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Discrete Mathematics CS 2610. September 26. Equal Boolean Functions. Two Boolean functions F and G of degree n are equal iff for all (x 1 ,..x n ) B n , F (x 1 ,..x n ) = G (x 1 ,..x n )
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Discrete Mathematics CS 2610 September 26
Equal Boolean Functions • Two Boolean functions F and G of degree n are equal iff for all (x1,..xn) Bn, F(x1,..xn) = G(x1,..xn) • Example: F(x,y,z) = x(y+z), G(x,y,z) = xy + xz, and F=G (recall the “truth” table we put on the board yesterday) • Also, note the distributive property: x(y+z) = xy + xz via the distributive law
x1 x2 x3 F(x1,x2,x3) 0 0 0 1 F(x1,x2,x3) = x1(x2+x3)+x1x2x3 0 0 1 0 F(x1,x2,x3) = x1x2+x1x3+x1x2x3 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 Boolean Functions • Two Boolean expressions e1and e2 that represent the exact same function F are called equivalent
Boolean Functions • More equivalent Boolean expressions: (x + y)z = xyz + xyz + xyz (x + y)z = xz + yz distributive = x1z + 1yz identity = x(y + y)z + (x + x)yz unit = xyz + xyz + xyz + xyz distributive = xyz + xyz + xyz idempotent We’ve expanded the initial expression into its sum of products form.
Boolean Functions • More equivalent Boolean expressions: xy + z = ?? = xy1 + 11z identity = xy(z + z) + (x + x)1z unit = xyz + xyz + x1z + x1z distributive = xyz + xyz + x(y + y)z + x(y + y)z unit = xyz + xyz + xyz + xyz + xyz + xyz distributive = xyz + xyz + xyz + xyz + xyz idempotent The example from yesterday.
x y z 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 Representing Boolean Functions • How to construct a Boolean expression that represents a Boolean Function ? F F(x, y, z) = 1 if and only if: (-x)(y)(-z) + (-x)yz + x(-y)z + xyz
x y z 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 Representing Boolean Functions • How to construct a Boolean expression that represents a Boolean Function ? F F(x, y, z) = x·y·z + x·y·z + x·y·z + x·y·z
Boolean Identities • Double complement: x = x • Idempotent laws: x + x = x, x · x = x • Identity laws: x + 0 = x, x · 1 = x • Domination laws: x + 1 = 1, x · 0 = 0 • Commutative laws: x + y = y + x, x · y = y · x • Associative laws: x + (y + z) = (x + y) + z x· (y · z) = (x· y) · z • Distributive laws: x + y·z = (x + y)·(x + z) x · (y + z) = x·y + x·z • De Morgan’s laws: (x · y) = x + y, (x + y) = x · y • Absorption laws: x + x·y = x, x · (x + y) = x the Unit Property: x + x = 1 and Zero Property: x·x =0
Boolean Identities • Absorption law: • Show that x·(x + y)=x • x·(x + y)= (x + 0) ·(x + y) identity • = x + 0 ·y distributive • = x + y· 0 commutative • = x + 0 domination • = x identity
Dual Expression (related to identity pairs) • The dualed of a Boolean expression e is obtained by exchanging + with , and 0 with 1 in e. Example: e = xy + zw e = x + y + 0 ed=(x + y)(z + w) ed= x · y · 1 Duality principle:e1e2iffe1de2d x (x + y) = x iff x + xy = x (absorption)
Dual Function • The dual of a Boolean function F represented by a Boolean expression is the function represented by the dual of this expression. • The dual function of F is denoted by Fd • The dual function, denoted by Fd, does not dependon the particular Boolean expression used to represent F.
Recall: Boolean Identities • Double complement: x = x • Idempotent laws: x + x = x, x · x = x • Identity laws: x + 0 = x, x · 1 = x • Domination laws: x + 1 = 1, x · 0 = 0 • Commutative laws: x + y = y + x, x · y = y · x • Associative laws: x + (y + z) = (x + y) + z x· (y · z) = (x· y) · z • Distributive laws: x + y·z = (x + y)·(x + z) x · (y + z) = x·y + x·z • De Morgan’s laws: (x · y) = x + y, (x + y) = x · y • Absorption laws: x + x·y = x, x · (x + y) = x the Unit Property: x + x = 1 and Zero Property: x·x =0
Boolean Expressions Sets Propositions • Identity Laws x 0 = x, x 1 = x • Complement Laws x x = 1, x x = 0 • Associative Laws (x y) z = x (y z) (x y) z = x (y z) • Commutative Laws xy = yx, xy = yx • Distributive Laws x(yz) = (x y) (x z) x(yz) = (x y) (x z) A Boolean Algebra is a set B with operations ,, and elements 0 and 1 such that the above properties hold for all x, y, and z in B.
minterms x y z + x y z + x y z DNF: Disjunctive Normal Form • A literal is a Boolean variable or its complement. • A minterm of Boolean variables x1,…,xn is a Boolean product of n literals y1…yn, where yi is either the literal xior its complement xi. Example: Disjunctive Normal Form: sum of products We have seen how to develop a DNF expression for a function if we’re given the function’s “truth” table.
maxterms (x + y + z) (x + y +z) (x +y + z) CNF: Conjunctive Normal Form • A literal is a Boolean variable or its complement. • A maxterm of Boolean variables x1,…,xn is a Boolean sum of n literals y1…yn, where yi is either the literal xior its complement xi. Example: Conjuctive Normal Form: product of sums
F x y z F F = x·y·z + x·y·z + x·y·z + x·y·z 0 0 0 1 0 F = (x·y·z) · (x·y·z) · (x·y·z) · (x·y·z) 0 0 1 0 1 = (x+y+z) · (x+y+z) · (x+y+z) · (x+y+z) 0 1 0 1 0 0 1 1 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0 Conjunctive Normal Form • To find the CNF representation of a Boolean function F • Find the DNF representation of its complement F • Then complement both sides and apply DeMorgan’s laws to get F: How do you build the CNF directly from the table ?
xy = (x+y) Functional Completeness • Since every Boolean function can be expressed in terms of ,+,¯, we say that the set of operators {,+,¯} is functionally complete. • {+, ¯} is functionally complete • {, ¯} is functionally complete (note: x + y = xy) • NAND | and NOR ↓ are also functionally complete, each by itself (as a singleton set). • Recall x NAND y is true iff x or y (or both) are false and x NOR y is true iff both x and y are false.
