Testing a new idea to solve the P = NP problem with mathematical induction

Transcription

1 Tesng a new dea o solve he P = NP problem wh mahemacal nducon Bacground P and NP are wo classes (ses) of languages n Compuer Scence An open problem s wheher P = NP Ths paper ess a new dea o compare he wo language ses and aemps o prove ha hese wo language ses conss of same languages by elemenary mahemacal mehods and basc nowledge of Turng machne ehods By nroducng a fler funcon C(,w) ha s he number of confguraons whch have more han one chldren (nondeermnsc moves) n he shores accep compuaon pah of a nondeermnsc Turng machne for npu w, for any language L() NP, we can defne a seres of s subses, L () = {w w L() C(,w) }, and a seres of he subses of NP as L = {L() L() NP} The nondeermnsc mul-ape Turng machne s used o brdge wo language ses L and L +1, by smulang he (+1)-h nondeermnsc move deermnscally n mulple wor apes, o reduce one (he las) nondeermnsc move Resuls The man resul s ha, wh he above mehods, he language se L +1, whch seems more powerful, can be proved o be a subse of L Ths resul collapses L P for all N Wh NP = N L, s clear ha NP P Because by defnon P NP, we have P = NP Dscusson There can be oher ways o defne he subses L and prove he same resul The resul can be exended o cover any ses of me funcons C, f f f C f 2 C, hen DTIE(C) = NTIE(C) Ths paper does no show any ways o fnd a soluon n P for he problem nown n NP PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

2 1 Tesng New Idea o Solve P=NP Problem wh ahemacal Inducon HUANG, YU BIN yubnhuang@yahoocom PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

3 P = NP 2 Absrac Bacground P and NP are wo classes (ses) of languages n Compuer Scence An open problem s wheher P NP Ths paper ess a new dea o compare he wo language ses and aemps o prove ha hese wo language ses conss of same languages by elemenary mahemacal mehods and basc nowledge of Turng machne ehods By nroducng a fler funcon w C, ha s he number of confguraons whch have more han one chldren (nondeermnsc moves) n he shores accep compuaon pah of a nondeermnsc Turng machne for npu w, for any language L NP, we can defne a seres of s subses, ( ) { w w L( ) C(, w) }, and a seres of he subses L of NP as L { L ( ) L( ) NP} The nondeermnsc mul-ape Turng machne s used o brdge wo language ses L and 1 L, by smulang he 1 -h nondeermnsc move deermnscally n mulple wor apes, o reduce one (he las) nondeermnsc move Resuls The man resul s ha, wh he above mehods, he language se L 1, whch seems more powerful, can be proved o be a subse of L Ths resul collapses L P for all N Wh N NP L, s clear ha NP P Because by defnon P NP, we have P NP Dscusson There can be oher ways o defne he subses L and prove he same resul The resul can be exended o cover any ses of me funcons C, f f f C f 2 C, hen C DTIE NTIE C Ths paper does no show any ways o fnd a soluon n P for he problem nown n NP Keywords: P versus NP, Compuaonal Complexy, nondeermnsc mul-ape Turng machne PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

4 P = NP 3 Tesng New Idea o Solve P=NP Problem wh ahemacal Inducon INTRODUCTION P and NP are wo classes (ses) of languages n Compuer Scence An open problem s wheher P NP Sephen Coo provded he formal descrpon of hs problem 1 For more nformaon please reference he exboo Theory of Compuaonal Complexy 2 To compare wo ses, he mahemacal way o prove P NP wll be provng L LP LNP (snce by defnon P NP, can also be LLNP LP ); and P NP equals LLP LNP However, a quc scan of lesones 3 of researches reveals ha: 1 os of he approaches o prove P NP aemp o prove L LNP LP The concern of hs approach s, for a ceran problem wh a well-nown soluon n NP, gvng a soluon n P does no prove LLNP LP Because LLP LNP, anyone can consruc unlmed dfferen L P ha sasfy L LNP LP Bu hs does no consruc a proof of LLNP LP 2 os of he approaches o declare LLP LNP only prove ha hey fal o fnd/prove a parcular language even hey prove ha s mpossble o fnd L NP ha s acually also n P Furhermore, L P, maybe he Turng machne searchng such language never hals or s undecdable, hs sll does no prove ha L P does no exs 3 Some oher papers are no wren wh well-esablshed compuaonal models and, hence, mae dffcul for he communy o undersand Ths paper ess a new dea 3 o solve LLNP LP usng elemenary mehods ha compare wo language ses, and prove P and NP are consss of same languages PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

5 P = NP 4 ETHOD P and NP are wo ses conss of nfne languages and each such language can have nfne acceped npus However, Turng machnes are all abou counable numbers And he lengh of he accep compuaon pahs of npus are also counable neger numbers These numbers of lenghs can be arbrary bg, bu by defnon never nfne These hn ha elemenary mehod such as mahemacal nducon can be helpful To apply mahemacal nducon, he languages or npus should be assocaed wh neger numbers To acheve hs goal, as he new dea, we can nroduce a fler funcon C, w ha s he number of confguraons whch have more han one chldren (nondeermnsc moves) n he shores accep compuaon pah of a nondeermnsc Turng machne for npu w Whou loss of generaly, assume s a one-ape Turng machne For any language L NP, we can hen defne: 1 A seres of subses of L, ( ) { w w L( ) C(, w) } for all N ; and 2 A seres of subses of NP as L L ) L( ) NP L ( for all N Apparenly, L ( ) P and L P To apply mahemacal nducon, he nondeermnsc mul-ape Turng machne s used o brdge wo languages L ( ) and ( ) L 1, and wo language ses L and L 1, defned by C(, w) and C (, w) 1 Gven any w L ( ) whose values of C (, w) are a mos 1 1, we can consruc a nondeermnsc mul-ape Turng machne o smulae he ha acceps all w L ( ) wh a mos nondeermnsc moves and same me complexy, by 1 smulang he 1 -h nondeermnsc (mulple possble) move deermnscally n mulple wor apes PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

6 P = NP 5 Because he square of a polynomal run me funcon s sll a polynomal run me funcon, we can consruc a one ape nondeermnsc Turng machne o smulae he above nondeermnsc mul-ape Turng machne whch sll has polynomal run me eans for any 1 ( NP, here exss, L ( ) L 1( ) Hence, we can ge L ( ) L L ) 1 Tha means L1 L P for any N Fnally, we can prove NP N L and NP P SYBOL & DEFINITIONS Empy se N s The se of non-negave neger numbers N operaor Language 1 le A se whose ems are ses The unon operaor of ses 1 s s for all s s N 1 1 s 2 s s s s s N { s, s, s } 1 N ; or Le be a fne alphabe (ha s, a fne nonempy se) wh a leas wo elemens, and * be he se of fne srngs over Then a language over s a subse L of Turng machne 1 A (one-ape) Turng machne consss of a fne sae conrol (e, a fne program) aached o a read/wre head movng on an nfne ape The ape s dvded no squares, each * PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

7 P = NP 6 capable of sorng one symbol from a fne alphabe ha ncludes he blan symbol b Each machne has a specfed npu alphabe, whch s a subse of, no ncludng he blan symbol b A each sep n a compuaon, s n some sae q n a specfed fne se Q of possble saes Inally, a fne npu over s wren on adacen squares of he ape, all oher squares are blan (conans b ), he head scans he lef-mos symbol of he npu, and s n he nal sae q A each sep s n some sae q and he head s scannng a ape square conanng some ape symbol s, and he acon performed depends on he par q, s and s specfed by he machne s ranson funcon (or program) The acon consss of prnng a symbol on he scanned square, movng he head lef or rgh one square, and assumng a new sae Formally, a Turng machne s a uple,, Q,, where,, Q are fne nonempy ses wh and b The sae se Q conans hree specal saes q, q accep, and q reec The ranson funcon sasfes If q, s q, s, h Q { q, q } Q { 1,1 } : accep reec, he nerpreaon s ha, f s n sae q scannng he symbol s, hen q s he new sae, s s he symbol prned, and he ape head moves lef or rgh one square dependng on wheher h s 1 or 1 We assume ha he ses Q and are dson ul-ape Turng machne 2[12] A mul-ape Turng machne s smlar o a one-ape Turng machne wh he followng excepons: PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

8 P = NP 7 1 I has a fne number of apes ha exends nfnely o he boh ends Each ape s equpped wh s own head All ape heads are conrolled by a common fne conrol 2 There are wo specal apes: an npu ape and an oupu ape a The npu ape s used o hold he npus only; s a read-only ape ha prohbs erasng and wrng b The oupu ape s used o hold he oupu srng when he compuaon of a funcon s concerned; s a wre-only ape 3 The oher apes are called he wor apes All wor apes are allowed o read, erase, and wre 4 For a -ape Turng machne, he ranson funcon sasfes Q { q, q } Q { 1,1 } : accep reec 5 The nal seng of he npu ape of he mul-ape Turng machne s he same as Confguraon 1 ha of he one-ape Turng machne, and all oher apes of he mul-apes Turng machne nally conan only blans A confguraon of s a srng xqy wh * x, y and q Q The nerpreaon of he confguraon xqy s ha s n sae q wh xy on s ape, wh s head scannng he lef-mos symbol of y If C and C are confguraons, hen C C f xqsy and C and q, s ( q, s, h) one of he followng holds: C xsqy and h 1 and y s nonempy C xsqb and h 1 and y s empy PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

9 P = NP 8 C xqasy and h 1 and x xa for some a C qbsy and h 1 and x s empy A confguraon xqy hals f q q accep, q } Noe ha for each non-halng { reec confguraon C here s a unque confguraon C ' such ha C C Dscusson The possble number of C s no more han he number of possble ransons of q,s, regardless of wha x or y s Compuaon 1 The compuaon of on npu * w s he unque sequence C, C 1, of confguraons such ha q w (or C q b f w s empy) and C C 1 for each C wh C 1 n he compuaon, and eher he sequence s nfne or ends n a halng confguraon If he compuaon s fne, hen he number of seps s one less han he number of confguraons; oherwse he number of seps s nfne Accep 1 We say ha acceps w f and only f he compuaon s fne and he fnal confguraon conans he sae q accep The language acceped by, denoed L ( ), has assocaed alphabe and s defned * by L( ) { w acceps w} Deermnsc Turng machne 2[14] Each confguraon of a machne here s a mos one move o mae, and hence here s a mos one nex confguraon, hs nd of he machnes are defned as deermnsc Turng machne PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

10 P = NP 9 Snce each confguraon of may a mos have one nex confguraon, he compuaon of a deermnsc Turng machne s a compuaon pah Such as: C nal C C fnal Nondeermnsc Turng machne 2[14] If allow more han one moves for some confguraons, and hence hose confguraons have more han one nex confguraons, he machne s called a nondeermnsc Turng machne Snce each confguraon of may have more han one nex confguraons, he compuaon of a nondeermnsc Turng machne on an npu w s, n general, a compuaon ree In he compuaon ree, each node s a confguraon and all s nex confguraons are s chldren The roo of he ree s he nal confguraon We can sll fnd he compuaon pah acceps he gven npu by cung all branches ha do no lead o he (shores) fnal accep confguraon Afer ha, loos le: C nal C C 1 C fnal All confguraons afer he nal confguraon can be one of he many possble confguraons For example, f C has more han one chldren, 1 C s he one of he chldren on he pah o C fnal Polynomal run me 1 We denoe by compuaon never hals, hen w w he number of seps n he compuaon of on npu w If hs PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

11 P = NP 1 Noe ha, for deermnsc Turng machne, he runme s he lengh of he compuaon pah For he nondeermnsc Turng machne, he runme s he shores compuaon pah n he compuaon ree whch acceps he gven npu For n N we denoe by T n he wors case run me of ; ha s, T n T ( n) max{ n w w }, where n s he se of all srngs over of lengh n We say ha runs n polynomal me f here exss such ha for all n, n DTIE 2[18] We defne DTIE o be he class of languages L ha are acceped by deermnsc Turng machnes wh n n DTIE C C for almos all n We le DTIE NTIE 2[19] We defne NTIE o be he class of languages L ha are acceped by nondeermnsc Turng machnes wh n n NTIE C C NTIE for almos all n We le P 2[21] P DTIE poly DTIE n N NP 2[21] NP NTIE poly NTIE n N PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

12 P = NP 11 C (, w) For any npu w L whch acceps, n he compuaon ree of he nondeermnsc Turng machne L, le C (, w) be he number of confguraons n he accep compuaon pah for w whch have more han one chldren n he compuaon ree ( ) L For any language NP L, we can defne a seres of subses of L ( ) as: By defnon we have: NP L ( ) { w w L( ) C(, w) }, N L, for any N Apparenly we have: L L1 L L and L P L Le L be he se of L of all NP L Apparenly we have: L P N Proposon 1 L L N PROPOSITIONS Proof Assume L L I means here mus exs a leas one L w L for all N w bu Accordng o he defnon of L, for hs w, C, w N So for npu w, he Turng machne wll never hal eans he Turng machne does no accep w w L Ths esablshes a conradcon Proposon 2 L { L N} PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

13 P = NP 12 Proof I s clear ha L { L N} have L { L N} N Accordng o Proposon 1, we Proposon 3 NP N L Proof By defnon, L N NP and NP L L NP Accordng o Proposon 2, L { L N} L NP N} NP { L By defnon, L L So L L Because L L L, we have NP NP N NP, we have NP N L Therefore NP N L LEA Lemma For any mul-ape Turng machne, here exss a one-ape Turng machne compung he same funcon as n me n O n Proof See Reference [2], Page 12, Page 23 THEORES 2 Theorem 1 L 1 L L Proof For any 1 1 assume on some sq L Le be he maxmum number of values ha a can q, of he one ape nondeermnsc Turng machne We can consruc a wor ape nondeermnsc Turng machne ha acceps he same language L ( ) wh a mos nondeermnsc moves 1 PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

14 P = NP 13 The dea s o rac and coun every nondeermnsc moves n he accep compuaon ree of npu w for Before he 1-h nondeermnsc move, he mul-ape Turng machne wors le A he me of he 1-h nondeermnsc move, he mul-ape Turng machne wll move deermnscally and all, a mos, possble moves wll be smulaed a he wor apes Because for every L, C, w 1 Afer he 1 w 1 nondeermnsc move, boh and wor deermnscally unl accep or reec he npu w -h Assume s a uple,, Q, and s anoher uple,, Q, We have: 1 Two Turng machnes have same ape symbols 2 For any sae q Q, creaes sae q Q for q s used by for saes afer he -h and before he 1-h nondeermnsc move 3 For any r and any r number of q Q, denoed as q Q for all r, creaes sae q q r Q ## 1 q # # q r1 s used by for saes a and afer he 1 -h nondeermnsc move 4 For any deermnsc move q, s q, s, l, creaes 1 dfferen deermnsc ransons for : q, s,, s q, s,, s, l,, l for q s he -h sae creaed for q Q for q s he -h sae creaed for q Q for 5 For each move of nondeermnsc moves q, s q, s, l, creaes 1 (nondeermnsc) ransons for : q, s,, s q 1, s,, s, l,, l for q s he -h sae creaed for q Q for 1 q s he 1 -h sae PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

15 P = NP 14 creaed for q Q for Noe he dfference, a nondeermnsc move wll swch usng he nex se of he saes 6 For nondeermnsc moves q s q, s, l,, q, s, l where r,, creaes one deermnsc ranson for : q, s,, s q # # qr, s,, sr, s,, l,, lr, l, Here, he s and l are wren repeaedly o show ha, f r 1, reuse s and l for he res of he wor apes and wor heads So all wor apes and wor heads have defnons (hngs o do) r r r 7 For any combnaon q # # qr, s,, sr, s, where r and q # # q r Q, a Creaes q # q r, s,, sr, s, q # # qr, s,, sr, s,, l,, lr, l, # f and only f here exss deermnsc move q, s q, s, l r for all b Creaes q # # qr, s,, sr, s, qreec, s,, sr, s,,1, f here exss a leas one r, deermnsc move q, s q, s, l does no exs (no defnon or only nondeermnsc moves avalable) 8 The nal sae of s q nal ha s he one creaed for q nal of PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

16 P = NP 15 9 For any q accep Q, he q accep Q, where q accep, s he -h sae creaed for for q accep 1 Sae q q # # r s an accep sae of f and only f here exss ha q s an accep sae of 11 For any q reec Q, he q reec Q, where q reec, s he -h sae creaed for for q reec 12 Sae q q # # r s a reec sae of f and only f for all ha q s a reec sae of 13 frs copy he npu ape o all he wor apes I s clear ha acceps same language as For every w L, he lengh of he shores compuaon pah of he accep compuaon ree are he same Regardless he nal O n wor o copy npu ape o all wor apes, and ' have same polynomal run me complexy T Because NP L r n T n n r 1 for, here exss r such ha for all n, Accordng o he Lemma, here exss an one ape nondeermnsc Turng machne me any L 1 acceps 2r L wh T n On, whch s also n polynomal run Noce ha only need a mos nondeermnsc moves o accep he npu So for, we have an one ape nondeermnsc Turng machne n polynomal run me acceps he same language wh a mos nondeermnsc moves eans L L By defnon of L, for any L NP, L L 1 1 PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

17 P = NP 16 Therefore, L 1 L Theorem 2 L P for all N Proof Because of L P, and Theorem 1, f L P, L1 L P, accordng o mahemacal nducon, L P for all N Theorem 3 NP P Proof Accordng o Theorem 2, we have L P for all N Hence we have N P N L Accordng o Proposon 3, we have NP L N, means NP N P Because N P P, NP P Theorem 4 P NP Proof I s rval ha P NP Accordng o Theorem 3, NP P Hence P NP DISCUSSION Noce ha here are unlmed ways o defne he subses of NP whch sasfy N NP L Wh he concluson of P NP, all of such L P Ths paper only gve one way o defne such L and prove all such L P s also provable L P There may be oher ways o defne L and The resul can be exended o cover any ses of me funcons C, f f f C f 2 C, hen C NTIE C DTIE Ths paper does no gve a way o fnd a soluon n P for a problem n NP PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215

18 P = NP 17 References [1] COOK, STEPHEN "THE P VERSUS NP PROBLE" Web <hp://wwwclaymahorg/ses/defaul/fles/pvsnppdf> [2] DU, DING-ZHU and KO, KER-I THEORY OF COPUTATIONAL COPLEXITY New Yor: JOHN WILEY & SONS, INC, 2 Prn [3] The P-versus-NP page Web hp://wwwwnuenl/~gwoeg/p-versus-nphm PeerJ PrePrns hps://dxdoorg/17287/peerpreprns1455v1 CC-BY 4 Open Access rec: 27 Oc 215, publ: 27 Oc 215