), a * (b + c) = (a * b) + (a * c)

Identity: a + 0 = a, a * 1 = a for some distinct unique 0 and 1

Complement: a + ã = 1, a * ã = 0

Page 30: Discrete Mathematics Lecture 4

Russell’s Paradox

• Set of all integers, set of all abstract ideas• Consider S = {A, A is a set and A A}• Is S an element of S?• Barber puzzle: a male barber shaves all those men who

do not shave themselves. Does the barber shave himself?

• Consider S = {A U, A A}. Is S S?• Godel: No way to rigorously prove that mathematics is

free of contradictions. (“This statement is not provable” is true but not provable) (consistency of an axiomatic system is not provable within that system)

Page 31: Discrete Mathematics Lecture 4

Halting Problem

• There is no computer algorithm that will accept any algorithm X and data set D as input and then will output “halts” or “loops forever” to indicate whether X terminates in a finite number of steps when X is run with data set D.

• Proof is by contradiction