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Labelled Graph Automorphisms, Cone Graphs, and P-Groups

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Labelled Graph Automorphisms, Cone Graphs, and P-Groups CHAPTER HI IAN~.r~n GRAPH AUTO-MORPHISMS, CONE GRAPHS, AND I'-GROUI~ Two results mark the origin of recent group-theoretic approaches to graph iso- morphism: The labelled graph automorphism problem, and the isomorphism problem for cone graphs. This is not to say that here group-theoretic techniques are applied to graph isomorphism for the first time. Rather, it appears that the impact of these results has been to convince researchers £hat a group-theoretic approach to graph isomorphism is a reasonable and practical line of attack. It is also true that cone graphs possess a topological structure which permits visualizing abstract properties of the automorphism group of graphs of bounded valence, and especially of trivalent graphs. In this chapter, we will develop both the polynomial time solution of the labelled graph automorphism problem and results concerning the complexity of an isomor- phism test for regular cone graphs. We will also discuss the relationship between cone graphs and Sylow p-subgroups of permutation groups, which seems to have trig- gered the isomorphism tests for graphs of fixed valence (Chapters IV and V). Finally, we will develop a number of basic algorithms for the class of p-groups. In Chapter IV we will develop further computational techniques for p-groups. L The Labelled Graph Automorphism Problem When designing an isomorphism test for graphs, it is a natural idea to attempt a vertex classification with the aim of reducing, to a manageable magnitude, the number of apr~o~ possible isomorphisms. That is, if X and X' are graphs to be tested for isomorphism, we wish to partition the vertices of X and of X' into classes such that an isomorphism can only map ~ vertex v of X into a vertex w of X' if v and w are in the same class, As a simple example illustrating this idea, consider classifying vertices by v~le~ce, i.e. by the number of edges incident to the vertex classified. Obviously, no isomorphism can map a vertex of valence k into a vertex of valence j~k, thus we have 61 here a sound classification criterion. Of course, if X and X' are regular graphs, i.e., if X and X' are graphs in which every vertex has the same valence, then this vertex classification yields no information. Over the two or so decades during which vertex classification has been the dom- inant style of approaching graph isomorphism, many elaborate criteria for classifying vertices have been proposed. Unfortunately, none of the proposed criteria has so far succeeded in solving the general problem. Therefore it is an interesting question to study how "good" a vertex classification scheme has to be in order to serve as basis for a polynomial time isomorphism test. This motivates Pl~B,.mm 1 (Labelled Graph Automorphism) Let X = (V,E) be a graph with n vertices, and assume that V has been partitioned into the classes C1 ..... Cs, forming the partition C, such that I Cit <- k, where k is a con- stant independent of n. Find all automorphisms of X which setwise stabilize the classes Ci, 1 -< i -< s. That is, find Auto(X) = I a e Aut(X) t (Vi-<s)(VxeCi)(xa c Ci) t, the subgroup of those automorphisms of X which respect the partition 6". In the following, we will view the partition C as the result of a vertem lab~Uing of the graph X with s distinct labels, The class Ci consists of the vertices in X which carry the i TM label. In order to demonstra~ the relationship of Problem 1 to vertex classification, we let X be the disjoint union of two connected graphs which are to be tested for isomor- phism, and we further assume that the partition C is the result of a correct vertex classification procedure. Then the two graphs are isomorphic iff every generating set for Autc(X) contains at least one permutation which exchanges the connected com- ponents of X. I. 1. A Deterministic Algorithm for Problem 1 We will show that Problem I has a deterministic polynomial time solution. Specifically, we will apply the techniques of Chapter If, Section 4, and demonstrate that we can make Autc(X ) (k,c)-accessible. 62 Recall the definition of (k,c)-accessibility (Chapter II, Definition 22). We will show that hutc(X) is (2,c)-accessible (for some constant c) by trapping it as the subgroup G (r) in a subgroup tower I = G (m) <: • " " < G (r) =Autc(X ) <: • -" <G (1) = G of a known group G. Intuitively, we let the groups G (ra), G(m-l) ..... G (r+i) be the pointwise stabilizers, in G (r), of the vertices in the classes Ci of X, stabilizing every vertex in an individual class at each step. Clearly we have simple membership tests for these groups. The groups G (r-l) ..... G(1) are obtained as the automorphism groups of certain labelled graphs Xj derived from X. Here, the trick is to define the graphs Xj such that Autc(Xj+l) is a subgroup of Autc(Xj). In particular, we define X1 to be the graph X stripped of all edges. Note that G(1) = Auto(X1) is the direct product of symmetric groups acting on the individual ver- tex classes of X, i.e., Auto(X1) = G (1) = rlSym(Ci) i=l Furthermore, since Aute(X1) respects the label classes, Auto(X) is a subgroup of Auto(X1). Note that we have generators for G O), and that we can easily test member- ship in this group. The graphs X2 ..... Xr are obtained by gradually adding back into X 1 the edges of X. Here it is crucial to add edges in batches, where each batch consists of all edges con- necting the vertices m two classes, Ch and Ci, This ensures that Autc(Xj+1) is a sub- group of Autc(Xj). We now formally specify the construction just outlined. Let X = (V,E) be the graph under consideration IVI = n, and let the partition of V be C = fC 1..... Csl, where I Cil -- k, I --- i-< s. Define El,i= t(v,w) EE vE Ci, wECj/, and letXl = (V,F1),where F1 =¢. Define the graphs Xj = (V,Fj), 1 < j ~ r, r = (~)+1, by Xu = (V,F~) = (V,F1uEI,1) X3 = (V,F~) = (V,F~)EI,~) Xs+1 = (V,Fs+I) = (V, FsUEI,s) 63 Xs+~ = (V,Fs+~) = (Y,Fs+z uE~,~) Xs+S = (V,Fs+3) = (V, Fs+euE2,3) Xr = (V,Fr) = (V,Fr-IUEs,,) = (¥,E) = X Furthermore, we define G(J) = Autc(Xi). ~I~ 1 Let X = (tl ..... 81, E) be a graph with E = I(1,2), (1,4), (2,3), (~,6), (2,7), (~,8), (3,5), (3,8), (4,5), (5,6), (6,7), (6,8)], and with the label classes C 1 = fI,2,8{, C2 = ~4,5,8{, and Ca = t7,81. X is shown in Figure I below. Then the graph sequence X1 ..... ?(7 is defined by the edge sets F1=¢ F~ = I(1,~), (~,a)~ Fa = Fe U t(1,4), (2,6), (3,5)t F4 = Fa u t(2,7), (~,8), (a,a)t F5 = F4 u 1(4,5), (5,6)1 F8 = F5 U 1(6,7), (6,a)l FT=Feu¢=E t f', I I '41 I I i i I I ! Cl 2* 151 C~ I ! ca / The graph X Figure i F6 = F7 since there is no edge (7,8) in E. Note that X7 = (V, FT) is the graph X. As illus- tration, the graph )(3 = (V,F3) is shown in Figure 2 below. Observe that the only edges 64 1 '- T(4 ]i C1 03 I I I'! ; I 3 ,-r"~ ~--t6, ;7 I i C3 18 The graph ]{3 Figure 2 present in X3 are those edges (v,w) of E such that v C C1 and w E C1LjCa. [] Having made these definitions, we need to verify that the resulting subgroup tower makes Autc(X) accessible, so that we can apply Theorem 12 of Chapter II, and determine Autc(X) in polynomial time. L~m~r~A 1 G (1) is the direct product of the symmetric groups Sym(Ci) acting on the individual s label classes, i.e., G (1) = ~-[Sym(Ci). i=1 Proof X1 has no edges, and G (0 = Autc(Xl), by definition. - L~XA 2 For 1 ~ j < r, G (j÷l) is a subgroup of G(J) of index at most (k!) 2. Proof Let Xj+ 1 = (V,FjL)Eha). Since Fj contains no edge connecting vertices in Ch with vertices in Ci, any automorphism of the labelled graph Xj+I is also an automor- phism of the labelled graph Xj, and so G (j+O < G (j). Furthermore, if ~, ~ E GO) such that ~r3~-I stabilizes the vertices of ChuC i pointwise, then ~ and ~ must be in the same right coset of G 0+l) in G 0). Since there are at most (k!) ~ distinct ways of permuting the vertices in Ch and in Ci, the index (G(J):GU+1)) cannot exceed this bound. " We now turn to the lower part of the subgroup tower. For 1 -- j -- s, we define c (r+j) = { ~ ~ G (r+j-~) i (Vxeq)(x" = x) 65 i That is, the group G (r+9 is the pointwise stabilizer in Auto(X) of the set uCi. ~ffil Recall that r = (~)+1, i.e., r is O(n~). Therefore, we have just defined a subgroup tower of height m = and so m is i.e.
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