DTIC ADA599937: Multiple-Valued Logic Operations with Universal Literals

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by  Defense Technical Information Center

Multiple-valued Logic Operations with Universal Literalst
Gerhard W. Dueck Jon T. Butler
Dept. of Math. and Comp. Sci. Dept. of Elec. and Comp. Eng.
St. Francis Xavier University Naval Postgraduate School
Antigonish, N. S. B2G 1CO Monterey, CA 93943-5121
However, in current-mode CMOS,w indow literals do not
Abstract
provide as efficient implementation as universal literals
We propose the use of universal literals as a means of
[DUE92b]. Also, the cost-table was shown to produce
reducing the cost of multiple-valued circuits. A universal
more cost effective implementations. Most minimization
literal is any function on one variable. The target
procedures using cost-tables in the past have been limited
architecture is a sum-of-products structure, where sum is
to one or two input variables kEI91, LEE83, ABD881.
the truncated sum and product terms consist of the
This restriction makes these minimization procedures inept
minimum of universal literals. A significant cost
for most practical functions. Dueck [DUE92b] proposed
reduction is demonstrated over the conventional window
an algorithm which combines cost-tables with direct cover
literal. The proposed synthesis method starts with a sum-
minimization that produces good results for functions with
of products expression. Simplification occurs as pairs of
up to four variables. Unfortunately, functions with more
product terms are merged and reshaped. We show under
than four variables require excessive CPU time.
what conditions such operations can be applied.
There are two serious limitations inherent in the direct
cover method. One is that it operates on minterms. This
1 Introduction implies that the function, even when it is given in a near
minimal form, has to be expanded into minterms.
Memory requirements become very large when the number
The goal of this paper is to develop the theoretical
of input variables increases. Also, the number of
framework needed to provide more efficient
implicants that have to be considered to cover a given
implementations of multiple valued logic circuits.
minterm may be very large. This occurs when a function
Towards this end, we propose the use of more complex
can be covered by a few large product terms.
literal functions than have been used in the past. Previous
Recently, Dueck et al. lDUE92aI proposed an algorithm
literal functions include the so-called window literal. We
that manipulates product terms directly without breaking
replace this with the universal literal. which is any logic
them into minterms. The algorithm makes use of the op-
function on a single variable. We propose a synthesis
erations: merge, sharp, and reshape. These operations are
method that combines and divides product terms. While
applied in a nondeterministic fashion, guided by the
this makes synthesis more complicated, it results in more
simulated annealing principle. Results from this
efficient realizations. There are two contributions of this
algorithm, which has been incorporated into HAMLET, are
paper 1. establishing a theoretical basis for the new
encouraging.
operations and 2. demonstrating their efficiency.
We want to be able to manipulate product terms in a
Many multiple-valued logic minimization algorithms
sum-of-product expression consisting of universal literals.
use the direct cover method pOM81, BES86, DUE871. In
In this paper, we redefine the primitive operations that
direct cover minimization, a minterm is selected according
have been successfully used with window literals and apply
to some criteria. From all the implicants that cover the
them to universal literals. This provides a basis for
selected minterm, the best one is chosen to be part of the
algorithms operating on universal literals.
solution. This process is iterated until all minterms are
covered. Several direct cover algorithms have been 2 Definitions and Notation
implemented in the PLA minimization tool HAMLET
[YUR90]. These algorithms use window literals.
Let xi be a variable that can assume any logic value in
the set R = (0,1;..,r - l), where r denotes the radix. Let
t Research supported by the Natural Sciences and Engineering X = (x~,x2,~..,xb,e) a set of n variables. An r-valued
RWeasesharinchgt Cono,u nDcCil othf rCouangahd ad iarnecdt bfyu nthdes Nata vtahle RNesaevaarlc hP oLsatbgroaradtuoartye, function is a mapping f:R n + R. The universal literal
School, Monterey, CA.
73
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Multiple-valued Logic Operations with Universal Literals
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13. SUPPLEMENTARY NOTES
14. ABSTRACT
We propose the use of universal literals as a means of reducing the cost of multiple-valued circuits. A
universal literal is any function on one variable. The target architecture is a sum-of-products structure,
where sum is the truncated sum and product terms consist of the minimum of universal literals. A
significant cost reduction is demonstrated over the conventional window literal. The proposed synthesis
method starts with a sumof products expression. Simplification occurs as pairs of product terms are
merged and reshaped. We show under what conditions such operations can be applied.
15. SUBJECT TERMS
16. SECURITY CLASSIFICATION OF: 17. LIMITATION OF 18. NUMBER 19a. NAME OF
ABSTRACT OF PAGES RESPONSIBLE PERSON

a. REPORT b. ABSTRACT c. THIS PAGE 7
unclassified unclassified unclassified
Standard Form 298 (Rev. 8-98)
Prescribed by ANSI Std Z39-18
e UOU~...U,-~ is a one-variable function f(xi), such Function fl covers f2, if fl@)2f2@),f or all
that f(j)= U,. For example, the identity function assignments of values p, where 2 is the greater-than-or-
f(x1) =e 0123 >%, has the property that x1 = 0 yields equal-to operator with logic values viewed as integers. For
f = O , xl=l yields f=1, x1=2 yields f =2, and example, the function e 1220 e 0221 >xz covers
xl = 3 yields f = 3. e 0120 >', e 0210 >x,.
Aproduct term of literals P(x1,x2,...,xn)= qoa11 The huncated sum operation, denoted +, is
h...aflsl,- l the p> ,,ropee raty20, ath21a-t *f-oar 2e,a-c1>h * ,a s.si*g.nemQ, OeUnnt1 a . 'o.fU nv,-a1l ue>Is. to where +f lo @n It +he f 2riVgh)t = i Ms aNrithrm- el,tifcl @ad)d+itif o2n@ w)1it.h logic
the variables xl,x2,.--,xn,P is the minimum of the values are viewed as integers.
values achieved by the literals for a. For example, We say that two universal literals e aioail...ai,-l> xi
eO123>,, <0123>,, is the minimum function; i.e. and e biobil* bir-l >xi intersect iff
r-1
P(xl,x2)=MIN(xl,q).
A product term representation is not unique. For Caijbij f 0.
example, for Pl(xl,x2) = < 3210 >,, e 1220 >,, and For example, literals e 1j0-03 0 and e 1010 intersect,
P2(x1,x2) =e2 210 >% , < 1230 > ,,, we can write while e 0223 and e lo00 do not. The distance
PI= P2. This product term is shown in Figure 1. There, between two product terms is the number of variables for
blank squares correspond to assignments where the
which the corresponding literals do not intersect. For
function is 0. Two product terms are said to be equivalent
example, the distance between product terms
if they realize the same function. Thus, <3210>,,
P4(x1,x2)=e1030>x<10 223>,, and Ps(xl,x2)=
e 1220 >% , and e 2210 >x, e 1230 ,>, are equivalent. e 1010 e lo00 >x, is 1, since there is one variable,
x2, where the literals do not intersect.
3 Product Term Operations
In this section, we describe operations that can be
performed on product terms. These operations have been
defined elsewhere for window literals [DUE92]. Here, they
3 t t m are extended to universal literals. Most operations are
Figure 1. A two-variable 4-valued function. intuitively easy to understand, but the conditions under
which they apply and their implementation in computer
Let aimax = MAX(aio,ail,...,a~-1), where 1, I i I n. programs are not trivial.
A product term is said to be normalized if all aya are
equal. For example, for Pl (xl ,x2) =e 3210 e 1220 >xz 3.1 The Merge Operation
and P2(x,,x2)=<2210>xle1230>x2, we have
a;lrx,ay = 3,2 and 2.3, respectively. Neither is A fundamental operation in our proposed method is the
normalized. However, P3(xl,x2)= e2 210 >+ e 1220 >", merging of two product terms on universal literals.
which is equivalent to P1(xlrx2)a nd P2(x1,x2), is Specifically, we ask under which conditions the truncated
normalized with a? = a y = 2. sum of two product terms, A and B, can be expressed as a
Let ahm= -- MIN(a;lrx,a~,...,a.~ )A product sincgl.e product C. The first result below specifies the form
term P can be converted into a normalized one equivalent of
to P by replacing all aii >ammPx with amma'. Note Lemma 1: If the truncated sum of two product terms, A
that the normalized product term is unique. For the and B, is expressible as a single product term
remainder of this paper, we will only consider normalized A+E=C=<C~OC~~...<C~~2~0-~~2 1***>~x22 ~-1 ...
product terms on universal literals. e c,oc,l~-c,-l then
Let a minterm be a function f(x1,x2,...,x,) that is c1.J . <- a1.J . + bIJ. .9 (1)
nonzero for exactly one assignment of values to the where aii and bij are components in the literals of A and
variables. A minterm can be represented by a product B,respectively,for l l i l na nd OS jIr-1.
term, where each literal consists of all 0 components Proof: Consider a cB, and choose an assignment
except one. a = (a1,a2,...,oaf ,v)a lues to variables, such that
The size of a product term is the number of assign- a,= j and a,,= j', where i'iti and ci7. = chmx.
ments of values to variables for which the product term is Because the product .term is normalized, we can always
nonzero. For example, the size of the product term choose ci7, as cmmax. It follows that C(a)=
e 2210 e 1220 >xz is 9. MIN(C~~,,C~,~,...,=C ,c~ii.) On the contrary, if
74
cy > ay + bV , then from A(a)I a,, and E(a)I bh, , Example 2. From Example 1, if A=<1030>,,
C(a)= cV > ay + by 2 A(a)+ E(a), which contradicts < 0223 >x2 and E = < 1010 >%, loo0 >%,, then
C=A+E. Q.E.D. C=A+B is expressible as a single product term
EEx =a<m 10p1l0e <1 lo. 00L >e+t . AS=e<e1 F0i3g0u>re, <20. 22C3 =>x A2 + Ea nids ecx c=e<p t1 f0o3r0 o ne <ca 1s2e2, 31 >% =,. c,, R<e ccaloll + th calot =Ck 1a,+ I 1 .U ,,S in+ce b hth,i s,
expressible as a single product term C=<1030>,, does satisfy (2) in Lemma 2, then it must not satisfy a) or
< 1223 >% , , which has the property Cka, = aka, + bh, , b) of Lemma 2. This can be seen as follows. Consider
except for k = 1 and a, = 0, where clo < qo+ blo. a = (0,2). Since A(a)+ E(a)= 1, b) is not satisfies.
But neither is a), as follows. For i = 1, A(a) f a,,, and
for i=2, E(a)*a,.
Definition 1: Two product terms, A and E, can be
merged iff there is a product term, C, such that C = A + E,
where + is the truncated sum.
For example, the product terms A and E in Example 2
can be merged into C. Note that Lemmas 1 and 2 give
A + B -- C conditions under which two product terms can be merged.
A third condition follows.
Figure 2. Merging of two product terms.
Lemma 3: If product terms A and B can be merged, then
the distance between them is no greater than 1.
Example 1 shows that the inequality of (1) cannot be
Prmk On the contrary, assume A and B are distance two
replaced by equality. For many examples, equality holds
or more apart. Thus, at least two of the universal literals
in (1). For such cases, we can show necessary and
in A do not intersect with their corresponding literals in E.
sufficient conditions for the merging of two product terms.
Let i andj denote the indices of these literals. That is, for
eLxepmremssaib 2le: asL aet s Ain galned p rBo dbuec ttw teorm p rCod, uwcht etreer ms. A + B is AE==.*....<< baiioobaiil,*....*b,~i- ,i ,->xl i .*e< bQjjoobajjll.*~*.*bQjjrr->-lxl j awnde
Cjaj = ajaj + bjaj 9 (2) have a,b,=O and at.db :.p =,.Op ,f,o r OIkIr-1. Let
for all 1 S j I n and 0 Ia i I r - 1 iff for all assignments a = a,a,...a,, and /3 be nonzero minterms
a = (a,,a,,...,oaf ,v,a)lu es to variables x,,~,,..., and in A and B, respectively. It follows that ala, >O,
x, either a,,, > 0, am, > 0 and blp, > 0, bZp,> 0, bnsm> 0.
a) There exists an i such that both A(a) = ai, and Assume A and E can be merged into one product term C.
E(a)= biai Thus, for C =...ec iOcil--.cir>-X,i < CjOCjl"'
or b) A( a)+ E(a)= r - 1. cjr-, >,.--., we have C=A +B, and it follows that
Proof: (if) Assume a) holds. Then, C(a)= A(a)+ Cia, >O', C2a2 >O, cmn >O and clpl >O, c2p2 >O,
E(a) = aiai + biai I uiaj + biaj for all 1 I J I n . The ... c > 0. Consider the assignment y = ala2--.
latter inequality is the condiuon for expressibility by a ai-lpai+l.--a,,I.t follows that C(y)> 0, since
single product term. Assume b) holds. We claim there is .C.i.a , >O, **** Ci-la,, cisi >Os Ci+lai+, >O,
noj such that ajaj+ bja, e r-1. On the contrary, if so, , cmm > 0. By the nonoverlap of A and E, we have
then A(a)+E(a)<r-f. But aja.+ bjaj 2r-1 for all uiaibiai= 0 and ujp.bjp=j 0. Since aiai > 0 and
1 I j I n implies C(a)= r - 1 and is the condition for bja. > 0, it follows thh biai = 0 and ujPj = 0. But, the
expressibility by a single product term. fodner implies that E(y)= 0, while the latter implies
(only if) Assume that neither a) nor b) hold for some A(y)=O. Thus, C(y)=A(y)+ E(y)=O, a contra-
assignment a'=(a;,a;;..,a:).W e show that A+E is diction. Q.E.D.
not expressible as a single product term such that (2) We now establish the conditions under which two
holds. Since A and B are product terms, product terms at distance 1 or less may be merged. We
A(a') = spa, I ajal and B(a') = bsa: I bjaj, for all consider three different cases.
1 I j I n. because a) does not hold, spa; < usa; or Case 1: Two product terms at distance 1 can be merged
bsa: bW;. Because b) does not hold, ;,a + bsa: < if the conditions in Lemma 4 are satisfied.
r - 1. Thus, for all 1 I i I n , aiai + biai > ups; + bsa; . Lemma 4: Let A and B be two product terms at distance
Further, r - 1 > C(a')= + bsa; < aiai + biarI cia:, 1, and let xi be the variable for which the corresponding
which contradicts the condioon that C is a single product literals do not intersect. A and E can be merged iff for all
term satisfying (2). Q.E.D. lIjIn,suchthat i#j and OSkIr-1,
75
a) if ajk > bjk, then bP = bminma,x cZa, ,-*-,c,~ ) = bjk. Consider the assignment a’ = (a;,
CX;,...,~:)su, ch that a; = aj,e xcept for j=i, in which
b) if bjk > ajk, then Ul.k = Uminmax . case as = bmmmx. It follows that C(a’)= A(a’)+
Example 3. Figure 3a below illustrates an A and B that B(a’) = bminmaaxn d that cjk = bjk.
satisfy the conditions in Lemma 4. In Figure 3a, Consider an assignment p = Cpl ,p2,.-.,p,,)o f values
A =c0 130 c 1300 and B =e 0120 to the variables, such that pi = k and, for lfj, pl = h,
>Xa .
e 0022 >+. Their truncated sum C =e 0130 where a,,, = ammx Because the product term is
e 1322 >xz. However, the functions D =e 0310 normalized, all a,?” are the same, and A@)= ajk. But,
e 1300 and E =c0 120 c 0022 shown in A and B do not intersect, and B(P) = 0. Thus, if A and B
>x2
Figure 3b, cannot be merged, because they don’t satisfy the can be merged into a single product term C, then
conditions in Lemma 4. C@)= A@)+ B@)= ajk. It follows that MIN(clal,
cZa, ,.-.,c,. ) = ajk. Consider the assignment p’ = @ ,;
p;,.-.,p:), such that = pi, except for j=i, in which
case p( = Ummsx. It follows that C@’) = A@’)+
BV’) = ammax and cjk = ajk. But ajk > bjk, and there
are contradictory requirements on cjk. It follows that A
and B cannot be merged. Q.E.D.
Case 2: Consider two product terms A and B at distance
0, such that B covers A. The merged product term C is
obtained as follows:
a b
Figure 3. Illustration for Lemma 4. C t B
for each minterm a included in A
Proof: (if) Assume that the conditions hold. Let modify C such that C(a)= A(a)+B (a)
cjk = MAX(ajk,bj6).W e show that for all assignments
a of values to vanables that C(a )= A( a)+ B( a). There Once C has been obtained we have to verify that it is
are two cases; either C(a)= 0 or C(a)f 0. If C(a)= 0, indeedequalto A+B.
then for at least one j and one k, cjk = 0, and thus for each minterm a included in C
Ujk ‘Ujk =o (Since C p= MM(ajk,bjk)). Thus, we must have C(a)= A(a)+ B(a)
C(a)= A(a)+ B(a). Now consider, C(a)f 0. Since A otherwise A and B cannot be merged.
and B are at distance 1 either A(a)= 0 or B(a)= 0. Example 4. Given A =c 0321 e 0323 and
>x2
Without loss of generality, assume that B(a)= 0. We B =c 0010 >x, e 0100 >x, we obtain C =e 0331
must show that C(a)=A(a). We have e 0323 >x2 . But C f A + B since C(2,3)f A(2,3)+
C(a)= MIN(cjaj ) = MIN(MAX(ajaj ,bjaj )) and biai = 0 B(2,3). The left hand side of the inequality evaluates to
since A(a)f 0. If aja. 2 bja,, then 3, whereas the right hand side is equal to 2.
MAX(ajaj,bjaj)=ajqjth is hold for i=j. If Finally, we consider product terms at distance 0 that do
ajaj c bjaj then condition b) applies, i.e. ajaj = aminmax, not fall into Case 2.
however aiai Saminmxe bjaj (in other words, it will Case 3: If A can be expressed as A = A, + A,, such that
never be the minimum) and this implies that the pair (Al,B) falls into Case 1 and (A2,B) falls into
MIN(MAX(ajaj *bjaj) I = MIN(ajajI - Case 2, then A and B can be merged if the following 2
(only if) Assume that A and B can be merged. That is, conditions hold
there exists a product term C such that C=A+B. Assume, 1) A, and B can be merged (let C, = A2 + B)
on the contrary, that the conditions are not satisfied. 2) A1 and C, can be merged (let C = Cl + A1)
Specifically, assume a) is not satisfied; the argument forb) If the decomposition A = A, + A, exists, then it is unique.
is similar. That is, suppose there is a j and a k such that
If such decomposition does not exist, then A and B cannot
ajk> bjk,b ut that bjk f bmmaX.S ince the product terms
be merged. The following condition must hold for A to be
are normalized, the latter inequality implies that expressed as A1 + A,, as described above
b.j-k. c, bam,oif,n )vma lauxe.sC toon tshide evra raianb laesss, isguncmh etnhat t aa i= = ( ak la,nad2, , [(Qik = 0) and (bik = o)] Or [(Qik f 0) and (bik f o)] ,
for l#j, al = h, where blh= bminma. xB ecause the product for all . Olker, lsicj, and JeiIn given
term is normalized, all b y a re the same, and B(a)= bjk. 1 I j In Essentially, we split the product term along the
j* variable.
But, A and B do not intersect, and A(a)= 0. Thus, if A
and B can be merged into a single product term C, then Example 5. Consider the product terms A =c 2100
C(a)= A(a)+B (a)= bjk. It follows that MIN(C~,,, c 1200 >x2 c 0112 >x3 and B =c 3200 e 0133 >*,
c 0223 >x3. We can express A as a sum of two product md
terms, split along x,, A = A1 + A, =< 1100 b ; =bjk, OIk<t-l, lIj<i,a ndic jln.
c lo00 >xz < 01 11 >x3 + c 2100 c 0200 >x2 < 01 12 The consensus term C = A’ + B”.
B can be merged with A, (Case 2) as We illustrate our definition of consensus with the
A2 + B =< 3200 < 0333 >x2 c 0223 >x3 = C1. following two examples.
Finally, Cl can be merged with A1 (Case 1) as
We Cco1n+c lAu1 d=e t<h a3t2 A00 a nd B< c1a3n3 b3e > xm, e<rg 0e2d2. 3 >’, = A + B. E<x 3a1m10p l>ex2 7.a nCd onBs =id<e r0 t0h2e3 p roducct t0e1rm23s >xA =.c W33e0 0h >axv, e
A’ =c 1100 c 0110 >x2 and B3=c 0022
3.2 The Consensus Operation c 0120 >x2. Since they can be merged, the consensus
term is C =< 1122 < 0120 > .,,
Informally, the consensus term C of two product terms Example 8. Consider the product terms A =e 3300 >xl
A and B is a largest product term that includes minterms 3210 >x, and B =c 0023 < 0123 >x shown in
from both, such that A + B covers C. The distance be- Figure 4. We have A’ =< 2200 > ,, c 6210 and
tween A and B must be either l or 0. We consider two B’ =c 0022 < 0120 >x2. Since they cannot be merged,
we find B” =< 001 1 < 01 10 The consensus term
CaSeS. >x2.
Case 1: First, we consider product terms A and B at is C=<2211>,1<0210~x2.
distance 1. Unfortunately, two product terms may have Case 2: We now consider product terms A and B at
more than one consensus term-according to the definition distance 0. The consensus term C of A and B includes all
given above. We illustrate this with the following minterms that are included in A and in B. We define A‘ as
example. follows;
ajk ifbjk #o
Example 6. Consider the product terms A =c 3300
u;k = io.
c 3210 >x2 and B =< 0023 c 0123 >x2 (see Figure 4.) otherwise’
We have the following seven normalized consensus terms for 0 I k I r - 1, 1 I j I n. Similarly, we define B’.
of size 8: c 1111 c 0110 >x2, c 1112 c 0120 >xz, Normalize A’ and B‘. If A’ and B‘ are mergable, then
< 1121> ,, < 0120 >x, , < 1122 c 0120 BX2, < 1211 the consensus C = A’ + B’. Otherwise, let B’’ be the
< 0210 >x2, c 2111 c 0210 >.+, and < 2211 >xl following product term;
< 0210 > .,, b!‘ P =m(bjk,q), forOlklr-landll jln,
where q is the minimum non-zero value of all bik. The
consensus C = A’ + B”.
Example 9. Consider the product terms A =c 0210
c1210>,, a n d B=<0133>, <0130>,,.
A‘ =c 0210 >x, < 0210 >xz a n d Bf =c 0130
c0130>,,. Since they cannot be merged we find
B” =c 01 10 >xl c 0110 >x2. The consensus term is
C =< 0320 < 0320 >x,.
Figure 4. Function used in Example 6. Example 10. Consider the product terms A =c 0210
< 1210 >=, and B =< 0313 >, < 0130
We give the following definition, which uniquely >x2.
A’ =c 0210 >xl 0210 >xz and Bf =< 0310
defines the consensus term of two product terms at distance
< 0130 >xz. A‘ and B’ can be merged and the consensus
1. Let xi be the variable for which the corresponding
term is C =< 0320 < 0330 >x2.
literals do not intersect. We define A‘ as follows;
u , ~ = uf~o,r 0lklr-1 3.3 The Sharp Operation
and
The sharp operation (denoted by #) has been used in
binary minimization algorithms [HON74]. In binary
logic, the sharp operation is defined as A# B = fi, where
for OSklr-1, lljci, andicjln. Similarly, we
A and B are product terms. Note that fi may not be
define B’. Normalize A’ and B‘. If A’ and B’ are
mergable, then the consensus C = A’+ B‘. If A‘ and B’ rseaatilsizfiaebs leA a =s Aa #s iBn g+l eA pBr.o dIun cat nt earnma.l ogTohues swhaayrp, wope edraetfiionne
are not mergable, then let B” be the following product
the sharp operation of two product terms A and B at
term distance 0 such that A + B = A# B + C + B# A, where C is
b; = MIN(U&,b;k), for 0 k 5 r - 1
I7
the consensus of A and B. If the distance between A and B This function requires at least five product terms when
is greater than zero, then A# B = A. A# B is a sum of window literals are used. Using window literals the
products expression that may not be unique. We illustrate function can be expressed as < 0300 < 3000
the sharp operation with the following two examples. + < om < 0222 > ,, + < 0011 < 0011 >x2
>12
Example 11. Given the product terms A =< 0330 +<0011>,<0001>,, +<OO01>,I<oo01>,2. AC-
< 3210 > ,, and B =< 0222 < 0020 >,?. A# B = cording to the cost estimates of Lei and Vranesic LEI911
<0330>,I<3200>,. In this case, there is a unique the implementation of this expression would cost 101.
solution.
Example 12. Given the product terms A=<0330>, xq2 0 1 2 3
< 3213 >X2andB =< 0022 < 0022 >%. Two possible
solutions of A# B are:
1) < 0330 < 3200 >xz + < 0300 < 0013 > ,,
2) < 0300 > ,, < 3213 > ,, + < 0030 >xl < 3200 >%
The consensus of A and B is C =< 0030 < 0033 > ,,
.
and B# A =< 0002 < 0022 > ,, We verify that the
+ +
following holds: A B = A# B C+ B# A.
Figure 5. A 2 variable function.
3.4 The Reshape Operation
Any minimization procedure must take into account
that normalized product .terms may not be cost effective.
The reshape operations of two product terms A and B is
defined as A# C + C+B # C, where C is the consensus of For example the product term < 01 11 < 0100 > ,, has a
cost of 20. The equivalent unnormalized product term
A and B. Note that when the distance between A and B is
+ < 0123 < 0100 >xz has a cost of 16. Therefore, a good
greater than 1 the reshape of A and B is A B.
minimizatlon procedure cannot be restricted to consider
only normalized product terms.
4 Comparison with Window Literals
In this paper, we have shown two operations, merge
and reshape. Merge produces one product term from two,
There is a significant increase in the complexity of the
while reshape produces two or more product terms from
analysis of sum-of-product expressions when window
two. Analogous operations have been successfully used in
literals are replaced by universal literals. However, there
a minimization algorithm with window literals PUE92aI.
also is a significant decrease in the number of product
terms. In a PLA, the space allotted to literals is large References
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[DUE871 G. W. Dueck and D. M. Miller, “A direct cover
In random logic, we can optimize space by
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Proceedings of the 18th International Symposium on
and Vranesic [LEI911 the cost of a universal literal
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and the MIN gate has a cost of 5. The following example
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minimization by simulated annealing,” Proceedings of
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Example 13. The function shown in Figure 5 can be Logic. May 1992, pp. 66-74.
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Below are 3 expressions with their corresponding costs. International Symposium on Multiple-Valued Logic,
May 1992, pp. 58-65.
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2) < 0300 < 3211 + < 0123 < 0013 > ,, 38 Journal of Research and Development, September
>x2 1974. pp. 443 - 458.
3) < 0300 e 3210 > ,, + < 0223 < 0013 > ,, 46
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79