Discrete Mathematics CS 2610

February 12, 2009

2

Agenda

Previously Finished functions Began Boolean algebras

And now Continue with Boolean algebras

3

But First

p q r, is NOT true when only one of p, q, or r is true. Why not?

It is true for (p Λ ¬q Λ ¬r)It is true for (¬p Λ q Λ ¬r)It is true for (¬p Λ ¬q Λ r)

So what’s wrong? Raise your hand when you know.

4

Injective Functions (one-to-one)

If function f : A B is 1-to-1 then every b B has 0 or 1 pre-image.Proof (bwoc): Say f is 1-to-1 and b B has 2 or more pre-images.Then a1, a2 st a1 A and a2 A, and a1 ≠ a2.

So f(a1) = b and f(a2) = b, meaning f(a1) = f(a2).

This contradicts the definition of an injection since when a1 ≠ a2 we know f(a1) ≠ f(a2).

5

Boolean Algebras (Chapter 11)

Boolean algebra provides the operations and the rules for working with the set {0, 1}.

These are the rules that underlie electronic and optical circuits, and the methods we will discuss are fundamental to VLSI design.

6

Boolean Algebra

The minimal Boolean algebra is the algebra formed over the set of truth values {0, 1} by using the operations functions +, ·, - (sum, product, complement).

The minimal Boolean algebra is equivalent to propositional logic where O corresponds to False 1 corresponds to True corresponds logical operator AND + corresponds logical operator OR - corresponds logical operator NOT

7

Boolean Algebra Tables

x

0

0

1

1

y

0

1

0

1

x + y

0

1

1

1

xy

0

0

0

1

x

0

1

x

1

0

x,y are Boolean variables – they assume values 0 or 1

8

Boolean n-Tuples

Let B = {0, 1}, the set of Boolean values.

Let Bn = { (x1,x2,…xn) | xi B, i=1,..,n}

.

B1= { (x1) | x1 B,}

B2= { (x1, x2), | xi B, i=1,2}

Bn= { ((x1,x2,…xn) | xi B, i=1,..,n,}

For all nZ+, any function f:Bn→B is called a Boolean function of degree n.

9

Example Boolean Function

x

0

0

0

0

1

1

1

1

y

0

0

1

1

0

0

1

1

z

0

1

0

1

0

1

0

1

F(x,y,z)=x(y+z)

F(x,y,z) =B3B

B3 has 8 triplets

0

0

0

0

1

1

0

1

10

Number of Boolean Functions

How many different Boolean functions of degree 1 are there?How many different Boolean functions of degree 2 are there?How many different functions of degree n are there ? There are 22ⁿ distinct Boolean functions of

degree n.

11

Combining Boolean Functions

Let F and G be two Booleans functions of degree n.

• Complement of F: F (x1,..xn) = F (x1,..xn)

• Boolean Sum : (F + G)(x1,..xn) = F (x1,..xn) + G (x1,..xn)

• Boolean Product: (F·G)(x1,..xn) = F(x1,..xn)·G(x1,..xn)

12

Equal Boolean Functions

Two Boolean functions F and G of degree n are equal iff for all (x1,..xn) Bn, F (x1,..xn) = G (x1,..xn)

Example: F(x,y,z) = x(y+z), G(x,y,z) = xy + zx

13

Boolean ExpressionsLet x1, …, xn be n different Boolean variables.

A Boolean expression is a string of one of the following forms (recursive definition): 0, 1, x1, …, or xn. are Boolean Expressions If E1 and E2 are Boolean expressions then -E1,

(E1E2), or (E1+E2) are Boolean expressions.

Example:E1 = x

E2 = y

E3 = z

E4 = E1 + E2= x + y

E5 = E1E2= x y

E6 = -E3 = -z

E7 = E6 + E4 = -z + x + y

E8 = E6 E4 = -z ( x + y)

Note: equivalent notation: -E = E for complement

14

Functions and Expressions

F(x1,x2,x3)

1110

1010

0100

1000

x3x2x1

0001

0101

0011

1111

A Boolean expression represents a Boolean function.

Furthermore, every Boolean function (of a given degree) can be represented by a Boolean expression with n variables.

F(x1,x2,x3) = x1(x2+x3)+x1x2x3

15

Boolean Functions

Two Boolean expressions e1 and e2 that represent the exact same function F are called equivalent

F(x1,x2,x3)

1110

1010

0100

1000

x3x2x1

0001

0101

0011

1111

F(x1,x2,x3) = x1(x2+x3)+x1x2x3

F(x1,x2,x3) = x1x2+x1x3+x1x2x3

16

Representing Boolean Functions

How to construct a Boolean expression that represents a Boolean Function ?

1110

1010

0100

0000

zyx

0001

1101

0011

1111

F

(-x)(y)(-z) + (-x)yz + x(-y)z + xyz

F(x, y, z) = 1 if and only if:

What about a 2-input multiplexer?

17

Boolean Identities

Double complement:x = x

Idempotent laws:x + x = x, x · x = x

Identity laws:x + 0 = x, x · 1 = x

Domination laws:x + 1 = 1, x · 0 = 0

Commutative laws:x + y = y + x, x · y = y ·

x

Associative laws:x + (y + z) = (x + y) + zx · (y · z) = (x · y) · z

Distributive laws:x + y ·z = (x + y)·(x + z)x · (y + z) = x ·y + x ·z

De Morgan’s laws:(x · y) = x + y, (x + y) =

x · yAbsorption laws:x + x ·y = x, x · (x + y) =

x

the Unit Property: x + x = 1 and Zero Property: x ·x = 0

18

Boolean Identities

Absorption law:

Show that x ·(x + y) = x

1) x ·(x + y) = (x + 0) ·(x + y) identity2) = x + 0 ·y distributive *3) = x + y · 0 commutative4) = x + 0 domination5) = x identity