Binary Time-Frame Expansion

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Elwin Boone
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1 Bary Tme-Frame Expaso Farza Fallah Fujsu Labs. o Amerca Suyvale, CA 9485 Absrac- Ths paper roduces a ew mehod or perormg me-rame expaso based o wrg he umber o me rames erms o powers o wo. I he proposed mehod, he behavor o a crcu or me rames, where < s modeled by urollg he crcu, 1,,, (log -1) mes ad combg hem. Ths ormulao o he problem makes possble o prue he search space quckly whe he problem s easble. To show he advaage o hs mehod, we have used o model he sae juscao problem ad solve he problem usg a SAT-solver. Expermeal resuls show several orders o magude speedup or some o-rval easble problems. Furhermore, mos cases he CPU me requreme grows learly erms o he umber o me rames. 1 Iroduco May exsg algorhms es ad vercao ca oly work o combaoal crcus where he oupu o he crcu s oly a uco o s pus, whch ca be easly corolled. Ths makes mpossble o use such algorhms o sequeal crcus drecly. The commo mehod o hadle sequeal crcus usg such algorhms s me-rame expaso whch was roduced several decades ago ad has bee wdely used whou ay mproveme sce he [1]. I me-rame expaso mehod a erave logc array s cosruced ad he problem s solved o. A erave logc array s a combaoal crcu ha mmcs he ucoaly o he sequeal crcu or a gve umber o me rames or clock cycles. The shorcomg o hs approach s ha he erave logc array models he behavor o he sequeal crcu or a exac umber o cycles. Ths makes ecessary o perorm me-rame expaso or oe, wo, ad more cycles ad solve he problem or each o he resulg erave logc arrays or sub-problems separaely. Solvg he sub-problems separaely does o perm o share he compuao o oe subproblem wh aoher oe. Ths paper preses a ew mehod or perormg me-rame expaso, amely bary me-rame expaso. I he proposed mehod, sead o urollg a sequeal crcu or a gve umber o mes, he behavor o he crcu s modeled or all me rames less ha a gve umber. Ths eecvely egraes he umber o me rames (.e., me) as a ew varable o he problem. Furhermore, eables searchg he me space (.e., he umber o me rames) a bary asho, resulg sgca speed-ups may cases. The res o hs paper s orgazed as ollows. I Seco we look a prevous work addressg hs problem. Seco 3 explas our bary me-rame expaso mehod. I Seco 4, we descrbe he ormulao o he sae juscao problem as a Boolea sasably problem. Expermeal resuls are preseed Seco 5. The coclusos ad uure work s preseed Seco 6. Prevous Work Ths seco explas coveoal me-rame expaso ad some o s varaos. I he coveoal mehod, he sequeal crcu s urolled o cosruc a erave logc array. Prmary Sae I Y Y 1 O Tme Frame Nex Sae I 1 O 1 Tme Frame 1 Y k I k Y k+1 O k Tme Frame k Fgure 1 A erave logc array. Fgure 1 shows a example o a erave logc array cosruced rom a sequeal crcu. A erave logc array mmcs he behavor o a sequeal crcu or exacly a gve umber o me rames. Uoruaely, may cases he umber o me rames s o kow a pror. As a example, suppose we are eresed dscoverg a codo o some sgals ever happes a crcu. Clearly, he umber o me rames s o kow. I //$17. IEEE

2 ! " # $ %! & ' I O I 1 O 1 I -1 O -1 ( ) Fgure The coveoal me-rame expaso algorhm. such a case, a erave algorhm s used o uroll he sequeal crcu 1,, 3,, mes. Assumg m deoes a sequeal crcu urolled m mes, he coveoal me-rame expaso 1,, 3,, are geeraed, where s a arbrary boud. 1 Fgure shows he erave algorhm used or perormg me-rame expaso. The algorhm eecvely perorms a lear search o he umber o me rames ad rus uco FUNC oce per erao or each geeraed erave logc array or moors he las me rame o he erave logc array. Ths lear search o he umber o me rames makes he algorhm ece. Furhermore, depedg o he problem, may be possble o skp some erao based o he resuls gahered he prevous eraos, bu he above algorhm does o provde such a opporuy. The algorhm preseed Fgure has bee used may applcaos o hadle sequeal crcus [,3,4,5,6,7]. I s possble o merge all eraos o he loop Fgure o oe by cosrucg a erave logc array or me rames ad solvg he problem or all me rames less ha ogeher. I oher words, sead o moorg oly he las me rame o he erave logc array, all me rames are moored. Ths ormulao wll help o decrease he CPU me o seg up he problem. I ca also solve he problem aser ha oher versos o he coveoal merame expaso, bu sll suers rom he ac ha uses a lear search o he umber o me rames. We wll gve a example o hs mehod Seco 4. As aoher varao o Fgure algorhm, each erao FUNC may be ru o me rames m-k o m, where k s a arbrary umber ad m s cremeed by k aer each erao. Burch e al., [13] roduced erave squarg or symbolc model checkg. Ierave squarg has some smlares wh our mehod ad ca 1 There s a expoeal upper boud erms o he umber o laches o a crcu or he requred umber o me rames, bu hs heorecal boud s oo loose o be o ay praccal use. I here s a rese le, wll suce o moor he las me-rame. Checkg all me rames wll be sll correc, bu wll crease he umber o soluos he problem s easble., + Fgure 3 Model o 1, + -1 * -1 + Fgure 4 The bary me-rame expaso o a sequeal crcu. expoeally decrease he umber o ecessary eraos o reach a xed-po, bu as hey ackowledge s mpraccal or large desgs due o he large sze o ermedae BDDs. 3 Bary Tme-Frame Expaso Our mehod s explaed hs seco. Le be a - b umber [ -1 ]. The, ca be wre as ollows, = =,.., 1 ad ca be wre as, = 1 < 1 - = where he cross operaor s deed as ollows, m+ m = We call he above ormula he expaso equao. Noe ha, = 1 = 1 = whch may be modeled usg copes o ad a mulplexor as depced Fgure 3. Hece, he behavor o he crcu or all me rames less ha or equal o may be modeled by modelg every erm he expaso equao as show Fgure 4. Ths ormulao has wo advaages, 1- I eables o search he umber o me rames a bary asho. I oher words, sead o learly searchg a value or, he search s doe o

3 /. Fgure 5 A Smple Crcu. he bs o. Ths makes he compuao aser may cases. - The exra clauses eeds o model he problem grows a logarhmc asho wh. Ths s because o he logarhmc growh o he umber o MUXes. I comparso, he ases verso o he coveoal me-rame expaso (.e., whe all me rames are moored), he growh o he exra clauses s lear. As a resul, he ew ormulao eeds smaller umber o clauses, bu he derece s o sgca. To llusrae he advaage o usg hs model cosder he problem o dg he umber o me rames ecessary o se he MSB o a 8-b couer o 1, assumg he al value o he couer s. The coveoal me-rame expaso proceeds by urollg he couer oce. Clearly, s o possble o se he MSB o 1 oe me rame. The algorhm coues wh urollg he couer ad more mes. Aer 18 rals, he algorhm ds a soluo o he problem. A each erao excep he las oe, he algorhm proves ha he problem s easble. O he coras, usg bary me-rame expaso, oe sep ca be proved ha or all values o whose egh b s zero, he problem s easble. Ths eables o prue a large umber o me rames rapdly. 3 Noe ha presece o a rese le, our ormulao o he problem creases he umber o soluos or a easble problem. The reaso s ha s possble o have more ha oe value or. I hs case, he acual umber o me rames ca be easly oud rom he value o ad he ormao abou he me-rame whose rese le has bee acvaed. As a example, cosder a problem ha ca be solved 5 me rames. I we use bary me-rame expaso ormulao o model all me rames less ha 8, oe soluo o he problem correspods o =5. Aoher soluo correspods o he case whe =7 ad he rese le o he secod me-rame s acve. I hs case he acual umber o me rames (.e., 5) ca be compued by dg how may me rames are 3 Ths example has bee used jus o llusrae he dea behd bary me-rame expaso mehod. I pracce, hs example s solved quckly by boh mehods. prese bewee he acvaed rese le ad he moor. 4 SAT Formulao o Sae Juscao Problem Gve a sequeal crcu, C, he problem o dg a sequece o vecors ha akes he sae o C rom a al value o a desred oe s called sae juscao problem. Sae juscao s a udameal problem es ad valdao. As aoher verso o he problem, oe mgh be eresed o prove ha a sae s o reachable wh a umber o clock cycles or s o reachable a all. Because sae juscao s a NP-Complee problem, s o possble o solve he problem or very large ad deep sequeal crcus. I hs seco we expla how sae juscao ca be ormulaed as a Boolea sasably problem. Frs, we gve a bre descrpo o how a crcu ca be modeled usg Boolea expressos. More complee descrpo may be oud [8]. Cosder he crcu o Fgure 5. Clauses may be wre or each gae he crcu o model he pu-oupu relaoshp o he gae. The AND gae ca be modeled by, (x z) (y z) (x y z) ad he Iverer ca be modeled usg, (x y) (x y). The above mehod was used by Larrabee o geerae suck-a aul es vecors usg a SAT solver [8]. We use hs mehod o model a sequeal crcu. We se he values o he laches he rs me rame o he al values ad check he values o he laches he las me rame agas he desred values. Isead o ermedae laches, buers are used. I he coveoal mehod, we use he erave algorhm depced Fgure o jusy a sae. As a vara o hs approach, we check he values o he laches o all me rames agas he desred values. Ths approach s o erave. Ye, as aoher vara, we creme he umber o me rames by k a each erao ad check he values o he laches o he las k me rames agas he desred values. I he ex seco, we compare he perormace o bary me-rame expaso agas he coveoal mehod ad s above varas. 5 Expermeal Resuls Ths seco provdes he expermeal resuls o usg our mehod o jusy a radom sae or sequeal crcus ISCAS89 bechmark sue [9]. We urolled he crcus ul he radom sae was jused or 496 rames were reached or he umber o varables or clauses he SAT problem was more ha he allowed lm by he SAT solver (we cascaded oe copy o he crcus wh he model Fgure 4 o search 1 space). Ths gave a seres

4 Perceage o Isaces 5.%.% 15.% 1.% 5.%.% Mehod # Problems # Sa. # Tme Ou CPU Tme (s) # Crcus # Lear Bary , Coveoal , Table 1 Comparso o wo approaches Perceage o Isaces : 7 8; 7 8< 7 8= 7 8> 7 8? 7 8@ 7 8A 9 : Rao o CPU Tmes Rao o CPU Tmes Fgure 6 Dsrbuo o rao o CPU mes or o-rval examples. o problems correspodg o dere umber o me rames or each crcu he bechmark. We used our bary me-rame expaso mehod ad he coveoal oe where he laches o all rames are compared agas he radom sae. To make easy o compare he resuls, we used equal umber o me rames boh cases. We used he CHAFF SAT solver [1] rug o a SUN ULTRA8 dual processor 45 MHz worksao wh GB memory o solve he sasably problems. For each problem, he me lm o he solver was se o oe hour. We dd o provde ay varable orderg or he SAT solver. a) Geeral Sascs ad Iormao abou Perormace o Bary Tme-Frame Expaso Fgure 7 Dsrbuo o rao o CPU mes or rval examples. Table 1 gves some sascs. Ou o 311 problems, 8 problems were sasable. Our bary me-rame expaso mehod aled 13 problems (correspodg o 4 crcus) gve he oe-hour me lm. I comparso, he ases verso o he old mehod aled 33 cases (correspodg o 17 crcus, some o hem havg small umber o laches, e.g., S89 wh 14 laches). Our mehod was able o hadle all problems ha he old oe could hadle. Ths showed ha our mehod was more robus ha he old oe. Toal CPU me o our approach or all examples excludg he me-ou oes were 7,934 secods. The correspodg umber or he old approach was 5,886 secods. Ou o 35 crcus ISCAS89 bechmark, or 6 crcus he CPU me grew learly whle creasg he umber o me rames. The correspodg umber or he old mehod was 7, all o hem were rval examples. We dvded he problems o wo ses, rval ad orval. 4 The rval se cluded all examples ha could be solved by jus usg mplcaos. The orval oes were he oes ha could be solved oly brachg was used. Abou 53% o he problems were o-rval ad he res were rval. Fgure 6 shows he dsrbuo o rao o CPU mes or orval examples. As oe ca see mos cases bary me-rame expaso ouperorms he coveoal oe. For example, or abou 19% o oal problems he ew mehod was bewee 1 o 1 mes aser ha he old oe. I some cases our mehod was several orders o magude aser ha he old oe. For o-rval problems, our mehod was a wors 3x slower ha he old approach. Fgure 7 shows he dsrbuo or rval problems. For hese problems, he old mehod s almos always beer ha our mehod, bu he derece speed s a mos 3x (.e., here s o expoeal derece). Noe ha or rval problems he easbly ca be proved by usg mplcaos; here s o eed o brach o varables. O he oher had whe we use bary me-rame expaso, we eed o brach o corol varables o MUXes. As a resul akes more me o prove easbly, bu he requred me does o grow expoeally. As oe ca see by usg bary me-rame expaso mehod, mos cases he easbly o problems ca be proved very quckly. b) Deals o Perormace or Sasable Problems There s o sgca derece bewee bary me-rame expaso ad he coveoal oe he problem s easble. Table gves CPU me ad some ormao abou 16 sasable saces. The examples edg wh leer b ad c have bee 4 We gored saces solved less ha.1 secod o emphasze hard saces.

5 Example Frames Tme Max-Dec- Level Decsos Orgal Clauses Orgal Lerals # Implcaos s1488b ,94 59,13 141,86 3,11,48 s1488c ,759 59,91 14,54 5,41,31 s1494b ,37 6,8 144,36 938, s1494c ,76 6, ,43 5,591,514 s8.1b ,46 9,785,34 57,78 135,48 8,444,346 s8.1c ,435 9,5,731 59, ,544 7,6,958 s7b s7c s386b ,54 8,61 8,996 s386c ,539 8,6 17,533 s51b ,7 11,164 1,85 s51c ,715 11,174 5,983 s8b ,98 9,86,89 s8c ,98 9,845 3,636 s83b ,84 1,16 3,94 s83c ,84 1,19 3,69 Table Comparso o wo mehods or sasable problems. ormulaed usg bary ad coveoal mehods, respecvely. As oe ca see he derece bewee wo ormulaos s eglgble. Ths may be due o he ac ha he brachg heursc SAT solvers usually gudes he search o pars o he space ha s lkely o have a soluo. Hece, he poros wh o soluo wll be by passed quckly boh ormulaos. c) Eec o Icreasg he Number o Tme Frames o CPU Tme Fgure 8 shows he CPU me growh or several crcus whe creasg he umber o me rames. For hree crcus amely S137, S3593, ad S38417, he CPU me grows learly erms o he umber o me rames. Ths resuled small CPU me or a large crcu lke S3593 wh 178 laches. For some crcus lke S98, he CPU me growh was o lear. As he las example, we have preseed S143 or whch he CPU me requreme grows rapdly ad surpasses oe-hour allocaed me. d) Dealed Comparso o Three Mehods Fally, Fgure 9 shows CPU me o usg hree dere mehods or me-rame expaso or crcu S1196. The lower curve correspods o bary me-rame expaso mehod. The curve ploed wh squares correspods o he coveoal me-rame expaso mehod whe all me rames are moored. The curve ploed wh ragles correspods o he coveoal me-rame expaso mehod whe oly he las 1 me rames are moored. Noe ha hs s a erave algorhm ad Fgure 9 shows he CPU me o each erao separaely. To compue he oal CPU me usg hs mehod, sum o he CPU me o all eraos (.e., pos o he graph) precedg he desred po has o be compued. As oe ca see, he CPU me o each erao hs mehod s almos equal o he CPU me or he case ha all me rames are moored. As a resul, he laer s much more ece. The oal CPU me o he ormer mehod decreases he umber o me rames moored each erao s creased. As oe ca see bary me-rame expaso mehod s abou wo orders o magude aser ha oher mehods. I ac, may pos o he wo op curves correspod o problems ha could o bee solved wh oe hour me lm. I he coveoal mehods, creasg he umber o me rames decreases he CPU me aer a cera po. Also, here s a sudde drop he CPU me whe usg he erave algorhm. These pheomea occur due o he heursc aure o he varable orderg algorhm used he SAT solver ad do o happe or all crcus. 6 Coclusos ad Fuure Work I hs paper we have preseed he dea o bary me-rame expaso ad used o prove ha a gve sae s o jusable wh a umber o me rames. The expermeal resuls showed ha our mehod ouperormed he coveoal me-rame expaso by several orders o magude may cases. Our expermes also showed ha or may crcus he CPU me requreme grew learly erms o he umber o me rames. We are currely ryg o dscover wha kds o crcus are beer sued or our mehod ad expla s lear behavor. I our expermes, we used a SAT solver o prove ha a sae s o jusable. I s eresg o see he

6 1. s s137 s98 1. s3593 s B C D E F G H I J K D F L G M D F N Fgure 8 CPU me growh versus he umber o me rames or several crcus whe usg bary me-rame expaso CPU Tme (s) Number o Tme Frames Fgure 9 Comparso o CPU me o hree dere mehods.

7 ! eec o usg bary me-rame expaso whle usg PODEM [11] or D-algorhm [1]. Fally, remas o bee see combg curre mehods o a hybrd oe makes possble o hadle all crcus ecely. 7 Reereces O P g h j Q k l R h S T U m V h U T j o W p X Y q r Z m [ j o \ j T s ] k U Y ^ u W X s [ h v _ ` a w b x c d y e z ` e { a } c ~ ƒ ƒ ƒ ˆ Š Œ ˆ Ž ˆ ± ² ³ ³ š ± œ µ ž µ ± Ÿ ¹ º» ¼ ³ ½ ¾ ¾ À Á ª Â Ã «Ä Á À À Å Â Æ Ç È É Ê Ë È É Ì Í Ì Î Ï Ð Ñ È Ñ Ê Ò Ì É Ï Ó Ï Ô È Õ Ï Ö Ñ Ì Õ Ø Ö Ê Î Ù Ú Û Ü Ý Þ ß àá â ã ä å æ ç è é å ê ë ç å ì í î Ú ë ï ð è ç å é ñ æ í å î æ å ò ó ô õ ö ø ù ù ô ú ú û ú ö ö ø ü ý þ ÿ þ ; < = " >? # A $ = > % B & C ' ( ) D E & * + A C, F -?. / < - G B H I 1 J 3 4 K 5 I L M 6 I 7 C 8 = 4 > A 9 : N O P Q R S Q T S U V W Q Q X Y X Z S T S Q O P [ Z \ T ] ^ P U _ S X ` R P a b c d e g g h j k l e m l o j p c o m q g l k j r s e o s e j u e j m v w r k r l s x y y z v z { } ~ { ƒ ˆ Š Œ Œ Ž Œ ˆ š š œ ž Ÿ ª «ª ± ² ³ µ ² ³ ¹ º» ¼ ½ ¾ À Á Â Ã ½ Ä Ä Á Å Æ Ç Æ È È É Æ Ç ½ Æ Å Ä Ä Ê È Æ Ë Ì Í Î Ì Ï Ð Ñ Ò Ó Ô Õ Í Ö Ø Ø Ù Ú Û Ü Ø Ý Û Ü Û Ô Ù Ù Ô Þ ß à á â ã ä å æ ç è é ê ë è ì è í î ç ï é î ð ñ í è ò ç ñ ë ó ò ð è í ô é õ ö ø ù ø ú û õ ø õ ø ü õ ü ý ö ú þ ö ÿ ù û õ ö ù û ö þ þ ú þ õ ü! " # $ $ % & ' & ( ) * +, -. ) / 1 * 3 4 ) 5 6 7, 8 9 * 6 : 5 ; : < = B C E F G H I J K L M N O P Q R S O T R O U P O V V W X Y Z [ \ ] ^ _ ` a b a c d e g h j k j l m e o p h q j r j s s u v v w x y w x z { } { ~ ƒ ƒ ˆ Š ˆ Œ ˆ Ž ƒ Œ ˆ ˆ ˆ ˆ š œ ž Ÿ ž š ª «ª ± ² ³ µ ¹ º» ¼ ¹ º ½ ¾ À Á Â Ã Ä Å Æ Ç Ç È Á { v q v } ~ ƒ ˆ Š Œ Ž š Ž š œ ž Ÿ š š š Ÿ Ÿ š Ž š Ž š Ÿ ª «± ² ³ ³ µ ª µ «µ µ ¹ º» ¼ ½ ¾ À Á Â ½ Â Ã ½ Ä Á Å Æ Ç È À Ã É Ê Ë Ë Ì Í Î Ï Ð Í Î Î Ê Ñ Ò Ó Ô Õ Ö Ö Ô Ø Ù Ú Û Ü Û Ý Þ ß à Þ á â Ý Û Ý Û ã ä å æ ç ä ä è é ã ê ë ê ì í í î ï ð ñ ò ó ó ô õ ô ò ö ø ù ú û ü ô ý þ ó û ü ÿ ò ò ö ô ø ù õ ô ü õ ô ò ÿ ú ò ö ò ü ø ô û ö! " # $ % & ' " ( " ) # * +, & -. / # " ( * : ; 3 < ; = = 5 A B C D A E A F G C H I E J I F K L M N L O P Q R S T Q U V N Q P W X Y Z [ [ [ \ ] ^ _ ` a b _ c b d e g h ] j k l k m l j o p q r s u v q p s u w x q y y u w x z w { w x } q ~ ƒ ˆ ˆ Š Œ Ž Ž Œ Ž Ž Ž Ž Ž š œ ž Ÿ œ Ÿ ž Ÿ ž œ ª œ œ Ÿ «± ² ³ ² ² µ µ ³ ² ¹ º» ¼ ± ½ ¾ À Á Â Ã Ä Å Æ Ç È É Ê Ë È Ì Í Î Ï Ð Í Î Ì Í Ñ Ò Ó Ô Õ Ö Ø Ù Ú Õ Ú Û Õ Ü Ù Ý Î Þ Ò Ò Ð ß à á ß â ã ä å æ ç è é ç ê ë ì ì ß í î ï ð ñ ò ó ô ô ó õ ö ö ø ö ù ú û ó ú ú ö ô ü ý ö ü ö ô ó ú þ ÿ ü ù þ ü! " # $ % &! & '! ( % ) * +,, -. #, / + 1 * 1 " : ; < = A B C ; < D E F A E F G H C I J K L M N O P M Q R S T U V W X Y Z W T X Y [ \ ] T ^ W [ _ ` T ^ ` _ a b _ X Z W Y [ c d e g h j k l j m l o p q r s j u v e u w x h y u e

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