Definitions. 03 Logic networks Boolean algebra. Boolean set: B 0,


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1 3. Boolean algebra 3 Logic networks 3. Boolean algebra Definitions Boolean functions Properties Canonical forms Synthesis and minimization alessandro bogliolo isti information science and technology institute /2 3. Boolean algebra Definitions Boolean set: B, Boolean constants:, Boolean variable:, Boolean functions: z=f(, 2,..., n ) Operations: and : B B (, y ) y y B z y y or : B B (, y ) y +y B z y f not : B : B n B B z ' ' alessandro bogliolo isti information science and technology institute 2/2
2 3. Boolean algebra Truth table: Boolean functions Table of 2 n rows that associates a Boolean value to each configuration of n independent variables There are 2 2n different functions of n variables Boolean epression: f : B n B a b c f Epression of Boolean variables, Boolean constants and operators f a b c' ab c ( ab ) c alessandro bogliolo isti information science and technology institute 3/2 3. Boolean algebra Properties Idempotent laws Distributive laws ( y z ) ( y ) ( z ) ( y z ) ( y ) ( z ) Associative laws yz ( yz ) ( y )z y z ( y z ) ( y ) z Commutative laws y y y y Identity elements Null laws (forcing elements) Complement laws ' ' Absorption laws y ( y ) De Morgan s laws (duality principle) ( y )' ' y' ( y )' ' y' alessandro bogliolo isti information science and technology institute 4/2
3 3. Boolean algebra Idempotent laws Idempotent laws By perfect induction: y y * * * y +y alessandro bogliolo isti information science and technology institute 5/2 3. Boolean algebra Absorption laws Absorption laws y ( y ) By Boolean manipulation: y ( y ) ( y ) y y alessandro bogliolo isti information science and technology institute 6/2
4 3. Boolean algebra De Morgan s laws De Morgan s laws ( y )' ' y' ( y )' ' y' By perfect induction: =, y= () = = + =+= =, y= () = = + =+= =, y= () = = + =+= =, y= () = = + =+= =, y= (+) = = == =, y= (+) = = == =, y= (+) = = == =, y= (+) = = == alessandro bogliolo isti information science and technology institute 7/2 3. Boolean algebra Canonical forms There are infinite equivalent Boolean epressions. The equivalence (i.e., identity) between two epressions can be demonstrated:. By perfect induction 2. By Boolean manipulation Canonical forms associate unique epressions to each function Checking the equivalence between two functions reduces to a comparison of their canonical representations alessandro bogliolo isti information science and technology institute 8/2
5 3. Boolean algebra Canonical forms (Sum of Products) Literal: independent variable taken either in true or complemented form (e.g.,, ) Minterm: Product of all independent variables taken either in true or complemented form A minterm represents a Boolean function that takes value corresponding to a unique configuration of input variables (e.g., f(a,b,c)=ab c takes vale for abc=) A Boolean function that takes value for M different configurations (that has M s in the truth table) can be epressed as the sum of the M minterms associated with the M s Any Boolean function can be epressed as a sum of products A sum of minterms, with fied variable order, is a canonical form alessandro bogliolo isti information science and technology institute 9/2 3. Boolean algebra From truth tables to SoPs a b c f = f = a b c + a bc + ab c + abc + abc alessandro bogliolo isti information science and technology institute /2
6 3. Boolean algebra Boolean minimization Given a Boolean function, find a Boolean epression that represents the function using a minimum number of literals f = a b c + a bc + ab c + abc + abc f = a c + ab c + abc + abc f = a c + ab c + ab 5 literals literals 7 literals In general, this is not an easy task There is a closedform solution for 2level SoP epressions There is no closedform solution for general multilevel epressions. Heuristic solutions found by Boolean manipulation alessandro bogliolo isti information science and technology institute /2 3. Boolean algebra Boolean minimization (eample) f = a b c + a bc + ab c + abc + abc 5 literals (SoP) = a c (b + b) + ab c + ab (c + c) literals (distributive) = a c + ab c + ab 7 literals (complement) = a c + a (b c + b) 6 literals (distributive) = a c + a (b c + b c + b) 8 literals (absorption) = a c + a ((b +b) c + b ) 7 literals (distributive) = a c + a (c + b) 5 literals (complement) = a c + a c + ab 6 literals (distributive) = (a + a) c + ab 5 literals (distributive) = c + ab 3 literals (complement) Remark: the number of literals doesn t decrease at every step. This makes the process non trivial alessandro bogliolo isti information science and technology institute 2/2
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