24 Nov Boolean Operations. Boolean Algebra. Boolean Functions and Expressions. Boolean Functions and Expressions


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1 24 Nov 25 Boolean Algebra Boolean algebra provides the operations and the rules for working with the set {, }. These are the rules that underlie electronic circuits, and the methods we will discuss are fundamental to VLSI design. We are going to focus on three operations: Boolean complementation, Boolean sum, and Boolean product Boolean Operations The complement is denoted b a bar (on the slides, we will use a minus sign). It is defined b  = and  =. The Boolean sum, denoted b + or b OR, has the following values: + =, + =, + =, + = The Boolean product, denoted b or b AND, has the following values: =, =, =, = 24 Nov 25 CS Nov 25 CS 32 2 Definition: Let B = {, }. The variable is called a Boolean variable if it assumes values onl from B. A function from B n, the set {(, 2,, n ) i B, i n}, to B is called a Boolean function of degree n. Boolean functions can be represented using epressions made up from the variables and Boolean operations. The Boolean epressions in the variables, 2,, n are defined recursivel as follows:,,, 2,, n are Boolean epressions. If E and E 2 are Boolean epressions, then (E ), (E E 2 ), and (E + E 2 ) are Boolean epressions. Each Boolean epression represents a Boolean function. The values of this function are obtained b substituting and for the variables in the epression. 24 Nov 25 CS Nov 25 CS 32 4 For eample, we can create Boolean epression in the variables,, and z using the building blocks,,,, and z, and the construction rules: Since and are Boolean epressions, so is. Since z is a Boolean epression, so is (z). Since and (z) are epressions, so is + (z). and so on Eample: Give a Boolean epression for the Boolean function F(, ) as defined b the following table: F(, ) Possible solution: F(, ) = () 24 Nov 25 CS Nov 25 CS 32 6
2 24 Nov 25 There is a simple method for deriving a Boolean epression for a function that is defined b a table. This method is based on minterms. Definition: A literal is a Boolean variable or its complement. A minterm of the Boolean variables, 2,, n is a Boolean product 2 n, where i = i or i =  i. Hence, a minterm is a product of n literals, with one literal for each variable. Definition: The Boolean functions F and G of n variables are equal if and onl if F(b, b 2,, b n ) = G(b, b 2,, b n ) whenever b, b 2,, b n belong to B. Two different Boolean epressions that represent the same function are called equivalent. For eample, the Boolean epressions, +, and are equivalent. 24 Nov 25 CS Nov 25 CS 32 8 The complement of the Boolean function F is the function F, where F(b, b 2,, b n ) = (F(b, b 2,, b n )). Let F and G be Boolean functions of degree n. The Boolean sum F+G and Boolean product FG are then defined b (F + G)(b, b 2,, b n ) = F(b, b 2,, b n ) + G(b, b 2,, b n ) (FG)(b, b 2,, b n ) = F(b, b 2,, b n ) G(b, b 2,, b n ) Question: How man different Boolean functions of degree 2 are there? Solution: There are 6 of them, F, F 2,, F 6 : F F 2 F 3 F 4 F 5 F 6 F 7 F 8 F 9 F F F 2 F 3 F 4 F 5 F 6 24 Nov 25 CS Nov 25 CS 32 Question: How man different Boolean functions of degree n are there? Solution: There are 2 n different ntuples of s and s. A Boolean function is an assignment of or to each of these 2 n different ntuples. Therefore, there are 2 2n different Boolean functions. Boolean Identities There are useful identities of Boolean epressions that can help us to transform an epression A into an equivalent epression B (see Table 5 on page 85 [6 th edition: page 753] in the tetbook). 24 Nov 25 CS Nov 25 CS
3 24 Nov =, law of double complement + =, idempotent laws = + =, identit laws = + =, domination laws = + = +, commutative laws = 24 Nov 25 CS (+z) = (+)+z, associative laws ( z) = ( ) z +z = (+)(+z), distributive laws (+z) = ( )+( z) () =  + , De Morgan s laws (+) = ()() + =, Absorption laws (+) = + =, unit propert () =, zero propert 24 Nov 25 CS 32 4 Dualit We can derive additional identities with the help of the dual of a Boolean epression. The dual of a Boolean epression is obtained b interchanging Boolean sums and Boolean products and interchanging s and s. 24 Nov 25 CS 32 5 Dualit Eamples: The dual of ( + z) is + z. The dual of  + ( + z) is ( + )(()z). The dual is essentiall the complement, but with an variable replaced b . (eercise 29, p. 88) The dual of a Boolean function F represented b a Boolean epression is the function represented b the dual of this epression. This dual function, denoted b F d, does not depend on the particular Boolean epression used to represent F. (eercise 3, page 88 [6 th ed. p.756]) 24 Nov 25 CS 32 6 Dualit Therefore, an identit between functions represented b Boolean epressions remains valid when the duals of both sides of the identit are taken. We can use this fact, called the dualit principle, to derive new identities. For eample, consider the absorption law ( + ) =. B taking the duals of both sides of this identit, we obtain the equation + =, which is also an identit (and also called an absorption law). Definition of a Boolean Algebra All the properties of Boolean functions and epressions that we have discovered also appl to other mathematical structures such as propositions and sets and the operations defined on them. If we can show that a particular structure is a Boolean algebra, then we know that all results established about Boolean algebras appl to this structure. For this purpose, we need an abstract definition of a Boolean algebra. 24 Nov 25 CS Nov 25 CS
4 24 Nov 25 Definition of a Boolean Algebra Definition: A Boolean algebra is a set B with two binar operations and, elements and, and a unar operation such that the following properties hold for all,, and z in B: = and = (identit laws) () = and () = (domination laws) ( ) z = ( z) and ( ) z = ( z) and (associative laws) = and = (commutative laws) ( z) = ( ) ( z) and ( z) = ( ) ( z) (distributive laws) 24 Nov 25 CS 32 9 Boolean Algebras Eamples of Boolean Algebras are:. The algebra of all subsets of a set U, with + =, =,  = complement, =, = U. 2. The algebra of propositions with smbols p, p 2,,p n, with + =, =,  =, = F, = T. 3. If B,, B n are Boolean Algebras, so is B.. B n, with operations defined coordinatewise. 24 Nov 25 CS 32 2 Logic Gates Electronic circuits consist of socalled gates. There are three basic tpes of gates. In each case the input is a Boolean epression and the output is another Boolean epression.  inverter + OR gate AND gate 24 Nov 25 CS 32 2 Logic Gates Eample: How can we build a circuit that computes the function + ()?  + () 24 Nov 25 CS () Multi switch light circuit Suppose we want a circuit for a light controlled b two switches, where changing the state of either switch changes the state of the light (on or off). If we let and be the states of the switches ( or ) then the boolean epression + ()() ( = complement) will do the job. 24 Nov 25 CS Multi switch light circuit This is because if both and are on () or off () + ()() will be, and otherwise will be. We can generalize this method. For three switches the Boolean epression z + ()(z) + ()(z) + ()()z will work. Can ou draw circuits implementing these epressions? (see pp. 825, 826 [6 th ed. pp. 763, 764]) 24 Nov 25 CS
5 24 Nov 25 Adding binar integers If we add two one bit integers and we get a sum for that bit position plus a carr bit. If we don t consider a carr bit from a lower bit addition we get what s called a half adder. If we do get consider an input carr bit we have a full adder. (see p. 827, 6 th ed.765) 24 Nov 25 CS Half Adder Given input bits and, the result bit will be + unless both and are, in which case the result is. This means that we can epress the result bit as (+)(()), or (+)( + ). The carr bit will be (we carr if both and are ) 24 Nov 25 CS Full Adder If we add a carr bit c from the previous order bit sum our result for this bit would be if one or three of c,, are, and otherwise. This means c +()(c )+()(c )+()()c would work, with carr bit c +(c )+()c +()c See p. 827 to check our implementation. A Boolean function can be implemented b man different Boolean epressions. Disjunctive normal form, the sumofproducts epansion we got from the table of values of the epression, is often not the most efficient. 24 Nov 25 CS Nov 25 CS For eample, the Boolean epression ( 2 ) ( ) 3 = (( 2 )+ 2 ) 3 + ( ) 3 = 3 + ( ) 3 = ( + ( )) 3 = 3 This last epression is a lot easier to compute. No gates required. Much simpler circuit. 24 Nov 25 CS Karnaugh Maps and the Quine McCluske Method are used for simplifing Boolean epressions. See section 2.4. We ll do some eamples on the board. 24 Nov 25 CS
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