(b) (b) f 1 = x 1 x 2 ( x 3 v x 4 )

Full text

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TsutomuSasao Jon T.Butler

DepartmentofComputerScience DepartmentofElectricaland

andElectronics ComputerEngineering

KyushuInstitute ofTechnology NavalPostgraduateScho ol

Iizuka820,Japan Monterey,CA93943-5121,U.S.A.

Abstract

A logic function f has a disjoint bi-decomposition i

f can b e represented as f = h(g 1 (X 1 );g 2 (X 2 )), where X 1 andX 2

aredisjointsetofvariables,andhisan

arbi-trarytwo-variablelogic fuction. f has a non-disjoint

bi-decompositionif canberepresentedasf(X

1 ;X 2 ;x)= h(g 1 (X 1 ;x);g 2 (X 2

;x)), where x is the common

vari-able. In this paper, we show a fast method to nd

bi-decompositions. Also,weenumeratethenumb erof

func-tionshavingbi-decomp ositions.

I Intro duction

Functionaldecompositionisabasictechniquetorealize

economical networks. If thefunction f is representedas

f(X 1 ;X 2 )=h(g(X 1 );X 2

),thenf canb e realizedbythe

networkshowninFig.1.1. Tondsuchadecomp osition,

1

X

2

X

g

h

f

Figure 1.1: A simple disjoint decomp osition.

1

X

2

X

g

h

g

1

2

f

Figure 1.2: A disjoint bi-decom-p osition.

1

X

2

X

g

h

g

1

2

x

f

Figure 1.3: A non-disjoint bi-decomp osition.

adecomp ositionchartwith2 n1 columnsand2 n2 rowsare used,wheren i

isthenumb erofvariablesinX

i

(i=1;2).

When n is large, the decomposition chart is to o large

to build. Recently, a method using BDDs has been

de-velop ed[13]. This greatly reduces memory requirements

andcomputation time. However,it isstill time

consum-ing, since we haveto check all the 0 n 1 +n 2 n1 1 partitions of n=n 1 +n 2

. Inthispaper,weconsiderbi-decompositions

oflogicfunctions,arestrictedclassoffunctional

decompo-sitionsthathavetheformf(X

1 ;X 2 )=h(g 1 (X 1 );g 2 (X 2 )).

Fig.1.2showstherealizationofthisdecomposition.

Thereasons weconsider bi-decomp ositions are as

fol-lows:

1) If f hasnobi-decomp osition, then thecomputation

timeisquitesmall.

2) Some programmable logic devices have two-input

logicelementsintheoutputs.

3) Iff hasabi-decomposition,thentheoptimizationof

theexpressionis relativelyeasy.

Arestictedclassofbi-decomp ositionshasb eenconsidered

by[8]. Thegoalsofthispap erare

1) Presentafastmetho dforndingbi-decomp ositions.

2) Enumerate the functions that have

bi-decom-positions.

Most of the proofs are omitted. They can be available

fromauthors. II Disjoint Bi-Decomposition Denition2.1 Let X = (X 1 ;X 2 ) be a partition of

the variables. A logic function f has a disjoint

bi-decomposition i f can be represented as f(X

1 ;X 2 ) = h(g 1 (X 1 );g 2 (X 2

)),wherehisanytwo-variablelogic

func-tion.

Iff hasadisjointbi-decomposition,thenf canb erealized

bythenetworkshownin Fig.1.2.

Denition2.2 Let X = (X 1 ;X 2 ) be a partition of the variables. Letn 1 andn 2

bethe numberofvariablesinX

1

andX

2

, respectively. A decompositionchart ofthe

func-tion f for a partition (X

1 ;X 2 ) consists of 2 n1 columns and2 n2

rowsof 0sand1s. The2 n1

distinctbinary

num-bers forX

1

are listedacrossthetop, andthe 2 n

2

distinct

binarynumbersforX

2

arelisteddowntheside. Theentry

forthechart correspondstothe valueof f(X

1 ;X

2 ).

Example 2.1 Twodecompositionchartsforthefunction

f(x 1 ;x 2 ;x 3 ;x 4 )=x 1 x 2 8x 3 x 4

areshown inFig. 2.1(a)

and(b). 2

Notethat thedecomposition chartis similarto the

Kar-naughmapwithadierentorderingforthecelllocations.

Denition2.3 Thenumberofdistinctcolumn(row)

pat-terns in the decomposition chart is called column (row)

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1

X

0

0

0

1

0

0

0

1

0

0

0

1

1

1

1

0

00

01

10

11

00

01

10

11

x

1

=(

,

x

2

)

X

1

0

0

0

0

0

1

0

1

0

0

1

1

0

1

1

0

00

01

10

11

00

01

10

11

x

1

=(

,

x

3

)

(a)

(b)

2

X

=(

x

2

,

x

4

)

2

X

=(

x

3

,

x

4

)

Figure2.1: Decomp osition chart.

x

y

x

y

z

z

V

Figure 3.1: A real-ization of f(x;y;z) = xyz_xyz.

x

y

z

Figure 3.2: Non-disjoint

bi-decomp osition for

f(x;y;z)=xyz_xyz.

Example 2.2 In Fig.2.1 (a), the row andcolumn

mul-tiplicities are two. In Fig. 2.1 (b), the row and column

multiplicitiesarefour. 2

Denition2.4 Let (f : X

1 ;X

2

) be the column

mul-tiplicities for f with respect to X

1 and X 2 . Let (f : X 2 ;X 1

)bethe rowmultiplicitiesforf with respecttoX

1

andX

2 .

Theorem 2.1 f has a disjointbi-decompositionof form

f(X 1 ;X 2 )=h(g 1 (X 1 );g 2 (X 2 )) i(f :X 1 ;X 2 )2 and (f:X 2 ;X 1 )2. I II Non-Disjoint Bi-Decomposition Denition3.1 Let X 1 and X 2

be disjoint sets of

vari-ables, and let x be disjoint from X

1

and X

2

. A logic

function f has anon-disjoint bi-decomposition i f can

be represented as f(X 1 ;X 2 ;x) = h(g 1 (X 1 ;x);g 2 (X 2 ;x)),

wherehisatwo-variablelogicfunction. Inthiscase,xis

calledthecommonvariable.

Afunctionf withanon-disjointbi-decompositioncanb e

realizedbythenetworkshowninFig.1.3.

Lemma3.1 Let X = (X

1 ;X

2

;x) be a partition of the

inputvariables. Leth(g

1 ;g

2

)beanarbitrarylogicfunction

oftwovariables. Then,

h(g 1 (X 1 ;x);g 2 (X 2 ;x))= xh(g 1 (X 1 ;0);g 2 (X 2 ;0))_xh(g 1 (X 1 ;1);g 2 (X 2 ;1)):

Denition3.2 Let x be the common variable of the

non-disjointbi-decomposition. Letf(X

1 ;X

2

;a)bea

sub-function,wherexissettoa0or 1.

1

X

1

0

0

0

1

0

0

0

1

0

0

0

0

1

1

1

00

01

10

11

00

01

10

11

x

1

=(

,

x

2

)

X

1

0

0

0

1

1

1

1

0

1

1

1

0

1

1

1

0

00

01

10

11

00

01

10

11

x

1

=(

,

x

2

)

(b)

f

1

=

x

1

x

2

(

x

3

v

x

4

)

(a)

f

0

=

x

1

x

2

x

3

x

4

2

X

=(

x

3

,

x

4

)

2

X

=(

x

3

,

x

4

)

Figure3.3: FunctionsinExample3.2

Theorem 3.1 f(X

1 ;X

2

;x)hasa non-disjoint

bi-decom-position of the form h(g

1 (X 1 ;x);g 2 (X 2 ;x)) i f(X 1 ;X 2 ;0) and f(X 1 ;X 2

;1) have disjoint

bi-decom-positions h(g 01 (X 1 );g 02 (X 2 )) and h(g 11 (X 1 );g 12 (X 2 )), respectively.

Example 3.1 Consider the three-variable function:

f(x;y;z) = xyz _ xyz. Suppose modules that

real-izes any function of two variables can be used. The

straightforwardrealizationshown inFig.3.1 requires ve

modules. The Shannon expansion with respect to x is

f(x;y;z) = xf(0; y;z) _xf(1;y;z); wheref(0;y;z) =

yz; andf(1;y;z) = yz: Note that both f(0;y;z) and

f(1;y;z)havebi-decompositionswithh(x;y)=xy. Since,

g 1 (x;y)=xg 01 (X 1 )_xg 11 (X 1 )=xy_xy; andg 2 (x;y )= xg 02 (X 2 )_xg 12 (X 2 ) = xz _xz: We have f(x;y;z) = g 1 (x;y)g 2

(x;z) = (xy_xy)(xz_xz): From this

expres-sion, we have the network in Fig. 3.2. This network

re-quiresonly threemodules. 2

Example 3.2 Consider the ve-variable function f =

x 5 f 0 _x 5 f 1 ,wheref 0 andf 1

areshowninFig.3.3. Since

bothf

0 andf

1

havedisjointbi-decompositionsoftheform

h(g 1 (X 1 );g 2 (X 2 )),f =x 5 f 0 _x 5 f 1

has anon-disjoint

bi-decompositionasfollows: f=x 5 fx 1 x 2 8x 3 x 4 g_x 5 fx 1 x 2 8(x 3 _x 4 )g =fx 5 (x 1 x 2 )_x 5 (x 1 x 2 )g8fx 5 (x 3 x 4 )_x 5 (x 3 _x 4 )g:

Theconverseistruealso. 2

Up to now, we only considered the case where there is

a singlecommon variable. However,the theoremcan b e

extenedtok commonvariables,where k2.

Denition3.3 Let X 1 , X 2 , and X 3 be disjoint sets of variables. Let f(X 1 ;X 2

;a) be the sub-functions, where

X

3

is set to a 2 f0;1g k

, and k denotes the number of

variablesinX 3 . Theorem 3.2 Let X 1 , X 2 , and X 3 be disjoint sets of

variables. Then, f has a non-disjoint bi-decomposition

of form f(X 1 ;X 2 ;X 3 ) = h(g 1 (X 1 ;X 3 );g 2 (X 2 ;X 3 )) i f(X 1 ;X 2

;a) has a decomposition of the form

h(g 1a (X 1 );g 2a (X 2 )) for all possible a 2 f0;1g k , where

kdenotes the numberof variablesinX

3 .

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Inthissection,weshownecessaryandsucient

condi-tionsfor a functionto havea disjointbi-decomposition.

Then, we show ecient algorithms to nd disjoint

bi-decompositions. In the previoussections, h(g

1 ;g

2 ) is an

arbitrarytwo-variable logic function. To nd a disjoint

bi-decomposition,weneedto consideronlythreetyp es:

1) ORtype: f =g 1 (X 1 )_g 2 (X 2 ), 2) ANDtype: f =g 1 (X 1 )g 2 (X 2 ),and 3) EXORtyp e: f =g 1 (X 1 )8g 2 (X 2 ).

Sincef hasan AND typ e disjointbi-decomposition i

f

hasORtypedisjointbi-decomp osition,weonlyconsider

theORtypeandEXORtypebi-decomp ositions.

Denition4.1 x and x are literals of a variable x. A

logicalproductwhichcontainsatmostoneliteralforeach

variable is called a product term or a product.

Prod-uct terms combined with OR operators form a

sum-of-productsexpression(SOP).

Denition4.2 Aprimeimplicant(PI)pofafunctionf

isaproductterm which impliesf,such that the deletion

of anyliteral fromp results in anew productwhich does

notimplyf.

Denition4.3 An irredundant sum-of-products

expres-sion(ISOP)is anSOP, whereeachproduct is aPI, and

no product can be deleted without changing the function

representedby the expression.

Denition4.4 Letf(X)beafunctionandpbeaproduct

of l iteral (s) in X. The restriction of f to p, denotedby

f(Xjp)isobtainedasfollows: If x

i

appearsinp, thenset

x

i

in1inf,andif x

i

appearsinp, thensetx

i in0inf. Example 4.1 Let f(x 1 ;x 2 ;x 3 ) = x 1 x 2 _x 2 x 3 and p = x 1 x 3

. f(Xjp) is obtained asfollows: Set x

1 =x 3 =1 in f,andwehavef(Xjx 1 x 3 )=f(1;x 2 ;1)=x 2 _x 2 =1. 2 Lemma4.1 pisanimplicantof f(X), if(Xjp)=1. Example 4.2 By Lemma 4.1, x 1 x 3 is an implicant of x 1 x 2 _x 2 x 3 ,showninExample 4.1. 2

Theorem 4.1 (ORtypedisjointbi-decomposition)f has

adisjointbi-decompositionof formf(X

1 ;X 2 )=g 1 (X 1 )_ g 2 (X 2

)i everyproductin anISOPforf consistsof

lit-eralsfromX 1 onlyorX 2 only. Denition4.5 x 0 =x. x 1 =x. Corollary4.1 If f(x 1 ;x 2 ;:::;x n

) has aPI of the form

x a1 1 x a2 2 111x a n n , wherea i

2f0;1g,then f has no OR type

disjointbi-decomposition. i 1 p 2 _111_p t

beanirredundantsum-of-productsexpression

forf,wherep

i

(i=1;2;:::;t)arePIsoff. Let5

0 bethe

trivialpartitionof f1;2;:::;ng,5

0

=[f1g;f2g;:::;fng].

Algorithm 4.1 (OR type disjoint bi-decomposition:

f(X 1 ;X 2 )=g 1 (X 1 )_g 2 (X 2 )). 1. For i = 1 to t, form 5 i from 5 i01 by merging two blocks 1 and 2 of 5 i01

if atleastone literal inp

i occursin both 1 and 2 . 2. If 5 t

has at least two blocks, then f(X

1 ;X

2 ) has a

disjoint bi-decomposition of the form f(X

1 ;X 2 ) = g 1 (X 1 )_g 2 (X 2 ), with X 1

the union of one or more

blocksof5

t andX

2

theunionoftheremainingblocks.

Example 4.3 Consider the ISOP: f(x

1 ;x 2 ;:::;x 6 ) = x 1 x 2 _x 2 x 3 _x 4 x 5 _x 5 x 6 . The productsx 1 x 2 andx 2 x 3 show that x 1 , x 2 , and x 3

are in the same block. Also,

the products x

4 x 5 and x 5 x 6 show that x 4 , x 5 , and x 6

are in the same block. Thus, we have the partition

[f1;2;3g;f4;5;6g]. The corresponding OR type disjoint

bi-decomposition is f(X 1 ;X 2 )=g 1 (X 1 )_g 2 (X 2 ), where X 1 =(x 1 ;x 2 ;x 3 )andX 2 =(x 4 ;x 5 ;x 6 ). 2

Example 4.4 Consider the function f with an ISOP:

f(x 1 ;x 2 ;x 3 ;x 4 ;x 5 )=x 1 x 2 x 3 _x 3 x 4 x 5 . 1) Theproductx 1 x 2 x 3 showsthatx 1 ,x 2 ,andx 3 belong

tothesame block.

2) Theproductx 3 x 4 x 5 showsthatx 3 ,x 4 ,andx 5 belong

tothesame block.

Thus, all the variables belong to the same block. From

this,itfollows thatf has no ORtypedecomposition. 2

Theorem 4.2 (AND type disjoint bi-decomposition) f

has a disjoint bi-decomposition of form f(X

1 ;X 2 ) = g 1 (X 1 )g 2 (X 2

)ieveryproduct inanISOPfor f consists ofliteralsfromX 1 onlyor X 2 only.

Lemma 4.2 [15]Anarbitraryn-variablefunctioncanbe

uniquelyrepresentedas f(x 1 ;x 2 ;:::;x n )=a 0 8(a 1 x 1 8a 2 x 2 81118a n x n ) 8(a 12 x 1 x 2 8a 13 x 1 x 3 81118a n01n x n01 x n ) 81118a 12111n x 1 x 2 111x n ; (4.1) wherea i

2f0;1g. The aboveexpression iscalleda

posi-tivepolarityReed-Mullerexpression(PPRM).

Foragivenfunctionf,theco ecientsa

0 ,a 1 ,a 2 ,:::,a 12111n

areuniquelydetermined. Thus,thePPRMisacanonical

representation. Thenumb erofpro ductsin(4.1)isatmost

2 n

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has a disjointbi-decomposition of the form f(X 1 ;X 2 )= g 1 (X 1 )8g 2 (X 2

)ieveryproductinthePPRMforf

con-sistsofliteralsfromX

1

onlyorX

2 only.

Corollary4.2 If the PPRM of an n-variable function

has the product x

1 x

2 111x

n

, then f has no EXOR type

disjointbi-decomposition.

Theorem 4.4 When f has an EXOR type disjoint

bi-decomposition, the number of true minterms of f is an

evennumber.

Corollary4.3 When the number of true minterms off

is an odd number, then f does not have an EXOR type

disjointbi-decomposition.

The signicance of this observation is that at least one

halfofthefunctionscanb equicklyrejectedascandidates

forEXORtyp edisjointbi-decomposition.

Let x

i

(i = 1;2;:::;n) b e the input variables of f.

Let p 1 8p 2 81118p t b e PPRM for f, where p i (i =

1;2;:::;t) are products. Let, 5

0

be thetrivial partition

off1;2;:::;ng,5

0

=[f1g;f2g;:::;fng].

Algorithm 4.2 (EXOR type disjoint bi-decomposition:

f(X 1 ;X 2 )=g 1 (X 1 )8g 2 (X 2 )). 1. For i =1 to t, form 5 i from 5 i01 by merging two blocks 1 and 2 of 5 i01

ifat leastone literalin p

i occursinboth 1 and 2 . 2. If5 t

hasatleasttwoblocks,thenf(X

1 ;X

2

)hasa

dis-jointbi-decompositionofformf(X

1 ;X 2 )=g 1 (X 1 )8 g 2 (X 2 ), with X 1

the union of one or more blocksof

5

t andX

2

the unionofthe remainingblocks.

Example 4.5 Consider the PPRM: f(x

1 ;x 2 ;:::;x 6 ) = x 1 x 2 8x 2 x 3 8x 4 x 5 8 x 5 x 6 . The products x 1 x 2 and x 2 x 3 show that x 1 , x 2 , and x 3

are in the same block.

Also, the productsx

4 x 5 and x 5 x 6 show that x 4 , x 5 , and x 6

are in the same block. Thus, we have the partition

[f1;2;3g;f4;5;6g]. The corresponding EXOR type

dis-joint bi-decomposition is f(X 1 ;X 2 ) = g 1 (X 1 )8g 2 (X 2 ), whereX 1 =(x 1 ;x 2 ;x 3 ) andX 2 =(x 4 ;x 5 ;x 6 ). 2

Algorithm 4.3 (Non-disjointbi-decomposition).

f(X 1 ;X 2 ;x i )=g 1 (X 1 ;x i )g 2 (X 2 ;x i ), wheredenotes

eitherOR,AND,orEXOR.Let(X

1 ;X 2 ;x i )beapartition ofthe variablesx 1 ,x 2 ,:::,andx n . Fori=1 ton,do i) Let f 0i = f(X 1 ;X 2 ;0). (Set x i to 0). Let f 1i = f(X 1 ;X 2 ;1). (Set x i to1). ii) If bothf 0i andf 1i

have the sametypeof disjoint

bi-decompositionswiththe samepartition,thenf has a

non-disjointbi-decomposition.

5.1 ORtype disjoint bi-decomposition

Weassumethat thefunctionis givenasanISOP with

t pro ducts. Note that t 2 n01

. The time to form the

partitionofvariablesisO(n1t).

5.2 EXORtyp edisjointbi-decomposition

APPRMcanb erepresentedbyafunctionaldecision

di-agram(FDD[5,15]). Eachpathfromtherootno detothe

constant1 node corresp ondsto aproduct in thePPRM.

Thus,thepartitionofthe inputvariablesisdirectly

gen-erated from the FDD. Thenumber of paths in an FDD

is O(2 n

), where n is the numb er of the input variables.

However,wecanavoidexhaustivegenerationofpaths as

follows: Letp

1 andp

2

b e productsina PPRM.Ifallthe

literalsinp

1

alsoapp earinp

2

,thenp

2

neednotb e

gener-atedintheAlgorithm,sincethepro ductp

1

thatcontains

more literals than p

2

is more important. By searching

thepathswithmoreliteralsrst,wecanecientlydetect

functionswithnodisjointbi-decomp osition.

Example 5.1 Consider the function f(X) given as a

PPRM:f(X)=x 1 8x 1 x 2 8x 3 x 4 8x 1 x 2 x 5 x 6 . In

construct-ingthe partitionof X,weneednotconsiderthe products

x 1 orx 1 x 2 ,sincex 1 x 2 x 5 x 6

hastheliteralsofx

1 andx 1 x 2 .

Inthis case,the product x

1 x 2 x 5 x 6 shows that x 1 ,x 2 ,x 5 , andx 6

belongto the samegroup. Also, the productx

3 x 4 showsthatx 3 andx 4

belongtothesamegroup. Thus,Xis

partitionedas X =(X 1 ;X 2 ), where X 1 =(x 1 ;x 2 ;x 5 ;x 6 ) andX 2 =(x 3 ;x 4 ). 2

Denition5.1 Let pbea product. The set of variables

inpisdenotedby V(p)=fx i jx i orx i appears inpg:For example,V(x 1 x 2 x 4 )=fx 1 ;x 2 ;x 4 g

Denition5.2 LetF beaPPRM.Aproductpissaidto

havemaximalvariablesetV(p)ifthereisnootherproduct

p 0

suchthatV(p)V(p 0

).

Example 5.2 For the PPRM, F = x

1 x 2 8 x 1 x 3 8 x 1 x 2 x 3 8 x 4 , V(x 1 x 2 ) = fx 1 ;x 2 g, V(x 1 x 3 ) = fx 1 ;x 3 g;V(x 1 x 2 x 3 ) = fx 1 ;x 2 ;x 3 g, and V(x 4 ) = fx 4 g. Thus,x 1 x 2 x 3 andx 4

havemaximal variablesets. 2

Theorem 5.1 A function f has an EXOR typedisjoint

bi-decompositionif afunction f 0

fromthe PPRMof f by

eliminating implicants not having maximal variable sets

hasan EXORtypedisjointbi-decomposition.

The following theorem says that if a function has an

EXOR type disjoint bi-decomp osition, then the number

ofproductsinthePPRMisrelativelysmall.

Theorem 5.2 If f has adisjointbi-decompositionofthe

form f(X 1 ;X 2 )= g 1 (X 1 )8g 2 (X 2

), then the number of

productsinthe PPRMisatmost2 n 1 +2 n 2 01,wheren i

isthe numberof variablesinX

i

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Bi-Decompositions

6.1 Functions with a small numb er ofvariables

In the previous sections, we showed that disjoint

bi-decompositions areeasy to nd. Inthis section, wewill

enumeratethefunctionswithdisjointbi-decomp ositions.

Denition6.1 Afunction f issaidtobenondegenerate

ifforallx i f(jx i )6=f(jx i ):

Denition6.2 Two functions f and g are

NP-equivalent,denoted by f NP

g,i g isderived fromf by

thefollowing operations:

1) Permutationof the inputvariables.

2) Negationsof the inputvariables.

Thefollowingiseasytoprove.

Lemma6.1 If f has a disjoint bi-decomposition and if

f NP

g, then g has also the same type of disjoint

bi-decomposition.

Lemma6.2 All the two-variable functionshave disjoint

bi-decompositions.

Example 6.1 There are 2 2

3

= 256 three-variable logic

functions of which 218 are nondegenerate. These

non-degenerate functionsare grouped into16 NP-equivalence

classes as shown in Table 6.1 [9]. In this table,

the column headed by N denotes the number of

func-tions in that equivalence class. Eight classes have

disjoint bi-decompositions, and three have non-disjoint

bi-decompositions. Note that 194 functions have

bi-decompositions. 2

The numb er of functions with AND type disjoint

bi-decompositions isequal to the numb er offunctions with

ORtyp edisjointbi-decompositions.

In the case of disjoint bi-decompositions, a function

has exactly one typ e of decomp osition (Lemma 6.4).

On the other hand, in the case of non-disjoint

bi-decompositions,afunctionmayhavemorethanonetype

ofbi-decompositions.

Example 6.2 Consider the three-variable function f =

x 1 x 3 _x 1 x 2

. Thisfunctionhasthreetypesofnon-disjoint

bi-decompositions: f=x 1 x 3 _x 1 x 2

(ORtypebi-decomp osition)

=x 1 x 3 8x 1 x 2

(EXORtyp ebi-decomp osition)

=(x 1 _x 3 )(x 1 _x 2

)(ANDtype bi-decomp osition) 2

Table6.1: NP-representativefunctionsofthreevariables.

Representativefunctions N Type Property

1x18x28x3 2EXOR 2x 1 x 2 x 3 8AND Disjoint 3x 1 _x 2 _x 3 8OR Bi-Decomposition 4x 1 (x 2 _x 3 ) 24AND 5x 1 _x 2 x 3 24OR 6x 1 (x 2 8x 3 ) 12AND 7x 1 _(x 2 8x 3 ) 12OR 8x 1 8x 2 x 3 24EXOR 9x1x2x3_x1 x2 x3 4 10(x1_x2_x3)(x1_x2 _x3) 4 Non-Disjoint 11x1x3 _x1x2 24 Bi-Decomposition 12x 1 x 2 x 3 _x 2 x 3 24 13(x 1 _x 2 _x 3 )(x 2 _x 3 ) 24 14x1x2_x2x3_x3x1 8 No 15x1x2_x2x3_x1x3_x1 x2 x3 8 Bi-Decomposition 16x1x2x3 _x1x2x3 _x1x2x3 8

N: Numberofthefunctionsintheclass.

Table6.2: Numberoffunctions.

n=2 n=3 n=4

Allthefunctions 16 256 65536

Nondegeneratefunctions 10 218 64594

Disjoint AND 4 44 1660

Functionswith OR 4 44 1660

bi-decomp osition EXOR 2 26 914

Non-disjont 0 80 3860

Total 10 194 8094

6.2 The numb erof functions with

bi-decomp ositions

Lemma 6.3 [4]: Let(n) bethe numberof

nondegener-ate functionsonnvariables. Then,

(n)= n X k =0 n k (01) n0k 2 2 k 2 2 n ;

wherea(n)b(n)means lim

n!1 a(n)

b(n) =1:

Lemma 6.4 Anondegeneratefunctionf hasatmostone

typeof disjointbi-decomposition:

1. f(X 1 ;X 2 )=g 1 (X 1 )1g 2 (X 2 ), 2. f(X 1 ;X 2 )=g 1 (X 1 )_g 2 (X 2 ), or 3. f(X 1 ;X 2 )=g 1 (X 1 )8g 2 (X 2 ), where g 1 and g 2

are nondegenerate functions on one or

morevariables.

Theorem 6.1 ThenumberoffunctionsN

disjoint (n)with disjoint bi-decompositions is N disjoint (n) = A dis (n)+ O dis (n)+E dis (n),where A dis (n)=n! X k 1 ;k 2 ;:::;k n 0 1k 1 +2k 2 +111+nkn=n n Y i=1 (i)0A dis (i) i! k i 1 ki!

(6)

Table7.1: Numb eroffunctions withbi-decompositions.

Decomp ositionType Numb erofFunctions

AND 853 Disjoint OR 264 EXOR 73 AND 162 Non-disjoint OR 91 EXOR 42 O dis (n)=n! X k 1 ;k 2 ;:::;k n 0 1k 1 +2k 2 +111+nk n =n n Y i=1 (i)0O dis (i) i! k i 1 k i ! E dis (n)=2n! X k 1 ;k 2 ;:::;kn0 1k 1 +2k 2 +111+nk n =n n Y i=1 (i)0E dis (i) i! ki 1 2 k i k i !

where the sums are over all partitions of n except the

trivialpartitionn=011+012+111+01(n01)+11n(i.e.

the sum is over all partitions where k

n =0), and where A dis (1)=O dis (1)=E dis (1)=0.

Table6.2showsthenumberoffunctionswithdisjoint

bi-decompositionsupto n=4.

VI I Experimental Results

Weanalyzedthebi-decomposabilityof136 b enchmark

functions. Over these multiple-output functions,the

to-talnumberofoutputs(functions)is1908. Foreach

func-tion, we determined whether there exists a disjoint

bi-decomposition. If none existed, we determined if there

existsanon-disjointbi-decomposition(withasingle

com-mon variable). Table 7.1 summarizes our results. It is

interesting that 1190 out of 1908 functions, or 62 p

er-cent, have disjoint bi-decompositions. Of the

remain-ing718 functions, 295 havenon-disjointdecompositions.

It should be noted that more than 295 functions have

non-disjointdecomp ositions, sincea functionwith a

dis-joint bi-decomposition may also have a non-disjoint

bi-decomposition.

VI II Conclusions and Comments

In this paper, we presented the bi-decomp osition, a

special case of functional decomp osition. Disjoint

bi-decompositionshavethefollowingfeatures:

1) They are easy to detect; we use ISOPs or PPRMs

ratherthandecomposition charts.

2) Programmable logic devices exist that realize

bi-decomp ositions.

3) If the function has an OR (AND) typ e

bi-decom-p osition, then we can optimize the expression

sep-arately.

Weenumeratedfunctionswithbi-decompositions. Among

218 nondegeneratefunctions of 4 variables, 194 have

bi-ofdisjointbi-decomp ositions.

Sincethefractionoffunctionswithdecomp ositions

ap-proaches to zero as n increase [4], the fraction of

func-tions with bi-decomp ositions also approaches to zero as

n increases. However, for 1908 functions we analyzed

ab out78%ofthemhadeither disjointornon-disjoint

bi-decomp ositions.

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