Notes for Lecture 17-18
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1 U.C. Berkeley CS278: Compuaional Complexiy Handou N7-8 Professor Luca Trevisan April 3-8, 2008 Noes for Lecure 7-8 In hese wo lecures we prove he firs half of he PCP Theorem, he Amplificaion Lemma, up o a saemen abou random walks on expanders ha will be proved laer. Lemma (Amplificaion) For all Σ 0, here is a δ 0 such ha c, here exiss Σ and a poly-ime reducion R, mapping Max-2-CSP-Σ 0 o Max-2-CSP-Σ, such ha he following hold for all insances C of Max-2-CSP-Σ 0. # of consrains of R (C) = O(#of consrains of R (C)) 2. op(c) = op(r (C)) = 3. op(c) δ op(r (C)) cδ, provided cδ δ 0 By a resul proved in he las lecure, wihou loss of generaliy we may assume he consrain graph G of C is a d-regular expander graph wih λ 2 (G) λ <, for an absolue consan λ indepeden of all he parameers of he lemma. Le C be he weighed consrain saisfacion problem over Σ = Σ +d+d2 + +d 0 (for some o be specified laer) defined as follows: Variables: For each variable v i, i =,..., n, of C we define a variable v i for C. Each variable v i of C can be seen o assign an elemen of Σ 0 o is associaed verex v i in G. The corresponding variable v i of C can be inerpreed o associae an elemen of Σ 0 o v i and any verex in G ha is a mos edges away from v i in G. Consrains: Firs we define a disribuion over pahs in G. Then we will associae a consrain wih each of hese pahs and will weigh hem by heir probabiliy. This will give us a weighed CSP. Use he following randomized procedure o pick a pah in G; his clearly defines a disribuion over pahs: Pick a saring verex v 0 a random. (D ) For i = o : Wih probabiliy /2 le v i+ v i. Wih probabiliy /2 le v i+ a random neighbor of v i. Remark This can be viewed as a random walk of lengh on he 2d-regular graph G which is he same as G excep ha G has addiionally d self loops a each verex. The consrain associaed wih a pah v 0 v v is saisfied in C if and only if he variables corresponding o v 0 and v in C give consisen assignmens o all he verices in he pah and hose assignmens saisfy he consrain associaed wih each edge in he pah. Lemma 2 (Main) The reducion described above saisfies he following:. op(c) = op(c ) =
2 2. op(c) δ op(c ) Ω λ, Σ0 ( δ ), provided δ < Proof:. The firs par is clear, because, if A is an assignmen saisfying C, hen A assigns an elemen of Σ 0 o each verex of he consrain graph G, so ha he consrains associaed wih each edge are saisfied. We can use hese same assignmens o generae an assignmen Ā for C in he naural way. Then Ā will saisfy he consrain corresponding o every pah p : v 0 v v of G because: ) The values Ā(v 0) and Ā(v ) are consisen by consrucion and 2) Every edge in he pah will be saisfied by he values ha Ā(v 0 ) and Ā(v ) assign o is endpoins, because each edge of he pah is eiher a self loop or an edge in G and A saisfies he consrain corresponding o every edge of G. 2. To show he second par, we will define a mapping from an assignmen Ā for C o an assignmen A for C and we will show ha value[a(c)] δ value[ā(c )] Ω λ, Σ0 () This proves he heorem, because, if op(c) δ, hen, for any Ā, he corresponding assignmen A will saisfy a mos a δ fracion of he consrains, and so, by (), he value of Ā can be a mos Ω( δ). The mapping is defined as follows. Le us fix an assignmen Ā for C. To define he assignmen A for C o which Ā is mapped we use he following disribuion over pahs saring a any given verex v 0 of G: (D 2 ) for i = o /2: Wih probabiliy /2 le v i+ v i. Wih probabiliy /2 le v i+ a random neighbor of v i. Le A(v 0 ) be he value mos likely o be given o v 0 by Ā(v /2) according o his disribuion. Noice ha Pr[A(v 0 ) = Ā(v /2)] Σ 0. We can pick a random consrain of C using process (D ). Equivalenly, we can use he following randomized procedure for some fixed b, b { /2 +,..., /2}: pick a (direced) edge (v /2 b, v /2 b+ ) from G a random (recall ha G is a 2d regular graph resuling from G by adding d self loops o each verex) (D 3 ) choose v /2 b,..., v 0 by aking a random walk in G saring a v /2 b choose v /2 b+2,..., v by aking a random walk in G saring a v /2 b+ Inuiion: Roughly speaking, if b = 0, he consrain picked by he above process will δ be violaed wih probabiliy a leas Σ The edge (v /2, v /2+ ) picked a he firs sep of he procedure will be violaed by A wih probabiliy a leas δ 2, because a leas a δ fracion of he edges of G are no saisfied by A and wih probabiliy 2 he firs sep of he above process will pick an edge of G (wih probabiliy 2 i will pick one of he self loops added o G o ge G ). Furhermore, as we commened above, A(v /2 ) will be consisen wih Ā(v 0) s assignmen for v /2 wih probabiliy a leas Σ 0. Similarly, A(v /2+) will be consisen 2
3 wih Ā(v ) s assignmen for v /2+ wih probabiliy a leas Σ 0 δ. Noe ha his Σ bound is only a lower bound on he probabiliy ha he middle edge of a random consrain is conradiced. If a similar bound holds for Ω( ) edges in he middle of a random pah and if he corresponding evens are close o being disjoin, we would ge our bound. Formally: Le us fix an assignmen Ā for C and he corresponding assignmen A for C. The following definiion is cenral o he remaining analysis. Definiion In a pah v 0 v v, a (direced) edge (v i, v i+ ) is fauly if A conradics consrain (v i, v i+ ). A(v i ) is consisen wih Ā(v 0). A(v i+ ) is consisen wih Ā(v ). I is easy o see ha, if a pah conains a fauly edge, hen he consrain corresponding o he pah is violaed by Ā. Moreover, if b, hen he probabiliy ( under ) (D 3 ) ha he iniially chosen edge is fauly should work ou as before o be Ω δ Σ So, if he corresponding evens for differen values of b were disjoin, we would be done. Wha we shall show nex is ha hey are close enough o being disjoin. Le F be he se of edges conradiced by A (or a subse of i of size δ E if he se of edges conradiced by A is bigger han ha). For every direced edge (u, v) F, le us define he random variable X (u,v),i for a randomly chosen pah v 0 v v in G : if X (u,v),i = (v i, v i+ ) = (u, v) Ā(v 0 ) is consisen wih A(v i ) Ā(v ) is consisan wih A(v i+ ) 0 oherwise Noice ha X (u,v),i is he indicaor variable ha, when picking a random pah of lengh in G, he i-h verex is u, he (i + )-h verex is v, Ā(v 0 ) is consisen wih A(v i ) and Ā(v ) is consisen wih A(v i+ ). If we define N = /2+ i=/2 (u,v) F X (u,v),i hen i is easy o see ha he (weighed) fracion of consrains conradiced by Ā is a leas Pr[N > 0]. The following proposiion complees he proof. Proposiion 3 Pr[N > 0] = Ω( δ) Proof: We prove his in hree pars: This is almos rue because his is walk has lengh insead of as is specified by our mapping and 2 2 (D2)., 3
4 . Pr[N > 0] E[N]2 E[N 2 ] 2. E[N] = Ω( δ) 3. E[N 2 ] = O( δ) (Par ) Claim 4 For every non-negaive random variable N: Pr[N > 0] (E[N])2 E[N 2 ] Proof: Le N>0 be he indicaor variable of he even {N > 0}, i.e. {, if N > 0 N>0 = 0, oherwise From he Cauchy-Schwarz Inequaliy i follows ha E[N] = E[N N>0 ] E[N 2 ] E[ 2 N>0 ] = E[N 2 ] E[ N>0 ] = E[N 2 ] Pr[N > 0] The claim follows. (Par 2) I is enough o show he following. Claim 5 For every (u, v) F and for every i {/2, /2 + }: Pr[X (u,v),i = ] = Ω[/ E ]. Proof: Pr[X (u,v),i = ] = Pr[v i = u and v i+ = v] Pr [ ] Ā(v0 ) is consisen wih A(v i ) Ā(v ) is consisan wih A(v i+ ) v i = u and v i+ = v Noe ha Pr[v i = u and v i+ = v] = 4 E, because G has 4 E direced edges. To lower bound he oher facor, we will need o compare he following experimens:. Pick a random direced edge (u, v) in G. Do a random walk in G of lengh i from u, and a random walk of lengh i from v. Call he end poins of hese wo walks a and b. 2. Pick a random direced edge (u, v) in G. Do a random walk in G of lengh /2 from u. Call he end poin a. Do a random walk in G of lengh /2 from v. Call he end poin b. Noe ha, by definiion of he mapping from Ā o A, Pr[Ā(a) is consisen wih A(u)] / Σ 0 and Pr[Ā(b) is consisen wih A(v)] / Σ 0. Once we condiion on he number of seps ha he walks ook in G (which is G wihou he self loops), he probabiliy of consisency is he same in he above experimens. The res of he analysis is he following: In experimen 2: 4
5 Pr[Ā(a) consisen wih A(u)] = Σ 0 = l Pr[Ā(a) consisen wih A(u) l seps aken in G] (S 2 ) Pr[l seps aken in G when a oal of /2 seps were aken in G ] While in experimen : Pr[Ā(a) consisen wih A(u)] = = l Pr[Ā(a) consisen wih A(u) l seps aken in G] (S ) Pr[l seps aken in G when a oal of i seps were aken in G ] The range of l in he above summaions is no he same when i 2. Ignoring his fac, since he firs facor of every erm in he above summaions is he same, o finish he claim i would be enough o show somehing of he flavor Pr[l seps aken in G when a oal of i seps were aken in G ] = Ω(Pr[l seps aken in G when a oal of /2 seps were aken in G ]) This is no always rue, bu i is rue when he value of i is close o 2 2 and he value of l is close o is expecaion, ha is, around /4. So, le us remove he ails from he above summaions! I is no hard o see ha here exiss some c = c(log Σ 0 ) such ha (S 2 ) implies 2 Σ 0 4 +c l= 4 c Pr[Ā(a) consisen wih A(u) l seps aken in G] Pr[l seps aken in G when a oal of /2 seps were aken in G ] 4 +c l= 4 c Pr[Ā(a) consisen wih A(u) l seps aken in G] O ( Pr[l seps aken in G when a oal of i seps were aken in G ] ) O(Pr[Ā(a) consisen wih A(u) in firs experimen]) The second inequaliy in he above derivaion relies on he fac ha, if we flip i fair coins i { 2,..., 2 + }, hen he probabiliy of geing l heads where l { 4 2 Recall ha in he saemen of he claim we asked i {/2, /2 + }. 5
6 c,..., 4 + c } is Θ( i ) = Θ( ). The hird inequaliy resuls from removing he ails from summaion (S ). Puing everyhing ogeher we ge E 2 + i= 2 X (u,v),i 2 ( ) 4 E Ω = Ω Σ 0 Σ0 ( ) 4 E. And so E[N] = Ω Σ0 ( ) δ E 4 E = Ω Σ 0 ( δ). (Par 3) Claim 6 E [ N 2] = O λ ( δ). Proof: For a pah v 0 v... v, randomly chosen as described above, we define random variables X (u,v),i and N as follows X (u,v),i = N = {, if vi = u and v i+ = v 0, oherwise 2 + i= 2 (u,v) F X (u,v),i Clearly, N N and, hus, i suffices o bound E [ N 2]. The bound follows from he following resul abou random walks in expanders, which we will prove in he nex lecure, applied o he l = 2 seps of he random walk beween sep 2 and sep 2 +. Lemma 7 Le G = (V, E) be a d regular graph wih λ 2 (G) λ <, le F E, and define δ = F / E. Pick a random walk of lengh l in G, and le M be he number of edges of F raversed in he walk. Then E[M 2 ] = O λ (δl + δ 2 l 2 ) The lemma above gives us a bound E[N 2 ] = O λ (δ + δ 2 ), bu recalling ha δ <, we have E[N 2 ] = O λ (δ ) as required. (end of proof of claim 6) (end of proof of proposiion 3) (end of proof of lemma 2) 6
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