## Pairwise independence and derandomization by Michael Luby, Avi Wigderson

By Michael Luby, Avi Wigderson

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M do: Compute Tj (v) = v j. For all i = 1, . . , n do: Compute Yi (β, v) = 1/m · m j=1 B(ei ⊕ Tj (v)) · β L ← L ∪ { bit(Y1 (β, v)), . . , bit(Yn (β, v)) }. j. From the above analysis, it follows that x ∈ L with probability at least 1/2, where this probability is over the random choice of v. As long as the running time T for computing B is large compared to n (which it is in our use of the Hidden Bit Technical Theorem to prove the Hidden Bit Theorem), the running time of S B is O(n3 T /δ 4 ).

4. , the graphs are not isomorphic, then P can tell which graph is sent by V and thus can compute c correctly, so that Pr[V accepts] = 1 = cyes . If the two graphs are isomorphic then, since the verifier sends a random permutation of the graph, the distribution on graphs received by the prover is the same whether b = 0 or b = 1, and since b is chosen randomly the prover can answer correctly with probability 1/2, and thus Pr[V accepts] = 1/2 = cno . Clearly, this protocol does not work if the coins used by the verifier to choose b and σ are public.

Proof. For the proof, we find it convenient to consider bits as being {1,−1}-valued instead of {0,1}-valued. For b ∈ {0, 1}, we let ¯b = (−1)b ∈ {1, −1}. Fix x ∈ {0, 1}n such that δxB ≥ δ. We can write δxB = Ez∈R {0,1}n B(z) · x z . For all i = 1, . . e. 0i−1 , 1, 0n−i , and let µi = δxB · xi . It follows that Ez∈R {0,1}n B(z) · x (ei ⊕ z) = µi . 2. Hidden Bit Theorem distributive over ⊕ implies that This is because (ei ⊕ z) = x x and because x 41 ei · x z ei = xi , and thus B(z) · x (ei ⊕ z) = B(z) · x z · xi .