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MOHAN LAL SUKHADIA UNIVERSITY

Seminar on “ Boolean Expression “

Presented to: Presented by: M.K. Jain Sir ---------------- Mrs. Priyanka Soni MCA 1st sem.

Introduction of Boolean Expression:

In computer science, a Boolean expression is an expression in a programing

language that produces a Boolean value when evaluated i.e. one of true or false.

Boolean expressions are used mostly with while loops, and conditional

statements. Boolean expressions can be contrasted with arithmetic expressions

which are expressions that evaluated to a number.

Boolean expressions are made up of the three logical operators (AND, OR, NOT),

the relational operator (> >= < = = <>) and functions that return true or false.

Logical Operators

• Electricity is the basic element of every digital device. When electricity flow in a device it is represented by 1. When flow is not available it is represented by 0.

• To produce the combine results of 2 or more inputs we use special type of operators called logical operators.

Use of Logic Operators/ Logic gates

• Logic gates are used in every digital electronic device.• The output depends on the gates to which input has been given.• Several applications are being processed by the help of logic gates,

such as computers, mobiles, USB drives, etc.

For Example:-1 way- like switches (2 inputs are connected to the plug than only

1 output is displayed)

There are 3 types of operators:

1. Logical AND operator

2. Logical OR operator

3. Logical NOT operator

Logical AND operators

• This AND operator is used to produce logical multiplication.

• This is represented by dot (.), and (&&).

• This operator produces true result only when both the member inputs are

in working states (1).

If any of the input member has false value than this will produce false

(0) output.

Truth Table

Symbolic Representation

INPUT OUTPUTA B Q=A.B0 0 0

0 1 0

1 0 0

1 1 1

Logical OR operator

• This OR operator is used to produce logical addition.

• This is represented by plus (+), or (| |).

• This operator produces true result only when one of the input is in

working states (1).

• If both of the inputs are false than this will produce false (0) output.

Truth Table


INPUT OUTPUTA B Q= A+B0 0 0

0 1 1

1 0 1

1 1 1

AB Q

Logical NOT operator

• It is also called Complement operator.

• This NOT operator is used to produce reverse result.

• This is represented by Desh (’) or Bar(‾).

• This operator produces true(1) result if input is false(0).

• If the inputs is true(1) than this will produce false(0) output.

Truth TableINPUT OUTPUT

A A’0 1

1 0


A A’

Basic Identities of Boolean Algebra

• The Boolean algebra contains different identities which are also

called theorems. These theorems are based on the logical operators.

• If we use logical OR in the identity, then the operation is called

‘Oring’ and if we use AND in an operation then the operation is called

‘Anding’.

Identities/ Theorems of Boolean Algebra

1. A + 0= AIf a variable is ORed with 0 then the result will be the variable.

2. A + 1= 1If a variable is ORed with 1 then the result will be 1 ever.

INPUT Theorem OUTPUTA A + 00 0 + 0 0

1 1 + 0 1

INPUT Theorem OUTPUTA A + 10 0 + 1 1

1 1 + 1 1

3. A . 0= 0If a variable is Anded with 0 then the result will be 0 ever.

4. A . 1= AIf a variable is ORed with 1 then the result will be 1 ever.

INPUT Theorem OUTPUTA A . 00 0 . 0 0

1 1 . 0 0

INPUT Theorem OUTPUTA A . 10 0 . 1 0

1 1 . 1 1

Identities

5. A + A= AIf a variable is ORed with itself then the result will be the same variable.

6. A . A= AIf a variable is Anded with itself then the result will be the same variable.

INPUT Theorem OUTPUT

A A + A0 0 + 0 0

1 1 + 1 1


A A . A0 0 . 0 0

1 1 . 1 1

Identities

7. A + A’= 1If a variable is ORed with its complement then the result will be 1 ever.

8. A . A’= 0If a variable is Anded with its complement then the result will be 0 ever.


A A’ A + A’0 1 0 + 1 1

1 0 1 + 0 1


A A’ A . A’0 1 0 . 1 0

1 0 1 . 0 0

Identities

Laws of Boolean Algebra

1. Idempotent Law:

(a) x + x= x (b) x .x= x

Proof:- Proof:-

= x+ x = x . x= (x + x).1 = x . x+ 0= (x + x).(x + x’) = x . x + x . x’= x + x + x . x’+ x . x’ = x . (x + x’)= x + 0 = x . 1= x = x

2. Absorption Law:

(a) x + x . y = x (b) x . (x + y)= x

Proof:- Proof:-

= x+ x . y = x . (x + y)= x . 1 + x . y = x . x + x . y= x . (1 + y) = x + x . y= x . (y + 1) = x . (1 + y)= x .1 = x . 1= x = x

3. Involution Law:-The inverse of an inversed variable produces the same variable as output.

(a) (x’)’ = x

Proof:-

Same

x x’ (x’)’0 1 01 0 1

4. Commutative Law:-The addition of 2 variables does not effect by their order. This law is called commutative law.

(a) x + y = y + x (b) x . y = y . xProof:- proof:-

Same Same

X Y X + Y Y + X0 0 0 00 1 1 11 0 1 11 1 1 1

X Y X . Y Y . X0 0 0 00 1 0 01 0 0 01 1 1 1

5. Distributive Law:-

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

Proof:- (a)

Same

X Y Z (Y + Z) X .(Y + Z) (X . Y) (X . Z) (X . Y) + (X . Z)0 0 0 0 0 0 0 00 0 1 1 0 0 0 00 1 0 1 0 0 0 00 1 1 1 0 0 0 01 0 0 0 0 0 0 01 0 1 1 1 0 1 11 1 0 1 1 1 0 11 1 1 1 1 1 1 1

6. Associative Law:-

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

Proof:- (a)

Same

X Y Z (Y . Z) X. (Y . Z) (X . Y) (X . Y) . Z0 0 0 0 0 0 00 0 1 0 0 0 00 1 0 0 0 0 00 1 1 1 0 0 01 0 0 0 0 0 01 0 1 0 0 0 01 1 0 0 0 1 01 1 1 1 1 1 1

5. De- Morgan’s Law:-

(a) (x + y)’ = x’ . y’ (b) (x . y)’ = x’ + y’

Proof:- (a)

Same

X Y X + Y (X + Y)’ X’ Y’ X’ . Y’0 0 0 1 1 1 10 1 1 0 1 0 01 0 1 0 0 1 01 1 1 0 0 0 0

Minimization Techniques

The techniques which are used minimize the variables or gates in a

logical expression are called minimization techniques.

The most commonly used minimization techniques are:-

1. Using Boolean Theorms

2. Using Karnough Map

3. Using Quine Macluskey Method(Tabular method)

4. Using Variable Mapping Method

Minterm• If a boolean function contains n variables than a product term which contains all

the variables once in either complemented or uncomplemented form is called minterm.• In the minterm main property of each minterm is that it will have value one only.• For n variable expression the no. of minterms are 2n like- if n=2 than minterms are

4, n=3 than minterms are 8.• All the minterms are can be represented by m0, m1, m2………..For example:-

X Y X . Y Minterms0 0 0 X’Y’0 1 0 X’Y1 0 0 XY’1 1 1 XY

mo

m1m2

m3

Sum Of Product (SOP)• When we perform the sum of logically multiplied inputs than the resultant

expression is called sum of product. The canonical or standard SOP is the sum of minterms.

Example:- Y(A,B,C)= m1 + m4 + m5 + m7= A’B’C + AB’C’ + AB’C + ABC

= A’B’C + AB’C + AB’C’ + ABC= B’C(A’ + A) + AB’C’ + ABC= B’C + AB’C’ + ABC

A B C Minterm0 0 0 A’B’C’ m00 0 1 A’B’C m10 1 0 A’BC’ m20 1 1 A’BC m31 0 0 AB’C’ m41 0 1 AB’C m51 1 0 ABC’ m61 1 1 ABC m7

Maxterm• If a Boolean function contains n variables and the sum term contains all the possible

combinations of n variables than the sum term is known as maxterm.• This term is opposite of minterm because the value of each maxterm is 0. The

maxterm can be represented as M0, M1,………..• The main property of each maxterm is that it has value 0 only for one comination of

n input variables.• Example:-

X Y X + Y Maxterms0 0 0 X+Y0 1 1 X+Y’1 0 1 X’+Y1 1 1 X’+Y’

M0

M1

M2

M3

Product Of Sum (POS)• The product of sum can be difined as the logical product of the maxterm for

which it has the value 0.

Example:- f(A,B,C)= (A’+B’+C) . (A+C) . (A’+B’)

= (A’+B’+C) . (A+C+BB’) . (A’+B’+CC’)

= (A’+B’+C) . ((A+C)+BB’) . ((A’+B’)+CC’)

By appling: x+y . z= (x+y) . (x+z)

= (A’+B’+C) . (A+C+B) . (A+C+B’) . (A’+B’+C) . (A’+B’+C’)

= M6, M0, M2, M6, M7

= M0, M2. M6, M7

Karnough Map (K- map)

• Some times the Boolean theorems and laws makes the simplification logics of Boolean

functions more complex. Than we have to use the k-map techniques.

• It is a graphical method which is used to simplify a Boolean function or to convert a truth

table into its equalent logic circuit.

• The k-map is designed by squares where each square represents a minterm or maxterm.

• We will determine these techniques by studying examples in order to establish the rules for

map manipulation.

Karnough Map

• 1 variable map

x f(x)0 f(0)1 f(1)

f(0) f(1)

x0 1

Truth Table

Literal

Binary Values

Function Values

Literal

Binary ValuesFunction Values 1x2 Karnaugh

Map

Karnough Map

• Case Study: 1 variable map

0 1

x0 1

Circle all 1 entries that, taken together, form a subcube (i.e. rectangle).

DEFINITION:When constructing SOP forms, a 2N -subcube is a rectangular region of a Karnaugh map consisting of 2N adjacent cells, each containing the same value 1 (or 0 for POS forms), and where N must be an integer greater or equal to zero.

Thus, the minimal expression of the function is: F(x) = x

Karnough Map

• Case Study: 1 variable map - Complementation

1 0

x0 1

The entry 1 in the first column corresponds to the prime implicant x’.Thus, the minimal expression of the function is: F(x) = x’

Karnough Map• 2 variable map

x y f(x,y) 0 0 f(0,0)0 1 f(0,1)1 0 f(1,0)1 1 f(1,1)

f(0,0)

y0 1

0x

1 f(1,0)

f(0,1)

f(1,1)

Truth Table

2x2 Karnaugh Map

Karnough Map

• 3 variable map

x y z f(x)0 0 0 f(000)0 0 1 f(001)0 1 0 f(010)0 1 1 f(011)1 0 0 f(100)1 0 1 f(101)1 1 0 f(110)1 1 1 f(111)

f(000) f(001) f(011) f(010)

yz00 01 11 10

0x

1 f(100) f(101) f(111) f(110)

2x4 Karnaugh Map

Note the way that the column indices change by only 1 bit at a time from left

to right.

Karnough Map

• 3 variable map

x y z f(x,y,z)0 0 0 f(000)0 0 1 f(001)0 1 0 f(010)0 1 1 f(011)1 0 0 f(100)1 0 1 f(101)1 1 0 f(110)1 1 1 f(111)

0 1 3 2

y00 01 11 10

0x

1 4 5 7 6

2x4 Karnaugh Map

One final alternative

labelling scheme replaces the

function value by the decimal minterm index

value.

z


0 1 1 0

yz00 01 11 10

0x

1 1 0 1 0

•Circle all 1 entries that, taken together, form a subcube (i.e. rectangle). •Start with the largest subcubes, then proceed to smaller subcubes•Generally speaking, there will be more than one independent subcube, each reflecting a different prime implicant.

Karnough Map

• 3 variable map

0 1 1 0

yz00 01 11 10

0x

1 1 0 1 0

Each subcube (rectangle) corresponds to a prime implicant term.

Gathering all terms in SOP form,

f = xy’z’ + x’z + yz


w x y z f(w,x,y,z)0 0 0 0 f(0000)0 0 0 1 f(0001)0 0 1 0 f(0010)0 0 1 1 f(0011)0 1 0 0 f(0100)0 1 0 1 f(0101)0 1 1 0 f(0110)0 1 1 1 f(0111)1 0 0 0 f(1000)1 0 0 1 f(1001)1 0 1 0 f(1010)1 0 1 1 f(1011)1 1 0 0 f(1100)1 1 0 1 f(1101)1 1 1 0 f(1110)1 1 1 1 f(1111)

f(0000) f(0001) f(0011) f(0010)

yz00 01 11 10

00

01wx

11

10 f(1000) f(1001) f(1011) f(1010)

f(0100) f(0101) f(0111) f(0110)

f(1100) f(1101) f(1111) f(1110)


w x y z f(w,x,y,z)0 0 0 0 f(0000)0 0 0 1 f(0001)0 0 1 0 f(0010)0 0 1 1 f(0011)0 1 0 0 f(0100)0 1 0 1 f(0101)0 1 1 0 f(0110)0 1 1 1 f(0111)1 0 0 0 f(1000)1 0 0 1 f(1001)1 0 1 0 f(1010)1 0 1 1 f(1011)1 1 0 0 f(1100)1 1 0 1 f(1101)1 1 1 0 f(1110)1 1 1 1 f(1111)

Note the way that both the

row and column indices change by only 1 bit at

a time.

0 1 3 2

yz00 01 11 10

00

01wx

11

10

4 5 7 6

12 13 15 14

8 9 11 10

This implies that two rows, or columns, whose indices differ by

only 1 bit value, are adjacent.

WRAP-AROUND!

Karnough Map• Case Study: 4 variable map

1 0 0 1

yz00 01 11 10

00

01wx

11

10

1 1 0 0

1 1 1 0

1 0 1 1Now to identify the

prime implicants:

f(w,x,y,z) = xy’ + x’z’ + wxz + wx’y

There are no subcubes of sizes: 24 =16 or 23 = 8.

We have a Subcubes of size: 22 =4. and 21 =2

Example of Boolean Expression

Y = A.B + A.B’

AND Gate

AND Gate

OR Gate

A.B

A.B’

A

B

A

B’A B A.B B’ A.B’ A.B + A.B’0 0 0 1 0 0

0 1 0 0 0 0

1 0 0 1 1 1

1 1 1 0 0 1

Conclusion

• We have studied and developed several techniques for simplifying Boolean expressions.

• These are based on the axioms, definitions and theorems of the Boolean Algebra, applied through the Boolean Calculus.

• Powerful tabular techniques have been developed for rapid reduction to some minimal cost forms using• Karnaugh maps

• An even more powerful technique has been developed by Quine and McCluskey (and Petrick).

Any

Query

Thank You

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