Class work 7 - The powering step
December 25, 2013
In this class work we will see a variant of the powering step. Recall that the powering step
is a reduction with the following properties.
The powering step: The input is a constraint graph G = (V, EG ) with alphabet Σ0 , and a
parameter t ∈ N. We assume that G is an (n, d, λ) expander for some λ bounded away from 1.
We construct a constraint graph H = (V, EH ) on the same vertex set as follows.
1.5t )
1. The alphabet of H is ΣH = (Σ0 )O(d
.
2. If unsat(G) = 0, then unsat(H) = 0.
3. If unsat(G) = ε for ε < 1/t, then unsat(H) > Ω(tε).
The construction of H: The graph H will be a weighted graph on the same set of vertices
V of G. We define the weight of the edge (u, v) according to the following random process:
Random process for choosing the edges from EH :
1. Choose j1 , j2 ∈ [−t/4, t/4] independently uniformly at random. Let ` = j1 + t + j2 .
2. Pick a random path v0 → v1 → · · · → v` of length ` in G.
3. Output (v0 , v` ).
Question 0. Let j1 , j2 ∈ [−t/4, t/4] be chosen independently uniformly at random. What is the
distribution of j1 + j2 ?
The assignment to the vertices of H is the opinion of each vertex about all vertices at
2
1.5t
distance at most 1.5t from it. Therefore, the alphabet is (Σ0 )1+d+d +···+d . The weight of an
edge (v0 , v` ) is defined according to the distribution above, and the constraint associated with
this edge checks that the opinions of v0 and of v` satisfy the constraints of G, and agree on the
common vertices.
The main part of the exercise is the following lemma.
Lemma 1 (Main lemma). If unsat(G) = ε for ε < 1/t, then unsat(H) > Ω(tε).
The proof outline is similar to the proof we saw in the reading that is defined with a random
stopping rule, but there are some differences to be worked out.
Given an assignment A : V → ΣH to H define an assignment M AJ : V → Σ0 for G as follows.
For u ∈ V define M AJ(u) as follows.
1. Pick a length s ∈ [t/2 − t/8, t/2 + t/8] uniformly at random.
2. Choosing a path of length s steps starting from u and terminating at some w.
3. Let M AJ(u) = arg maxσ∈Σ0 Pr[A(w)u = σ].
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Denote by F the edges edges of G that are not satisfied by M AJ. By the assumption that
unsat(G) = ε, it follows that |F | ≥ ε|E|.
Next we define a random variable N that counts the number of “faulty edges” in a random
path chosen from EH , and prove that unsat(H) ≥ Pr[N > 0] ≥ Ω(εt).
Defining N : Let j1 , j2 ∈ [−t/4, t/4], let i ∈ [−t/8, t/8], and let P = v0 → v1 → · · · → v` be a
path of length ` = j1 + t + j2 . We say that an edge (u, v) ∈ EG is i’th faulty for the path P if
1. vj1 +t/2+i = u and vj1 +t/2+i+1 = v.
2. (u, v) ∈ F .
3. A(v0 )u = M AJ(u) and A(v` )v = M AJ(v).
Clearly if an edge (u, v) is i’th faulty for a path P , then the corresponding edge in H is not
satisfied.
For all i ∈ [−t/8, t/8] define a random variable Xi ∈ {0, 1} that is equal to 1 if the random
Pt/8
path chosen by the verifier contains i’th faulty edge, and let N = i=−t/8 Xi . We clearly have
unsat(H) ≥ Pr[N > 0], and so it is enough to show that Pr[N > 0] ≥ Ω(εt).
Claim 2. The following holds.
1. E[N ] ≥ Ω(εt).
2. E[N 2 ] ≤ O(εt).
3. Conclude that Pr[N > 0] ≥ Ω(εt).
Items 2 and 3 are very similar to those appeared in the reading. Below we prove Item 1.
Proof of Item 1 The key step in the proof is the following. For each i ∈ [−t/8, t/8] and for
(u,v)
each (u, v) ∈ F define the distribution Bi
.
1. Choose j1 , j2 ∈ [−t/4, t/4] independently uniformly at random. Let ` = j1 + t + j2 .
2. Pick a random path v0 → · · · → vj1 +t/2+i = u of length j1 + t/2 + i ending at u.
3. Pick a random path v = vj1 +t/2+i+1 → · · · → v` of length t/2 − i + j2 − 1 starting at v.
4. Output (v0 , v` ).
Note that this is equivalent to picking a random path v0 → v1 → · · · → v` according to the
distribution of EH conditioned on the event that vj1 +t/2+i = u and vj1 +t/2+i+1 = v.
Question 1. For every edge (u, v) ∈ EG , and for every i ∈ [−t/8, t/8] it holds that
Pr
(u,v)
(v0 ,v` )∼Bi
[A(v0 )u = M AJ(u) and A(v` )v = M AJ(v)] ≥
1
.
4|Σ0 |2
(u,v)
(Hint: use independence of v0 and v` when given (u, v) and i, and relate Bi
we used to define M AJ.)
to the distribution
Question 2. Prove that for all i ∈ [−t/8, t/8] it holds that Pr[Xi = 1] ≥ Ω(ε). Conclude Item 1.
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