CS286.2 Lecture 4: Dinur s Proof of the PCP Theorem

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1 CS86. Lecture 4: Dur s Proof of the PCP Theorem Scrbe: Thom Bohdaowcz Prevously, we have prove a weak verso of the PCP theorem: NP PCP 1,1/ (r = poly, q = O(1)). Wth ths result we have the desred costat completeess ad soudess, but the proof s expoetally log. It takes dffcult work to shorte the proof to get the full PCP theorem: NP PCP 1,1/ (r = O(log ), q = O(1)). Now we wll see a dfferet approach to the proof of the PCP theorem whch was recetly dscovered (006) by Irt Dur. The approach here wll be to take a result wth a small soudess gap, NP PCP 1,1 1 m (r = O(log ), q = O(1)), ad gradually amplfy the soudess to get the desred gap whle keepg the legth of the proof costat. The proof ca be uderstood usg the CSP varat of the PCP theorem: Theorem 1. For ay L NP there s a polyomal-tme mappg x {0, 1} ϕ x where ϕ x s a q-csp over m = poly() clauses, q O(1), such that: x L = ω(ϕ x ) = 1 x / L = ω(ϕ x ) 1/ The ma Lemma facltatg Dur s proof s the followg: Lemma. There exsts a effcet polyomal-tme procedure mappg a -CSP ϕ = ϕ () to -CSP ϕ = ϕ (+1) as follows ϕ = ϕ () over Σ = {0, 1} = ϕ = ϕ (+1) over Σ = {0, 1} ϕ () has m costrats = ϕ (+1) has m Cm costrats where C s some uversal costat ω(ϕ () ) = 1 = ω(ϕ (+1) ) = 1 ω(ϕ () ) 1 ɛ = ω(ϕ (+1) ) 1 ɛ To see why ths lemma mples the PCP theorem, here s the dea: take some NP-complete laguage lke 3 COLORING. By defto t s NP-hard to decde for the correspodg - CSP ϕ (0) whether ω(ϕ (0) ) = 1 or ω(ϕ (0) ) 1 m 1, where m s the umber of costrats (whch the case of 3 COLORING s the umber of edges the graph). Applyg the above Lemma oce brgs us from {ω(ϕ (0) ) = 1 or ω(ϕ (0) ) 1 m 1 } to {ω(ϕ(1) ) = 1 or ω(ϕ (1) ) 1 m }, 1

2 ad so applyg t t tmes brgs us to {ω(ϕ (t) ) = 1 or ω(ϕ (t) ) 1 t m }. Settg t log m, we re doe, otg that ϕ (t) C t m m log C+1 remas polyomal the sze of the orgal stace ϕ (0). The proof of the Lemma has two ma steps: gap amplfcato (GA) ad alphabet reducto (AR). These steps operate as follows o our orgal -CSP stace: CSP: ϕ = ϕ () a -CSP GA ϕ = ϕ () a -CSP AR ϕ = ϕ (+1) a -CSP Clauses: m GA m Cm AR m C m Alphabet: Σ = {0, 1} GA Σ = [w] = {0, 1,..., w 1} where w s some larger teger AR Σ = {0, 1} Completeess: ω(ϕ = ϕ () ) = 1 GA ω(ϕ = ϕ () ) = 1 AR ω(ϕ = ϕ (+1) ) = 1 Soudess: ω(ϕ = ϕ () ) 1 ɛ GA ω(ϕ = ϕ () ) 1 6ɛ AR ω(ϕ = ϕ (+1) ) 1 ɛ As oted above, repeatg these steps O(log m) tmes gves a costat gap wth oly a polyomal blow-up the umber of costrats (sce all clauses volve a costat umber of varables, the same automatcally holds for the umber of varables). The harder part s costructg the gap amplfcato procedure, so that s what we wll look at ths lecture. It requres the use of expader graphs. Expader Graphs A expader graph s a graph G = (V, E) whch has good expaso propertes. A loose defto would be that there s ever a small subset of vertces that has a lot of edges sde, but coectg the subset to the rest of the graph wth oly a few edges. We focus o -vertex d-regular graphs, ad gve the followg defto. Defto 3. Let, d be tegers ad 0 ρ < 1. A -vertex d-regular graph G s called a (, d, ρ)-expader f S [] s.t. S, E(S, S) ρd S. (1) Note that d S s the maxmum umber of edges that we could talk about. Also, otce that we always have ρ 1.1 ). The followg theorem states the exstece of good expaders. Theorem 4. ɛ > 0, d = d(ɛ), N = N(ɛ) s.t. N, G whch s a (, d, 1 ɛ)- expader. 1 To see ths, for ay graph G cosder a radomly chose S obtaed by cludg every vertex depedetly wth probablty 1/. I expectato the set S wll have exactly d/4 edges gog out, ad so there must exst a set S havg as may edges leavg t.

3 We wll ot prove t, but t s ot too hard to show usg the probablstc method: a radom graph works. Explct costructos, o the other had, are much harder. The followg lemma about expaders states the key property that wll be used the aalyss of Dur s gap amplfcato procedure. Lemma 5. Suppose that G s a (, d, ρ)-expader. The for ay set S [] wth S, t, Pr u t v(u S, v S) S ( ( S 1 ρ + ) t ) () where u t v deotes choosg a uformly radom u V, makg a t-step radom walk startg there, ad arrvg at v V. Proof. We ll prove t for t = 1. By defto of expaso, Pr u t v(u S, v S) = Pr(u S)Pr(v S u S) = S S S (1 ρ) ( S + 1 ) ρ ( S + 1 ρ where the frst equalty uses S / 1/ ad the last that 1 ρ 1 ρ for ρ. Note ths s a bt of a wasteful boud. Alteratvely there s a algebrac techque: ρ s related to the secod largest egevalue λ (G) of the adjacecy matrx A(G) as λ (G) 1 ρ (ths s called Cheeger s equalty). The use λ (G t ) = λ (G) t ad the use ρ 1 λ. G t s a graph whose adjacecy matrx specfes the paths of legth t G as edges, rather tha the paths of legth 1; applyg the t = 1 case prove above to G t would yeld the lemma. We omt the detals. Proof/Costructo of Gap Amplfcato Step We just wat to get the ma dea of ths proof, ad so we assume that the costrat graph G of ϕ s a (, d, ρ)-expader for some ρ < 1. (A addtoal step Dur s proof shows that ths ca always be the case, wthout loss of geeralty.) Suppose ω(ϕ) 1 ɛ. The ay assgmet to the vertces volates a fracto of at least ɛ of the costrats. These correspod to a bad set of edges of G, whch we ca call S. We wat to fgure out how to defe ϕ as specfed for the gap amplfcato step. Fx a parameter t ad defe ϕ as follows: ϕ s varables are the same as ϕ s ϕ s alphabet s Σ = {0, 1} 1+d+d +...+d t. The dea s that each varable holds a value for tself, ts earest eghbors ad etc... all the way to the t-th earest eghbors. 3 ),

4 The costrats are weghted costrats; we ca always go back to uweghted costrats by duplcatg the heavy (hgher probablty) costrats as may tmes as requred. The costrats are geerated accordg to the followg dstrbuto: 1. Choose a radom vertex u.. Make a t-step radom walk G from u to v. 3. Look at the values of the varables of ϕ, x u ad x v, that are assocated to the vertces u ad v. Each of these varables provdes a assgmet of values {0, 1} to all vertces at dstace at most t from them. Check that the values provded by x u ad x v agree o all vertces the path u v, ad that the commo values satsfy all of ϕ s costrats alog the path. The whole checkg procedure makes a sgle costrat that s placed o a (u, v) edge for ϕ. Why does ths work? The frst thg to check would be that ω(ϕ) = 1 = ω(ϕ ) = 1, whch s clear: all varables x ϕ ca be assged values cosstet wth a sgle satsfyg assgmet x for ϕ. Secod ad more mportatly s to check that ω(ϕ) 1 ɛ = ω(ϕ ) 1 6ɛ. Let s take a assgmet y to the varables for ϕ, such that y volates a fracto (1 ωϕ ) = 1 δ of the clauses. What we ll show: there exsts some x that volates a fracto at most 6 δ of the costrats of ϕ. Sce by assumpto ω(ϕ) 1 ε, we must have 6 δ coclude the proof. ɛ ad so δ 6ɛ, whch wll We make a mportat smplfcato. Assume that y s obtaed the correct way from some x for ϕ: whe two y overlap (that s, they provde values for the same vertces of G) the ther values match. If ths smplfyg assumpto s ot true, the the proof volves a addtoal step of amoutg to a decodg procedure whch defes x at ay vertex to be the majorty value amog those provded by all eghbourg y; t s the possble to show that there s always a strog majorty. Our smplfyg assumpto thus amouts to assumg that majorty s uamty! I So, our goal s to show that δ 6ɛ. We ll show that f y volates at most a fracto δ of the clauses, the x volates at most a fracto δ t of clauses. But the best x volates at least a fracto ɛ of clauses.. so, δ t ɛ ad so δ tɛ ad we ca take t 6 to get what we wat. (Note that we could get better for larger t but remember that the alphabet sze for ϕ creases very quckly wth t...) We eed to show that a radom costrat s volated wth probablty of at least tɛ. A radom costrat s a whole buch of costrats from ϕ. If these were chose depedetly the the probablty we completely avod the bad set S would be (1 ɛ) t 1 tɛ, ad so wth probablty at least tɛ we see a volato. However, costrats are ot depedet: they are selected from the radom walk. The followg lemma wll let us complete the proof. Lemma 6. Let G be a (, d, ρ) expader. Let S be a subset of edges of G such that S /m = ε, where m s the total umber of edges. Let t < 1/(dε) be a teger. The ρ Pr (at least oe edge of the t-walk s S) u t v 8d tε. 4

5 Applyg the lemma wth S the set of edges correspodg to costrats that are ot satsfed by the assgmet x, we obta that δ ρ /(8d)tε. Choosg t = 48d/ρ (where ρ s the expaso factor of a good expader used the costructo, ad ca be take to be a small costat, e.g. ρ = 1/4) the gves us δ 6ε (as log as ε s ot too large, case we d be doe already), as desred. Proof. For = 1,..., t defe dcator varables X = 1 f the -th edge of the t-step radom walk from u to v s S, ad X = 0 otherwse. The E[X ] = S /m = ε: sce the frst vertex u of the walk s chose uformly at radom, ay dvdual edge of the walk s also a uformly radom edge G. By learty of expectato, we deduce [ E[X] = E ] X = E[X ] = tε, where X s the umber of edges alog the walk that are S. The probablty the lemma s Pr(X > 0). We wll use a secod momet boud Pr(X > 0) E[X] E[X ], (3) whch holds for ay radom varable X 0. For ths t oly remas to lower boud the secod momet E[X ]. Let S be the set of vertces G that touch a edge S. Wrte ] E[X ] = E[ X X j 1,j t = E[X X j ] 1,j t = E [ X ] + E[X X j ] = tε + Pr(-th edge S ad j-th edge S) tε + tε + tε + t (dε) + tdε/ρ (3 + 4d/ρ )tε, Pr(-th vertex S ad j-th vertex S) dε (dε + ( (1 ρ )/ ) ) j where the here the key step s gve the secod equalty, whch we appled Lemma 5 (the factor dε comes from the fact that S d S ). I the last equalty we used tdε < 1. Together wth (3) ths cocludes the proof of the lemma. 5