Lec15 Boolean Algebra

8/3/2019 Lec15 Boolean Algebra

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[Lecture 15][Boolean Algebra]

Boolean Algebra


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Boolean Algebra

Invented by and named after George Boole.

Commonly, and especially in computer science and digital electronics, this

term is used to mean two-value logic.

Provide the operations and rules for working with the set {0,1}.

Map logical propositions to symbols.

Permit manipulation of logic statements using Mathematics.

Invented by and named after George Boole.

Commonly, and especially in computer science and digital electronics, this

term is used to mean two-value logic.

Provide the operations and rules for working with the set {0,1}.

Map logical propositions to symbols.

Permit manipulation of logic statements using Mathematics.

Definition:

ABoolean algebra written as (B, +, . , d , 0, 1), consists of a setB

containing distinct elements 0 and 1, binary operators + and . onB, and a

unary operatord onB, such that the identity laws, complement laws,associative laws, commutative laws and distributive laws holds.

Definition:

ABoolean algebra written as (B, +, . , d , 0, 1), consists of a setB

containing distinct elements 0 and 1, binary operators + and . onB, and a

unary operatord onB, such that the identity laws, complement laws,associative laws, commutative laws and distributive laws holds.


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Identity Law

x+ 0 =x,

x. 1 =x

Complement Law

x+xd = 1

x.xd = 0

Associative Law

(x+ y) + z=x+ (y+z)

(x. y) . z=x. (y. z)

Identity Law

x+ 0 =x,

x. 1 =x

Complement Law

x+xd = 1

x.xd = 0

Associative Law

(x+ y) + z=x+ (y+z)

(x. y) . z=x. (y. z)

Commutative Law

x + y = y + x

x . y = y . x

Distributive Law

x + (y . z) = (x + y) . (x + z)

x . (y + z) = x . y + x . z

Commutative Law

x + y = y + x

x . y = y . x

Distributive Law

x + (y . z) = (x + y) . (x + z)

x . (y + z) = x . y + x . z


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Note:

0 and 1 in the previous definition are merely symbolic names.

In general, they have nothing to do with the numbers 0 and 1. + and are merely binary operators. In general, they have

nothing to do with the ordinary addition and multiplication.

Some examples of Boolean algebras are

(a) ({0, 1}, , , d, 0, 1)

(b) (P(U), , ,D , , U)

where (i) and plays the role of +.

(ii) and plays the role of.

(iii)d

andD

plays the role of complement.(iv)plays the role of 0 and the universal set

U plays the role of 1.

Note:

0 and 1 in the previous definition are merely symbolic names.

In general, they have nothing to do with the numbers 0 and 1. + and are merely binary operators. In general, they have

nothing to do with the ordinary addition and multiplication.

Some examples of Boolean algebras are

(a) ({0, 1}, , , d, 0, 1)

(b) (P(U), , ,D , , U)

where (i) and plays the role of +.

(ii) and plays the role of.

(iii)d

andD

plays the role of complement.(iv)plays the role of 0 and the universal set

U plays the role of 1.


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5

Boolean Expressions and Boolean Functions

LetZ2 = {0, 1}, a variable x is called a Boolean variable if itassumes only values from Z2.

A function from toZ2 is called a Boolean function of

degree n, where = {(x1, , xn) | xi Z2 , 1 e i e n}.

Boolean functions can be represented using expressions

made up from variables and Boolean operations.

LetZ2 = {0, 1}, a variable x is called a Boolean variable if itassumes only values from Z2.

A function from toZ2 is called a Boolean function of

degree n, where = {(x1, , xn) | xi Z2 , 1 e i e n}.

Boolean functions can be represented using expressions

made up from variables and Boolean operations.

n

Z2

n

Z2

rprqprqpF !),,(

zxyzxzyxF !),,(

Example:


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Example:

F(x, y, z) = xy + z is a Boolean function wherexy + z

is a Boolean expression.

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

1 1 1 1 0 1

1 1 0 1 1 1

1 0 1 0 0 0

1 0 0 0 1 1

0 1 1 0 0 0

0 1 0 0 1 1

0 0 1 0 0 00 0 0 0 1 1

The truth table forF(x, y, z) = xy + z is:


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Example:How many Boolean functionsof degree 2 are there?

Soluti

on:A Boolean function of degree 2 is a function fromZ2

2= {(0,0), (0,1),(1,0), (1,1)} toZ2={0,1}. Hence, there are 16 different Booleanfunctionsof degree 2. The following table displays those functions,labeled F1, , F16.

x y F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 F16

0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

Question: How many different Boolean Functions of degree n are there?

Answer:n22


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Identity Name Identity Name

(x + y) + z = x + (y + z)(x.y) z = x (y. z)

AssociativeLaws

x + 1 = 1x. 0 = 0

Domination/Bound Laws

x + y = y +x

x y =y x

Commutative

Laws

x + (x. y) =x

x. (x + y) =x

Absorption

Laws

x (y +z) = (x y) + (x z)

x+ (y z) = (x+ y) (x+ z)

Distributive

Laws

(xd)d = x Double

ComplementLaws

x + 0 =x

x 1 = x

Identity Laws 0 d = 1

1d = 0

0 and 1 Laws

x + xd = 1

x xd = 0 UnitP

ropertyZero Property

(x + y)d

= xd

.yd

(x. y)d = xd +

yd

D

eM

organsLaws

x + x = x

x x =x

Idempotent

Laws


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ExampleLet x, y, z be the Boolean variables. Determine

whether x + x.y.z = xistrue.

Solution:

x + x.y.z = x.1 + x.y.z Identity

= x.(1 + y.z) Distributive= x. 1 Domination

= x Identity

ExampleLet x, y, z be the Boolean variables. Determine

whether x + x.y.z = xistrue.

Solution:

x + x.y.z = x.1 + x.y.z Identity

= x.(1 + y.z) Distributive= x. 1 Domination

= x Identity


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Boolean identitiescome in pairs. Such pairs are called

duals. The dualof a Boolean expression isobtained

by interchanging Boolean sums (+) and Boolean product(.), and

by interchanging 0s and 1s.

Boolean identitiescome in pairs. Such pairs are called

duals. The dualof a Boolean expression isobtained

by interchanging Boolean sums (+) and Boolean product(.), and

by interchanging 0s and 1s.

Duality

Example

The dual of (x + y)d = xdyd is (xy)d = xd + yd

The dual ofxyd = 0 if and only ifxy = x is

x + yd = 1 if and only ifx + y = x

Example

The dual of (x + y)d = xdyd is (xy)d = xd + yd

The dual ofxyd = 0 if and only ifxy = x is

x + yd = 1 if and only ifx + y = x


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A literalis a Boolean variable or its complement.A literalis a Boolean variable or its complement.

Minterm and MaxtermMinterm and Maxterm

A minterm of the Boolean variables x1, x2, , xn is a Boolean

producty1y2yn, where yi = xi oryi = xid.

A minterm of the Boolean variables x1, x2, , xn is a Boolean

producty1y2yn, where yi = xi oryi = xid.

A maxterm is defined as a Boolean sum of n literals, with

one literal for each variable.

A maxterm is defined as a Boolean sum of n literals, with

one literal for each variable.

Hence, a minterm is a product ofn literals, with one literal foreach variable.


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x1 x2 x3 Minterm Maxterm

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

Example: 3 variables

321xxx 321

xxx

321xxx 321 xxx

1 2 3x x x

1 2 3x x x

1 2 3 x x x

31 2 x x x

Complete the rest!


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Also known as sum-of-products expansion.

Sum of minterms corresponding to input combinations

for which the function produces a 1 as output.

Also known as sum-of-products expansion.

Sum of minterms corresponding to input combinations


A B C F1 F20 0 0 0 10 0 1 1 00 1 0 0 10 1 1 1 01 0 0 0 11 0 1 1 01 1 0 1 01 1 1 1 0

F1 = + ABC+ A'BC + AB'C + ABC'A'B'C

Disjunctive Normal Form (DNF)


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Discrete Structures 14

Also known asproduct-of-sums expansion. Sum of maxterms corresponding to input combinations


Also known asproduct-of-sums expansion. Sum of maxterms corresponding to input combinations


Conjunctive Normal Form (CNF)

Any Boolean Expression has a DNF and a CNF.

Given a Boolean function, we can obtain the DNF andCNF either by Truth table or Boolean identity laws.


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Logic Gates

Input

x

NOT GateOutput

x

Two inputs AND Gate

Inputs

x

y

Output

xy

x

0 1

1 0

x

x y

0 0 0

0 1 0

1 0 0

1 1 1

xy


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Inputs

x

y

Output

x + y

Two Inputs ORGate x y x+y

0 0 0

0 1 1

1 0 1

1 1 1

Logic Gates


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NORGate

NAND Gate


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Combination of Gates


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Examples:Construct circuits that produce the following outputs:

a) b) c)xyx )( )( zyx )()( zyxzyx

Solution:

b) )( zyx

xyx )( a)


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c) )()( zyxzyx


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Example: Given a function F(x, y, z) with the truth table below. Write

the sum-of-products (minterm expression) representing F(x,y, z) and

draw the corresponding logic diagram.


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Example

A function f(x, y, z) has truth table as below.

Write the sum-of-products (minterm expression) representing f(x, y,z) .

Simplify the expression and draw a logic diagram using as few gates as possible.

x y z f(x, y, z)

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 01 1 1 1

Discrete Structures 23


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zyxzyxzyxf !),,(

z

yyz

Solution:

Sum-of-product

Simplify the expression:

zy

zyxx

zyxzyxzyxf

!

!

!

)(

),,(

Logic diagram:


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Boolean expression for a logic circuit may be reduced

to a simpler form.

The simplified expression forms a circuit uses less logic

gates but performs the same function.

We can use Boolean identity laws to simplify a Boolean

function. Another method to simplify a Boolean function is by

using Karnaugh map.

Boolean expression for a logic circuit may be reduced

to a simpler form.

The simplified expression forms a circuit uses less logic

gates but performs the same function.

We can use Boolean identity laws to simplify a Boolean

function. Another method to simplify a Boolean function is by

using Karnaugh map.

Simplify this Boolean expression using

identity laws:F(x, y, z) = zyxzyxyzxzyx

Minimization ofCircuits

Answer: yxzx


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K-map for short

Invented in 1950 by Maurice Karnaugh, a telecommunication

engineer in Bell Labs.

An excellent aid for simplification of Boolean expression in

DNF form.

Make use of the human brain's excellent pattern-matchingcapability to decide which terms should be combined to get

the simplest expression.

K-map for short

Invented in 1950 by Maurice Karnaugh, a telecommunication

engineer in Bell Labs.

An excellent aid for simplification of Boolean expression in

DNF form.

Make use of the human brain's excellent pattern-matchingcapability to decide which terms should be combined to get

the simplest expression.

Karnaugh Map


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Karnaugh map in 2 variables

xy yx

yx yx

x

x

y y


xyz

yzx zyx

zxy zyx zyx

zyx zyx

x

x

yz zy zy zy


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wx

xw

yz zy zy zy

xw

xw xyzw

zw xy zywx zywx

yzxw zyxw zyxw zyxw

yzxw zyxw zyxw zyxw

wxyz

zxyw zyxw zyxw


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Important points:

1 isplaced in the cell representing a minterm if

thisminterm ispresent in the expansion.

Cells are said to be adjacent if the minterms

differ in exactly one literal.

Identify thelargest possib

le b

locks and cover allthe 1s with the fewest blocks.

Circle blocksof cells in the K-map that represent

minterms that can be combined and find the

corresponding

sum-of-product

s.

Important points:

1 isplaced in the cell representing a minterm if

thisminterm ispresent in the expansion.

Cells are said to be adjacent if the minterms

differ in exactly one literal.

Identify thelargest possib

le b

locks and cover allthe 1s with the fewest blocks.

Circle blocksof cells in the K-map that represent

minterms that can be combined and find the

corresponding

sum-of-product

s.


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Example:

Use Karnaugh map to simplify the following Boolean expressions:

1)

2)

3)

4)

yxyx

zyxyzxzyxzyxxyz

zyxzyxzxyzyxxyz

zyxwzyxwzyxwzxywzywxzwxy

Given a truth table of a Boolean expression,we can also simplify the expression by first

constructing the DNF from the truth table.


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Solution:

xyxyx !

zyyxyzzyxyzxzyxzyxxyz !

1 1

x

x

y y

1 1

1 1 1

x

x

yz zy zy zy

2.

1.

xyyxyxyx !! )(


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zxyzyxzyxzxyzyxxyz !

zyxwzywzxwzwxzyxwzyxwzyxwzxywzywxzwxy !

3.

1 1 1

1 1

x

x

yz zy zy zy

4.

wx

xw

yz zy zy zy

1 1

1 1

1 1

xw

xw


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Answer:

1. The duals are x + (y.1) and (xd + 0)(ydz).

2.

Question:

1. Find the duals ofx(y + 0) and xd.1 + (yd + z).

2. Use K-map to simplify .yxyxyx

1

1 1

x

x

y y

yxyxyxyx !

Lec15 Boolean Algebra

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