+ ɛ Hardness for Max-E3Lin

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1 Advanced Approximation Algorithms (CMU 5-854B, Spring 008) Lecture : δ vs. + ɛ Hardness for Max-E3Lin April 8, 008 Lecturer: Ryan O Donnell Scribe: Jonah Sherman Recap In this lecture, we finish the hardness reduction for Max-E3Lin]. Recall we began with a Label- Cover(K,L) instance, with left vertices U and right vertices V. We associate K variables u,..., u K with each u U and L variables v,..., v L for each v V. Furthermore, we identify these variables with functions f u : {, } K {, } and g v : {, } L {, } respectively. U V u K L v K] L] Also recall that for the completeness direction of our proof, we will encode the assignment that u U gets key a K] by taking f u to be the a th dictator function, f(x) = x a (and similarly for V, L). Thus, for soundness, our test should also have the property that if f u passes with sufficiently high probability, we can decode f u into a small set of suggested keys. Last time, we saw how to come up with such a test. However, it s not enough to merely test that the functions suggest a labelling we need them to suggest a good labelling. That is, we could have all of the f u and g v be perfectly decodable dictator functions, and they would pass all the tests proposed last time with high probability, even if that assignment doesn t actually satisfy any of the constraints. Thus, our test needs to somehow also simultaneously check that the suggested key/label pairs satsify the edge constraints.

2 The New Test We alter the test from last time to take the constraints into account. Choose b {, }, x {, } K, y {, } L, independently and uniformly at random. Choose λ {, } L from the δ-biased distribution (i.e., each λ i is independently w.p. δ). Define x π {, } L by (x π ) i := x π(i). Set z := x π y λ (b,..., b). Check f(x)g(y)g(z) = b. x K π x π L Next we analyze the probability that a given pair of functions f, g pass. Prf, g pass] = E + bf(x)g(y)g(z)] b,x,y,λ = + ˆf(S)ĝ(T )ĝ(u)e bχ S (x)χ T (y)χ U (z)] = + = + = + S K] T,U L] S,T,U S,T,U S K] ˆf(S)ĝ(T )ĝ(u)e bχ S (x)χ T (y)χ U (x π )χ U (y)χ U (λ)χ U (b,..., b)] ˆf(S)ĝ(T )ĝ(u) E ] b U + E χ S (x)χ U (x π )] E χ T U (y)] E χ U (λ)] } b {{} x y λ }{{}}{{} U odd] T =U] ( δ) U ( δ) T ĝ(t ) ) ˆf(S) E x χ S (x)χ T (x π )] }{{} ( )

3 Now, note that, ] ( ) = E x i x π(j) x i S j T = E x i i i S i K] = E i K] = i K] x π (i) T x π (i) T +i S] i i S π (i) T odd ] = S = π odd (T )] ] where, for each T L], we define, π(t ) := {a K] : π (a) T } π odd (T ) := {a K] : π (a) T is odd} Thus, we have, Prf, g pass] = + ( δ) T ĝ(t ) ˆf(πodd (T )) The following two facts about π(t ) and π odd (T ) will be useful. Fact.. π odd (T ) π(t ). Fact.. If T is odd, then π odd (T ). We re now ready to prove the completeness and soundness of the test.. Completeness Suppose f : {, } K {, }, g : {, } L {, } are matching dictators (i.e., f(x) = x a and g(x) = x α where π(α) = a). Then, g has Fourier support {{α}}, f has Fourier support {{a}}, and π odd ({α}) = {a}, so, Prf, g pass] = + ( δ)ĝ({α}) ˆf(πodd ({α})) = δ 3

4 . Soundness As usual, we ll prove soundness using the probabalistic method. Specifically, we want to come up with random decoding functions Dec f : f u a K] and Dec g : g v α L], such that if f u, g u pass with sufficiently high probability, then π vu (Dec g (g v )) = Dec f (f u ) holds with sufficiently high probability. As before, we ll decode f u, g v in two steps, by first decoding to some suggestion sets Sugg(f u ) K], Sugg(g v ) L], and then choosing the actual key/label uniformly at random from those suggestions. To do this, we want Sugg(f u ), Sugg(g v ) to be small (independent of K, L) and to have the property that if f u, g v pass with probability at least + ɛ, then there exist a Sugg(f u), α Sugg(f v ) such that π vu (α) = a. Suppose f, g pass with probability at least + ɛ. Then, ɛ ( δ) T ĝ(t ) ˆf(πodd (T )) ( δ) T ĝ(t ) ˆf(πodd (T )) Recall from Parseval s theorem that T L] ĝ(t ) =. Thus, we can think of ĝ as a probability distribution on sets T L]. Then, the last line is equivalent to, ] ɛ E ] ( δ) T ˆf(πodd (T )) T ĝ Note the expression inside the expectation is just a RV on T taking values in 0, ], so by a simple averaging argument, ] ɛ Pr ] ( δ) T ˆf(πodd (T )) ɛ T ĝ }{{} GOOD T Now, suppose some set T L] is good (i.e., GOOD T happen. Specifically, occurs). Then, many good things T is odd. T ln(/ɛ) δ =: B. π odd (T ) 0, by Fact.. π odd (T ) B, since π odd (T ) π(t ) and π(t ) T B. ˆf(π odd ) ɛ The last point suggests a good way to decode f. Define, Sugg(f) := {S K] : S B, ˆf(S) ɛ } Note that since each ˆf(S) ɛ, there can be at most /ɛ such S, and each S has size at most B. Thus, Sugg(f) B/ɛ. 4

5 The only remaining question is how to decode g. One immediate but naive idea would be to take the union of all good T. However, we don t have any kind of bound on the number of such good T. That is, it could be the case that there are many small good sets T, in which case the union could be very large. So, we re going to need to decode g in a different manner. Now, in the past, our random decoding has essentially consisted of deterministically decoding g to some small suggestion set, and then choosing the actual label uniformly at random from that. Of course, there s no need for the first step to be deterministic; we re free to have Sugg(g) be a random function too. So, we can simply let Sugg(g) = T, where T is drawn from the distribution given by ĝ. We ve shown how to decode f and g to suggestion sets of small size. Now, suppose T is good. Then, the union in the definition of Sugg(f) includes π odd (T ), so π odd (T ) Sugg(f). Since π odd (T ) is non-empty, there exists some a π odd (T ). But then a π(t ), so there exists some α T with π(α) = a. That is, for any good T, there exists a Sugg(f) and α T = Sugg(g) such that π(α) = a. Now, putting it all together, suppose f, g pass with probability + ɛ. Then, we know Pr T ĝ GOOD T ] ɛ. Thus, Pr T ĝ T B, a Sugg(f), α T : π(α) = a] ɛ Since Dec(f) is chosen uniformly from Sugg(f) and Dec(g) is chosen uniformly from T, we have, References Prπ(Dec(g)) = Dec(f)] Pr T ĝ GOOD T ] Sugg(g) Sugg(f) ɛ B ɛ B = ɛ3 B = ɛ 3 4δ ln (/ɛ) ] J. Håstad. Some optimal inapproximability results. J. ACM, 48(4): , 00. 5