Boolean Algebra. Sungho Kang. Yonsei University

Transcription

1 Boolean Algebra Sungho Kang Yonsei University

2 Outline Set, Relations, and Functions Partial Orders Boolean Functions Don t Care Conditions Incomplete Specifications 2

3 Set Notation $09,3/#0,9 438 v V Element is a member of set v V v V S V S V Element is not a member of set v Cardinality (number of members) of set V Set is a subset of S The empty set (a member of all sets) V The complement of set S U The universe: S = U S U V S V 3

4 Set Notation $09,3/#0,9 438 Inclusion ( ) Proper Inclusion ( ) Complementation Intersection ( ) Union ( ) Difference 4

5 Power Sets $09,3/#0, V = { S S V } 2 V The power set of set V (the set of all subsets of set V V V 2 = 2 The cardinality of a power set is a power of 2 5

6 Power Sets $09,3/#0,9 438 V = {0,1,2} 3-member set 2 V = {, {},{},{}, { 01, },{ 02, },{ 12, }, {,,} 012 } 1 subset with 0 members 3 subsets with 1 members 3 subsets with 2 members 1 subset with 3 members V V 3 2 = 2 = 2 = 8 Power sets are Boolean Algebras 6

7 Cartesian Products $09,3/#0,9 438 The Cartesian Product of sets A and B is denoted A x B Suppose A = { 012,,}, B = { ab, }, then A B ={( 0, a),( 0, b),( 1, a),( 1, b),( 2, a),( 2, b)} A = {,,} 012 Set A is unordered (, 1 b) ( ) denotes Ordered Set 7

8 Binary Relations $09,3/#0,9 438 The Cartesian Product of sets A and B is denoted A x B A x B consists of all possible ordered pairs (a,b) such that a A and b B A subset R A Bis called a Binary Relation Graphs, Matrices, and Boolean Algebras can be viewed as binary relations 8

9 Binary Relations as Graphs or Matrices $09,3/#0,9 438 E = { ab, cb, cd, cf} A B A = {,}B ac = { bd,, f} a c b d f a b c d f Bipartite Graph Rectangular Matrix 9

10 Binary Relations as Graphs or Matrices $09,3/#0,9 438 E = { ab, ac, ad, bd, cd} { aa, bb, cc, dd} V V V = {, a b,, c d} b a d c a b c d Directed Graph Square Matrix 10

11 Properties of Binary Relations $09,3/#0,9 438 R V V A binary relation can be reflexive, and/or transitive, and/or symmetric, and/or antisymmetric We illustrate these properties on the next few slides 11

12 Notation for Binary Relations $09,3/#0,9 438 If R A B, we say that A is the domain of the relation R, and that B is the range. If ( ab, ) R, we say that the pair is in the relation R, or arb. 12

13 Missing Figure 3.2, Page $09,3/#0,9 438 = {, 1234,, } A Graph View 13

14 Example: Less than or Equal $09,3/#0,9 438 A = B = {,,, 012}, R A B= " "

15 Example: a times b = 12 $09,3/#0,9 438 V = {,,, 012f}, R V V = {( u, v) u v= 12} R = {( 112, ),( 2, 6),( 3, 4),( 4, 3),( 6, 2),( 121, )}

16 Reflexive Binary Relations V $09,3/#0,9 438 = {, a b,, c d} R V V v V vrv b a c d A binary relation R V V is reflexive if and only if (,) vv R for every vertex v V a b c d

17 Non-Reflexive Binary Relations V $09,3/#0,9 438 = {, a b,, c d} R V V a v V vrv b c d Non-Reflexivity implies that there exists v V such that (,) vv R. Here d is such a v. a b c d

18 Transitive Binary Relations $09,3/#0,9 438 If uvw,, V, and urv, and vrw, then urw. b a d c e A binary relation is transitive if and only if every (u,v,w) path is triangulated by a direct (u,w) edge. This is the case here, so R is transitive. 18

19 Non-Transitive Binary Relations $09,3/#0,9 438 R V V uvw,, V, such that urv, vrw, but urw b a d c e A binary relation is not transitive if there exists a path from u to w,, through v that is not triangulated by a direct (u,w) edge Here (a,c,e) is such a path. Edge (a,e) is missing 19

20 Symmetric Binary Relations $09,3/#0,9 438 A binary relation is symmetric if and only if every ( uv, ) edge is reciprocated by a (, vu) edge R V V (,) uv R (,) vu R ( urv vru)

21 Non-Symmetric Binary Relations $09,3/#0,9 438 R V V (,) uv Rsuch that (,) vu R A binary relation is non-symmetric if there exists an edge ( uv, ) not reciprocated by an edge (, vu) b a d Edge (c,a) is missing c e 21

22 Antisymmetric Binary Relations $09,3/#0,9 438 ( u, v) R, ( urv, vru) ( v = u) A binary relation is antisymmetric if and only if no ( uv, ) edge is reciprocated by a (, vu) edge v = u b a d c e Not antisymmetric if any such edge is reciprocated : here b d, d b 22

23 Functions $09,3/#0,9 438 A function f are binary relations from set A (called the domain) to B (called the range) But, it is required that each a in A be associated with exactly 1 b in B For functions, it cannot be true that both (a,b) in R and (a,c) in R, b different from c 23

24 The Binary Relation $09,3/#0,9 438 The Binary Relation of Relations to Synthesis/Verification Partial Orders Equivalence Relations Compatibility Relations *DC=don t care Combinational Logic (no DCs) => (0,1) Boolean Algebra Combinational Logic (with DCs) => Big Boolean Algebras Sequential Logic (no DCs) Sequential Logic (with DCs) 24

25 The Path Relation * $09,3/#0,9 438 b G a d =( V, E) This graph defines path relation c e a b c d e * * * * * {,,, abcd,} e {, cd,} e {,} ce {,} ce {,} ce Relation * is sometimes called Reachability 25

26 Equivalence Relations $09,3/#0,9 438 b a d c e Note R is reflexive, symmetric, and transitive 26

27 Equivalence Relations b G a d =( V, E) This graph defines path relation * c e NOT an equivalence relation: E b a d $09,3/#0,9 438 R= {( uv, ) v u u v *, * } G= ( V, R) c e This graph defines R An equivalence relati on: R 27

28 The Cycle Relation b G a d =( V, E) This graph defines path relation * c e C= {( uv, ) v u} * G= ( V, R) a c b d e $09,3/#0,9 438 These are called the Strongly Connected Components of G 28

29 Refinement $09,3/#0,9 438 Given any set B a partition of B is a set of subsets B i B with two properties Bi B j = φ for all i j i B i = B Given two partitions P 1 and P 2 of a set S, P 1 is a refinement of P 2 if each block B 1 i of P 1 is a subset of some block of π2 29

30 Other Binary Relations $09,3/#0,9 438 Partial Orders (Includes Lattices, Boolean Algebras) Reflexive Transitive Antisymmetric Compatibility Relations Reflexive Not Transitive--Almost an equivalence relation Symmetric 30

31 Functions $09,3/#0,9 438 A function f from A to B written f : A B is a rule that associates exactly one element of B to each element of A A relation from A to B is a function if it is right-unique and if every element of A appears in one pair of the relation A is called the domain of the function B is called the co-domain (range) If y=f(x) : A B, y is image of x Given a domain subset C A IMG(f,C) = { y B x C y = f(x) } preimage of C under f PRE(f,C) = { x A y C y = f(x) } A function f is one-to-one (injective) if x y implies f(x) f(y) A function f is onto (surjective) if for every y B, there exists an element x A, such that f(x) = y 31

32 Algebraic Systems!,79, 7/078 ( V, ) The pair is called an algebraic system V is a set, called the carrier of the system is a relation on V V = V 2 (, are similar to ) This algebraic system is called a partially ordered set, or poset for short 32

33 Partially Ordered Sets!,79, 7/078 A poset has two operations, and +, called meet and join (like AND and OR) Sometimes written or ( V,, + ) ( V, +,, ) ( V 2, +,, ) even, since is implied 33

34 Posets and Hasse Diagrams!,79, 7/078 Integers (Totally Ordered): Missing Figure = {, 1234,, } A Matrix View Graph View Hasse Diagram Hasse Diagram obtained deleting arrowheads and redundant edges 34

35 Posets and Hasse Diagrams!,79, 7/078 ( V, +,, ) V = = { a, b, c, d} {( aa, ),( ab, ),( ac, ),( ad, ), ( bb, ),( bd, ), ( cc, ),( cd, ),( dd, )} :distance from top is refelxive, antisymmetric, and transitive: a partial order b a d b c, c b c \ \ Relation Hasse Diagram b d c a 35

36 Meet!,79, 7/078 An element m of a poset P is a lower bound of elements a and b of P, if and. m a m b m is the greatest lower bound or meet of elements a and b if m is a lower bound of a and b and, for any m such that and also,. m a m b m m a m m b 36

37 Join: the Dual of Meet!,79, 7/078 m P b a m b m An element of a poset is a upper bound of elements and of P, if and. a m is the least upper bound or join of elements and if m is an upper bound of a and b and, for any m such that and also,. a m b m m m m m a b a b 37

38 Meet ( ) and Join (+)!,79, 7/078 posets: d d d b b c b c b d a a a a c ab = a cb = a cb = a c a=? a+ b= b c+ b=? c+ b= d d+ b=? 38

39 POSET Properties!,79, 7/078 Theorem If x and y have a greatest lower bound (meet), then x x y x j y Similarly, if x and y have a least upper bound (join), then m x x + y Proof: Since meet exists, x x y by definition. Also, since join exists, by definition. x x+ y 39

40 More POSET Properties!,79, 7/078 Theorem x y x y = x Proof ( ): This means assume x y. x is a lower bound of x and y (by Def.) x y m x is also the meet of x and y Proof: by contradiction. Suppose x x y. Then m x such that x m where m= x y. But since m was a lower bound of x and y, m x as well. Thus m = x, by the anti-symmetry of posets. Proof ( ): From Definition of meet 40

41 Well-Ordered!,79, 7/078 If all pairs of elements of a poset are comparable, then the set is totally ordered If every non-empty subset of a totally ordered set has a smallest element, then the set is well-ordered e.g.) Natural numbers Mathematical Induction Given, for all n N, propositions P(n), if P(0) is true for all n>0, if P(n-1) is true then P(n) is true then, for all n N, P(n) is true 41

42 Lattices!,79, 7/078 Lattice: a poset with both meet and join for every pair of elements of the carrier set Boolean Algebra: a distributed and complemented lattice Every lattice has a unique minimum element and a unique maximum element 42

43 Lattices and Not Lattices!,79, 7/078 ab = a cb = a a+ b= b c (lattice) d d d b b c b c b d a a a a c + b=? cb = a c a=? c+ b= d d+ b=? (lattice) (Boolean algebra) 43

44 Examples of Lattices!,79, 7/078 d = 1 b c a = 0 g = 1 e f d b c a = 0 e = 1 b c d a = 0 44

45 More Notation!,79, 7/078 There Exists a vin set V is denoted by v V The following are equivalent: a b ( ab + a b ) a b and a b ( a + b)( a+ b )) a is true if and only if b is true Does ( a + b)( a+ b ) make sense in a poset? No--Complement is defined for lattices but not for posets 45

46 Properties of Lattices!,79, 7/078 Meet, Join, Unique maximum (1), mi nimum (0) element are always defin ed Idempotent: Commutative: Associative: Absorptive: x + x = x x x = x x+ y= y+ x x y= y x x+ ( y+ z) = ( x+ y) + z x ( y z) = ( x y) z x ( x+ y) = x x+ ( x y) = x Absorptive properties are fundamental to op timization 46

47 Duality!,79, 7/078 Every lattice identity is transformed into another identity by interchanging: + and and 0and 1 Example: x ( x+ y) = x x+ ( x y) = x 47

48 Complementation If x+y=1 and xy=0 then x is the complement of y and vice versa A lattice is complemented if all elements have a complement 48

49 Examples of Lattices!,79, 7/078 Complemented? yes no yes d = 1 b c a = 0 bc = a= 0 b+ c= d = 1 g = 1 e f d b c a = 0 bc = a= 0 b+ c= d 1 e = 1 b c d a = 0 bc = a= 0 b+ c= e= 1 49

50 Distributivity!,79, 7/078 Semi-distributivity: x ( y+ z) ( x y) + ( x z) x+ ( y z) ( x+ y) ( x+ z) Proof : 1. x y x (def. of meet) 2. x y y y+ z (def. of meet, join) 3. x y x ( y+ z) (def. of meet) 4. x z x ( y+ z) (mutatis mutandis: y z) 5. x ( y+ z) ( x y) + ( x z) (def. of join) 50

51 Distributivity!,79, 7/078 Boolean Algebras have full distributivity: x ( y + z) = ( x y) + ( x z) x+ ( y z) = ( x+ y) ( x+ z) Boolean Algebras are complemented. That is, x = y ( x y = 0) and ( x + y = 1) must hold for every x in the carrier of the poset 51

52 Definition of Boolean Algebra!,79, 7/078 A complemented, distributive lattice is a Boolean lattice or Boolean algebra Idempotent x+x=x xx=x Commutative x+y=y+x xy = yx Associative x+(y+z)=(x+y)+z x(yz) = (xy)z Absorptive x(x+y)=x x+(xy) = x Distributive x+(yz)=(x+y)(x+z) x(y+z) = xy +xz Existence of the complement 52

53 Are these Lattices Boolean Algebras?!,79, 7/078 Complemented and distributed? yes no no d = 1 g = 1 e = 1 e f b c d b c d b c a = 0 a = 0 a = 0 bc = a= 0 b ( c+ d) = b b+ c= d= 1 bc = a= 0 ab ( + c) = ( bc ) + ( bd ) = a ab + = ac a= 0 b+ c= d 1 53

54 Fundamental Theorem of Boolean Algebras!,79, 7/078 Every poset which is a Boolean Algebra has a power of 2 elements in its carrier All Boolean Algebras are isomorphic to the p ower set of the carrier. Example: V V = {}, 1 2 = {,} 1 = { 01, } 1 ( = 1) 0 ( = ) 54

55 Examples of Boolean Algebras!,79, 7/078 V {}, 1 2 {,} 1 { 01,} = V = = 1 ( = 1) 0 ( = ) 1-cube V = {,}, 01 2V = {, 01,,{ 01,}} = {,,,} ( = { 01, } = V) ( = ) 2-cube 55

56 Atoms of a Boolean Algebra!,79, 7/078 A Boolean Algebra is a Distributive, Comple mented Lattice The minimal non-zero elements of a Boolean Algebra are called atoms 1 1-atom V n = 2 n atoms 0 0 Can a Boolean Algebra have 0 atoms? atoms NO! 56

57 Examples of Boolean Algebras!,79, 7/078 A V = {, abc,} = 2 A = { A, { ab, },{ ac, },{ bc, }, abc,,, 3 atoms } 1 ( = V) {,} ab {,} ac {,} bc {} a {} b {} c 0 ( = ) 57

58 Examples of Boolean Algebras!,79, 7/078 n= A = 4 atoms, n+ 1= 5levels, 2 = 16 elts n A V Level 4--1 elt* Level 3--4 elts Level 2--6 elts Level 1--4 elts Level 0--1 elts = {, abcd,, } = 2 A = { A, {, abc,},{, abd, },{,, acd},{,, bcd}, { ab, },{ ac, },{ ad, },{ bc, },{ bd, },{ cd, } { a},{ b},{ c},{ d} } * elt = element C 4 4 C 4 3 C 4 2 C 1 4 C

59 Boolean Algebras!,79, 7/078 Theorem Complementation is unique. proof: Suppose x and y are both complements of ( ). Hence x x + y = 1, xy = 0 y = y( x+ x ) = x y + xy = x y = xy + xx = x ( y+ x) = x Note we used distributivity. Similarly, we have Theorem (Involution): ( x = ) x 59

60 Properties of Boolean Algebra!,79, 7/078 x + x y = x+y x(x +y) = xy 60

61 Lemma: The Isotone Property!,79, 7/078 x y xz yz Proof: By Theorem 3.2.2, x y x = xy, so we get xz = xyz = xyzz =( xz)( yz). Note we used i dempotence, commutativity, and associativity. Then we use Theorem again to prove the lemma. y x xz yz xz yz z 61

62 Duality for Boolean Algebras!,79, 7/078 Every Boolean Algebra identity is transformed into another valid identity by interchanging: + and and 0 and 1 and () () This rule not valid for lattices Example: xx = 0 x + x = 1 62

63 Theorem!,79, 7/078 x y xy = x + y= 1 0 y x Proof: By the isotone property we have x y xy yy xy 0 xy = 0 The second identity follows by duality. 63

64 Other Theorems!,79, 7/078 DeMorgan s Laws Consensus ( x + y) = x y ( xy) = x + y xy + x z + yz = xy + x z ( x+ y)( x + z)( y+ z) = ( x+ y)( x + z) 64

65 Consensus Examples!,79, 7/ a bc + abd + bcd = abc + a bd abe + bce + bde + ac' d = abe + be( c + d) + a( c + d) = be( c + d) + a( c + d) Note use of DeMorgan s Law = bce + bde + ac' d 65

66 Consensus Example!,79, 7/ Is wrong. Replace by a c d+ b c d+ acd+ bcd = ( abd ) + acd + bcd + acd+ bcd = abd + bcd + acd Uphill Move: Note addition of redundant consensus term enables deletion of two other terms by consensus This avoids local minima--a crucial part of the logic minimization paradigm 66

67 Functions and Boolean Functions 44 0,3: Ordinary functions of 1 variable: f ( x): D R f D R Ordinary functions of 2 real variables: f ( xy, ): D D R f ( D D ) R x y x y 67

68 Functions and Boolean Functions 44 0,3: Boolean functions of n variables: a b f ( x,, x ): B B, B= { 0,, 1} 1 n n f ( x,, x ) ( B B) B 1 n 0 68

69 Boolean Formulae 44 0,3: Boolean Formulae: Meets and Joins of Variables and Constants F 1 = 0 F = x x + x x F = xxx + xx F = ax + b??? 3 1 a 1 0 b 69

70 n-variable Boolean Formulae 44 0,3: B = { 0, l, 1} The elements of B are Boolean formulae. Each variable x 1, l, x n is a Boolean formula. If g and h are Boolean formulae, then so are g + h g h g A string is a Boolean formula if and only if it derives from finitely many applications of these rules. 70

71 Distinct Formulas, Same Function 44 0,3: B = {, 0 a, b,} 1 x x F = x + x G = x + x x Truth Table for f: B 2 B 0000 aaaa bbbb ab1 0ab1 0ab1 0ab1 0ab1 aa11 b1b ab1 aa11 b1b

72 Boolean Functions on Boolean Algebra 44 0,3: B = { 0, l, 1} f ( x1, l, xn) = xi is a Boolean function f( x1, l, xn ) = e B is also If g and h are Boolean, functions then so are g+ h g h g A function is Boolean if and only if it derives from finitely many applications of these rules. 72

73 Cofactors 44 0,3: Positive Cofactor WRT x 1 : Negative Cofactor WRT x 1 : f = f (, 1 x,, x ) x 1 f = f( 0, x,, x ) x 1 Note: prime denotes complement 2 2 n n f = f (, 1 x,, x ) = f x 2 n x 1 1 f = f ( 0, x,, x ) = f x 2 n x = = 1 73

74 Cofactors 44 0,3: Positive Cofactor WRT x 1 : Negative Cofactor WRT x 1 : Example: f = abc d + a cd + bc f = bc d + bc a f = + cd + bc a f = f (, 1 x,, x ) x 1 f = f( 0, x,, x ) x This term drops out when a is replaced by 1 This term is unaffected n n 74

75 Cofactors 44 0,3: B= {, 0 a, b,}, 1 f( x): B B f = f( a, x,, x ) x = a 2 n f = f(, 1 x,, x ) x 1 1 Example: 2 f = ax + bx f = aa + bx = 0 + bx = bx x 1 = a 1 2 n n

76 Boole s Expansion Theorem 44 0,3: f = ax + bx f = a0+ bx = bx, f = a1+ bx = a + bx x Example: f ( x1, x2,, xn ) = x f + x f = [ x + f ] [ x + f ] i x i x i x i x i i i i x Sum form: f = x1( bx2) + x1 ( a + bx2) Product form: f = [ x + ( a+ bx )] [ x + ( bx )],

77 Variables and Constants 44 0,3: In the previous slide a and b were constants, and the x 1, l, x n were variables. But we can also use letters like a and b as variables, without explicitly stating what the elements of the Bo olean Algebra are. 77

78 Boole s Expansion (Sum Form) 44 0,3: f = abc d + a cd + bc f = bc d + bc a f = + cd + bc a f = afa + a fa = a( bc d + bc) + a ( cd + bc) 78

79 Boole s Expansion (Product Form) 44 0,3: f = [ a+ f ][ a + f ] a f = abc d + a cd + bc = [ a+ ( cd + bc)][ a + ( bc d + bc)] = aa + a( bc d + bc) + a ( cd + bc) + ( cd + bc)( bc d + bc) Note application = abc d + a cd + bcd + bc = abc d + a cd + bc a of absorptive law 79

80 Minterm Canonical Form 44 0,3: The minterm canonical form is a canonical, or st andard way of representing functions. From p10 0, f = x + y + is z represented by millions of distinct Boolean formulas, but just 1 minterm canonical form. Note f = x+ y + z= ( x yz ) Thus some texts refer to f as {0,1,3,4,5,6,7} f = x + y z x + y z x yz+ xy z+ ( 0, 1, 3, 4, ) 80

81 Minterm Canonical Form 44 0,3: f ( x1, x2, l, xn) xf + xf 1 x 1 x 1 1 xx f + xx f + xx f + xx f 1 2 xx 1 2 xx 1 2 xx 1 2 xx r x mx f + x mx x f + m 1 n x mx 1 n 1 n x mx x 1 n 1 n 1 n + xx mx f + xmx f 1 2 n x x mx 1 n x mx 1 2 n 1 n levels of recursive cofactoring create constants 2 n These elementary functions are called minterms n 81

82 Minterm Canonical Form 44 0,3: Thus a Boolean function is uniquely determined by its values at the corner points 0m00, m01, l, 1m1 f ( x1, x2, l, xn) = xf + xf 1 x 1 x 1 1 xx f + xx f + xx f + xx f 1 2 xx 1 2 xx 1 2 xx 1 2 xx r = x mx f + x mx x f + m 1 n x mx 1 n 1 n x mx x 1 n 1 n 1 n + xx mx f + xmx f 1 2 n x x mx 1 n x mx These 2 n constants are aptly called discrimants 1 2 n 1 n 82

83 Minterm Canonical Form 44 0,3: f = ax + bx, f = a+ bx, f = bx 1 2 x 2 x = xxa + xx ( a+ b) + xx 0 + xxb x x2 F = ax1 + bx2 xxa + xx + xxb f: B 2 v B B= {, 0 a, b,} This function is Boolean (from Boolean Formula) aaaa bbbb ab1 0ab1 0ab1 0ab1 aa11 00bb aa11 00bb aa11 00bb aa11 00bb Thus all 16 cofactors match--not just discriminants 83

84 Now Suppose Truth Table is Given 44 0,3: Here we change 15th cofactor, but leave the 4 discriminants unchanged Since the given function doesn t match at all 16 cofactors, is not Boolean F x x2 F xxa + xx + xxb aaaa bbbb ab1 0ab1 0ab1 0ab1 aa11 00bb aa11 001b aa11 00bb aa11 00bb 84

85 2-Variable Boolean Functions 44 0,3: Minterms 2-variable functions 15 4 f = f = x + y, x + y, x+ y, x+ y f = x, y, xyxy,, + xyxy, + xy f = x y, xy, x y, xy 0 0 f = 0 Note that the minterms just depend on the number and names of the variables, independent of the particular Boolean Algebra 85

86 Boolean Function Algebras 44 0,3: The notation Fn(B) means the Boolean Algebra whose ca rrier is the set of all n-variable Boolean Functions which map B into Fn(B): Bn -> B minterms n B = { 01, } If the atoms of Fn(B) are its n-variable B is called the Base Algebra of the Boolean Function al gebra The atoms of B are called Base Atoms F ( B): B B n 86

87 Large Boolean Algebras? 44 0,3: When you design an optimal circuit, each gate must be optimized with respect to its Don t Cares Because of Don t Cares, 4 functions of ( xy, ) are equivalence preserving replacements for gate g Optimal Design: pick best such replacement 87

88 Boolean Algebra F1({ 0, ab,, 1}) Level (= # of atoms) ,3: atoms of F1{ 0, ab,, 1}) are base atoms minterms Each element has 3 Atoms Each element has 2 Atoms 1 Each element has 1 Atom Note: base atoms act like 2 extra literals a y, a = b y 88

89 Boolean Algebra F ({ 0, 1}) 2 Level (= # of atoms) ,3: poset: { f,, f } i j f f i j minterms(f ) minterms(f ) 3 minterms (3 atoms) 2 minterms (2 atoms) 1 minterm (1 atom) 89

90 Boolean Algebra 44 0,3: F1({ 0, ab,, 1}) Boolean Algebra of 1-Variable Functions atoms of F1{ 0, ab,,}) 1 base atoms minterms 3 Atoms are 2 Atoms 1 Atom Subalgebras are incompletely specified Boolean Functions (most important) 90

91 Boolean Subalgebra 44 0,3: Boolean Subalgebra: Interval of 2-Variable Function Lattice The 3-Atom element x + y = xy + xy + x y is ONE of subalgebra 2-Atom elements are atoms of subalgebra 1-Atom element xy is ZERO of subalgebra 91

92 More On Counting 44 0,3: Level (= # of atoms) 4 3 An algebra (or subalgebra) with n + 1 levels has exactly 2 n elements, because n + 1 levels implies n atoms

93 A larger Subalgebra 44 0,3: Interval of 2-Variable Function Lattice ONE of algebra (4-Atoms) is also one of this subalgebra 2-Atom elements are atoms of subalgebra 1-Atom element is ZERO of subalgebra 93

94 Counting Elements of Fn ( B) 44 0,3: For each of the 2 n minterms, the discriminant can be chosen as any of the B elements of B. Therefore, the number of elements of is Examples ( 2 ) A( B) ( 2 ) ( A( B) 2 ) = ( 2 ) = 2 B = { 01, }, A( B) = 1 B n n n n n ( 2 ) 2 4 = 2 B = 2 = 2 = 16 2 F n ( B) n n ( 2 ) 2 8 = 3 B = 2 = 2 =

95 Counting Elements of Fn(B) 44 0,3: Examples with B = {, 0 a,,}, b1 A( B) = {,} a b n n ( A( B) 2 ) = 2 2 = 2 = 2 3 n Examples with ( A( B) 2 ) = 3 2 = 2 = 2 Note 2(4) base atoms act like 1(2) extra variables n A( B ) = 4 4 n n n ( A( B) 2 ) = 2 2 = 2 = 2 n ( A( B) 2 ) = 3 2 = 2 =

96 Boolean Difference (Sensitivity) 44 0,3: A Boolean function f depends on a if and only if f f. Thus a a f / a = fa fa is called the Boolean Difference, or Sensitivity of f with respect to a 96

97 Boolean Difference (Sensitivity) 44 0,3: Example: f = abc + a bc f / a = f f = ( bc) ( bc) = a a f / b= f f = ( ac) ( a c) = c b b Note the formula depends on a, but the implied function does not 0 97

98 Don t Care 43 9,708 An interval [L, U] in a Boolean algebra B is the subset of B defined by [L,U] = { x B : L x U } Satisfiability don t cares Observability don t cares 98

99 Intervals and Don t Cares $50. 1.,9 438 D = D Obs + D Sat = xy L = g D = gd = ( xy + xy )( x+ y) = ( xy + xy ) = g U = g + D = ( x y + xy ) + x' y' = x' + y' 99

100 Don t Cares 43 9,708 Input Filter Output Filter xy if and only if x y f = x, f = y z/ w= 0 for all possible xy. D Obs xy if and only if local input xy never occurs D Sat The complete don t care set for gate g is g Sat Obs D = D + D 100

101 Don t Cares Local inputs 43 9,708 x = u + v, y= v ( y= 0) ( x = 1) For this circuit, local input combinations x y ( x = 0, y = 0) do not occur. That is, the local minterm xy is don t care. 101

102 Don t Cares 43 9,708 z= w + t z/ w= z z = t 1 = t For this circuit, global input combination 10 sets z t = uv = 1 which makes insensitive to w. However, local input pair 10 ( xy ) is NOT don t care, since uv also gives xy, and in this case. t u v = 1 w w 102

103 Computing ALL Don t Cares 43 9,708 u v x y w t z z/ w xy D Sat if and only f x (,)= u v x and f y (,)= u v y does not occur for any row uv, in the truth table. Here, D Sat = x y ( 00 does not occur) 103

104 Computing ALL Don t Cares 43 9,708 Note these differ! u v x y w t z z/ w Similarly xy if and only for every row uv, such that f x (,)= u v x and f y (,)= u v y, z/ w= zw zw = 0 Here 10 ( xy ) is NOT don t care since z/ w= 1 in the first row. D Obs 104

105 Computing ALL Don t Cares 43 9,708 Note these differ! u v x y w t z z/ w g Sat Obs D = D + D = x y g = xy + xy ( + xy ) x + y Thus the exclusive OR gate can be replaced by a NAND 105

106 Don t Cares and Function Subalgebras 43 9,708 Suppose we are given a Boolean Function f and a don t care set D. Then any function in the interval (subalgebra) [ fl, fu ] = [ fd, f + D] is an acceptable replacement for f in the environment that produced D. Here fd is the 0 of the subalgebra and f + D is the

107 Incompletely Specified Functions $50. 1.,9 438 Suppose we are given a Boolean Function g and a don t care set D. Then the triple ( f, dr, ) where f = gd, d = D, and r = ( f + D) is called an Incompletely specified function. Note f + d+ r = gd + D+ ( g + D) =