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Counting Inequivalent Colorings Concepts: Graph Automorphism

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Counting Inequivalent Colorings Concepts: Graph Automorphism Chapter 18: Counting Inequivalent Colorings Concepts: Graph automorphism, vertex and edge coloring, equivalent color- ings, equivalence class of colorings, number of inequivalent colorings, color- fixing automorphisms, fixed points of an automorphism, formula for number of inequivalent colorings, formula for number of restricted colorings. 1. Automorphisms of a graph: Permutations of the vertices that don't change the edge set. Example 1.1: There are 8 automorphisms of C4: 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 ; ; ; ; 1 2 3 4 1 4 3 2 2 1 4 3 2 3 4 1 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 ; ; ; : 3 2 1 4 3 4 1 2 4 1 2 3 4 3 2 1 To find these, we can either work out all 24 permutations of C4 and see which ones fix C4, or we can construct them as follows: any permutation must produce the cycle 1-2-3-4-1. There are four permutations which produce this cycle in the clockwise orientation and there are four permutations which produce this cycle in the counter-clockwise direction. Illustration 1.2: Permutations applied to C4. 2. C-coloring of a graph: A function c : V ! C to a finite set of colors C. The color of the vertex v is c(v). The number of C-colorings of an n-vertex graph is njCj. 2 Example 2.1: There are 4 = 16 fR; Bg-colorings of C4. Illustration 2.2: fR,Bg colorings of C4 3. Action of a graph automorphism on a coloring: Given a graph automorphism f of a graph G, and a coloring c of the vertices of G, permute the vertices of G using f, leaving the colors in place, obtaining a coloring 1 2 ··· n 1 2 ··· n f ∗c of G. If c = and f = c(1) c(2) ··· c(n) f(1) f(2) ··· f(n) then f(1) f(2) ··· f(n) f ∗ c = : c(1) c(2) ··· c(n) 1 1 2 3 4 1 2 3 4 Example 3.1: Let c = and f = . Then BRRR 2 1 4 3 2 1 4 3 1 2 3 4 f ∗ c = = : BRRR RBRR 4. Equivalence classes of colorings: We say that two colorings c1 and c2 are equivalent if there is a graph automorphism f which transforms c1 into c2: f ∗ c1 = c2. If f ∗ c1 6= c2 for any graph automorphism f, then we say 0 0 that c1 and c2 are inequivalent. Given a coloring c, we set [c] = fc : c ∼ c g. The distinct sets [c] form a partition of the set of all colorings. Terminology in textbook: [c] is the orbit of c under the action of automor- phisms and is denoted by cG. See page 457, Definition 18.15. Example 4.1: Example 3.1 implies 1 2 3 4 1 2 3 4 ∼ : BRRR RBRR In fact, all colorings with 3 Rs and 1 B are equivalent, and we have 1 2 3 4 = BRRR 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 ; ; ; : BRRR RBRR RRBR RRRB 1 2 3 4 1 2 3 4 Example 4.2: Let c = and c = . There 1 BBRR 2 BRBR is no automorphism that sends the subset f1; 2g to the subset f1; 3g, there- fore f ∗ c1 6= c2 for any f. Hence c1 and c2 are inequivalent colorings. Geometrically, every time we rotate or flip c1 we obtain a coloring in which the two blue vertices are joined by an edge. Since c2 doesn't have two blue vertices joined by an edge, we never never obtain c2 by acting on c1 by an automorphism. Illustration 4.3: Equivalence classes of fR,Bg-colorings of C4. 5. Partitioning automorphisms: Let c be a coloring. Then for each automorphism f, f ∗ c is a coloring equivalent to c. If [c] = fc1; c2; : : : ; ckg 2 where c1 = c, then we can partition the elements of the automorphism group via G = G1 [ G2 [···[ Gk; where Gi = ff 2 G : f ∗ c1 = cig: 1 2 3 4 f(1) f(2) f(3) f(4) Example 5.1: Let c = . Then f∗c = . BRRR BRRR If we denote by ci the coloring in which the vertex i is colored blue and the other vertices are colored red, then we have f ∗ c1 = cf(1) for all f. Therefore Gi = ff 2 G : f(1) = ig and G = G1 [ G2 [ G3 [ G4: Looking at G in Part 1 above we see that jG1j = jG2j = jG3j = jG4j = 2. 1 2 3 4 f(1) f(2) f(3) f(4) Example 5.2: Let c = . Then f∗c = . BRBR BRBR If we denote by cfi;jg the coloring in which vertices i and j are colored blue and the other two vertices are colored red, then we have f ∗ c = cff(1);f(3)g for all f. Therefore Gfi;jg = ff 2 G : ff(1); f(3)g = fi; jgg and G = Gf1;3g [ Gf2;4g: Looking at G in Part 1 above we see that jGf1;3gj = jGf2;4gj = 4. We will see in Theorem 8 below that whenever we partition the automor- phism this way, each set in the partition contains the same number of auto- morphisms. 6. Color-fixing automorphisms: Let c be a coloring and let f be an automorphism. We say that f fixes c if f ∗ c = c. See the illustrations for examples. The set of all color-fixing automorphisms of c is G(c) = ff 2 G : f ∗ c = cg: Terminology in textbook: G(c) is the stabilizer of c, denoted Gc. See page 456, Definition 18.11. 3 7. Fixed-points of an automorphism: Let c be a coloring and let f be a automorphism. We say that c is a fixed-point of f is f ∗ c = c. See the illustrations for examples. The set of all fixed-points of an automorphism f is C(f) = fc : f ∗ c = cg: Terminology in textbook: C(f) is the set of fixed points of the automorphism f, denoted by Ff . See page 458, Definition 18.19. Theorem 8: Let c be a coloring, let [c] = fc1; : : : ; ckg, and let Gi = ff 2 G : f ∗ c1 = cig: Then jGij = jGjj for all i and j. Proof: It suffices to prove that jGij ≤ jGjj for any i and j. Choose a fixed automorphism gij that satisfies gij ∗ ci = cj. We will use this to construct a mapping from Gi to Gj. Given f 2 Gi we have c1 7! ci using f, therefore c1 7! ci 7! cj using gij ◦ f, therefore gij ◦ fi 2 Gj. The mapping f 7! gij ◦ f 0 −1 −1 0 0 is injective: gij ◦ f = gij ◦ f =) gij ◦ gij ◦ f = gij ◦ gij ◦ f =) f = f . Since the mapping is injective, jGij ≤ jGjj. Theorem 9: For each coloring c, j[c]j · jG(c)j = jGj. This is equivalent to Lemma 18.16, p. 457. Proof: Write [c] = fc1; : : : ; ckg where c1 = c. Then by Theorem 8, jGj = jG1j + ··· + jGkj = k · jG1j = j[c]j · jG(c)j: Theorem 10: Assume there are m different equivalence classes of colorings. Then X jG(c)j = mjGj: c Proof: By Theorem 2 we have X X jGj jG(c)j = : j[c]j c c 4 However, given [c] = fc1; : : : ; ckg, we have X jGj X jGj jGj = = k = jGj; j[ci]j j[c]j j[c]j ci2[c] ci2[c] so given that there are m different equivalent equivalence classes of colorings, the entire sum is jGj + jGj + ::: jGj = mjGj. Theorem 11: Assume there are m different equivalence classes of colorings. Then 1 X m = jC(f)j: jGj f This is equivalent to Theorem 18.20, page 458. Proof: Using Theorem 10, we have X X X X X X mjGj = jG(c)j = 1 = 1 = jC(f)j: c c f f c f f∗c=c f∗c=c Theorem 12: Let f be an automorphism with k cycles. Then jC(f)j = jCjk. 1 2 3 4 Example 12.1: Let f = be an automorphism of C . We 2 1 4 3 4 count the number of colorings fR; Bg-colorings c that f fixes using a counting 1 2 3 4 argument. Let c = be a coloring fixed by f. Then c(1) c(2) c(3) c(4) c = f ∗ c implies 1 2 3 4 2 1 4 3 = : c(1) c(2) c(3) c(4) c(1) c(2) c(3) c(4) It is required that c(1) = c(2) and c(3) = c(4). There are 2 ways to choose the common color of c(1) and c(2) and 2 ways to choose the common color of c(3) and c(4), for a total of 22 colorings fixed by f. Proof of Theorem 12: A coloring c fixed by f must satisfy c(i) = c(f(i)) for each i. This implies c(1) = c(f(1)) = c(f(f(1))) = ··· , c(2) = c(f(2)) = c(f(f(2))) = ··· , etc. Decompose f into disjoint cycles and choose a color arbitrarily for all the vertices in a given cycle. If f has k different cycles and there are jCj colors to choose from, then jC(f)j = jCjk. 5 Theorem 13: Assume there are m different equivalence classes of colorings. Then 1 X m = jCjcycles(f): jGj f Proof: Substitute Theorem 12 into Theorem 11.
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