1. BOOLEAN ALGEBRA
Boolean Algebra

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

3. Boolean Algebra

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15. Min Terms are those which occupy
Min Terms and Max Terms
Min Terms are those which occupy
minimum area on Venn Diagram
Max Terms are those which occupy
maximum area on Venn diagram.
Boolean Algebra

16. Boolean Algebra

17. Boolean Algebra

18. Boolean Algebra

19. Nand and Nor gates are called Universal gates
LOGIC GATES
Nand and Nor gates are called Universal gates
as any Boolean function can be realized with the
help of Nand and Nor gates only
Boolean Algebra

20. For example, NOT, OR, AND gates can be realized by only Nand gates.
Boolean Algebra

21. Boolean Algebra

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24. Boolean Algebra

25. SIMPLIFICATION OF BOOLEAN FUNCTIONS
Algebraic Method
Tabular Method
K-Map Method
Schienman Method
Boolean Algebra

26. ALGEBRAIC METHOD Advantage: Disadvantage:
Simplify using algebraic theorems
Advantage:
First Method based on Boolean Algebra theorems
Disadvantage:
No Suitable algorithm to apply (Trial type of method)
Boolean Algebra

27. TABULAR METHOD Advantage: Disadvantage:
Also called Quine McClusky Method
Advantage:
It may work for any no. of variables
Disadvantage:
Simplification from table is quite involved
Boolean Algebra

28. K-MAP METHOD Karnaugh Method. Advantage: Disadvantage:
Karnaugh Map. Also called Vietch
Karnaugh Method.
Advantage:
Simplest and Widely accepted
Disadvantage:
Applicable for only upto Six variables
Boolean Algebra

29. SCHIENMAN METHOD decimal numbers and their simplification Advantage:
Columnwise writing of minterms as
decimal numbers and their simplification
Advantage:
Very suitable for computerization. Applicable for any
number of variables. Parallel Processing
Disadvantage:
May not result in most simplified answer for some
problems
Boolean Algebra

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38. 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.
Boolean Algebra

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59. SYMMETRIC FUNCTIONS DEFINITION PROPERTIES IDENTIFICATION
Boolean Algebra

60. 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.
Boolean Algebra

61. f(x,y,z) = x’y’z+xy’z’+x’yz’ If we substitute x = y and y = x
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)
Boolean Algebra

62. is Not a Symmetric function
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 symmetric
w.r.t x and y. So Partially Symmetric
f(x1,x2,x3) = x1’x2’x3’ + x1x2’x3+ x1’x2x3
is 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’
Boolean Algebra

63. The variables in which a function is symmetric are called the
VARIABLES OF SYMMETRY
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64. Necessary and Sufficient condition for function f(x1,x2…
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 it
assumes 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
Boolean Algebra

65. A Symmetric function is denoted by
Sa1,a2…ak (x1,x2….xn), where S designates the
property of symmetry, the subscripts designate
the a numbers, and (x1,x2….xn) designate the
variables of symmetry.
Boolean Algebra

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70. IDENTIFICATION The switching function to be tested for
symmetry is written as a table in which all the
minterms contained in the function are listed
by their binary representation
Boolean Algebra

71. For example, the function f(x,y,z) = (1,2,4,7) is written as shown:
x y z a# The arithmetic sum of each
column in the table is computed
written under the column. This
sum is referred to as a column-
sum. The number of 1’s in each
row is written in the corresp. position in column a#. This no. is called
ROW SUM.
Boolean Algebra

72. If an n-variable function is symmetric and one
of 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
Boolean Algebra

73. For the example, all column sums equal 2,
and there are two row sums, 1 and 3, that must
be checked for “Sufficient Occurrence”.
>> 3!/(3-1)! = 3 ; 3!/(3-3)! = 1
Both row sums occur the required number of
times. Therefore, the function is symmetric
and can be expressed by S1,3(x,y,z).
Boolean Algebra

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75. Since the column sums are not all the same,
further tests must be performed to determine
if the function is symmetric, and if it is, to
find its variables of symmetry.
The column sums can be made the same by
complementing the columns corresponding
to variables x and y.
Boolean Algebra

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77. The new column sums are now computed and are
found identical. The row sums are determined
next and entered as a#. Each row sum is tested
by the binomial co-efficient occurrence.
4!/(4-2)! = 6 ; 4! /(4-3)!3! = 4
Since, all row sums occur the required number
of times, the function is symmetric,its variables
of summetry are w,x’,y’,z and its a numbers are 2
and 3. ( f = S2,3(w,x’,y’,z))
Boolean Algebra

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

79. Boolean Algebra

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

81. The function can be expanded about w. w x y z a# Column Sums :
The column sums can be made by complementing
the columns corresponding to variables x and y.
Boolean Algebra

82. Each row is tested by the binomial coefficients
x’ y’ z a# Column Sums:
x y z
Each row is tested by the binomial coefficients
for sufficient occurrence.
3!/(3-2)!2! = 3
Symmetry: S2(x’,y’,z)
Boolean Algebra

83. 1 0 1 0 1 The column sums can be made
w x y z a#
The column sums can be made
the same by complementing
the columns corresponding
to variable z.
x y z’ a#
Each row is tested by the
binomial coefficients for
sufficient occurrence.
Boolean Algebra

84. So the function f is written as f=w’S2(x’,y’,z) + wS2(x,y,z’)
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’)
Boolean Algebra

85. questions?
Boolean Algebra