cs104 : discrete structures chapter ii fundamental structures

CS104 : Discrete Structures

Chapter IIFundamental Structures

04/19/23 Prepared by Dr. Zakir H. Ahmed 2

Set Theory

Set Theory - Set


Set: A set is an unordered collection of well-defined objects. The objects in a set are also called the elements or members of the set. A set is said to contain its elements

Some examples:• A={1, 2, 3} is the set containing “1” and “2”

and “3”. So, 1, 2, 3 A, but 5 A. • {1, 1, 2, 3, 3} = {1, 2, 3}, since repetition is

irrelevant.• {1, 2, 3} = {3, 2, 1}, since sets are

unordered.• V={a, e, i, o, u} is the set of all vowels in

English alphabet

Set Theory - Set


Some more examples: = {} is the empty set, or the set

containing no elements.• B = {b} is the singleton set, or the set

containing only one element.• N = {1, 2, 3, …} is a way we denote an

infinite set, set of natural numbers• Z = {….., -2, -1, 0, 1, 2, …….} is the set of

integers• Q = {p/q | pZ, qZ, q≠0} is the set of

rational numbers• R = Q U Qc, set of real numbers• C = {a + ib | a, b R, i = √-1}, set of

complex numbers

Note: {}

Set Theory – Subset and Superset


Subset: The set A is said to be a subset of the set B if and only if every element of A is also an element of B, and B is said to be superset of A.

A B means “A is a subset of B.”or, “B contains A.”or, “every element of A is also in B.”or, x ((x A) => (x B)).and B A means “B is a superset of A.”Example: Let A = {a, b, c} B = {a, b, c, d},

C = { a, c, d, e}then A B and B A, but A is not subset of C.

Venn Diagram

A

B

Set Theory – Equal sets

Equal sets: Two sets A = B if and only if A and B have exactly the same elements.

A = B iff, A B and B Aiff, x ((x A) <=> (x B)).

Example: If A = {a, b, c, d} and B = {a, b, c, d} , then AB and BA, so A = B.

Note: For any set S,(i) Ø S and (ii) S S


Set Theory – Proper subset

Proper subset: A B means “A is a proper subset of B”, i.e., A B, and A B.

Examples:

•{1, 2, 3} {1, 2, 3, 4, 5}

•{1, 2, 3} {1, 2, 3, 4, 5}Questions:

•Is {1, 2, 3}?

•Is {x} {x}?

•Is {x} {x}?

•Is {x} {x, {x}}?

•Is {x} {x, {x}}?04/19/23 Prepared by Dr. Zakir H. Ahmed 7

Set Theory - Power set


Power set: If S is a set, then the power set of S is the set of all subsets of the set S. The power set of S is denoted by P(S). i.e., P(S) = { x | x S }.

Examples:

•If S={a}, then P(S)={, {a}}.

•If S = {a, b}, then P(S)={, {a}, {b}, {a, b}}.

Questions:

•If S = , then P(S)=?

•If S = {,{}}, then P(S)=?Fact: If S is finite, |P(S)| = 2|S|. (if |S|=n, |P(S)|=2n)

Set Theory - Cartesian Product

Cartesian Product: Let A and B be two sets. The Cartesian Product of A and B, denoted by AXB, is the set of all ordered pairs (a, b), where a A and b B. i.e.,A x B = { (a, b) | a A and b B}

Examples:• If A={1, 2}, B={a, b, c} then AXB={(1, a),

(1, b), (1, c), (2, a), (2, b), (2, c)}Questions:

• If A and B be two finite sets, the |AXB|=?• |A|+|B|, |A+B|, or |A||B|

Facts: AXB ≠ BXA• A1 x A2 x … x An = {(a1, a2,…, an) | ai Ai, for

i=1, 2, …, n}


Set Theory – Union and Intersection


Union: The union of two sets A and B, denoted by AB, is the set that contains those elements that are either in A or in B, or in both:A B = { x | x A or x B}

Intersection: The intersection of two sets A and B, denoted by AB, is the set containing those elements in both A and B:A B = { x | x A and x B}

AB AB

Set Theory – Union and Intersection

Aii1

n

A1 A2 An

Aii1

n

A1 A2 An


Example: If A={1, 2, 3}, B={1, 3, 5} then A B ={1, 2, 3, 5} and AB={1, 3}

Questions: If A = {x : x is a CS major student}, and B = {x : x is a MS major student}, then AB and AB=?

• If A = {x : x is a US president}, and B = {x : x is deceased}, then AB=?

• If A = {x : x is a US president}, and B = {x : x is in this room}, then AB=?

Generalized Unions and Intersections:

Set Theory – Disjoint Set


Disjoint sets: Two sets are said to be disjoint if their intersection is the empty set.

Examples:

•If A = {1, 3, 5, 7, 9}, and B = {2, 4, 6, 8, 10}, then AB=Ø

•If A = {x | x is a CS major student} and B = {x | x is a PS major student}, then AB=Ø

Questions:

•Give some examples of two sets A and B such that AB=Ø

Set Theory – Difference of sets


Difference: The difference of two sets A and B, denoted by A-B, is the set containing those elements that are in A but not in B: A – B = { x | xA and xB}

Examples:

•If A = {1, 3, 5}, and B = {1, 2, 3}, then A-B = {5}, and B-A = {2}. So, A-B B-A.

AB

Set Theory - Complement

Complement: Let U be the universal set. The complement of the set A, denoted by Ac or Ā, is set containing those elements that are not in A:Ac = { x | x A}

Examples:

•If U={1, 2, …, 10}, A={1, 3, 5, 7, 9}, then Ac={2, 4, 6, 8, 10}.

•If A = {x : x is bored}, then Ac = {x : x is not bored}

Facts:

•Uc = Ø and Øc = U

•A - B = A Bc 04/19/23 Prepared by Dr. Zakir H. Ahmed 14

AU

Set Theory – Set Identities

A B C

B CA (B C)A BA C(A B) (A C)

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

01110111

00000111

00000011

00000101

00000111


Prove the following law using Membership table:A (B C) = (A B) (A C)

Proof:

Set Theory – Set Identities


Prove the following law using Venn diagram:A (B C) = (A B) (A C)

Proof:

A B AU

C C

UB

A (B C) (A B) (A C)

Set Theory - Famous Identities

IdentitiesName

A U = AA U = A

Identity laws

A U U = UA = Domination laws

A U A = AA A = A

Idempotent laws

(Ac)c = AComplementation

law

A U B = B U AA B = B A

Commutative laws

A U (B U C) = (A U B) U CA (B C) = (A B)

CAssociative laws


Set Theory - Famous Identities

IdentitiesName

A U (B C) = (A U B) (A U C)

A (B U C) = (A B) U (A U C)

Distributive laws

(A U B)c = Ac Bc

(A B)c = Ac U BcDe Morgan’s laws

A U (A B) = AA (A U B) = A

Absorption laws

A U Ac = UA Ac = Complement laws


Question:Prove the above identities by using Venn diagram and Membership tables ??

Cardinality of a set

Cardinality of a set: Let S be a set. If there are exactly n distinct elements in S, where n is a non-negative integer, we say that S is a finite set and that n is the cardinality of S. The cardinality of S is denoted by |S|.

Example: Let A be the set of English alphabets, then |A|=26.

Questions:

•|Ø|=?

•If B={1, 1, 1}, then |B|=?

•If S = { , {}, {,{}} }, then |S|=?


Set Theory – Inclusion-exclusion

Inclusion-exclusion theory: We are interested in finding cardinality of the union sets.

Question Example:|A| = How many people are wearing a watch?|B| = How many people are wearing jackets?|A B| = How many people are wearing a

watch OR jackets?Answer:


AB Wrong or right?


Answer:•Note that |A|+|B| counts each element that is

in A but not in B, or in B not in A, exactly once.

•Each element that is in both A and B will be counted twice

•So, elements in A B will be subtracted the result, i.e.,

|A B| = |A| + |B| - |A B|Question: Generalize the formula for 3 sets



Example:There are 150 CS majors100 are taking CS53070 are taking CS52030 are taking both

Question:How many are taking neither?

Answer:


150 – (100 + 70 - 30) = 10

CS530CS520

Set Theory – Computer representation

• Let U be a finite universal set. Let a1, a2,…, an be an arbitrary ordering of the elements of U.

• Represent a subset A of U with the bit string of length n, where

Aaif

Aaifbiti

i

ith

0

1


Examples:

• Let U={1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and

• Then the set A={1, 3, 5, 7, 9} can be represented by the string of bits: 10 1010 1010

• The set B={1, 2, 4, 9} can be represented as: 11 0100 0010

Set Theory – Computer representation


Examples:

• A : 10 1010 1010

• B : 11 0100 0010

• The set A U B can be represented as: 11 1110 1010

• The set A B can be represented as:10 0000 0010

Questions:

•If C={1, 6, 8, 10}, express following sets with bit strings

A-B, Ac, A U (B C) and A (B U C)


Functions

Functions – Definition

Definition: Let A and B be two sets. A function f : A B is an assignment of exactly one element of B to each element of A. We write f(a)=b if b is the unique element of B assigned by the function f to the element a of A.

Example 1:Let A = {Hussain,

Muhammad, Hassan, Eisa} B = {Amina, Fatima,

Khadiza, Mariyam}Also, let f: A B be defined as

f(a) = mother(a).


a

A

b=f(a)

B

f(a)

f

Hussain

Muhammad

Hassan

Eisa

Amina

Fatima

Khadiza

MariyamA B

Functions - Examples


Example 2: Let S={Ahmed, Hussain, Muhammad, Musa, Badr} be a set of students enrolled in CS100 course. Each student is assigned a letter grade from the set G={A,B,C,D,F} as follows:

Ahmed AHussain BMuhammad CMusa DBadr F

This assignment is an example of a function.Questions: Give some examples of functions?

Functions are sometimes called

mappings or transformations

Functions – Examples

Example 3: Suppose we have following graph:And I ask you to describe the red function.What’s the function?

Notation: f: RR,

f(x) = -(1/2)x - 25


f(x) = -(1/2)x - 25domai

n

codomain

Functions – Image and Preimage

Definition: If the function f : A B, then A is the domain and B is the codomain of f. If f(a)=b, we say that b is the image of a and a is a preimage of b. The range of f is the set of all images of elements of A. Also, if f is a function from A to B, we say that f maps A to B.

Example 4: - image({Hussain, Hassan}) = {Fatima} image(A) = B – {Khadiza}, range of f is the set

{Amina, Fatima, Mariyam}


HussainMuhammad

Hassan Eisa

AminaFatimaKhadizaMariyam

A B

Functions – Image and Pre-image

Example 5: - preimage({Fatima}) = {Hussain, Hassan}

preimage (B) = AFor any set P A, image(P) = {b : a P, f(a) = b}For any Q B, preimage(Q) = {a: b Q, f(a) =

b}


Hussain

Muhammad

Hassan

Eisa

Amina

Fatima

Khadiza

Mariyam

pre-image(Q) = f-1(Q)image(P) = f(P)

A B

Functions – Image and Pre-image

Example 6: - Let f be the function that assigns the last two bits of a bit string of length 2 or greater to that string. For example, f(11010) = 10. then, the domain of f is the set of all bits of length 2 or greater, and both the codomain and range are the set {00, 01, 10, 11}.

Example 7: - The domain and codomain of functions are specified in programming languages. E.g., the Java statement: int floor (float real) { ….. }, and Pascal statement function floor (x: real): integerstate that the domain of floor function is the set of real numbers, and codomain is set of the integers.


Functions – On Real Number

Definition: Let f1 and f2 be two functions from A to R. Then f1+f2 and f1f2 are also functions from A to R defined by

(f1+f2)(x) = f1(x) + f2(x),

(f1f2)(x) = f1(x)f2(x).

Example 8: Let f1 and f2 be two functions from R to R such that f1(x) = x2 and f2(x) = x – x2. What are the functions f1+f2, f1f2?

Solution: From the definitions,(f1+f2)(x) = f1(x) + f2(x) = x2 + (x – x2) = x,

(f1f2)(x) = f1(x)f2(x) = x2(x – x2) = x3 – x4.04/19/23 Prepared by Dr. Zakir H. Ahmed 32

Functions – Injection and Surjection

Injection: A function f from the set A to the set B is said to be one-to-one (injective, an injection), if and only if f(a) = f(b) implies that a = b for all a and b in the domain of f. Every x B has at most 1 preimage.

Surjection: A function f from A to B is said to be onto (surjective, an surjection), if and only every element b B, there is an element a A with f(a) = b. Every b B has at least 1 preimage.


Neither injection nor surjectionHussain

MuhammadHassan Eisa

AminaFatimaKhadizaMariyam

A B

Functions – Bijection

Bijection: A function f from the set A to the set B is said to be one-to-one correspondence (bijective, an bijection), if it is both injection and surjection. Every b B has exactly 1 preimage.


Muhammad

Hassan

Eisa

Amina

Fatima

Mariyam

An important implication of this characteristic:The preimage (f-1) is a function!

A B


Example 9: Determine whether the function f: {a, b, c, d} {1, 2, 3, 4, 5} with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is injective.

Sol: The function f is one-to-one since f takes on different values at the four elements of its domain.

Example 10: Determine whether the function f: Z Z, f(x) = x2 is injective.

Sol: The function f is not one-to-one since, for instance, f(1) = f(-1) = 1, but 1 ≠ -1.

Example 11: Determine whether the function f: {a, b, c, d} {1, 2, 3} with f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3 is surjective.

Sol: The function f is onto, since three elements of the codomain are images of elements in the domain.



Example 12: Determine whether the function f: Z Z, f(x) = x2 is surjective.

Sol: The function f is not onto since, for instance, there is no integer x with x2 = -1.

Example 13: Determine whether the function f: {a, b, c, d} {1, 2, 3, 4} with f(a) = 4, f(b) = 2, f(c) = 1, and f(d) = 3 is bijective.

Sol: The function f is one-to-one since no two values in the domain are assigned the same function value. It is also onto because all four elements of the codomain are images of elements in the domain. Hence, f is a bijection.




Example 14:

Functions – Questions

Q1. Suppose f: R+ R+, f(x) = x2.Is f one-to-one?

Is f onto?

Is f bijective?


Q2. Suppose f: R R+, f(x) = x2.Is f one-to-one?

Is f onto?

Is f bijective?

Functions – Questions

Q3. Suppose f: R R, f(x) = x2.Is f one-to-one?Is f onto?Is f bijective?

Q4. Let f be a function from {a, b, c, d} to {1, 2, 3, 4} with f(a)=4, f(a)=3, f(b)=2, f(c)=1 and f(d)=3.Is f one-to-one?Is f onto?Is f bijective?


Functions – Inverse Functions

Definition: Let f be an one-to-one correspondence from the set A to the set B. The inverse function of f is the function that assigns to an element b є B the unique element a є A such that f(a) = b. The inverse function of f is denoted by f-1. Hence, f-1(b) = a when f(a) = b.

A one-to-one correspondence is called invertible, since we can define an inverse of this function.

A function is not invertible, if it is not one-to-one correspondence


a=f-1(b) .

A

. b=f(a)

B

f-1(b)

f(a)

f-1

f


Example 15: Find out whether the function f: {a, b, c} {1, 2, 3} with f(a) = 2, f(b) = 3, and f(c) = 1 is invertible. And if yes, what is its inverse?

Sol: The function f is invertible, since it is bijective. The inverse function f-1 reverses the correspondence given by f, so f-1(1) = c, f-1(2) = a, and f-1(3) = b

Example 16: Determine whether the function f: R R, with f(x) = x2 is invertible.

Sol: The function f is not one-to-one, since for instance, f(-2) = f(2) = 4, but 2 ≠ -2.

Q5: find out whether the function f: Z Z, with f(x) = x+1 is invertible. And if yes, what is its inverse?


Functions – Compositions of Functions

Definition: - Let g:AB, and f:BC be functions. Then the composition of f and g, denoted by f o g, is defined by (f o g)(x) = f(g(x))

Note that the composition f o g can not be defined unless the range of g is a subset of the domain of f.


.a

A

.g(a)

B

.f(g(a))

Cg f

g(a) f(g(a))

(f o g)(a)

f o g


Example 17: Let g:{a, b, c}{a, b, c} with g(a)=b, g(b)=c, g(c)=a. Also, let f:{a, b, c}{1, 2, 3} with f(a)=3, f(b)=2, f(c)=1. What are the compositions of f and g, i.e., (f o g), and g and f, i.e., (g o f)?

Sol: The composition f o g, is defined by(f o g)(a) = f(g(a)) = f(b) = 2,(f o g)(b) = f(g(b)) = f(c) = 1, and(f o g)(c) = f(g(c)) = f(a) = 3.Note that g o f is not defined, because the range of f is not subset of the domain of g.

Q6: Let g:{a, b, c}{a, b, c} with g(a)=a, g(b)=b, g(c)=c. Also, let f:{a, b, c}{1, 2, 3} with f(a)=1, f(b)=2, f(c)=3. What are the compositions (f o g) & (g o f)?



Example 18: Let f:ZZ with f(x) = 2x+3, and g:ZZ with g(x) = 3x+2. What are the compositions (f o g) and (g o f)?

Sol: Both compositions (f o g) and (g o f) are defined. (f o g)(x) = f(g(x)) = f(3x+2) = 2(3x+2)+3 = 6x+7, and(g o f)(x) = g(f(x)) = g(2x+3) = 3(2x+3)+2 = 6x+11.

Definition: Let A and B be two sets and f: AB be a function. The graph of the function f is the set of ordered pairs

{(a, b) | , a A and f(a) = b}.

Note: The graph of f:AB is a subset of AXB.04/19/23 Prepared by Dr. Zakir H. Ahmed 44



Graph of f(n)=2n+1 from Z to Z

Graph of f(n)=n2 from Z to Z

Example 19:

Functions – Properties

Some properties:• f(Ø) = Ø• f({a}) = {f(a)}• f(A U B) = f(A) U f(B)• f(A B) f(A) f(B)• f-1() = • f-1(A U B) = f-1(A) U f-1(B)• f-1(A B) = f-1(A) f-1(B)


Functions – Familiar functions

Polynomials: f(x) = a0xn + a1xn-1 + … + an-1x1 + anx0

Example: f(x) = x3 - 2x2 + 15

Exponentials: f(x) = cdx

Example: f(x) = 310x, f(x) = ex

Logarithms: log2 x = y, where 2y = x.

Ceiling: f(x) = x the least integer y so that x y.Example: 1.2 = 2; -1.2 = -1; 1 = 1Floor: f(x) = x the greatest integer y so that x y.Example: 1.8 = 1; -1.8 = -2; -5 = -5Question: what is -1.2 + 1.1 ?



Relations

Relations – Definition

Relation: Let A and B be two sets. A binary relation (R) from A to B is a subset of A x B, i.e., R AxB.

Example 1: Let A = Set of students; B = Set of courses.

R = {(a,b) | student a is enrolled in course b}Example 2: Let A = Set of cities; B = Set of

countries. Define the relation R by specifying that (a, b) belongs to R if city a is the capital of b. For instance, (Riyadh, Saudi Arabia), (Delhi, India), (Washington, USA) are in R.

Example 3: Let A={0, 1, 2} and B={a, b}. {(0, a), (0, b), (1, a), (2, b)} is a relation from A to B. This means, 0Ra, but 1Rb.


Relations – On a Set

Relation: A relation on the set A is a relation from A to A. That is, a relation on a set A is a subset of A x A.

Example 4: Let A = {1, 2, 3, 4}. Which ordered pairs are in the relation R={(a, b) | a divides b}?

Sol: (a, b)єR iff. a and b are positive integers not exceeding 4 such that a divides b, we see thatR={(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)}The pairs in R are displayed graphically and in tabular form:


1. .1

2. .2

3. .3

4. .4

R1 2 3 41234

X X X X X X X X

Relations – Examples

Example 5: Consider the relations on the set of integers:R1= {(a, b) | a ≤ b},

R2={(a, b) | a > b},

R3={(a, b) | a = b or a = -b},

R4={(a, b) | a = b},

R5={(a, b) | a = b+1},

R6={(a, b) | a+b ≤ 3},Which of these relations contain each of the pairs (1, 1), (1, 2), (2, 1), (1, -1) and (2, 2)?

Sol: The pair (1, 1) is in R1, R3, R4 and R6; (1, 2) is in R1and R6; (2, 1) is in R2, R5 and R6; (1, -1) is in R2, R3

and R6; and finally, (2, 2) is in R1, R3 and R4.


Relations – Properties

Reflexivity: A relation R on a set A is called reflexive if for all a A, (a, a) R.

Symmetry: A relation R on a set A is called symmetric if (b, a) R whenever (a, b) R, for all a, b A.

Antisymmetry: A relation R on A is called antisymmetric if for all a, b A, if (a, b) R and (b, a) R, then a = b.

Transitivity: A relation on A is called transitive if (a, b) R and (b, c) R imply (a, c) R, for all a, b, c A.



Example 6: Which of the relations from Example 5 are reflexive and symmetric?

Sol: The reflexive relations from Example 5 are R1

(because a ≤ a, for all integer a), R3 and R4.For each of the other relations in this example it is easy to find a pair of the form (a, a) that is not in the relation.The symmetric relations are R3, R4 and R6.

R3 is symmetric, for if a=b or a=-b, then b=a or b=-a.

R4 is symmetric, since a=b implies b=a.

R6 is symmetric, since a+b ≤ 3 implies b+a ≤ 3.None of the other relations is symmetric.



Example 7: Which of the relations from Example 5 are antisymmetric?

Sol: The antisymmetric relations from Example 5 are R1, R2, R4 and R5. R1 is antisymmetric, since the inequalities a ≤ b and b ≤ a imply that a = b. R2 is antisymmetric, since it is impossible for a>b and b>a. R4 is antisymmetric because two elements are related with respect to R4 iff. they are equal. R2 is also antisymmetric, since it is impossible that a = b+1 and b = a+1. None of the other relations is antisymmetric.



Example 8: Which of the relations from Example 5 are transitive?

Sol: The transitive relations from Example 5 are R1, R2, R3 and R4. R1 is transitive, since a ≤ b and b ≤ c imply a ≤ c. R2 is transitive, since a > b and b > c imply a > c. R3 is transitive, since a = ±b and b = ±c imply a = ±c. R4 is transitive, since a = b and b = c imply a = c. R5 and R6 are not transitive.


Relations – Question

Q1: Consider following relations on {1, 2, 3}:R1= {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3)},

R2={(1, 1), (1, 2), (2, 2), (3, 2), (3, 3)},

R3={(2, 1), (2, 3), (3, 1)}

R4={(2, 3)},

Which of the relations are reflexive, symmetric, antisymmetric and transitive?


Relations – Equivalence relations

Definition: A relation on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive.1. Reflexive ( a A, aRa)2. Symmetric (aRb => bRa)3. Transitive (aRb and bRc => aRc)

Example 13: Let R be the relation on the set of real numbers such that aRb iff. a-b is an integer. Is R an equivalence relation?

Sol: As a-a = 0 is an integer for all real numbers a. So, aRa for all real numbers a. Hence R is reflexive. Let aRb, then a-b is an integer, so b-a also an integer. Hence bRa, i.e., R is symmetric.If aRb and bRc, then a-b and b-c are integers. So, a-c = (a-b) + (b-c) is also an integer. Hence, aRc. Thus R is transitive. Consequently, R is an equivalence relation.


Relations – Example

Example 14: Is the relation “divides” on the set of positive integers equivalence relation?

Sol: As a | a , whenever a is a positive integer, the “divides” relation is reflexive.Let a | b and b | c. Then there are positive integers k and l such that b = ak and c = bl. Hence, c = a(kl). So, a | c. It follows that the relation is transitive.This relation is not symmetric, as 1 | 2, but 2 ∤ 1. Hence, the relation is not an equivalence relation.

Q2: Let R be the relation on the set of integers such that aRb iff. a=b or a=-b. Is R an equivalence relation?

Q3: Is the relation “≤” on the set of real numbers equivalence relation?



End of Chapter-II

cs104 : discrete structures chapter ii fundamental structures

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