Lecture 18: Graph Isomorphisms

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1 INFR11102: Computational Complexity 22/11/2018 Lectue: Heng Guo Lectue 18: Gaph Isomophisms 1 An Athu-Melin potocol fo GNI Last time we gave a simple inteactive potocol fo GNI with pivate coins. We will show that it can also be achieved using only public coins. Theoem 1. GNI AM. We will take a moe quantitative appoach. Fo any gaph G with n vetices, let aut(g) = {π π(g) = G} be its automophism goup. Let iso(g) = {π(g) π S n } be the set of gaphs isomophic to G. Conside the set {(G, π) π S n }. Clealy π(g) iso(g), and each one appeas exactly aut(g) times. Namely, Now, fo G 1 and G 2, define n! = {(G, π) π S n } = aut(g) iso(g). S := {(G, σ) σ aut(g ), G = G1 o G = G2 }. Thus, if G 1 = G2, then S = n!, and othewise S = 2n!. To distinguish these two cases, once again we will use paiwise independent hash family. Let H be such a family fom S to T whee T is some abitay set of size 4n!. Fix a paticula element α T. Ou potocol is the following: 1. Athu picks a andom function H H and pesent it to Melin; 2. Melin etuns an element (G, σ) S and a pemutation τ; 3. Athu accepts if (1) τ(g ) = G 1 o G 2 ; (2) σ(g ) = G ; and (3) H(G, σ) = α. Note that the last veification step can be done easily in deteministic polynomial time. Conditions (1) and (2) veify that (G, σ) is indeed a element of S, and (3) assets a fact whose pobability to happen distinguishes the two scenaios of S. If S = n!, then by the definition of paiwise independent hash function, S P [ s S, H(s) = α] H H T =

2 Othewise S = 2n!, then by inclusion-exclusion, P H H [ s S, H(s) = α] s S = S T P [H(s) = α] H H s,s S ) ( S 2 1 T S 2 2 T 2 = = 3 8. P [H(s) = H H H(s ) = α] Thus, we have ceated a constant gap in the accepting pobability between the two cases. We can employ the standad epeat and vote tick to amplify such a gap. Details of the amplification ae omitted. 2 Evidence against NP-completeness of gaph isomophisms Recall the gaph isomophism (GI) poblem. Since thee was no efficient algoithm, it is natual to wonde whethe the poblem is NP-complete. Howeve, this is also unlikely to be the case, unless the polynomial hieachy collapses. Theoem 2. If GI is NP-complete, then Σ p 2 = Π p 2. Poof. It is sufficient to show that if GI is NP-complete, then Σ p 2 Π p 2. Conside the QBF 2 poblem, which is complete fo Σ p 2 and whose input ae fomulas of the following fom: ψ = x y φ(x, y). Since by assumption, GI is NP-complete, GNI is conp-complete. Thus, thee is a eduction R( ) such that fixing x, ψ (x) := y φ(x, y) is valid if and only if R(ψ (x)) GNI. Last time, we showed that GNI AM and AM = AM 1 whee AM 1 is the one-sided eo vesion. Let P (x) := R(ψ (x)). Thus, by appopiate amplification, thee is a poly-time TM M such that P (x) GNI P [ z M(P (x),, z) = 1] = 1; P (x) GNI P [ z M(P (x),, z) = 1] 2 n 1, whee n = x and both and z all have length bounded by a polynomial in n. We claim that ψ is valid x z M(P (x),, z) = 1. (1) This implies the theoem. To veify (1), we have two cases: 2

3 1. If ψ is valid, then x such that P (x) GNI which implies that x z, M(P (x),, z) = 1. This implies that x z, M(R(ψ (x)),, z) = If ψ is not valid, then x, P (x) GNI. Thus, which implies, via the union bound, x P [ z M(P (x),, z) = 1] 2 n 1, P [ x z M(P (x),, z) = 1] x {0,1} n P [ z M(P (x),, z) = 1] 2 n 2 n 1 = 1/2 < 1. In othe wods, P [ x z M(P (x),, z) = 0] > 0. The pobabilistic method implies that x z M(P (x),, z) = 0 ( x z M(P (x),, z) = 1). If we look moe caefully at the poof of Theoem 2, the only cucial popety of GNI we used is that GNI AM. Coollay 3. If conp AM, then Σ p 2 = Π p 2. Recall that NP AM Π p 2. Coollay 3 implies that AM sits in an inteesting position at the complexity landscape. 3 Counting gaph isomophisms We have seen that some decision poblems in P have #P-complete counting countepats. One natual question is that whethe the counting vesion of GI is easy o had. Name: #GI Input: Two gaphs G 1 and G 2. Output: How many pemutations ae thee to make G 1 identical to G 2? Clealy #GI is no easie than GI. complexity. Next we show that they actually have the same Theoem 4. #GI t GI. To show Theoem 4, we need an intemediate poblem. Recall that aut(g) is the automophism goup of a gaph G. 3

4 Name: #Aut Input: A gaph G. Output: aut(g) Lemma 5. #Aut t GI. Poof. Let G = (V, E) be a gaph with V = n. Conside a paticula vetex v V. Let C v (G) := {π(v) π aut(g)} be the set of vetices that v can map to via an automophism, and let S v (G) := {π π aut(g) and π(v) = v} be the set of automophisms fixing v. Basic goup theoy implies that aut(g) = C v (G) S v (G). One way to undestand this fact is by choosing a π u fo each u C v (G) such that π u aut(g) and π u (v) = u. Evey π aut(g) can be uniquely decomposed into π u σ whee σ S v. The claim follows. Next we will compute C v (G) and S v (G) sepaately. We go though evey vetex u V using the GI oacle to detemine whethe an automophism exists mapping v to u. To do so, let H be a igid gaph with n + 1 vetices such that aut(h) contains only the identity. 1 Constuct G v by taking a copy of G and a copy of H, and then gluing v G to an abitay vetex w H. Similaly, constuct G u by gluing u to w. We ask the GI oacle whethe G v = Gu. Since vetices in H must map to vetices in H (H has one moe vetex than G), such an isomophism exists if and only if v is mapped to u. Namely G v = Gu if and only if u C v (G). We still need to count S v (G). The idea is to use self-educibility. Namely we want to tansfom it into a smalle instance of #Aut itself. In fact, we claim that S v (G) = aut(g v ). The eason is the same as above, namely that all vetices in the copy of H can only map to vetices in H, and H has only one automophism. Although G v contains 2n vetices, n + 1 of them can only map to themselves. Hence, the numbe of fee vetices in G v is n 1. Let v 1 := v, and to continue, we pick an abitay fee vetex. Call it v 2, and we poceed to compute C v2 (G v1 ). Namely, we attach a igid gaph H of size 2n + 1 to v 2 to get G v1,v 2 and go though all vetices in V \ {v 1, v 2 } to detemine thei membeship in C v2 (G v1 ) using the GI oacle. Then we ecusively compute S v2 (G v1 ). This ecusion can only go down n steps. In fact, we constuct a sequence of gaphs G v1, G v1,v 2,, G v1,...,v n 1, each one fixing one moe vetex and having polynomial size. It can be veified that This finishes the poof. aut(g) = C v1 (G) C v2 (G v1 ) Cvn (G v1,...,v n 1 ). Now we ae eady to pove Theoem 4. Poof of Theoem 4. We fist use the GI oacle to test whethe G 1 = G2. If not, then we etun 0. Othewise, we compute the numbe of automophisms of G 1 using Lemma 5. We claim that this is also the numbe of isomophisms fom G 1 to G 2. 1 Such gaphs do exist! 4

5 To be moe specific, let iso(g 1, G 2 ) := {π π(g 1 ) = G 2 }. Ou claim is that iso(g 1, G 2 ) = aut(g 1 ) if G 1 = G2. Fix an abitay pemutation π 0 iso(g 1, G 2 ). Fo any σ iso(g 1 ), it is easy to see that π 0 σ(g 1 ) = π(g 1 ) = G 2. Thus, π 0 σ iso(g 1, G 2 ). It implies that π 0 aut(g 1 ) iso(g 1, G 2 ). On the othe hand, fo each π iso(g 1, G 2 ), we have that π0 1 π (G 1 ) = π0 1 (G 2 ) = G 1. Thus, π0 1 π aut(g 1 ), namely π = π 0 σ fo some σ aut(g 1 ). It implies that iso(g 1, G 2 ) π 0 aut(g 1 ). To summaize, we have that iso(g 1, G 2 ) = π 0 aut(g 1 ). Taking the cadinality on the both sides yields the claim. Remak (Bibliogaphic). Theoem 2 was fist shown by Boppana, Håstad, and Zachos [BHZ87]. Relevant chaptes ae [AB09, Chapte 8.2]. Refeences [AB09] Sanjeev Aoa and Boaz Baak. Computational Complexity - A Moden Appoach. Cambidge Univesity Pess, [BHZ87] Ravi B. Boppana, Johan Håstad, and Stathis Zachos. Does co-np have shot inteactive poofs? Inf. Pocess. Lett., 25(2): ,