TASK 3

MATH DISCRETETAF 3023

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1. Sets reprisented via computer

Computer Representation of Sets

There are two methods to represent sets using a computer :

•Unordered fashion.•Arbitrary ordering.

a. Store the elements of the set in an unordered fashion: -Computing operation are union, intersection and differ entence of two set. -Each operation require large amount of searching elements.

b. Store the elements using an arbitrary ordering of the elements of the universal set. 

• Easy to make compute set.

1. Let's assume that set U is a finite and of reasonable size.

2. First, we'll specify an arbitrary ordering of elements of U, for instance a1, a2,

a3, ..., an. 3.Then, we'll represent a subset A of U with the

bit string of length n, where the i th bit in this string is:                                                       if ai ∈ A,

and 0, if ai ∉ A.

A. Store the elements of the set in an unordered fashion:

-Computing operation are union, intersection and differ entence of

two set. -Each operation require large

amount of searching elements.

Example 1:    Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, and the ordering of elements of U has the elements in

increasing order; that is, ai = i . What bit strings represent the subset of all odd integers in U, the subset of all even integers in U, and the subset of integers not exceeding 5 in U?

                         Solution:   The bit string that represents the set of odd integers in U, namely, {1, 3, 5, 7, 9}, has a one

bit in the first, third, fifth, seventh, and ninth positions, and a zero elsewhere. It is

10 1010 1010. Similarly, we represent the subset of all even integers in U, namely, {2, 4, 6, 8, 10}, by the

string  01 0101 0101.

 The set of all integers in U that do not exceed 5, namely, {1, 2, 3, 4, 5}, is represented by

the string  11 1110 0000.

1)      Ā (Complement)Solution:

Bit string for A: 0101 0111,Then Ā = 1010 1000

(Therefore, Ā = {0,1,2,4}) 

2)      A ∩ B (Intersection)Solution:

Bit string for A: 0101 0111 Bit string for B: 1111 1100

 Then A ∩ B =0101 0111 ˄  1111 1100 = 0101 0100                                        

(A ∩ B = {1,3,5} ) 

3)      A U B (Union)Solution:

Bit string for A: 0101 0111 Bit string for B: 1111 1100

 Then A U B = 0101 0111 ˅  1111 1100 = 1111 1111

(A U B = U = {0,1,2,3,4,5,6,7}) 

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2. RELATION

Relation between Two Sets

• A relation is a set of ordered pairs(x,y) , -the first component are the input values(Domain).- second component are the output values(Range).

• A relation R defined by ( , {(x,y):x  }), is the set of ordered pairs of all Input Values (x) and all Output Values(y) such that each element x is related to the corresponding element y.

• Here, x=Domain of the Relation y=Range of the Relation x & y no values, it will repeated

x is called Independent variable and y is called Dependent variable because its value depends on the x-value chosen.

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3. function

DEFINITION OR CONCEPT OF FUNTION ON GENERAL SETS

 Function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.

As an example:-

The function f maps to A to B

INJUCTIve

•f (a) = f (b) implies that

a = b for all a and b in the domain of f.

•Involves one to one function

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BOOLEAN FUNCTION§ A Boolean function describes the ways to gain the value of

boolean output via logical calculation from Boolean inputs. The functions play a basic role in questions of complexity theory

also the design of circuits and chips for digital computers.

The properties of Boolean functions play a critical role in cryptography, particularly in the design of symmetric key

algorithms Boolean functions are often represented by sentences in

propositional logic, and sometimes as multivariate(involving 2 or more variables)

A literal is a Boolean variable or its complement. A minterm of the Boolean variables x1,x2,...,xn is a Boolean product y1y2 ···yn,

where yi = xi or yi = xi. Hence, a minterm is a product of n literals, with one literal for each variable.

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example:

§Find the sum-of-products expansion for the function F (x, y, z) = (x + y)z.

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Solution: We will find the sum-of-products expansion of F (x, y, z) in two ways.

First, we will use Boolean identities to expand the product and simplify. We find that

F (x, y, z) = (x + y)z =xz+yz=x1z+1yz =x(y+y)z+(x+x)yz =xyz+xyz+xyz+xyz

=xyz+xyz+xyz.

Distributive law Identity law Unit property Distributive law Idempotent law

Second, we can construct the sum-of-products expansion by determining the values of F for all possible values of the variables x, y, and z. These values are found in Table 2. The sum-of- products expansion of F is the Boolean sum of three minterms corresponding to the three rows of this table that give the value 1 for the function. This gives

F (x, y, z) = xyz + xy z + xyz.

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It is also possible to find a Boolean expression that represents a Boolean function by taking a Boolean product of Boolean sums. The resulting expansion is called the conjunctive normal form or product-of-sums expansion of the function. These expansions can be found from sum-of-products expansions by taking duals.

INJUCTIve

•f (a) = f (b) implies that

a = b for all a and b in the domain of f.

•Involves one to one function

SURJECTIVE

•Onto function

•every element

b ∈ B there is an element a ∈ A with f (a) = b.

BIJECTION

•one-to-one function AND

onto function

Inverse FunctionDefinition

Let f be the bijection from set A to the set B The inverse function of f is the function that

assigns to an element b belonging to B the unique element a in A such that f (a) = b

The inverse function of f is denoted by fˉˡ Hence fˉ (b) = a ˡ when f(a) = b

Example of Inverse Function

Example let f : R R be the bijection given by the

formula: f(x)= 4x + 1 Write f(x) = y ,that is 4x + 1 = y Then solve for x 4x = y + 1 x = y – 1 4Answer: fˉˡ (y)= = y – 1 4

Definition of Composition of Functions

Process to combine two functions . One function is to be done first then its

result is substitute in place of each x to another function.

notation for a composite function : ( ƒº g) (x)= ƒ( g(x))

Example

 Given that ƒ (x) =5x and g (x) =x²+2.

Find ( ƒºg )(x) and (gºƒ)(x)

Solution 1: (ƒºg)(x) Solution 2: -ƒ( g (x)) - g(ƒ(x)) ƒ (x²+2) - g(5x) 5( x²+2) -(5x)²+2 = 5x²+10 -5x²+2

Example :

Given that ρ(x)=x+4 and h(x) = x² get the value of

a)(hºρ)(x) b) (ρºh)(3) -h(ρ(x)) -ρ(h(x)) -h(x+4)² -ρ(3)² -x²+8x+8 -9+4 =11