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BOOLEAN ALGEBRA

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BOOLEAN ALGEBRA -REVIEW

Boolean Algebra was proposed by George Boole in 1853.Basically AND,OR NOT can be expressed as Venn Diagrams

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Min Terms and Max Terms

Min Terms are those which occupyminimum area on Venn Diagram

Max Terms are those which occupymaximum area on Venn diagram.

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LOGIC GATES

Nand and Nor gates are called Universal gatesas any Boolean function can be realized with thehelp of Nand and Nor gates only

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For example, NOT, OR, AND gates can be realizedby only Nand gates.

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SIMPLIFICATION OF BOOLEAN FUNCTIONS

• Algebraic Method

• Tabular Method

• K-Map Method

• Schienman Method

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ALGEBRAIC METHOD

Simplify using algebraic theorems

Advantage:

First Method based on Boolean Algebra theorems

Disadvantage:

No Suitable algorithm to apply (Trial type of method)

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TABULAR METHOD

Also called Quine McClusky Method

Advantage:

It may work for any no. of variables

Disadvantage:

Simplification from table is quite involved

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K-MAP METHOD

Karnaugh Map. Also called VietchKarnaugh Method.

Advantage:

Simplest and Widely accepted

Disadvantage:

Applicable for only upto Six variables

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SCHIENMAN METHOD

Columnwise writing of minterms asdecimal numbers and their simplification

Advantage:

Very suitable for computerization. Applicable for anynumber of variables. Parallel Processing

Disadvantage:

May not result in most simplified answer for someproblems

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Steps for simplification:

• Try to find single one’s– 2 one’s– 4 one’s– 8 one’s

Always see is a higher combination exists. If a higher combination exists, wait. Be sure that you have managed the lower combination first.

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SYMMETRIC FUNCTIONS

• DEFINITION

• PROPERTIES

• IDENTIFICATION

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Definition

A switching function of n variables

f(X1,X2….Xn) is called a symmetric (or

totally symmetric), if and only if it is invariant

under any permutation of its variables. It is

partially symmetric in the variables Xi,Xj

where {Xi,Xj} is a subset of {X1,X2…Xn} if

and only if the interchange of the variables

Xi,Xj leaves the function unchanged.

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EXAMPLES

• f(x,y,z) = x’y’z+xy’z’+x’yz’If we substitute x = y and y = x

x = z and z = x y = z and z = y

TOTALLY SYMMETRIC with respect to x,y,z

• f(x,y,z) = x’y’z + xy’z’is Prettily Symmetric in the variables x and z.(x = z and z = x)

• f(x,y,z) = z’y’x + zy’x’is a Symmetric function (x = y and y = x)

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• f(x,y,z) = y’x’z +yx’z’)is Not a Symmetric function

This function is symmetric w.r.t x and z, but not symmetricw.r.t x and y. So Partially Symmetric

•f(x1,x2,x3) = x1’x2’x3’ + x1x2’x3+ x1’x2x3is not symmetric w.r.t. the variables x1,x2,x3, but is

symmetric w.r.t the variables x1,x2,x3’ >> f is not invariant under an interchange of variables

x1,x3. That is, x3’x2’x1’+x3x2’x1 +x3’x2x1 != f

>> But f is invariant under an interchange of variables x1,x3’That is, x3x2’x1 + x3’x2’x1 + x3x2x1’ = f

So f is symmetric w.r.t the variables x1,x2 and x3’

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The variables in which a function is symmetric are called the

VARIABLES OF SYMMETRY

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Necessary and Sufficient condition for function f(x1,x2….xn) to be symmetric is that it may be specified by a set of numbers {a1,a2…ak} where 0<an<n,such that itassumes the value 1 when and only when ai of the variables are equal to 1. The numbers in the set are called the a-numbers

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A Symmetric function is denoted bySa1,a2…ak (x1,x2….xn), where S designates theproperty of symmetry, the subscripts designatethe a numbers, and (x1,x2….xn) designate thevariables of symmetry.

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IDENTIFICATION

The switching function to be tested forsymmetry is written as a table in which all theminterms contained in the function are listed by their binary representation

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For example, the function f(x,y,z) = (1,2,4,7) is written as shown:

x y z a# The arithmetic sum of each0 0 1 1 column in the table is computed0 1 0 1 written under the column. This1 0 0 1 sum is referred to as a column-1 1 1 3 sum. The number of 1’s in each 2 2 2 row is written in the corresp. position in column a#. This no. is called ROW SUM.

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If an n-variable function is symmetric and oneof its row sums is equal to some number a, then,by definition, there must exist n!/(n-a)!a! rows which have the same row sum.

If all the rows occur the required number of times,then all colums sums are identical

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For the example, all column sums equal 2,and there are two row sums, 1 and 3, that mustbe checked for “Sufficient Occurrence”.>> 3!/(3-1)! = 3 ; 3!/(3-3)! = 1

Both row sums occur the required number oftimes. Therefore, the function is symmetricand can be expressed by S1,3(x,y,z).

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The column sums are 6 4 4 6Since the column sums are not all the same,further tests must be performed to determineif the function is symmetric, and if it is, tofind its variables of symmetry.

The column sums can be made the same bycomplementing the columns correspondingto variables x and y.

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The new column sums are now computed and arefound identical. The row sums are determinednext and entered as a#. Each row sum is testedby the binomial co-efficient occurrence.

4!/(4-2)! = 6 ; 4! /(4-3)!3! = 4Since, all row sums occur the required number of times, the function is symmetric,its variablesof summetry are w,x’,y’,z and its a numbers are 2 and 3. ( f = S2,3(w,x’,y’,z))

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If columns w and z are complimented, insteadof x and y, the table shown below results and \since all its row sums occur the required no. oftimes, f can be written as f = S1,2(w’,x,y,z’)

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The column sums are all identical, but row sum 2 does not occur six times as required.One way to overcome this difficulty is byexpanding the function about any one of its variables

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The function can be expanded about w.

w x y z a# Column Sums :0 0 0 0 0 x y z0 0 1 1 2 1 1 20 1 0 1 2

The column sums can be made by complementingthe columns corresponding to variables x and y.

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x’ y’ z a# Column Sums:1 1 0 2 x y z1 0 1 2 2 2 20 1 1 2

Each row is tested by the binomial coefficientsfor sufficient occurrence.

3!/(3-2)!2! = 3

Symmetry: S2(x’,y’,z)

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w x y z a#1 0 1 0 1 The column sums can be made1 1 0 0 1 the same by complementing1 1 1 1 3 the columns corresponding 2 2 1 to variable z.

x y z’ a#0 1 1 2 Each row is tested by the1 0 1 2 binomial coefficients for1 1 0 2 sufficient occurrence.2 2 2

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3!/(3-2)! 2! = 3

Symmetry: S2(x,y,z’)

So the function f is written as

f=w’S2(x’,y’,z) + wS2(x,y,z’)

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questions?