Discrete Mathematics CS 2610

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Discrete Mathematics CS 2610. February 12, 2009. Agenda. Previously Finished functions Began Boolean algebras And now Continue with Boolean algebras. 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)

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### Discrete Mathematics CS 2610

February 12, 2009

Agenda
• Previously
• Finished functions
• Began Boolean algebras
• And now
• Continue with Boolean algebras
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.
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).
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.
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
• Ocorresponds to False
• 1 corresponds to True
•  corresponds logical operator AND
• + corresponds logical operator OR
• - corresponds logical operator NOT

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Boolean Algebra Tables

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

Boolean n-Tuples
• Let B = {0, 1}, the set of Boolean values.
• LetBn = { (x1,x2,…xn) | xi B, i=1,..,n}

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B1= { (x1) | x1 B,}

B2= { (x1, x2), | xi B, i=1,2}

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

• For all nZ+, any function f:Bn→B is called a Boolean function of degree n.
Example Boolean Function

F(x,y,z) =B3B

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

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B3 has 8 triplets

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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.
Combining Boolean Functions
• Let Fand G be two Booleans functions of degree n.
• Complement ofF: 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)
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
Boolean Expressions
• Let 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 E2are 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

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

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Functions and Expressions
• 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.

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F(x1,x2,x3)

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

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

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Boolean Functions
• Two Boolean expressions e1and e2 that represent the exact same function F are called equivalent

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Representing Boolean Functions
• How to construct a Boolean expression that represents a Boolean Function ?

F

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

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

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) + z

x· (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 · y

• Absorption laws:

x + x ·y = x, x · (x + y) = x

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

Boolean Identities
• Absorption law:
• Show that x ·(x + y)=x
• x ·(x + y)= (x + 0) ·(x + y) identity
• = x + 0 ·y distributive *
• = x + y · 0 commutative
• = x + 0 domination
• = x identity