A criterion for easiness of certain SAT-problems

Transcription

1 A criterio for eaie of certai SAT-problem Berd R. Schuh Dr. Berd Schuh, D Köl, Germay; keyword: compleity, atifiability, propoitioal logic, P, NP, -i-3sat, eay/hard itace Abtract. A geeralized -i-3sat problem i defied ad foud to be i compleity cla P whe retricted to a certai ubet of CN epreio. I particular, -i-ksat with o retrictio o the umber of literal per claue ca be decided i polyomial time whe retricted to eact READ-3 formula with equal umber of claue (m) ad variable (), ad o pure literal. Alo idividual itace ca be checked for eaie with repect to a give SAT problem. By idetifyig whole clae of formula a beig olvable efficietly the approach might be of iteret alo i the complemetary earch for hard itace. Itroductio. May problem i propoitioal logic are varietie of the deciio problem SAT? ad are i compleity cla NP. Eample are -i-3sat which i the problem of decidig whether for a give 3- CN formula there eit a aigmet which evaluate eactly oe literal per claue to true, or NOT-ALL-EQUAL-3SAT which ak for a aigmet with at leat oe true ad oe fale literal per claue. Other, like e.g. HORN-SAT or 2-SAT, are kow to be decidable i liear time ad thu belog to compleity cla P. or thee ad other eample ee e.g. []. A eteio of ome of thee NP problem to a more geeral requiremet o the umber of true literal ad to itace where the umber of literal i each claue i ot retricted to eactly 3 will

2 2 i geeral ehace the compleity of the problem. Oe ca idetify retricted CN epreio, however, for which the problem lie i compleity cla P. The baic idea i a follow. Give a SAT problem which typically ak the quetio Doe a truth aigmet eit with property X?. If you maage to fid i time polyomial a limited et of aigmet which are the oly oe to fulfill a eceary coditio of property X the they form the oly certificate which eed to be teted o property X, ad the whole proce i doe efficietly. I the followig I will formulate coditio which allow to determie uch a limited et for certai SAT problem. The criterio ca be evaluated i time polyomial ad be ued to determie a give itace a eay with repect to the problem coidered. Notatio. A Boolea formula i Couctive ormal form CN by defiitio i a couctio of claue, where each claue i a diuctio of literal. A literal i a occurrece of a Boolea variable (atom, baic variable) or it ivere/egative/egated. The Table lit ome parameter by which a geeral CN formula may be characterized, though ot completely. m umber of claue umber of variable k umber of literal i claue C p p p occurrece of atom a (alo called degree, frequecy or appearace ) = um of egative (egated) ad poitive (uegated) literal of variety m umber of claue C, for which k = umber of atom a, for which p = N N + N - total umber of literal umber of poitive literal umber of egative literal The followig relatio hold for CN epreio i term of the above quatitie. m m ; N N N p k m m

3 3 I the followig we ue the term { k, p} m SAT itace or { k, p} m CN for a Boolea CN epreio with m claue, variable ad o more tha k literal per claue ad o more tha p occurrece per variable. By droppig the prefi we idicate that the epreio ha eactly k literal /p occurrece per claue/variable. Itace without pure literal, i.e. p p 0 for all N will be called completely mied. Note that eact epreio, i.e. { r, r} or { r, r} ( m, m ) ( m, m) - CN automatically are. { rr, } mm - CN, due to m= ad the relatio m The followig commoly ued term are pecial cae: A 3SAT itace i a Boolea epreio i CN with k =3 for all claue,,2,..., m, i.e. a {3, } m p SAT itace. A READ-3 SAT itace i a Boolea CN epreio i which all variable have degree 3 or le, i.e. a {, 3} m k SAT itace. Ay { k, p} m ( ', ') SAT itace ca be traformed to a {3, p} m SAT itace without lo of atifiability ad i polyomial time. Likewie it i poible to traform ay { k, p} m SAT itace to a eact READ-3 CN, i.e. a {,3} m ( ', ') k SAT itace. Combiig both reductio lead to a ( ", ") { 3,3} m ( ", ") SAT or a {3, 4} m SAT itace i the bet cae. No way from a arbitrary CN formula lead to a CN with eactly 3 literal per claue ad o more tha 3 occurrece of each variable. Tovey oticed that i thi ee the {3, 4} m SAT problem i the mallet NP- complete atifiability problem. I fact, ay {r, r} m SAT itace i atifiable ad thu trivial i a way [2]. The proof ue Hall theorem [3]. The PART-SAT problem. We ow defie a cla of atifiability problem by Defiitio PART-SAT: Let k p {, } m - CN, ad let i {0,.,..,m} i {0,,..., k} be a partitio of m, i.e. m 0... k. Doe a truth aigmet eit uch that may claue cotai eactly true literal each? To relate the problem to the pecific partitio we will alo ue the otatio { } SAT. A a eample et 0 for all ecept, i.e. m, {0,m,0,0} SAT coicide with -i-3sat. ad retrict to {3, p} m - CN. The

4 4 A a further eample let agai be {3, p} m ad et 0 for all ecept ad 2. The decidig {0,, 2,0} SAT for all pair 2 m i equivalet to decidig NOT-ALL-EQUAL-3SAT. Oe ca alo ue PART-SAT to ivetigate the quetio whether certai CN epreio have aigmet which leave a give umber of claue uatified, 0 0. The criterio. The cetral criterio for idetifyig p. t. SAT problem i the followig Theorem If { } SAT i retricted to oe of two ubet of CN epreio, either k ( m, ) { k, p} mi( p, p ) or k ( m, ) { k, p} ma( p, p ), it i decidable i time polyomial time 2, where i the umber of variable with p p. A a corollary to the theorem we ca tate, that - SAT i decidable i time polyomial, if itace are retricted to formula for which equal either the miimum or the maimum umber of literal which ca be aiged true, ad for which o variable occur i equal umber of poitive ad egative literal. Sum atifiability ad proof of theorem. or ay CN itace with m claue, atom ad literal l, ad ay truth aigmet T :{atom,atom,...,atom } {0,}, umbered by {, } a defied i [4], we defie the um 2 atifiability a the total of true literal uder aigmet T : m ( ) T ( l ). A a double um it ca be evaluated either by ummig over claue or over variable firt: ( ) ( ) ( ) var iable claue i a obviou otatio. A a ide remark we tate, that for a eact READ-3 CN, { k,3} m, i a particularly imple quatity, becaue i thi cae the characteritic fuctio implifie to () p p ( ) : 2 e 2 (e e ) { }

5 5 where ( ) e {(e ) / 2} (2coh ) p p i the umber of pure variable. Thu for completely mied eact READ-3 formula ( p 0 ) follow a biomial ditributio, ad coequetly the umber of aigmet T uder which () evaluate to k i. k We ow retur to the geeral cae. To prove the theorem coider the circumtace of { }-SAT. If there i a aigmet 0 with the deired property the it mut belog to a et of aigmet which fulfill ( ) accordig to the defiitio of the um atifiability. Quite geerally ha a miimum ad maimum value with repect to all aigmet: : mi ( ) mi( p, p ) mi : ma ( ) ma( p, p ) ma The miimum ad maimum tate are degeerate if there are variable with p p, the degeeracy beig 2, if deote the umber of variable with p p. If p p for all variable the there i eactly oe aigmet mi/ma for which ( mi/ma ) mi/ma hold, amely mi/ma, ( / )g( p p), repectively. If there were more tha oe uch aigmet it ecearily would lead to a reductio of (i cae of maimum) or a ehacemet (i cae of miimum). Therefore, if i retricted to the ubet mi( p, p ), the there i ut oe mi aigmet with thi property, ecept for the p p degeeracy which lead to a factor 2 for the allowed aigmet. Determiig mi/ma i a imple coutig procedure, workig through the variable. Thu the problem ca be decided i the tated umber of tep., of The ame lie of argumet work for { k, p} m retricted to ma( p, p ) coure. Illutrative eample. The -i-3sat problem wa proved to be NP-complete by Schaefer a a pecial cae of Schaefer' dichotomy theorem [5]. Similarly we ca argue that the -i-ksat problem, i.e. the ame problem without retrictio o the umber of literal per claue, i NP-hard, a well. With the help of the theorem it i poible to idetify a ubcla of CN itace for which the problem ca be decided i

6 6 polyomial time. or -i-3sat, or more geerally -i-ksat m mut be fulfilled, becaue eactly oe true literal per claue i required. Thu, accordig to the theorem, for { k, p} m with either m mi( p, p) or m ma( p, p) ad p p compleity cla P. for all -i-ksat i i Take a a illutratio the formula ( a, c)( a, b, c,e)( a, b, e)( b, d,e)( b,d, e )(b,d). Determiig the miimum of the um atifiability i a imple coutig proce. To make thi proce more clearly arraged we ue a otatio i term of the adacecy matri cheme, ee [4] for detail. I thi otatio row repreet claue ad colum variable. A cro (ot to be cofued with the aigmet ide!) at poitio (,) tad for a poitive literal of variety i claue, for a egative. 0 at poitio (,) i the matri cheme idicate that variable doe ot appear i claue. The aforemetioed eample the read I thi repreetatio it i immediately clear that variable c ha oe poitive ad oe egative literal. Thu there are two aigmet which miimize, amely (-,,,-,-) ad (-,,-,-,-). Each mut be checked agait each claue to determie whether it lead to eactly true literal. Oly the firt aigmet pae thi tet, the ecod coflict already with the firt claue. Oe may retrict the allowed epreio further to completely mied eact READ-3 formula. Thi way oe get rid of degeerate variable with p p. The mi ad the oly o-trivial cadidate for -i-ksat epreio which are i P are itace with m 3. We call uch epreio quare for obviou reao. The followig itace are illutratio of uch quare CN =

7 7 All three itace are atifiable. But ot all are -i-3sat epreio. The oly aigmet which miimize i =(-,,,-,-). But it doe ot atify claue 2. Thu there i o aigmet that give oe true literal per claue. Sice 0 2m alo {0,0,m,0} - SAT ca be checked with ma ut oe aigmet, amely (,-,-,,). It fail to achieve 2 true literal i the firt claue. or 2, mi (,,,,,, ) which violate claue 2. Wherea the ame mi obviouly lead to eactly oe true aigmet i each claue for itace 3. Alo the awer to the {0,0,m,0} - SAT problem i poitive ow. Agai, m ad ma (,,...,) i the oly aigmet to 2 3 ma be checked. A look at the matri cheme reveal that ettig all aigmet to true ideed lead to the deired reult. A a further illutratio coider the PART-SAT problem {0, m,0,...,0} SAT ad {0,0, m,...,0} SAT retricted to {3, p} m, ad write mi/ma a follow ( N p p ) mi 2 ( N p p ). ma 2 or eact 3-CN N 3m hold ad both PART-SAT problem are efficietly olvable for all which fulfill m p p ad have o degeerate variable. Obviouly all quare completely mied {3,3} mm a dicued before belog to thi cla (ee formula 2 ad 3 ). A le ymmetric eample would be The aigmet to be checked are (-,,-,) for the -i-3sat problem, ad (,-,,-) for the 2-i 3-SAT problem. Both fail to meet the requiremet. A eample where the criterio doe ot idetify ay eay itace at all i the 2/2/4-SAT problem decribed i []. Oe earche for aigmet which have eactly two true literal i each claue of a quare {4, 4} mm - CN, where variable with occurrece 4 have two poitive ad two egative literal. Accordig to [] the problem i NP-complete. The followig two itace erve a a illutratio for m=5 ad m=6:

8 Thi i a cae where miimum ad maimum value of the um atifiability coicide: 2 2m. O the other had we are dealig with the PART-SAT problem mi ma {0,0,2,0,0} SAT ad two true literal per claue mea 2m. Thu the aumptio of the theorem are fulfilled. But the degeeracy i maimal ow, m. So all 2 aigmet are cadidate to be teted i priciple. I fact, for the m=5 itace there are two aigmet with the required property, amely (-,,,,)=(fale, true, true, true, true) ad it egatio. Alo NOT-ALL-Equal-SAT (NAE-SAT) ca ot be implified with the help of the theorem. NAE-SAT ak for aigmet which lead to at leat oe true ad oe fale literal i each claue. If the et of allowed itace i retricted to {3,3} m - CN, the earched for aigmet mut deliver either oe or two true literal per claue, ad thu NAE-SAT i equivalet to decidig {0,, m,0 } SAT for all 0,,...,m. Thi i priciple i a O( 2 m ) tak. Thi problem ca alo be put the followig way. Sice for eact 3-CN the umber of true literal i claue ca oly take value or 2 for NAE-aigmet, which i equivalet to (3 ) 2, the equatio 2 3 ( ) ( ) 2 m i a eceary coditio for aigmet which olve the NAE-3SAT problem. I term of adacecy matri elemet the coditio may be writte: m 0 with f f, ' ' ' ' ' Ay NAE aigmet i to be foud amog the olutio of thi equatio. Although the ' are eaily calculated i p.t. for ay give 3-CN, there i i geeral o efficiet way to determie the allowed. Note that the imilar equatio which determie the atifyig aigmet of the regular 3SAT problem cotai additioal term, liear ad triliear i, [4]. Though 3SAT ad NAE-SAT belog to the ame compleity cla NP, oe i tempted to ay that NAE-3SAT - although it impoe troger coditio tha 3SAT - i omewhat eaier tha 3SAT, ice it lack the triliear term.

9 9 Cocluio. I have derived criteria for Boolea CN formula to be eay itace for a cla of SAT-problem, termed PART-SAT. PART-SAT ak for aigmet which geerate eactly true literal i claue, ad m, the total umber of claue. The criterio tate that a itace i decidable i time polyomial (time a degeeracy factor, if there are variable with p p ) i.e. eay with repect to the problem poed provided maimum value of the total of true literal, mi/ma equal either the miimum or. Thi latter quatity ca be determied i liear time due to the additivity of i both claue ad variable. I geeral, it i difficult to ue thi electio criterio to igle out a imply defiable cla of epreio a PART-SAT- eay, i.e. a a cadidate for P-compleity. I cae of -i-3sat or more geerally l-i-ksat with l k which are pecial cae of PART-SAT, uch a imple cla could be idetified, amely the cla of quare, completely mied READ-3 formula. Neverthele, oe ca alway check idividual itace o eaie with repect to a give PART-SAT problem. The hope i to eae the earch for hard itace via thi complemetary tool, too. Referece. [] ee e.g. Ha Kleie Büig, Theodor Lettma: Propoitioal Logic: Deductio ad Algorithm. [2] Craig A. Tovey: A Simplified NP-Complete Satifiability Problem; Dicrete Applied Mathematic 8 (984) [3] P. Hall: O repreetatio of ubet, J. Lodo Math. Sot. 0 (935) [4] Schuh, Berd R.: SAT for pedetria, arxiv: [c.cc]. [5] Schaefer, Thoma J. (978): The compleity of atifiability problem (PD). Proceedig of the 0th Aual ACM Sympoium o Theory of Computig. Sa Diego, Califoria. pp