PCP Theorem Proof: Powering Stage

1. PCP Theorem Proof: Powering Stage Part II Irit Dinur Presentation by Boris Danilovich and Jonathan Heimann

2. The Story So Far • Input: • G=(V,E) an (n,d,λ)-expander with alphabet Σ. • A set of constraints CΣXΣ. • Construction of new G’=(V’,E’): • V’= V • E’– Any A.S.R.W from a to b => (a,b)E’. • w(a,b) = P[A.S.R.W begins on a and ends on b] • Σ’ - All binary strings up to length dt • C’ = {(ε1, ε2)Σ’XΣ’ | ε1 and ε2 agree about neighbors, • derived plurality assignment satisfies C}

3. A.S.R.W – After Stopping Random Walk B.S.R.W – Before Stopping Random Walk Let G=(V,E) be a regular graph. Back to Basics

4. Back to Basics σ’:V’→Σ’ – An assignment which minimizes UNSAT(G’). σ:V→Σ – σ(v) = the “popular” σ’(w)v, for w’s that are neighbors of v. F’E is the set of all edges in E which are not satisfied by assignment σ. FF’E is produced from F’ by removing edges until |F|/|E| = UNSAT(G). Make an A.S.R.W from a to b in a graph G. • An edge (a,b)E’ is badif it contains a step u→v for which: • For every wV a t-neighbor of a and b in graph G, ‘(a)w = ‘(b)w • distG(u,a)  t AND distG(v,b)  t • The assignment ) ‘(a)u,‘(b)v) is a violation of the constraint on (u,v).

5. Back to Basics Make an A.S.R.W from a to b in a graph G. • A step u→v within such a walk is called faulty if: • (u,v)F • distG(u,a)  t AND ‘(a)u = (u) • distG(v,b)  t AND ‘(b)v = (v) N = # number of faulty steps in an A.S.R.W.

6. t O(1) Back to Basics UNSAT(G’) = P[A bad edge inG’ will be generated by an A.S.R.W in G] ≥ P[The chosen A.S.R.W walk in G contains a faulty step] = P]N>0] ≥ E[N]2 E[N2] Second moment method ≥? min(UNSAT(G),1/t)

7. Setting immediate goals E[N]2 We want to prove: P[N>0] ≥ = [t/O(1)]ּ[|F|/|E|] We will prove: 1) E[N]2 ([t/O(1)]ּ[|F|/|E|])2 2) E[N2]  [t/O(1)] ּ[|F|/|E|] E[N2]

8. Lemma 1:E[N] [t/4|∑|2]ּ[|F|/|E|] Remember we talk about A.S.R.W לפי הגדרה זה קבוע... E[N] = E[#faulty steps] = = E[over all uv E: #faulty uv steps] = =E[over all uvF: #fauly uv steps]  [#possible steps  F]ּminuvF{E[#faulty uv stepF]}= = 2|F|ּminuvF{E[#faulty uv step  F]} If we prove that: minuvF {E[#faulty uv step  F]} [t/(8|∑|2)ּ[(1/|E|) We get that: E[N][t/4|∑|2]ּ[|F|/|E|]as required

9. Lemma 1:1) E[N] [t/4|∑|2]ּ[|F|/|E|]minuvF{E[#faulty uv step  F]} [t/(8|∑|2)ּ[(1/|E|) [Note: if we prove this bound to be a lower bound of every {u→vF faulty step}, we are done] For any step uvF: E[#faulty uv steps] = A fact about expectancy: E[A] = ∑ E[A|B(k)]ּPr[B(k)] =∑E[#faulty uv steps |exactly k uv steps]ּPr[exactly k uv steps] Remember that in an specific A.S.R.W, if a u→v step is faulty once, it will is faulty each time in that walk. E[#faulty uv steps | exactly k uv steps] = = k * Pr[uv steps are faulty | exactly k uv steps] We’ll set this value in and receive: =∑k*Pr[uv steps are faulty |exactly k uv steps]ּPr[exactly k uv steps] 1 k 1

10. Lemma 1:1) E[N] [t/4|∑|2]ּ[|F|/|E|] minuvF{E[#faulty uv step  F]} [t/(8|∑|2)ּ[(1/|E|)Continue – 2 out of 3 Overall, we got (from previous slide): E[#faulty uv steps]= =∑k * Pr[uv steps are faulty|exactly k uv steps] ּPr[exactly k uv steps] We’ll claim and soon prove: P[uv steps are faulty | exactly k uv steps]  1/(4|∑|2) We’ll set this value in and receive: =∑1/(4|∑|2) * k ּPr[exactly k uv steps] = we take a constant value out of the sum = 1/(4|∑|2) * ∑ kּPr[exactly k uv steps] = = 1/(4|∑|2) * E[#of uv steps (Not only faulty)] = 1 2 1

11. Lemma 1:1) E[N] [t/4|∑|2]ּ[|F|/|E|] minuvF{E[#faulty uv step  F]} [t/(8|∑|2)ּ[(1/|E|)Continue – 3 out of 3 מציין(בהסתברות) = 1/(4|∑|2) * E[#of uv steps(Not only faulty)] = = 1/(4|∑|2) * E[∑[uv is the ith step]] = 1) The chances of each ith step to be uv are equal 2) The expected number of steps = t = 1/(4|∑|2) *t ּ E[ [uv is the ith step]] = Regular G  the probability of going over a particular edge in an A.S.R.W = 1/|E|  the probability of making that uv step = 1/(2|E|) = 1/(4|∑|2) * t /(2|E|) = = 1/(8|∑|2) * t /|E| A.S.R.W d-Regular 1 i 1

12. Lemma 1:1) E[N] [t/4|∑|2]ּ[|F|/|E|] minuvF{E[#faulty uv step  F]} [t/(8|∑|2)ּ[(1/|E|)Debt from last week:E[#uv steps] = t/(2|E|) E[#of uv steps(Not only faulty)] = = ∑E[#uv steps | exactly k steps] ּ Pr[exactly k steps] = = ∑ k/(2|E|)ּ Pr[exactly k steps] = = 1/2|E| ּ ∑ k ּ Pr[exactly k steps] = = 1/2|E| ּ t = = t/2|E| Each step has Pr=1/2|E| A Geometric Expectation

13. Yet more debts Suppose we did an A.S.R.W in a graph G=(V,E), conditioned on making exactly k u→vsteps, where k≥1 is some constant and (u,v) a fixed edge in E. Then: • Distribution of the final vertex b ~ Distribution of a B.S.R.W from v. • Distribution of the initial vertex a ~ Distribution of a B.S.R.W. from u. • a and b are independent.

14. Proof • Change the condition: instead of taking exactly k u→v steps, we’ll take at • least k. • Take kth u→v step. A.S.R.W turns into B.S.R.W (choice of stopping on v • taken yet) and there are no further restrictions. Denote Y = # of u→v steps. Pw = P[A B.S.R.W from v stops on vertex w] = P[b=w | Y≥k]=? P[b=w | Y=k] 1 P[(b=w) ۸ (Y≥k)] P[(b=w) ۸ (Y=k)] + P[(b=w) ۸ (Y≥k+1)] P[b = w | Y≥k]= = P[Y≥k] P[Y≥k] Pu→v = P[u→v taken in A.S.R.W] Pu→v’ = P[u→v taken in A.S.R.W | begin on v]

15. Proof P[(b=w) ۸ (Y=k)] + P[(b=w) ۸ (Y≥k+1)] 1 P[b = w | Y≥k]== Pw P[Y≥k] P[Y≥k] = Pu→v∙ (Pu→v’)k-1 P[(b=w) ۸ (Y≥k+1)] = P[b = w | Y≥k+1] = Pw- B.S.R.W is memoryless P[Y≥k+1] => P[(b=w) ۸ (Y≥k+1)] = P[b = w | Y≥k+1]∙P[Y≥k+1] = Pw ∙ Pu→v ∙ (Pu→v’)k Set these values in : 1 P[(b=w) ۸ (Y=k)] + Pw∙ Pu→v ∙ (Pu→v’)k Pw = => Pu→v ∙ (Pu→v’) k-1 P[(b=w) ۸ (Y=k)] = Pw ∙ Pu→v ∙ (Pu→v’)k-1 ∙ (1 - Pu→v’) Finally: P[(b=w) ۸ (Y=k)] Pw ∙ Pu→v ∙ (Pu→v’)k-1 ∙ (1 - Pu→v’) P[b=w | Y=k] = = = Pw P[Y=k] Pu→v ∙ (Pu→v’)k-1 ∙ (1 - Pu→v’)

16. Where are we standing? E[N]2 Remember: UNSAT(G’) ≥ E[N2] • After bounding E[N]2, we are one step away from completing the • analysis and put a constant-growth lower-bound on UNSAT(G’). • Something is rotten in the state of Denmark: • Edges of E’ are weighted. • G’ is too large – Ω(n2).

17. Recalling G’’ • In the previous lecture, G’’ was created out of G’, by: • Splicing a single edge in E’, into several unweighted parallel edges in E, according to its weight. Growth: by a constant factor. • Removing all edges in E’ which were generated by a walk in G that was longer than B=(10ln|Σ|)t steps.

18. Doesn’t this hurt the credibility of our argument? Note that G’’G’. Couldn’t the removal of edges hurt the UNSAT? EB = E’ – E’’ α = |EB|/|E’| = P[Edge eE’ generated by walk longer than B] = P[A.S.R.W takes more than B steps] = (1-1/t)B ≤ exp(-B/t) = exp(-10ln|Σ|t/t) = 1/(|Σ|10) - Constant! Worst case – All edges of EB were unsatisfied UNSAT(G’’) = (|F’|- α|E’|)/[|E’|(1 – α)] = |F’|/[|E’|(1-α)] – α|E’|/[|E’|(1-α)] = 1/(1-α)∙UNSAT(G’) – α/(1-α) = O(UNSAT(G’)) Change is by small constant factor only! Note: might be a problem in the first iterations, if UNSAT is too small

19. A faulty* step A faulty step: Let there be a walk from a to b chosen by an A.S.R.W from a in graph G. • A step u→v within such a walk is called faulty if: • (u,v)F • distG(u,a)  t AND ‘(a)u = (u) • distG(v,b)  t AND ‘(b)v = (v) • A step is faulty* if: • it is faulty. • the number of steps in the entire a→b walk is at most B. Denote the random variable N* = # of faulty* steps in an A.S.R.W

20. A faulty* step u→v is a faulty* step in an A.S.R.W from a to b => (a,b) is a bad edge in G’’ This is because σ’(a) and σ’(b) agree with the popular opinion, and according to it (u,v)F (meaning, the constraint on the edge is not satisfied by σ). Reminder: Let there be a walk from a to b chosen by an A.S.R.W from a in graph G. • An edge (a,b)E’ is badif it contains a step u→v for which: • For every wV a t-neighbor of a and b in graph G, ‘(a)w = ‘(b)w • distG(u,a)  t AND distG(v,b)  t • ) ‘(a)u,‘(b)v) is a violation of the constraint on (u,v).

21. t O(1) Bounding UNSAT(G’’) UNSAT(G’’) = P[A bad edge inG’’ will be chosen by an A.S.R.W in G] ≥ P[The chosen A.S.R.W walk in G contains a faulty* step] = P]N*>0] Each walk containing a faulty* step generates a bad edge. ≥ E[N*]2 E[N*2] Second Moment Method ≥? min(UNSAT(G),1/t)

22. t |F| 8|Σ|2 |E| If we could just prove this… |F|/|E| = UNSAT(G) Lemma 1: E[N*] ≥ |F| Lemma 2: E[(N*)2] ≤ O(1)∙t∙ |E|

23. t |F| 8|Σ|2 |E| t |F| 4|Σ|2 |E| Proof Lemma 1: E[N*] ≥ S – # of steps in walk E[N*] = E[N ∙ 1{S≤B}] = E[N ∙ (1 – 1{S>B})] = E[N] – E[N ∙ 1{S>B}] ≥ – E[N ∙ 1{S>B}] An indicator Lower bound on E[N], shown before.

24. t |F| t |F| 8|Σ|2 |E| 8|Σ|2 |E| t |F| t |F| 8|Σ|2 |E| 4|Σ|2 |E| Proof Lemma 1: E[N*] ≥ We want to find an upper-bound on E[N ∙ 1{S>B}]. NF – # of steps that were in F E[N ∙ 1{S>B}] = P[S>B] ∙ E[N | S>B] = (1-1/t)B ∙ E[N | S>B] ≤ exp(-B/t) ∙ E[NF | S>B] = exp(-B/t) ∙ E[S | S>B] ∙ (|F|/|E|) = exp(-B/t) ∙ (B+t) ∙ |F|/|E| = exp(-10ln|Σ|) ∙ t(10ln|Σ| + 1) ∙ |F|/|E| = (10ln|Σ| + 1)/|Σ|10 ∙ t(|F|/|E|) ≤ 1/(8|Σ|2) ∙ t(|F|/|E|) = => E[N*] ≥– E[N ∙ 1{S>B}] ≥ Proof follows.

25. t |F| 8|Σ|2 |E| Proof Lemma 1: E[N*] ≥ Why E[NF | S>B] = E[S | S>B] ∙ (|F|/|E|) ? NF = # of steps of walk in F S = # of steps in walk Xi = 1[ith step is in F] We’ll show E[NF] = E[S] ∙ (|F|/|E|). This also holds with the condition. E[NF] = E[Σi=1Xi] = E[X1 +…+ XN] = E[X1∙P[S≥1] + X2∙P[S≥2] +…] E[Σi=1Xi∙P[S≥i]] = Σi=1(P[S≥i]∙E[Xi]) = Σi=1P[S≥i]∙P[Xi] = |F|/|E| ∙ ΣPi=1[S≥i] = |F|/|E| ∙ Σi=1(P[S=i] + P[S=i+1] + …) = (|F|/|E|) ∙ Σi=1Σj=iP[S=j] = (|F|/|E|) ∙ Σi=1i∙P[S=i] = (|F|/|E|) ∙ E[S] N ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞

26. Time to get the job done |F| Lemma 2: E[(N*)2] ≤ t ∙ O(1) ∙ |E| Each faulty* step is in F, so: E[(N*)2] ≤ E[N2] ≤ E[(NF)2] Denote Χi = 1[ith step is in F]. Then: NF ≤ Σi=1Xi ∞

27. Proof |F| Linearity of expectancy Lemma 2: E[(N*)2] ≤ t ∙ O(1) ∙ |E| E[(NF)2] ≤ E[(ΣXi)2] = E[Σi=1Xi2 + Σi≠jXiXj] = Σi=1E[Xi2] + Σi≠jE[XiXj] = Σi,j=1E[XiXj] = Σi,j=1P[XiXj] = Σi,j=1P[(Xi=1) ۸ (Xj=1)] ≤ 2Σi=1Σj≥i(P[Xi=1] ∙ P[Xj=1 | Xi=1]) = 2Σi=1(P[Xi=1] ∙ Σj≥iP[Xj=1 | Xi=1]) P[Xj=1 | Xi=1] = ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ XiXj is an indicator ∞ ∞ { 1 i=j P[walk takes at least j-i more steps] X P[The (j-i)th step of a walk beginning in F, is in F] i≠j

28. Proof |F| Lemma 2: E[(N*)2] ≤ t ∙ O(1) ∙ |E| P[walk takes at least j-i more steps] = (1-1/t)j-i -Walk didn’t stop j-i times P[The (j-i)th step of a walk beginning in F, is in F] = ? Therefore, P[The (j-i)th step of a walk beginning in F, is in F] = |F|/|E| + (λ/d)j-i-1 A Blast from the Past! G=(V,E)is an (n,d,λ)-expander and FE. Then, the probability that a random walk which starts on an edge from F will pass on an edge from F on its tth step is bounded by |F|/|E| + (λ/d)t-1.

29. Rejoice!!! Yay, calculations! |F| Lemma 2: E[(N*)2] ≤ O(1) ∙ t ∙ |E| E[(N*)2] ≤ E[(NF)2] ≤ 2Σi=1(P[Xi=1] ∙ [1 + Σk=1(1-1/t)k ∙ (|F|/|E| + (λ/d)k-1)]) = 2Σi=1(P[Xi=1] ∙ [1 + Σk=1(1-1/t)k ∙ |F|/|E| + Σk=1(1-1/t)k(λ/d)k-1]) ≤ 2Σi=1(P[Xi=1] ∙ [1 + Σk=1(1-1/t)k ∙ |F|/|E| + Σk=1(λ/d)k-1]) ≤ Reminder: if |a|<1, then Σt=0at converges to 1/(1-a). 2Σi=1(P[Xi=1] ∙ [1 + (t-1) ∙ |F|/|E| + O(1)]) ≤ Remember: we bound the growth of the UNSAT by a constant 1/t. Therefore, |F|/|E| ≤ 1/t. O(1) ∙ Σi=1P[Xi=1] = O(1) ∙ E[NF] = O(1) ∙ t ∙ ∞ ∞ ∞ ∞ ∞ (1-1/t) < 1 λ<d, because G is an expander ∞ |F| ∞ |E|

30. Summary Given an (n,d,λ)-expander G, we manage to build a new graph G’’ with a considerably larger UNSAT, in polynomial time. We’ve done this by allowing unsatisfied edges to “spread the disease”, using an expander’s properties. Some of the graph properties were hurt. This will be dealt with in the next lecture.

31. The end Questions?