Boolean Algebra. Introduction. Soitltion: Using the definitions of complementation, the Boolean sum, and the Boolean product, it follows that =0+0 =0.

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1 C H A P T E R Boolean Algebra. Boolean Functions.2 Representing Boolean Functions.3 Logic Gates.4 Minimization of Circuits T he circuits in computers an other electronic evices have inputs, each of which is either.ll. a or a I, an prouce outputs that are also Os an s. Circuits can be constructe using an basic element that has two ifferent states. Such elements inclue switches that can be in either the on or the off position an optical evices that can either be lit or unlit. In 938 Claue Shannon showe how the basic rules oflogic, first given b George Boole in 854 in his The Laws of Thought, coul be use to esign circuits. These rules form the basis for Boolean algebra. In this chapter we evelop the basic properties of Boolean algebra. The operation of a circuit is efine b a Boolean function that specifies the value of an output for each set of inputs. The first step in constructing a circuit is to represent its Boolean function b an epres sion built up using the basic operations of Boolean algebra. We will provie an algorithm for proucing such epressions. The epression that we obtain ma contain man more operations than are necessar to represent the function. Later in the chapter we will escribe methos for fining an epression with the minimum number of sums an proucts that represents a Boolean function. The proceures that we will evelop, Karnaugh maps an the Quine-McCluske metho, are important in the esign of effi cient circuits.. Boolean Functions Introuction Boolean algebra provies the operations an the rules for working with the set to, I}. Electronic an optical switches can be stuie using this set an the rules of Boolean algebra. The three operations in Boolean algebra that we will use most are complementation, the Boolean sum, an the Boolean prouct. The complement of an element, enote with a bar, is efine b = an T = O. The Boolean sum, enote b + or b OR, has the following values: + =, + =, + =, + = O. The Boolean prouct, enote b. or b AND, has the following values: =, =, =, =. When there is no anger of confusion, the smbol can be elete, just as in writing algebraic proucts. Unless parentheses are use, the rules of preceence for Boolean operators are: first, all complements are compute, followe b Boolean proucts, followe b all Boolean sums. This is illustrate in Eample. EXAMPLE Fin the value of. + ( + ). Soitltion: Using the efinitions of complementation, the Boolean sum, an the Boolean prouct, it follows that + ( + ) = + T =+ =. II-I 749

2 75 II/Boolean Algebra / -2 The complement, Boolean sum, an Boolean prouct correspon to the logical operators, -., v, an /\, respectivel, where correspons to F (false) an correspons to T (true). Equal ities in Boolean algebra can be irectl translate into equivalences of compoun propositions. Conversel, equivalences of compoun propositions can be translate into equalities in Boolean algebra. We will see later in this section wh these translations iel vali logical equivalences an ientities in Boolean algebra. Eample 2 illustrates the translation from Boolean algebra to propos itional logic. Translate + ( + ) EXAMPLE 2 =, the equalit foun in Eample, into a logical equivalence. Solution: We obtain a logical equivalence when we translate each into a T, each into a F, each Boolean sum into a isjunction, each Boolean prouct into a conjunction, an each complementation into a negation. We obtain (T /\ F) v -. (T v F) == F. Eample 3 illustrates the translation from propositional logic to Boolean algebra. EXAMPLE 3 Translate the logical equivalence (T /\ T ) v -.F == T into an ientit in Boolean algebra. Solution: We obtain an ientit in Boolean algebra when we translate each T into ai, each F into a, each isjunction into a Boolean sum, each conjunction into a Boolean prouct, an each negation into a complementation. We obtain (. I) + =. Boolean Epressions an Boolean Functions Let B = {O, I }. Then B n = {(" X2,, n ) I Xi E B for i n } is the set of all possible n -tuples of Os an I s. The variable is calle a Boolean variable if it assumes values onl from B, that is, if its onl possible values are an. A function from B n to B is calle a Boolean function of egree n. unks CLAUDE ELWOOD SHANNON ( I). Claue Shannon was born in Petoske, Michigan, an grew up in Galor, Michigan. His father was a businessman an a probate juge, an his mother was a language teacher an a high school principal. Shannon attene the Universit of Michigan, grauating in 936. He continue his stuies at M.LT., where he took the job of maintaining the ifferential analer, a mechanical computing evice consisting of shafts an gears built b his professor, Vannevar Bush. Shannon's master's thesis, written in 936, stuie the logical aspects of the ifferential analer. T his master's thesis presents the first application of Boolean algebra to the esign of switching circuits; it is perhaps the most famous master's thesis of the. twentieth centur. He receive his Ph.D. from M.LT. in 94. Shannon joine Bell Laboratories in 94, where he worke on transmitting ata efficientl. He was one of the first people to use bits to represent information. At Bell Laboratories he worke on etermining the amount of traffic that telephone lines can carr. Shannon mae man funamental contributions to information theor. In the earl 95s he was one of the founers of the stu of artificial intelligence. He joine the M.LT. facult in 956, where he continue his stu of information theor. Shannon ha an unconventional sie. He is creite with inventing the rocket-powere Frisbee. He is also famous for riing a uniccle own the hallwas of Bell Laboratories while juggling four balls. Shannon retire when he was 5 ears ol, publishing papers sporaicall over the following ears. In his later ears he concentrate on some pet projects, such as builing a motorize pogo stick. One interesting quote from Shannon, publishe in Omni Magazine in 987, robots what ogs are to humans. An I am rooting for the machines." is "I visualize a time when we will be to

3 -3. Boolean Functions 75 EXAMPLE 4 The function F(, ) = from the set of orere pairs of Boolean variables to the set to, I} is a Boolean function of egree 2 with F(I, ) =, F(I, ) = I, F(O, ) =, an F(O, ) =. We ispla these values of F in Table... TABLE I I F(,) I I I Boolean functions can be represente using epressions mae up from variables an Boolean operations. The Boolean epressions in the variables XJ, X2,,Xn are efine recursivel as,, XJ, X2,, Xn are Boolean epressions; if EJ an E2 are Boolean epressions, then E. (EJE2), an (EJ + E2) are Boolean epressions. Each Boolean epression represents a Boolean function. The values of this function are obtaine b substituting an I for the variables in the epression. In Section.2 we will show that ever Boolean function can be represente b a Boolean epression. EXAMPLE 5 Fin the values of the Boolean function represente b F(,, z) = + z. Solution: The values of this function are isplae in Table 2. Note that we can represent a Boolean function graphicall b istinguishing the vertices of the n-cube that correspon to the n-tuples of bits where the function has value. EXAMPLE 6 IlO III The function F(,, z) = + z from B 3 to B from Eample 3 can be represente b istingu ishing the vertices that correspon to the five 3-tuples (,, ), (,,), (,, ), (,,), an (,, ), where F(,, z) =, as shown in Figure. These vertices are isplae using soli black circles... Boolean functions F an G of n variables are equal if an onl if F(bl. b2,., bn) = G(bJ, b2,, bn) whenever bj, b2,, bn belong to B. Two ifferent Boolean epressions that OIO--+- OIl represent the same function are calle equivalent. For instance, the Boolean epressions, +, an. are equivalent. The complement of the Boolean function F is the function F, where F(J,...,n) = F(J,..., n). Let F an G be Boolean functions of egree n. The Boolean sum F + G an the Boolean prouct F G are efine b FIGURE (F + G)(J,..., n) = F(J,..., n) + G(J,..., n), (F G)(J,..., n) = F(J,..., n)g(j,..., n). TABLE 2 z z F(,,z) = + Z I I I I I I I I I I I I I I I I I I I I I

4 752 II/Boolean Algebra -4 TABLE 3 The Boolean Functions of Degree Two. FI F2 F3 F4 Fs F6 F, Fa F, FlO FII FI2 FI3 FI4 FI5 FI6. A Boolean function of egree two is a function from a set with four elements, namel, pairs of elements from B = {O, I}, to B, a set with two elements. Hence, there are 6 ifferent Boolean functions of egree two. In Table 3 we ispla the values of the 6 ifferent Boolean functions of egree two, labele F, F2,, F6 EXAMPLE 7 How man ifferent Boolean functions of egree n are there? Solution: From the prouct rule for counting, it follows that there are 2 n ifferent n-tuples of Os an s. Because a Boolean function is an assignment of or to each of these 2 n ifferent n-tuples, the prouct rule shows that there are 2 2" ifferent Boolean functions of egree n... Table 4 isplas the number of ifferent Boolean functions of egrees one through si. The number of such functions grows etremel rapil. Ientities of Boolean Algebra There are man ientities in Boolean algebra. The most important of these are isplae in Table 5. These ientities are particularl useful in simplifing the esign of circuits. Each of the ientities in Table 5 can be prove using a table. We will prove one of the istributive laws in this wa in Eample 8. The proofs of the remaining properties are left as eercises for the reaer. EXAMPLE 8 Show that the istributive law CV + z) = + z is vali. Solution: The verification of this ientit is shown in Table 6. The ientit hols because the last two columns of the table agree... TABLE4 The Number of Boolean Functions of Degree ". Degree Number , ,294,967, ,446,744,73,79,55,66

5 -5. Boolean Functions 753 TABLE 5 Boolean Ientities. Ientit = += = +O= l = +l= = +=+ = +(+z)=(+)+z () = () + = ( + )( + z) (+z)=+z ( ) :;:::+ ( + ) = + = ( + ) = +=l =O Name Law of the ouble complement Iempotent laws Ientit laws Domination laws Commutative laws Associative laws Distributive laws M9f ' laws Absorp Pfll ws Unit propert Zero propert The reaer shoul compare the Boolean ientities in Table 5 to the logical equivalences in Table 6 of Section.2 an the set ientities in Table in Section 2.2. All are special cases of the same set of ientities in a more abstract structure. Each collection of ientities can be obtaine b making the appropriate translations. For eample, we can transform each of the ientities in Table 5 iqto a IQgical equivalence b changing each Boolean variable into a propositional variable,!lch nto a F, each into a T, each Boolean sum into a isjunction, each Boolean prouct,nto a conjunction, an each complementation into a negation, as we illustrate in Eample 9. TABLE 6 Verifing One of the Distributive Laws. z + z z ( +z) +z

6 754 II / Boolean Algebra EXAMPLE 9-6 Translate the istributive law + = ( + )( + z) in Table 5 into a logical equivalence. Solution: To translate a Boolean ientit into a logical equivalence, we change each Boolean variable into a propositional variable. Here we will change the Boolean variables,, an z into the propositional variables p, q, an r. Net, we change each Boolean sum to a isjunction an each Boolean prouct into a conjunction. (Note that an o not appear in this ientit an complementation also oes not appear.) This transforms the Boolean ientit into the logical equivalence p v (q /\ r ) == (p V q) /\ (p V r). This logical equivalence is one of the istributive laws for propositional logic in Table 6 in Section.2. Ientities in Boolean algebra can be use to prove further ientities. We emonstrate this in Eample. EXAMPLE Eam: Prove the absorption law ( + ) = using the other ientities of Boolean algebra shown in Table 5. (This is calle an absorption law because absorbing + into leaves unchange.) Solution: The steps use to erive this ientit an the law use in each step follow: ( + ) = = ( + O)( + ) X+. = +Y = = Ientit law for the Boolean sum Distributive law of the Boolean sum over the Boolean prouct. Commutative law for the Boolean prouct X+ Domination law for the Boolean prouct. Ientit law for the Boolean sum Dualit The ientities in Table 5 come in pairs (ecept for the law of the ouble complement an the unit an zero properties). To eplain the relationship between the two ientities in each pair we use the concept of a ual. The ual of a Boolean epression is obtaine b interchanging Boolean sums an Boolean proucts an interchanging Os an s. EXAMPLE Fin the uals of ( + ) an + ( + z). Solution: Interchanging. signs an + signs an interchanging Os an s in these epressions prouces their uals. The uals are + ( ) an (X + O)(), respectivel. The ual of a Boolean function F represente b a Boolean epression is the function represente b the ual of this epression. This ual function, enote b F, oes not epen on the particular Boolean epression use to represent F. An ientit between functions represente b Boolean epressions remains vali when the uals of both sies of the ientit are taken. (See Eercise 3 for the reason this is true.) This result, calle the ualit principle, is useful for obtaining new ientities. EXAMPLE 2 Construct an ientit from the absorption law ( + ) = b taking uals. Solution: Taking the uals of both sies of this ientit prouces the ientit + is also calle an absorption law an is shown in Table 5. =, which

7 /-7. Boolean Functions 755 The Abstract Definition of a Boolean Algebra In this section we have focuse on Boolean functions an epressions. However, the results we have establishe can be translate into results about propositions or results about sets. Because of this, it is useful to efine Boolean algebras abstractl. Once it is shown that a particular structure is a Boolean algebra, then all results establishe about Boolean algebras in general appl to this particular structure. Boolean algebras can be efine in several was. The most common wa is to specif the properties that operations must satisf, as is one in Definition. DEFINITION A Boolean algebra is a set B with two binar operations V an /\, elements an, an a unar operation - such that these properties hol for all,, an z in B: vo= /\ = V= I /\ = } } Ientit laws ( V ) V Z= V ( V Z) (/\ )/\ z=/\ ( /\ z) V = V /\ = /\ }. Complement laws } V (/\ z)= ( V ) /\ ( V Z ) /\ ( V z)= (/\ ) V (/\ z) Associative laws } Commutative laws Distributive laws Using the laws given in Definition, it is possible to prove man other laws that hol for ever Boolean algebra, such as iempotent an omination laws. (See Eercises ) From our previous iscussion, B = {O, I} with the OR an AND operations an the com plement operator, satisfies all these properties. The set of propositions in n variables, with the V an/\ operators, F an T, an the negation operator, also satisfies all the properties of a Boolean algebra, as can be seen from Table 6 in Section.2. Similarl, the set of subsets of a universal set U with the union an intersection operations, the empt set an the universal set, an the set complementation operator, is a Boolean algebra as can be seen b consulting Table in Section 2.2. So, to establish results about each of Boolean epressions, propositions, an sets, we nee onl prove results about abstract Boolean algebras. Boolean algebras ma also be efine using the notion of a lattice, iscusse in Chapter 8. Recall that a lattice L is a partiall orere set in which ever pair of elements, has a least upper boun, enote b lub(, ) an a greatest lower boun enote b glb(, ). Given two elements an of L, we can efine two' operations V an /\ on pairs of elements of L b V = lub(, ) an /\ = glb(, ). For a lattice L to be a Boolean algebra as specifie in Definition I, it must have two properties. First, it must be complemente. For a lattice to be complemente it must have a least element an a greatest element an for ever element of the lattice there must eist an element such that V = an /\ = O. Secon, it must be istributive. This means that for ever,, anz inl, V ( /\ z) = ( V ) /\ ( V z) an/\ ( V z)= (/\ ) V (/\ z). Showing that a complemente, istributive lattice is a Boolean algebra is left as Eercise 39 at the en of this section.

8 756 II/Boolean Algebra -8 Eercises. Fin the values of these epressions. a) b) + c ) 2. Fin the values, if an, of the Boolean variable satisf these equations. a) l = O c ) X = beginning of this section. In each case, use a table as in Eample 8. ) ( + ) that b) + = O a) b) 7. Verif the omination laws. Translate the equation in part (a) into a propositional equivalence b changing each to a F, each to a T, each Boolean sum into a isjunction, each Boolean prouct into a conjunction, each complementation into a negation, an the equals sign to a propositional equivalence sign. Show that (. ) + ( ) =. Translate the equation in part (a) into a propositional equivalence b changing each to a F, each I to a T, each Boolean sum into a isjunction, each Boolean prouct into a conjunction, each complementation into a negation, an the equals sign to a propositional equivalence sign. 5. Use a table to epress the values of each of these Boolean functions. a) F(,, z) = b) F(,, z) = + 6. Use a table to epress the values of each of these Boolean functions. verte that correspons to a 3-tuple where this function has the value. 8. U se a 3-cube Q 3 to represent each of the Boolean func tions in Eercise 6 b (tisplaing a black circle at each verte that correspons to a 3-tuple where this function has the value. 9. What values of the Boolean variables an satisf = +?. How man ifferent Boolean functions are there of e + = using the other laws in Table 5. ts"2. Show that F(,, z ) = + z + z has the value if an onl if at least two of the variables,, an z have the value. 3. Show that 2. Verif D e Morgan's laws Verif the unit propert. 23. Verif the zero propert. The Boolean operatoreb, calle the XOR operator, is efine b EB =, EB =, EB =, an EB = o. 24. Simplif these epressions. b) EB a) EB c ) EB ) EB 25. Show that these ientities hol. a) EB = ( + )() b) EBY = (Y) + (X) = EB. 27. Prove or isprove these equalities. a) EB (EBz) = ( EB)EBz b) + (EBz) = ( + )EB ( + z) 28. Fin the uals of these Boolean epressions. c ) F(,, z) = + () ) F(,, z) = (z + Z) U se a 3-cube Q 3 to represent each of the Boolean func tions in Eercise 5 b isplaing a black circle at each gree 7? 9. Verif the associative laws. 2. Verif the first istributive law in Table c ) EB ( + z) = ( EB) + ( EBz) a) F(,, z) = z b) F(,, z) = +. Prove the absorption law 8. Verif the commutative laws. 26. Show that EB c ) F(,, z) = + () ) F(,, z) = ( + Z) Verif the iempotent laws. 6. Verif the ientit laws. ) X = a) Show that (. ) + (. I + ) =. b) 4. Verif the law of the ouble complement. + + z = + + z. Eercises 4-23 eal with the B oolean algebra to, I} with aition, multiplication, an complement efine at the a) + c ) + z *29. Suppose that b) ) z F is a Boolean function represente b a Boolean epression in the variables XI,, n Show that. F(XI,..., n) = F( X I,..., n ). *3. Show that if F an G are Boolean functions represente b Boolean epressions in n variables an F = G, then F = G, where F an G are the Boolean functions represente b the uals of the Boolean epressions rep resenting F an G, respectivel. [Hint: Use the result of Eercise 29.] *3. How man ifferent Boolean functions F(,, z) are there such that F(X,, Z) = F(,, z) for all values of the Boolean variables,, an z? *32. How man ifferent Boolean functions F(,, z) are there such that F(,, z) = F(,, z) = F(,, z ) for all values of the Boolean variables,, an z? 33. Show that ou obtain De Morgan's laws for proposi tions (in Table 5 in Section.2) when ou transform De Morgan' s l aws for Boolean al gebra in Table 6 into logical equivalences.

9 .2 Representing Boolean Functions Show that ou obtain the absorption laws for proposi tions (in Table 5 in Section.2) when ou transform the absorption laws for Boolean algebra in Table 5 into logical equivalences. In Eercises , use the laws in Definition to show that the state properties hol in ever Boolean algebra. 35. Show that in a Boolean algebra, the iempotent laws 39. Show that De Morgan's laws hol in a Boolean algebra. That is, show that for all (XA) = v. an, ( v) = A an 4. Show that in a Boolean algebra, the moular properties hol. That is, show that A ( V ( Az» = ( A) V ( Az ) anv (A ( V z» = ( v )A ( V z ). v = ana = hol for ever element. 4. Show that i n a Boolean algebra, i f v unique complement such that v = ana = o. 42. Show that in a Boolean algebra, the 36. Show that in a Boolean algebra, ever element has a 37. Show that in a Boolean algebra, the complement of the element is the element an vice versa. 38. Prove that in a Boolean algebra, the law of the ouble complement hols; that is, = for ever element =, then = an =, an that if A =, then = an =. ual of an ientit, obtaine b interchanging thev ana operators an in terchanging the elements an, is also a vali ientit. 43. Show that a complemente, istributive lattice i s a Boolean algebra. Representing Boolean Functions Two important problems of Boolean algebra will be stuie in this section. The first problem is: Given the values of a Boolean function, how can a Boolean epression that represents this function be foun? This problem will be solve b showing that an Boolean function can be represente b a Boolean sum of Boolean proucts of the variables an their complements. The solution of this problem shows that ever Boolean function can be represente using the three Boolean operators., +, an -. The secon problem is: Is there a smaller set of operators that can be use to represent all Boolean functions? We will answer this problem b showing that all Boolean functions can be represente using onl one operator. Both of these problems have practical importance in circuit esign. Sum-of-Proucts Epansions We will use eamples to illustrate one important wa to fin a Boolean epression that represents a Boolean function. EXAMPLE TABLE z F G Fin Boolean epressions that represent the functions F (,, z) an G (,, z), which are given in Table. Solution: An epression that has the value when = z = an =, an the value otherwise, is neee to represent F. Such an epression can be forme b taking the Boolean prouct of,, an z. This prouct,, has the value if an onl if = Y = z =, which hols if an onl if = z = an = O. To represent G, we nee an epression that equals when = = an z =, or when = z = an =. We can form an epression with these values b taking the Boolean sum of two ifferent Boolean proucts. The Boolean prouct has the value if an onl if = = an z = O. Similarl, the prouct has the value if an onl if = z = an =. The Boolean sum of these two proucts, +, represents G, because it has the... value if an onl if = = I an z = or = z = an =. Eample illustrates a proceure for constructing a Boolean epression representing a function with given values. Each combination of values of the variables for which the function has the value leas to a Boolean prouct of the variables or their complements.

10 758 / Boolean Algebra - DEFINITION A literal is a Boolean variable or its complement. A minterm of the Boolean variables Xl, X2,, Xn is a Boolean prouct YIY2 Yn, where Yi = Xi or Yi = Xi. Hence, a minterm is a prouct of n literals, with one literal for each variable. A minterm has the value for one an onl one combination of values of its variables. More precisel, the minterm YI Y2. Yn is I if an onl if each Yi is I, an this occurs if an onl if Xi = I whenyi = Xi an Xi = whenyi = Xi' EXAMPLE 2 Fin a minterm that equals I if Xl = X3 = an X2 = X4 = X5 = I, an equals otherwise. Solution: The minterm XIX2X3X4X5 has the correct set of values. B taking Boolean sums of istinct minterms we can buil up a Boolean epression with a specifie set of values. In particular, a Boolean sum of minterms has the value I when eactl one of the minterms in the sum has the value I. It has the value for all other combinations of values of the variables. Consequentl, given a Boolean function, a Boolean sum of minterms can be forme that has the value I when this Boolean function has the value I, an has the value when the function has the value O. The minterms in this Boolean sum correspon to those combinations of values for which the function has the value I. The sum of minterms that represents the function is calle the sum-of-proucts epansion or the isjunctive normal form of the Boolean function. (See Eercise 42 in Section.2 for the evelopment of isjunctive normal form in propositional calculus.) EXAMPLE 3 Etra Eamples""" Fin the sum-of-proucts epansion for the function F(,, z) = ( + )z. Solution: We will fin the sum-of-proucts epansion of F(,, z) in two was. First, we will use Boolean ientities to epan the prouct an simplif. We fin that F(,, z) = ( + )z = z + Distributive law = X lz + I Ientit law = ( + Y)z + ( + X) Unit propert = Distributive law = + + z. Iempotent law Secon, we can construct the sum-of-proucts epansion b etermining the values of F for all possible values of the variables,, an z. These values are foun in Table 2. The sum-ofproucts epansion of F is the Boolean sum of three minterms corresponing to the three rows of this table that give the value I for the function. This gives F(,, z) = + +. It is also possible to fin a Boolean epression that represents a Boolean function b taking a Boolean prouct of Boolean sums. The resulting epansion is calle the conjunctive normal form or prouct-of-sums epansion of the function. These epansions can be foun from sum-of-proucts epansions b taking uals. How to fin such epansions irectl is escribe in Eercise at the en of this section.

11 JJ-JJ.2 Representing Boolean Functions 759 TABLE 2 z + z I I I I I I I I I I I I I I I I I I I I I I (+)z I I I Functional Completeness Ever Boolean function can be epresse as a Boolean sum of mintenns. Each mintenn is the Boolean prouct of Boolean variables or their complements. This shows that ever Boolean function can be represente using the Boolean operators., +, an -. Because ever Boolean function can be represente using these operators we sa that the set {., +, - } is functionall complete. Can we fin a smaller set of functionall complete operators? We can o so if one of the three operators of this set can be epresse in tenns of the other two. This can be one using one of De Morgan's laws. We can eliminate all Boolean sums using the ientit +=, which is obtaine b taking complements of both sies in the secon De Morgan law, given in Table 5 in Section., an then appling the ouble complementation law. This means that the set {., - } is functionall complete. Similarl, we coul eliminate all Boolean proucts using the ientit = +, which is obtaine b taking complements of both sies in the first De Morgan law, given in Table 5 in Section., an then appling the ouble complementation law. Consequentl {+, - } is functionall complete. Note that the set {+,.} is not functionall complete, because it is impossible to epress the Boolean function F() = using these operators (see Eercise 9). We have foun sets containing two operators that are functionall complete. Can we fin a smaller set of functionall complete operators, namel, a set containing just one operator? Such sets eist. Define two operators, the I or NAND operator, efine b I = an I = o I = I = ; an the -I- or NOR operator, efine b -I- = -I- = -I- = an o -I- =. Both of the sets { I } an { -I- } are functionall complete. To see that { I } is functionall complete, because {., -} is functionall complete, all that we have to o is show that both of the operators an - can be epresse using just the I operator. This can be one as = l, = ( I ) I ( I ). The reaer shoul verif these ientities (see Eercise 4). We leave the emonstration that { -I- } is functionall complete for the reaer (see Eercises 5 an 6).

12 76 II/Boolean Algebra - 2 Eercises. Fin a Boolean prouct of the Boolean variables a) ==, Z=. c) =, =z=. Boolean prouct o f the Boolean sums foun i n parts (a), (b), an (c) in Eercise 7.],, an z, or their complements, that has the value if an onl if b) =O,= l,z=o. ) ==z=o. 9. Show that the Boolean sum Yl nation of the values of the variables, namel, when Xi = if Yi =Xi an Xi = if Yi =Xi. This Boolean sum is calle a materm. 2. Fin the sum-of-proucts epansions of these Boolean functions. b) F(,)= a) F(,)= + c) F(,)=. Show that a Boolean function can be represente as a B oolean prouct of materms. This representation is calle the prouct-of-sums epansion or conjunctive normal form of the function. [Hint: Inclue one ma term in this prouct for each combination of the variables where the function has the value.] )F(,)= 3. Fin the sum-of-proucts epansions of these Boolean functions. a) F(,,z)= + + z b) F(,,z)=( + z) c) F(,,z)=X ) F(,,z)=. Fin the prouct-of-sums epansion of each of the Boolean functions in Eercise Fin the sum-of-proucts epansions of the B oolean func 2. Epress each o f these Boolean functions using the oper ators. an -. tionf(,,z) that equals if an onl if b) =O. ) =O. a) =O. c) + =O. 5. Fin the sum-of-proucts epansion of the Boolean func tionf (w,,,z) that has the value if an onl if an o number of w,,,an z have the value. 6. Fin the sum-of-proucts epansion of the Boolean func tion F(XI' X2,X3,X4,X5) that has the value if an onl if three or more of the variables Xl, 2, 3, X4, an X5 have the value. Another wa to fin a Boolean epression that represents a Boolean function is to form a B oolean prouct of Boolean sums of literals. Eercises 7- are concerne with represen tations of this kin. 7. Fin a Boolean sum containing either or X, either or, an either z or that has the value if an onl if z a) ==, z=o. c) =z=, =. b) ==z=o. 8. Fin a Boolean prouct of B oolean sums of literals that has the value if an onl if either == an z=, =z= an=, : ==z=o. [Hint: Take the.3 + Y Yn, where Yi =Xi or Yi =Xi, has the value for eactl one combi b) + (X + z) ) (+ +Z) a) + + z c) (+)i) 3. Epress each of the Boolean functions in Eercise using the operators+ an Show that b) =( I ) I ( I ). a) X= I. c) + =( I ) I ( I ). 5. Show that a) = -!-. b) =( -!- )..!- ( -!- ). c) + =( -!- )..!- ( -!- ). 6. Show that { -!- } is functionall complete using Eercise Epress each of the Boolean functions in Eercise 3 using. the operator I. 8. Epress each of the Boolean functions in Eercise 3 using the operator -!-. 9. Show that the set o f operators complete. {+,. } i s not functionall 2. Are these sets of operators functionall complete?.){ +,EB} b) CEB} c) { " EB} Logic Gates Introuction udaic Boolean algebra is use to moel the circuitr of electronic evices. Each input an each output of such a evice can be thought of as a member of the set to, I }. A computer, or other electronic evice, is mae up of a number of circuits. Each circuit can be esigne using the rules of Boolean algebra that were stuie in Sections. an.2. The basic elements of circuits are calle gates. Each tpe of gate implements a Boolean operation. In this section we efine several tpes of gates. Using these gates, we will appl the rules of Boolean algebra to esign circuits that perform a variet of tasks. The circuits that we will stu in this chapter give output

13 -3.3 Logic Gates 76 (a) Inverter FIGURE Basic Tpes of Gates. :--.: + (b) OR gate (c) AND gate FIGURE 2 Gates with n Inputs. that epens onl on the input, an not on the current state of the circuit. In other wors, these circuits have no memor capabilities. Such circuits are calle combinational circuits or gating networks. We will construct combinational circuits using three tpes of elements. The first is an inverter, which accepts the value of one Boolean variable as input an prouces the complement of this value as its output. The smbol use for an inverter is shown in Figure l( a). The input to the inverter is shown on the left sie entering the element, an the output is shown on the right sie leaving the element. The net tpe of element we will use is the OR gate. The inputs to this gate are the values of two or more Boolean variables. The output is the Boolean sum of their values. The smbol use for an OR gate is shown in Figure I (b). The inputs to the OR gate are shown on the left sie entering the element, an the output is shown on the right sie leaving the element. The thir tpe of element we will use is the AND gate. The inputs to this gate are the values of two or more Boolean variables. The output is the Boolean prouct of their values. The smbol use for an AND gate is shown in Figure (c). The inputs to the AND gate are shown on the left sie entering the element, an the output is shown on the right sie leaving the element. We will permit multiple inputs to AND an OR gates. The inputs to each of these gates are shown on the left sie entering the element, an the output is shown on the right sie. Eamples of AND an OR gates with n inputs are shown in Figure 2. Combinations of Gates Combinational circuits can be constructe using a combination of inverters, OR gates, an AND gates. When combinations of circuits are forme, some gates ma share inputs. This is shown in one of two was in epictions of circuits. One metho is to use branchings that inicate all the gates that use a given input. The other metho is to inicate this input separatel for each gate. Figure 3 illustrates the two was of showing gates with the same input values. Note also that output from a gate ma be use as input b one or more other elements, as shown in Figure 3. Both rawings in Figure 3 epict the circuit that prouces the output +. EXAMPLE Construct circuits that prouce the following outputs: (a) ( + ), (b) ( +Z), an (c) ( + + z)(xz). Solution: Circuits that prouce these outputs are shown in Figure 4.

14 762 II/Boolean Algebra ---.! Y ----.t ! ---.! FIGURE 3 Two Was to Draw the Same Circuit L- ( + ) (a) --- (b) ---.r t z---./ + +z (c) ---pj ( + + z ) ----.! z ---.t FIGURE 4 Circuits that Prouce the Outputs Specifie in Eample.

15 -5.3 Logic Gates I --,.--. +z+ FIGURE 5 A Circuit for Majorit Voting. Eamples of Circuits We will give some eamples of circuits that perform some useful functions. EXAMPLE 2 A committee of three iniviuals ecies issues for an organization. Each iniviual votes either es or no for each proposal that arises. A proposal is passe if it receives at least two es votes. Design a circuit that etermines whether a proposal passes. Solution: Let = I if the first iniviual votes es, an = if this iniviual votes no; let = if the secon iniviual votes es, an = if this iniviual votes no; let z = if the thir iniviual votes es, an z = if this iniviual votes no. Then a circuit must be esigne that prouces the output from the inputs,, an z when two or more of,, an z are. One representation of the Boolean function that has these output values is + z + (see Eercise 2 in Section.). The circuit that implements this function is shown in Figure EXAMPLE 3 TABLE F(,) I I I I I I Sometimes light fitures are controlle b more than one switch. Circuits nee to be esigne so that flipping an one of the switches for the fiture turns the light on when it is off an turns the light off when it is on. Design circuits that accomplish this when there are two switches an when there are three switches. Solution: We will begin b esigning the circuit that controls the light fiture when two ifferent switches are use. Let = when the first switch is close an = when it is open, an let = when the secon switch is close an = when it is open. Let F (, ) = when the light is on an F(, ) = when it is off. We can arbitraril ecie that the light will be on when both switches are close, so that F (, I) =. This etermines all the other values of F. When one of the two switches is opene, the light goes off, so F( I, ) = F(O, ) = O. When the other switch is also opene, the light goes on, so F(O, ) =. Table isplas these values. We see that F(, ) = +. This function is implemente b the circuit shown in Figure r-""". + ' FIGURE 6 A Circuit for a Light Controlle b Two Switches.

16 764 II /Boolean Algebra , z ----/ ----, z ----' FIGURE 7 A Circuit for a Fiture Controlle b Three Switches. TABLE 2 z F(,,z) I I I I I I I I I I I I I I I I We will now esign a circuit for three switches. Let,, an z be the Boolean variables that inicate whether each of the three switches is close. We let = when the first switch is close, an = when it is open; = when the secon switch is close, an = when it is open; an z = when the thir switch is close, anz = Owhen it isopen. LetF(,, z) = when the light is on an F (,, z) = when the light is off. We can arbitraril specif that the light be on when all three switches are close, so that F (,, ) =. This etermines all other values of F. When one switch is opene, the light goes off, so F(I,, ) = F(I,, ) = F(O,, ) = O. When a secon switch is opene, the light goes on, sof(i,, ) = F(O,, ) = F(O,, ) =. Finall, when the thir switch is opene, the light goes off again, so F(O,, ) =. Table 2 shows the values of this function. The function F can be represente b its sum-of-proucts epansion as F (,, z) = The circuit shown in Figure 7 implements this function.... udale TABLE 3 Input an Output for the Half Aer. Input Output s c I I I I I I I Aers We will illustrate how logic circuits can be use to carr out aition of two positive integers from their binar epansions. We will buil up the circuitr to o this aition from some component circuits. First, we will buil a circuit that can be use to fin +, where an are two bits. The input to our circuit will be an, because these each have the value or the value. The output will consist of two bits, namel, s an c, where s is the sum bit an c is the carr bit. This circuit is calle a multiple output circuit because it has more than one output. The circuit that we are esigning is calle the half aer, because it as two bits, without consiering a carr from a previous aition. We show the input an output for the half aer in Table 3. From Table 3 we see that c = an that s = + = ( + )(). Hence, the circuit shown in Figure 8 computes the sum bit s..an the carr bit c from the bits an. We use the full aer to compute the sum bit an the carr bit when two bits an a carr are ae. The inputs to the full aer are the bits an an the carr Cj. The outputs are the sum bit s an the new carr Ci+. The inputs an outputs for the full aer are shown in Table 4.

17 .3 Logic Gates -7 X+ Y S Ci = 765 XY Ci +XYCi + XY Ci + XYCi (X+Y )(X ) J Ci+ =XY Ci +XY Ci+ Carr = FIGURE 8 The Half Aer Xo TABLE 4 Yo Input an Output for the Full Aer. I Output CI S CI+ FIGURE 9 A Full Aer. So Full aer YI Input XY Y ci 2 Y2 FIGURE Aing Two Three-Bit Integers with Full an Half Aers. The two outputs of the full aer, the sum bit s an the carrci + J. are given b the sum of-proucts epansions Cj + Cj + Cj + Cj an Cj + Cj + Cj + Cj, respec tivel. However, instea of esigning the full aer from scratch, we will use half aers to prouce the esire output. A full aer circuit using half aers is shown in Figure 9. Finall, in Figure l O we show how full an half aers can be use to a the two three-bit integers (X2XIXoh an (Y2YIYoh to prouce the sum (S3S2SISoh. Note that S3, the highest-orer bit in the sum, is given b the carr C2. Eercises In Eercises -5 fin the output of the given circuit.. Y ---I 3. X ---' ---"" Y " z X Y---.J --.! ' X -- z--_--' Y -- Y---- z

18 II/Boolean Algebra (loh Two gates that are often use in circuits are NAND an NOR gates. When NAND or NOR gates are use to represent cir cuits, no other tpes of gates are neee. The notation for these gates is as follows: --"'::""---+ z L- z 6. Construct circuits from inverters, AND gates, an OR gates to prouce these outputs. a) c) b) + + ) (X + z)( + Z) 7. Design a circuit that implements majorit voting for five iniviuals. 8. Design a circuit for a light fiture controlle b four switches, where flipping one of the switches turns the light on when it is off an turns it off when it is on. 9. Show how the sum of two five-bit integers can be foun using full an half aers.. Construct a circuit for a half subtractor using AND gates, OR gates, an inverters. A half subtractor has two bits as input an prouces as output a ifference bit an a borrow.. Construct a circuit for a full subtractor using AND gates, OR gates, an inverters. A full subtractor has two bits an a borrow as input, an prouces as output a ifference bit an a borrow. 2. Use the circuits from Eercises an to fin the if ference of two four-bit integers, where the first integer is greater than the secon integer. *3. Construct a circuit that compares the two-bit integers (loh (YlYoh, an returning an output of when the first of these numbers is larger an an output of other wise. *4. Construct a circuit that computes the prouct of the two-.4 (YlYO )2. bit integers an The circuit shoul have four output bits for the bits in the prouct I Z ---- L- *5. Use NAND gates to construct circuits with these outputs. a) b) + Y c) ) B *6. Use NOR gates to construct circuits for the outputs given in Eercise 5. *7. Construct a half aer using NAND gates. *8. Construct a half aer using NOR gates. A multipleer is a switching circuit that prouces as output one of a set of input bits base on the value of control bits. 9. Construct a multipleer using AND gates, OR gates, an inverters that has as input the four bits an an the two control bits an Set up the circuit so that is the output, where i is the value of the two-bit integer Xi Co Cl. o, Xl, X2, X3 (Cl CO )2. The epth of a combinatorial circuit can be efine b spec ifing that the epth of the initial input is an if a gate has n ifferent inputs at epths respectivel, then its outputs have epth equal to ma ; this value is also efine to be the epth of the gate. The epth of a combinatorial circuit is the maimum epth of the gates in the circuit. l 2 n, (l 2... n ) + 2. Fin the epth of a) b) c) ) the circuit constructe in Eample 2 for majorit voting among three people. the circuit constructe in Eample 3 for a light con trolle b two switches. the half aer shown in Figure 8. the full aer shown in Figure 9. Minimization of Circuits Introuction The efficienc of a combinational circuit epens on the number an arrangement of its gates. The process of esigning a combinational circuit begins with the table specifing the output for each combination of input values. We can alwas use the sum-of-proucts epansion of a circuit to fin a set of logic gates that will implement this circuit. However, the sum-of proucts epansion ma contain man more terms than are necessar. Terms in a sum-of proucts epansion that iffer in just one variable, so that in one term this variable occurs an in the other term the complement of this variable occurs, can be combine. For instance, consier the circuit that has output I if an onl if = = z = I or = z = I an = o. The

19 -9.4 Minimization of Circuits r ! z ---I--'" + ---, z ----' : ----.!:O z FIGURE Two Circuits with the Same Output. sum-of-proucts epansion of this circuit is +. The two proucts in this epansion iffer in eactl one variable, namel,. The can be combine as + = ( + Y)(z) =. (z) =z. Hence, z is a Boolean epression with fewer operators that represents the circuit. We show two ifferent implementations of this circuit in Figure. The secon circuit uses onl one gate, whereas the first circuit uses three gates an an inverter. This eample shows that combining terms in the sum-of-proucts epansion of a circuit leas to a simpler epression for the circuit. We will escribe two proceures that simplif sum-of-proucts epansions. The goal of both proceures is to prouce Boolean sums of Boolean proucts that represent a Boolean function with the fewest proucts of literals such that these proucts contain the fewest literals possible among all sums of proucts that represent a Boolean function. Fining such a sum of proucts is calle minimization ofthe Boolean function. Minimizing a Boolean function makes it possible to construct a circuit for this function that uses the fewest gates an fewest inputs to the AND gates an OR gates in the circuit, among all circuits for the Boolean epression we are minimizing. Until the earl 96s logic gates were iniviual components. To reuce costs it was important to use the fewest gates to prouce a esire output. However, in the mi-96s, integrate circuit technolog was evelope that mae it possible to combine gates on a single chip. Even though it is now possible to buil increasingl comple integrate circuits on chips at low cost, minimization of Boolean functions remains important. Reucing the number of gates on a chip can lea to a more reliable circuit an can reuce the cost to prouce the chip. Also, minimization makes it possible to fit more circuits on the same chip. Furthermore, minimization reuces the number of inputs to gates in a circuit. This reuces the time use b a circuit to compute its output. Moreover, the number of inputs to a gate ma be limite because of the particular technolog use to buil logic gates. The first proceure we will introuce, known as Karnaugh maps (or K-maps), was esigne in the 95s to help minimize circuits b han. K-maps are useful in minimizing circuits with up to si variables, although the become rather comple even for five or si variables. The secon proceure we will escribe, the Quine-McCluske metho, was invente in the 96s. It automates the process of minimizing combinatorial circuits an can be implemente as a computer program.

20 II / Boolean Algebra Unfortunatel, minimizing Boolean functions with man variables is a computationall intensive problem. It has been shown that this problem is an NP-complete problem (see Section 3. 3 an [Ka93]), so the eistence of a polnomial-time algorithm for minimizing Boolean circuits is unlikel. The Quine-McCluske metho has eponential compleit. In practice, it can be use onl when the number of literals oes not ecee ten. Since the 97s a number of newer algorithms have been evelope for minimizing combinatorial circuits (see [Ha93] an [Ka93]). However, with the best algorithms et evise, onl circuits with no more than 25 variables can be minimize. Also, heuristic (or rule-of-thumb) methos can be use to substantiall simplif, but not necessaril minimize, Boolean epressions with a larger number of literals. Kamaugh Maps lidksi:::i : L:I:J FIGURE 2 K-maps in Two Variables. EXAMPLE l To reuce the number of terms in a Boolean epression representing a circuit, it is necessar to fin terms to combine. There is a graphical metho, calle a Karnaugh map or K-map, for fining terms to combine for Boolean functions involving a relativel small number of variables. The metho we will escribe was introuce b Maurice Karnaugh in 953. His metho is base on earlier work b E. W. Veitch. (This metho is usuall applie onl when the function involves si or fewer variables.) K-maps give us a visual metho for simplifing sum-of-proucts epansions; the are not suite for mechanizing this process. We will first illustrate how K-maps are use to simplif epansions of Boolean functions in two variables. We will continue b showing how K-maps can be use to minimize Boolean functions in three variables an then in four variables. Then we will escribe the concepts that can be use to eten K-maps to minimize Boolean functions in more than four variables. There are four possible minterms in the sum-of-proucts epansion of a Boolean function in the two variables an. A K-map for a Boolean function in these two variables consists of four cells, where a is place in the cell representing a minterm if this minterm is present in the epansion. Cells are sai to be ajacent if the minterms that the represent iffer in eactl one literal. For instance, the cell representing is ajacent to the cells representing an. The four cells an the terms that the represent are shown in Figure 2. Fin the K-maps for (a) +, (b) +, an (c) + +. Solution: We inclue a in a cell when the minterm represente b this cell is present in the <IIII sum-of-proucts epansion. The three K-maps are shown in Figure 3. We can ientif minterms that can be combine from the K-map. Whenever there are I s in two ajacent cells in the K-map, the minterms represente b these cells can be combine int.o a prouct involving just o e f. e r.ari ble : f EJR ar.et: :X n e represente b aja ent cells an can be com? p.to ' c e fr t t ;f t X) =. Moreover, If I s are m all four cells, the fo r f.fnte can R ppmrm 4 Hl.J:g Rn- term, namel, the Boolean epression that involves none of the variables. We circle' SlocKs of cells in the K-map that represent minterms that can be combine an then fin the corresponing sum of proucts. The :f. MAURICE KARNAUGH (BORN 924) Maurice Kamaugh, born in New York Cit, receive his B.S. from the Cit College of New York an his M.S. an Ph.D. from Yale Universit. He was a member of the technical staff at Bell Laboratories from 952 until 966 an Manager of Research an Development at the Feeral Sstems Division of AT&T from 966 to 97. In 97 he joine IBM as a member of the research staff. Kamaugh has mae funamental contributions to the application of igital techniques in both computing an telecommunications. His current interests inclue knowlege-base sstems in computers an heuristic search methos.

21 -2.4 Minimization of Circuits 769 :EE :EE :EE (a) (b) (c) FIGURE 3 K-maps for the Sum-of-Proucts Epansions in Eample. :EE :58 :B8] (a) (b) (c) FIGURE 4 Simplifing the Sum-of-Proucts Epansion from Eample 2. goal is to ientif the largest possible blocks, an to cover all the I s with the fewest blocks using the largest blocks first an alwas using the largest possible blocks. EXAMPLE 2 Simplif the sum-of-proucts epansions given in Eample. Solution: The grouping of minterms is shown in Figure 4 using the K-maps for these epansions. Minimal epansions for these sums-of-proucts are (a), (b) +, an (c) +... A K-map in three variables is a rectangle ivie into eight cells. The cells represent the eight possible minterms in three variables. Two cells are sai to be ajacent if the minterms that the represent iffer in eactl one literal. One of the was to form a K-map in three variables is shown in Figure 5(a). This K-map can be thought of as ling on a cliner, as shown in Figure 5(b). On the cliner, two cells have a common borer if an onl if the are ajacent. To simplif a sum-of-proucts epansion in three variables, we use the K-map to ientif blocks of minterms that can be combine. Blocks of two ajacent cells represent pairs of minterms that can be combine into a prouct of two literals; 2 2 an 4 blocks of cells represent minterms that can be combine into a single literal; an the block of all eight cells ji jiz i i i (a) (b) FIGURE 5 K-maps in Three Variables.