PCP Theorem by Gap Amplification

Transcription

1 PCP Theorem by Gap Amplificaion Bernhard Vesenmayer JASS 2006 Absrac The PCP Theorem provides a new classificaion of NP. Since he original proof by [AS98], several new proofs occured. While he firs proof used highly sophisicaed mehods, he new approaches ry o use simpler ones. In his paper he proof by [Din05] is presened. I uses he equivalence of his problem o he NP-hardness of gap-3sat, i.e. i is his hardness ha is shown. The saisfiabiliy gap of a se of consrains is he smalles fracion of unsaisfied consrains over all assignmens for he variables. Gap-3SAT is he problem of deciding wheher his gap is 0 or greaer han a posiive consan. In his paper herefore 3SAT is polynomially reduced o gap-3sat via gap amplificaion, i.e. he gap is blown up o a consan fracion. 1 Inroducion There are by now many srucural differen proofs of he famous PCP-Theorem. Mos of hem are quie echnical and use highly sophisicaed mehods. Originally i has been saed via ineracive proofs. In [FGL + 96] a surprising connecion has been found beween his heorem and a formulaion via gap-3sat. Tha means hey have shown, ha he PCP-Theorem is equivalen o saing ha gap-3sat is NP-hard. In oher words, i is NP-hard o disinguish beween UNSAT(C ) = 0 and UNSAT(C ) = α > 0 for a consrain sysem C. This is he basis of he proof by [Din05] ha is presened in his paper. Naurally, if he consrain sysem is no saisfiable, hen UNSAT(C ) 1, where n is he number of consrains in C. n Via a polynomial amplificaion algorihm his gap is blown up o a consan α which renders his problem equivalen o gap-3sat. This paper is par of a alk a JASS 06. I is herefore no self-conaining, bu relies on he preparaive paper [Bul06] in which underlying srucures and basic resuls are presened. Only he exisence of assignmen eser is used wihou proof. The proofs of he oher resuls are mosly aken ou of [Din05] and can be found here or in [Bul06]. 2 Preprocessing In order o apply he main amplificaion lemma, we need a nicely srucured graph. In his secion is shown ha any consrain graph can be polynomially ransformed ino such TU München, Bolzmannsraße 3, Garching bei München, Germany, ( vesenmay@ma.um.de). 1

2 2 2 PREPROCESSING a nice graph. By nicely srucured is mean ha he graph is regular, of consan-degree and expanding. The ransformaion includes wo seps which are formalized in he nex wo proposiions. Proposiion 2.1 (Consan degree) Any consrain graph G = (V, E), Σ, C can be ransformed ino a (d 0 + 1)-regular graph G = (V, E ), Σ, C such ha V = 2 and for some global consans d 0, c > 0. c UNSAT(G) UNSAT(G ) UNSAT(G) Proof. Fix d 0 and define d := d For each n le X n be a d 0 -regular expander on n verices wih h(x n ) h 0 as is guaraneed by [Bul06, Lemma 17]. Le d v be he degree of v V. Replace each v by X dv and pu equaliy consrains on hese edges. Denoe his wih [v]. Le [V ] := v V [v] and E 1 be he union of he edges. For every (v, w) E pu an edge beween one of he verices of [v] and [w], such ha each such verice only sees one such exernal edge. The consrains on hese edges are given by he consrains of he original graph. Denoe his se of edges by E 2. Tha means we have consruced a d-regular graph G : G := ([V ], E = E 1 E 2 ) Le us now show ha his graph mees he requiremens. We sar wih he upper bound. Le σ : V Σ be an assignmen of G. Define σ : [V ] Σ by: v V, x [v] : σ (x) = σ(v) Since he consrains of he addiional edges of E 1 are me in his case, we see ha he fracion of unsaisfied edges decreases or says consan. Tha means: UNSAT(G ) UNSAT(G) The oher bound is more complex. Le σ : [V ] Σ be a bes assignmen. Define σ : V Σ by: v V : σ(v) := maxarg a Σ (P x [v] (σ (x) = a)) Tha means σ(v) is he mos popular value of [v]. Le F E be he se of edges ha rejec σ, F E be he se of edges ha rejec σ. Le S be he se of hose verices in [V ] ha were ouvoed in he definiion of σ, i.e. S := v V {x [v]; σ (x) σ(v)}. Every exernal edge in E 2 corresponding o an edge in F eiher rejecs σ or has a leas one endpoin in S. This gives us: F + S F = α, where α := F. Firs recall ha = d. We now have wo cases: 1. F α = α : Therefore we have he desired lower bound: 2 2d UNSAT(G ) = UNSAT σ (G ) 1 2d UNSAT σ(g) 1 2d UNSAT(G).

3 3 2. F < α: Then S α. Fix v V and define: 2 2 S v := [v] S S v a := {x S v ; σ (x) = a} Then we have S v a [v] 2. Therefore he edge expansion propery of X d v yields: E(S v a, [v] \ S v a) h 0 S v a. All of hose edges rejec σ, because of heir equaliy consrains. Tha means: F h 0 Sa v = h 0 S v α = h 0 S h 0 2 = h 0α 2d v V a Σ v V And herefore he lower bound follows as above: UNSAT(G ) h 0 2d UNSAT(G) Recalling ha d is independen of G, he proposiion follows wih c := min( 1 2d, h 0 2d ). Proposiion 2.2 (Expanderizing) Le d 0, h 0 > 0 be some global consans. Any d-regular consrain graph G can be ransformed ino G such ha 1. G is (d+d 0 +1)-regular, has self-loops, and λ(g ) d+d 0 +1 h2 0 d+d 0 +1 < deg(g ), 2. size(g ) = O(size(G)), and 3. d d+d 0 +1 UNSAT(G) UNSAT(G ) UNSAT(G). Proof. This ransformaion will be done by adding edges wih rivial consrains. Le E loop := {(v, v); v V } and X = (V, E ) be a d 0 -regular expander on V verices wih h(x) h 0 (again, he exisence is guaraneed by [Bul06, Lemma 17]). Define: G = (V, E E E loop ) This clearly is a (d + d 0 + 1)-regular graph of he same size as G. We can now apply he general resul for expanders (see [Bul06, Lemma 19]) and ge: λ(g ) d + d h(g ) 2 d(g ) d + d h 2 0 d + d < d + d As he new edges are always saisfied and he number of edges is increased by a facor d+d 0 +1 d, he fracion of unsaisfied consrains drops a mos by his facor. Now we can finally sae he lemma ha allows us o ransform a given consrain graph ino a nicely-srucured one. Lemma 2.3 (Preprocessing) There exis consans 0 < λ < d and β 1 > 0 such ha any consrain graph G can be ransformed ino a consrain graph G, denoed G = prep(g), such ha 1. G is d-regular wih self-loops, and λ(g ) λ < d. 2. G has he same alphabe as G, and size(g ) = O(size(G)). 3. β 1 UNSAT(G) UNSAT(G ) UNSAT(G). Proof. Apply Proposiion 2.1 on G, hen apply Proposiion 2.2 on he resul. The lemma is proven wih β 1 = c d. d+d 0 +1

4 4 3 POWERING 3 Powering In his secion we consruc ou of a given consrain graph G = (V, E), Σ, C a new graph G = (V, E), Σ d, C for every N as follows: 1. The verices of G are he same as he verices of G 2. u and v are conneced by k edges in E iff he number of -sep pahs from u o v in G is exacly k. 3. The alphabe of G is Σ d, where every verex specifies values for all is neighbours reachable in seps. 4. The consrain associaed wih an edge e = (u, v) E is saisfied iff he assignmens for u and v are consisen wih an assignmen ha saisfies all of he consrains of he pah, induced by he neighbourhoods of u and v. (In case his resuls in wo differen values for a verex, hen he consrain is auomaically no saisfied.) In he sequel e E is called edge and e E is called pah. This consrucion yields he desired amplificaion of he saisfiabiliy-gap. I is based on he fac, ha he number of edges in G increases by d 1, bu each edge in G is possibly included in d 1 pahs. Therefore, if a consrain corresponding o an edge e is no saisfied his yields o possibly d 1 rejecions of he corresponding assignmen for G. Firs we sar wih a echnical proposiion: Proposiion 3.1 For every p [0, 1] and c > 0 here exiss some 0 < τ 1 such ha if l 1 l 2 ( l 1 l 2 ), hen k, k pl 1 k pl 2 c( l 1 l 2 ) : τ P (B l 1,p = k) P (B l2,p = k) 1 τ Proof. The proposiion is perfecly symmeric in l 1 and l 2. Tha means wihou loss of generaliy we can assume l 1 l 2. Wrie herefore l 1 = l 2 + r for some 0 r l 1. We have P (B l1,p = k) = = = ( ) l2 + r p k (1 p) l 2+r k k l l k l l k l 2 + r l 2 + r k l l k l l k ( ) l2 p k (1 p) l2 k (1 p) r k l 2 + r l 2 + r k P (B l 2,p = k) For all a r l 1 we have wih l 2 k l 2 p + c l 1 (1 p)l 2 + c l 1 : l 2 + a l 2 + a k l 2 (1 p)l 2 + (c + 1) 1 ( 1 c + 1 ) l 1 1 p (1 p) l 1 l 2 + a l 2 + a k l 2 + l1 (1 p)l 2 c 1 ( ) 4c 1 + l 1 1 p (1 p) l 1

5 5 This yields he proposiion wih τ = e 4c+1 1 p. Clearly we have ha if UNSAT(G) = 0, hen UNSAT(G ) = 0. In he oher case we have he amplificaion of he saisfiabiliy gap: Lemma 3.2 (Powering) Le λ < d, and Σ be arbirary consans. There exiss a consan β 2 = β 2 (λ, d, Σ ) > 0, such ha for every N and for every d-regular consrain graph G = (V, E), Σ, C wih self-loops and λ(g) λ ( UNSAT(G ) β 2 min UNSAT (G), 1 ). Proof. Le σ be a bes assignmen for G, i.e. UNSAT σ (G ) = UNSAT(G ). By definiion, σ(v) specifies values for every w in is -neighbourhood. Le σ(v) w denoe his value. For every 1 j and v V le X v,j be a random variable wih disribuion a Σ : P (X v,j = a) := # j-sep pahs saring from v and ending a w wih σ(w) v = a # j-sep pahs saring from v Tha means X v,j represens he value assigned o v by a random poin wihin j-disance. Wih his define he corresponding assignmen σ for G as follows: σ(v) := maxarg a Σ (P (X v, 2 = a)) Le F E be a subse of he edges ha rejec σ, such ha UNSAT(G) = F, and I r := { r < i + r}. For each e E define he following random variable 2 2 { 1 if e = (v 0,..., v ) where (v i 1, v i ) F σ(v 0 ) vi 1 = σ(v i 1 ) σ(v ) vi = σ(v i ) N i (e) := 0 oherwise N(e) := i I r N i If N(e) > 0, hen his pah has go a rejecing edge "in he middle", i.e. i clearly rejecs σ. Tha means P (N > 0) UNSAT σ (G ). In he sequel we will esimae his probabiliy. To his end we make use of [Bul06, Lemma 23], i.e. P (N > 0) E2 (N). Therefore we E(N 2 ) have o cope wih E(N) and E(N 2 ). 1. E(N): From he definiion we can gain he probabiliy of P (N i > 0) as follows: P (N i > 0) = P ((u, v) F ) P (X u,i 1 = σ(u))p (X v, i = σ(v)) = F P (X u,i 1 = σ(u))p (X v, i = σ(v)) For any l I r we can decompose as follows. Disinguish beween he loops in a pah and he res. Tha means a pah is deermined by he number of loops and heir posiion and he res of he pah wihou loops. This consideraions yield: P (X u,l = σ(u)) = l P (B l,p = k)p (X u,k = σ(u)), k=0

6 6 3 POWERING where B l,p is binomially disribued wih p = 1 1 and d X u,k is defined like X u,k bu wihou loops. In he sequel we will use 3.1 o esimae ha value. Firs se r := (hen l l) and M := max{ N; +r 2 2 r}. Now 2 choose c > 0 such ha he following is me: p(1 p) c 2 M < 1 2 Σ. Se J := {k N; k pl k p 2 c( 2 l)}. Then we have wih Chebyshev P (B 2,p J) ( P B l,p pl > c( 2 ) ( l) + P B 2,p p 2 > c( 2 ) l) lp(1 p) c 2 ( l) + p(1 p) 2 c 2 ( l) 2 p(1 p) + r c Σ Remember, by consrucion of σ we have P (X u, = σ(u)) 1. We can now apply 2 Σ 3.1 wih l 1 =, l 2 2 = i 1: P (X u,i 1 = σ(u)) k J P (B i 1,p = k)p (X u,k = σ(u)) τ k J P (B 2,p = k)p (X u,k = σ(u)) τ(p (X u, = σ(u)) Σ ) τ 2 Σ The same holds for P (X v, i = σ(v)). Therefore we finally have: E(N) = i I r E(N i ) = i I r P (N i > 0) = i I r F P (X u,i 1 = σ(u))p (X v, i = σ(v)) τ 2 F I r 4 Σ 2 Ω( ) F 2. E(N 2 ): To his end we define he following random variables: { 1 if e i F Z i (e) := 0 oherwise Z(e) := i I r Z i (e)

7 7 Then we clearly have N(e) Z(e) and we can sar o calculae. E(Z 2 ) = E(Zi 2 ) + 2 E(Z i Z j ) = I r F + 2 i I r i<j;i,j I r i<j;i,j I r E(Z i Z j ) E(Z i Z j ) = P (Z i Z j > 0) = P (Z i > 0)P (Z j > 0 Z i > 0) = F P (Z j > 0 Z i > 0) P e = ( Z j (e) > 0 Z i (e) > 0) = P e = i+1(z j i+1 (e ) > 0 Z 1 (e ) > 0) F + The las inequaliy is he reason why he regulariy condiion on G had o be posulaed. There we used he ( resul ) given in [Bul06, Theorem 21], saing ha P (Z s (e) > 0 Z 1 (e) > 0) F + λ(g) s, d if G is d-regular. Having ha λ < d, we know ha K := ( λ i i=0 d) <. Gahering all hose resuls, we can esimae E(N 2 ): ( E(N 2 ) E(Z 2 ) I r F + 2 F ( ) ) j i F λ + d i<j;i,j I r O( ( ) ) F 2 ) ( I + 2 r 2 F + I r F K = O( ( ) ) F 2 + O() F. Finally we can show he gap amplificaion. Recall ha F = UNSAT(G) and disinguish beween hese wo cases: 1. UNSAT(G) 1 : In his case we have This finally yields: E(N 2 ) O( ( ) F + O() F ) 2 O( ) F. ( ) j i λ d UNSAT(G ) P (N > 0) Ω( ) 2 F 2 2 O( ) F Ω( ) F Ω( ) min(unsat(g), 1 ) 2. UNSAT(G) > 1 : We know ha E(N 2 ) O() F. This yields he desired resul in his case, oo: UNSAT(G ) P (N > 0) Ω( ) 2 F 2 2 O() F Ω(1) F Ω(1) 1 = Ω( ) 1 Ω( ) min(unsat(g), 1 ) Therefore he proposiion follows.

8 8 4 COMPOSITION 4 Composiion The previous sep amplified he saisfiabiliy gap, bu has blown up he alphabe. I remains o reduce he size of he alphabe. This will be done making use of composiion. We will ake a consrain of he given consrain graph. This will be pu ino an assignmen eser P, which produces a consrain graph on an alphabe Σ 0 wih Σ 0 = O(1). The se of all such graphs ha were produced his way is pu ogeher ino he resuling graph. Le for he sequel SAT(Φ) {0, 1} n denoe he se of assignmens ha saisfy a given boolean circui Φ. Definiion 4.1 [Assignmen Teser] An Assignmen Teser wih alphabe Σ 0 and rejecion probabiliy ε > 0 is a polynomial-ime ransformaion P whose inpu is a circui Φ over Boolean variables X, and whose oupu is a consrain graph G = (V, E), Σ 0, C such ha X V, and such ha he following holds. Le V = V \ X, and le a : X {0, 1} be an assignmen. 1. If a SAT(Φ), here exiss b : V Σ 0 such ha UNSAT a b (G) = If a SAT(Φ) hen for all b : V Σ 0 holds UNSAT a b (G) ε dis(a, SAT(Φ)). Such an algorihm exiss, see for example [Din05, 6.2]. Fixing such an assignmen eser P we can now formulae and prove he following lemma: Lemma 4.2 (Composiion) Le P an assignmen eser wih consan rejecion probabiliy ε > 0, and alphabe Σ 0, Σ 0 = O(1). There exiss β 3 > 0 ha depends only on P, such ha any consrain graph G = (V, E), Σ, C can be ransformed ino a consrain graph G = (V, E ), Σ 0, C, denoed by G P, such ha size(g ) = M( Σ ) size(g), and β 3 UNSAT(G) UNSAT(G ) UNSAT(G) Proof. Assignmen eser are only defined for consrains over Boolean variables. Therefore we firs prepare he graph wih ha respec. Le e : Σ {0, 1} l be an encoding wih relaive Hamming disance ϱ > 0, l = O(log Σ ). Replace each v V by l Boolean variables [v]. Replace each consrain c over he wo variables v, w by c over [v] [w], such ha c is saisfied iff he assignmen for [v] [w] is a legal encoding via e of an assignmen for v and w ha saisfies c. Now run P on any such c and ge consrain graphs G c = (V c, E c ), Σ 0, C c. Wihou loss of generaliy we can assume E c = E c for any consrains c, c (oherwise add edges wih rivial consrains and se ε := ε min c C E c max c C E c ). Now define he resuling graph G as follows: G = (V, E ), Σ 0, C, where V = c C V c, E = c C E c, C = c C C c. Le us now check, if G has go he desired properies. We know ha any c is ransformed ino a consrain c : {0, 1} 2l {T, F }. There are only finiely many such consrains possible. Se M as he maximal size of he oupu graph of P for an inpu c. Then M only depends on Σ and P, and we have size(g ) M size(g). Le σ : V Σ 0 be a bes assignmen for G. Define σ : V Σ by: σ(v) := minarg a Σ (dis(e(a), σ ([v]))).

9 9 Tha means σ(v) is he value whose encoding via e is closes o σ ([v]). Le c be a consrain over he variables u, v ha rejecs σ. Now he propery of he error correcing code comes in: A leas a fracion of ϱ of he bis of 2 σ ([u]) or of σ ([v]) (or of boh of hem) has o be changed in order o lead o a saisfying value. Tha means dis(σ [u] [v], SAT( c)) ϱ. Since 4 P is an assignmen eser wih rejecion probabiliy of ε, we have ha a leas a fracion of ε ϱ of he consrains in C 4 c rejec σ. The assumpion E c = E c for any consrains c, c and E = c C E c herefore guaranees:unsat(g ) εϱ UNSAT(G). 4 5 Main heorem Now we have everyhing a hand o finally sae our main heorem. Theorem 5.1 (Main) For any Σ, Σ = O(1), here exis consans C > 0 and 0 < α < 1, such ha given a consrain graph G = (V, E), Σ, C one can consruc, in polynomial ime, a consrain graph G = (V, E ), Σ 0, C such ha 1. size(g ) C size(g) and Σ 0 = O(1). 2. If UNSAT(G) = 0 hen UNSAT(G ) = 0 3. UNSAT(G ) min(2 UNSAT(G), α). Proof. We consruc G using he lemmas of he previous secions: G = (prep(g)) P Firs we noice ha each lemma only incurs a linear blowup of he size. More precisely, he number of edges increases by a consan facor during Preprocessing and Powering. In he Composiion sep he size grows by a facor ha depends only on Σ d and on P which do no depend on G. Le β 1, β 2, β 3 be he consans in he Lemmas 2.3,3.2,4.2. Now choose ( ) 2 2 = β 1 β 2 β 3 and α = β 3β 2. Then we have alogeher: UNSAT(G ) 4.2 β 3 UNSAT((prep(G)) ) β 3 β 2 min(unsat(prep(g)), ) 2.3 β 3 β 2 min(β1 UNSAT(G), 1 ) min(2 UNSAT(G), α). This proves he heorem. A las we can now prove he PCP Theorem. [Bul06, Lemma 9] has already shown ha he PCP Theorem is equivalen o showing ha Gap-3SAT is NP-hard. Corollary 5.2 Gap-3SAT is NP-hard.

10 10 REFERENCES Proof. According o [Bul06, Theorem 14], i is NP-hard o decide if UNSAT(G) = 0 or no for a given consrain graph G wih Σ = 7. Le G 0 be such an insance and G i be he oucome of applying he main heorem on G i 1. Se k := log(α E 0 ) = O(log n). This way we have for all i k : size(g i ) C i size(g 0 ) = poly(n). If UNSAT(G 0 ) = 0 hen UNSAT(G i ) = 0. If no, hen we have by inducion: UNSAT(G i ) min(2 i UNSAT(G 0 ), α). If UNSAT(G 0 ) > 0, hen UNSAT(G 0 ) 1, so surely we have E 0 2k UNSAT(G 0 ) > α and herefore UNSAT(G k ) α. Now a local gadge reducion akes G k o 3SAT form, while mainaining he saisfiabiliy gap up o some consan. This proves he corollary. References [AS98] S. Arora and S. Safra. Probabilisic checking of proofs: A new characerisaion of NP. J.ACM, 45(1):70-122, [Bul06] Lukas Bulwahn. The PCP Theorem: An Inroducion. Munich, [Din05] Iri Dinur. The PCP Theorem via Gap Amplificaion. Technical Repor TR05-46, Elecronic Colloquium on Compuaional Complexiy, [FGL + 96] U. Feige, S. Goldwasser, L. Lovász, S. Safra, and M. Szegedy. Approximaing clique is almos NP-complee. Journal of he ACM, 43(2): , 1996.