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Boolean Algebra1 BOOLEAN ALGEBRA Boolean Algebra2 BOOLEAN ALGEBRA -REVIEW Boolean Algebra was proposed by George Boole in 1853. Basically AND,OR NOT.

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Presentation on theme: "Boolean Algebra1 BOOLEAN ALGEBRA Boolean Algebra2 BOOLEAN ALGEBRA -REVIEW Boolean Algebra was proposed by George Boole in 1853. Basically AND,OR NOT."— Presentation transcript:

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2 Boolean Algebra1 BOOLEAN ALGEBRA

3 Boolean Algebra2 BOOLEAN ALGEBRA -REVIEW Boolean Algebra was proposed by George Boole in 1853. Basically AND,OR NOT can be expressed as Venn Diagrams

4 Boolean Algebra3

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12 Boolean Algebra11

13 Boolean Algebra12

14 Boolean Algebra13

15 Boolean Algebra14

16 Boolean Algebra15 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.

17 Boolean Algebra16

18 Boolean Algebra17

19 Boolean Algebra18

20 Boolean Algebra19 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

21 Boolean Algebra20 For example, NOT, OR, AND gates can be realized by only Nand gates.

22 Boolean Algebra21

23 Boolean Algebra22

24 Boolean Algebra23

25 Boolean Algebra24

26 Boolean Algebra25 SIMPLIFICATION OF BOOLEAN FUNCTIONS Algebraic Method Tabular Method K-Map Method Schienman Method

27 Boolean Algebra26 ALGEBRAIC METHOD Simplify using algebraic theorems Advantage: First Method based on Boolean Algebra theorems Disadvantage: No Suitable algorithm to apply (Trial type of method)

28 Boolean Algebra27 TABULAR METHOD Also called Quine McClusky Method Advantage: It may work for any no. of variables Disadvantage: Simplification from table is quite involved

29 Boolean Algebra28 K-MAP METHOD Karnaugh Map. Also called Vietch Karnaugh Method. Advantage: Simplest and Widely accepted Disadvantage: Applicable for only upto Six variables

30 Boolean Algebra29 SCHIENMAN METHOD 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

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32 Boolean Algebra31

33 Boolean Algebra32

34 Boolean Algebra33

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38 Boolean Algebra37

39 Boolean Algebra38 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.

40 Boolean Algebra39

41 Boolean Algebra40

42 Boolean Algebra41

43 Boolean Algebra42

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45 Boolean Algebra44

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50 Boolean Algebra49

51 Boolean Algebra50

52 Boolean Algebra51

53 Boolean Algebra52

54 Boolean Algebra53

55 Boolean Algebra54

56 Boolean Algebra55

57 Boolean Algebra56

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59 Boolean Algebra58

60 Boolean Algebra59 SYMMETRIC FUNCTIONS DEFINITION PROPERTIES IDENTIFICATION

61 Boolean Algebra60 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.

62 Boolean Algebra61 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)

63 Boolean Algebra62 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’

64 Boolean Algebra63 The variables in which a function is symmetric are called the VARIABLES OF SYMMETRY

65 Boolean Algebra64 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/17/3387847/slides/slide_65.jpg", "name": "Boolean Algebra64 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

66 Boolean Algebra65 A Symmetric function is denoted by S a1,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.

67 Boolean Algebra66

68 Boolean Algebra67

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70 Boolean Algebra69

71 Boolean Algebra70 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

72 Boolean Algebra71 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 0 0 1 1column in the table is computed 0 1 0 1written under the column. This 1 0 0 1sum is referred to as a column- 1 1 1 3sum. The number of 1’s in each 2 2 2row is written in the corresp. position in column a#. This no. is called ROW SUM.

73 Boolean Algebra72 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

74 Boolean Algebra73 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 S 1,3 (x,y,z).

75 Boolean Algebra74

76 Boolean Algebra75 The column sums are 6 4 4 6 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.

77 Boolean Algebra76

78 Boolean Algebra77 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 = S 2,3 (w,x’,y’,z))

79 Boolean Algebra78 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 = S 1,2 (w’,x,y,z’)

80 Boolean Algebra79

81 Boolean Algebra80 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

82 Boolean Algebra81 The function can be expanded about w. w x y z a# Column Sums : 0 0 0 0 0x y z 0 0 1 1 21 1 2 0 1 0 1 2 The column sums can be made by complementing the columns corresponding to variables x and y.

83 Boolean Algebra82 x’ y’ z a# Column Sums: 1 1 0 2x y z 1 0 1 22 2 2 0 1 1 2 Each row is tested by the binomial coefficients for sufficient occurrence. 3!/(3-2)!2! = 3 Symmetry: S 2 (x’,y’,z)

84 Boolean Algebra83 w x y z a# 1 0 1 0 1The column sums can be made 1 1 0 0 1the same by complementing 1 1 1 1 3the columns corresponding 2 2 1to variable z. x y z’ a# 0 1 1 2Each row is tested by the 1 0 1 2binomial coefficients for 1 1 0 2sufficient occurrence. 2 2 2

85 Boolean Algebra84 3!/(3-2)! 2! = 3 Symmetry:S 2 (x,y,z’) So the function f is written as f=w’S 2 (x’,y’,z) + wS 2 (x,y,z’)

86 Boolean Algebra85 questions?


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