Ringing the Changes II

Cayley Color Graphs Let G be a group and H a subset of G. We define a Cayley digraph (directed graph) by: Each element of G is a vertex of the digraph Γ and an arc is drawn from vertex a to vertex b if there is an element h H so that b = ha in G. If H = H -1 (the inverse of every element in H is also in H), then for every arc from a to b there is also an arc from b to a. In this situation it is common to replace the pair of arcs by an undirected edge and the Cayley digraph is called a Cayley graph. When each element of H is its own inverse, we may label the edges of the Cayley graph by the elements of H which determine the edge. This labeled graph is called a Cayley color graph.

Cayley Color Graphs Let G be the symmetric group S 3 whose elements are (written in cyclic notation): G = {I, (12), (13), (23), (123), (132)}. Let H = {(12), (132)}. The Cayley digraph C G (H) is: I (12) (132) (13) = (132)(12) (23) = (12)(132) (123)

Cayley Color Graphs Let H' = {(123), (132)}. The Cayley graph C G (H') is: I (12) (132) (13) (23) (123)

Cayley Color Graphs Let H'' = {(12), (13)}. The Cayley color graph C G (H'') is: I (12) (123) (23) (12) (13) (13) (132)

Ringing the Changes We can use Cayley color graphs to model change ringing. For n bells, we will use the symmetric group S n which consists of all the permutations on n symbols. The transitions (permutations which either fix or interchange adjacent bells) form the set H. Since each transition is an involution (has order 2), we can form the Cayley color graph C Sn (H). To obtain an extent on n bells which satisfies rules 1, 2 and 3 we need only find a Hamiltonian cycle (which visits each vertex exactly once before returning to the starting vertex) in this color graph and starting at the identity vertex, trace the cycle using the edge labels to obtain each change.

Hamiltonian Cycles As an example consider an extent of 3-bells (singles). The Cayley color graph is on S 3 and H = {(12),(23)}. I (12) (123) (23) (12) = a (23) = b (13) (132) Hamiltonian: ababab = (ab) 3 quick six bababa = (ba) 3 slow six Quick Six 123 = I 213 = (12) 231 = (123) 321 = (13) 312 = (132) 132 = (23) 123 = I

Another Example Consider the Cayley color graph of S 4 with H = {(12), (23), (34)}. Starting with (12) this is Double Court Minimus.

Another Example Consider the Cayley color graph of S 4 with H = {(12), (23), (34)}. Starting at JT gives the Johnson-Trotter sequence. JT Starting with (12) this is Double Canterbury Minimus.

Finding Hamiltonian Cycles Graphs containing a Hamiltonian cycle are called Hamiltonian graphs. Deciding whether or not a graph is Hamiltonian is an NP-problem, meaning that there is no known fast algorithm that will answer this question for all graphs. There are some theorems which give sufficient (but not necessary) conditions for a graph to be Hamiltonian (Ore, Dirac, Chvátal) but none of these are applicable in our situation. For small graphs we can use a basic backtracking algorithm to find Hamiltonian cycles, but in our case there is a theoretical result which is useful.

Rapaport's Construction In Cayley colour groups and Hamilton lines [Scripta Mathematica 24(1959), pp.51-58], Elvira Strasser Rapaport proved that the Cayley color graphs we are interested in are in fact Hamiltonian. Her proof is based on the following lemma: Lemma: A connected graph, regular of degree three, has a Hamiltonian cycle if there exists a set P of cycles and a set Q of 4-cycles each partitioning the vertex set of the graph and such that no member of P contains every vertex of a member of Q.

Rapaport's Lemma Lemma: A connected graph, regular of degree three, has a Hamiltonian cycle if there exists a set P of cycles and a set Q of 4-cycles each partitioning the vertex set of the graph and such that no member of P contains every vertex of a member of Q. Proof (cont.): If E is in some Q i, then the two other edges through v 1 must be in C 1 and one of them, say v 1 p 1 is in Q i. Similarly, an edge v 2 p 2 in C 2 is also in Q i. We could thus carry out the above operation again, a contradiction. Thus, E is in no Q i. The edges v 1 p 1 and v 1 p 3 are in C 1. As v 1 is in some Q 1, these two edges must also be in Q 1 (as E is not and valency is 3). C 1 thus contains 3 vertices and 2 consecutive edges of Q 1. The consecutive edges could not have arisen from the operation, so these edges must have been in some P i to start with. The 4 th vertex of Q 1 could not be in this P i by hypothesis.

Rapaport's Theorem Theorem: For n 3, the Cayley color graph C sn (H) is Hamiltonian when H = {g 1 =(12), g 2 =(12)(34)(56)..., g 3 = (23)(45)(67)... }. Pf: The permutations (12) and (1 3 5 7...) {= g 2 g 3 } are known to generate S n, so the three permutations of H certainly do. This means that the Cayley color graph is connected. For n > 4, start at any vertex and follow the path g 1 g 2 g 1 g 2. Since this product is the identity transformation, the path determines a 4-cycle. If two of the 4-cycles generated this way had a vertex in common, they would have to be identical. Let Q be the set of all such 4- cycles. There are n!/4 such 4-cycles in Q.

Rapaport's Theorem Theorem: For n 3, the Cayley color graph C sn (H) is Hamiltonian when H = {g 1 =(12), g 2 =(12)(34)(56)..., g 3 = (23)(45)(67)... }. Pf(cont): The set P consists of the 12-cycles that are formed from the relation (g 1 g 3 ) 6 = id. There are n!/12 disjoint 12-cycles in P. As no product of alternating g 3 and g 1 can equal g 2 (consider what happens to 4 and 5) none of these 12-cycles can contain all (or even 3) vertices of one of Q's 4-cycles. Thus, the conditions of the lemma are satisfied and a Hamiltonian cycle exists in this graph. If n = 3, H = {g 1 = g 2 = (12), g 3 = (23)} and the quick six singles gives the result. If n = 4, replace the 12- cycles of P by the 6-cycles generated by (g 1 g 3 ) 3 =id.

Example when n = 4 α I β (12) (12)(34) (23) η λ η λ β α

Other Choices Although Rapaport did not use them, P could also be made up of 2n-cycles which come from (g 2 g 3 ) n = id. There are n!/2n = ½(n-1)! of these 2n-cycles. For n = 4 we would have 3 8-cycles forming P, and the algorithm of the lemma will still work.

Example when n = 4 α I β (12) (12)(34) (23) η λ η λ β α

Other Choices However, as the following examples show, not all Hamiltonian cycles come from this construction of Rapaport.

Example when n = 4 α I β (12) (12)(34) (23) η λ η λ β α

Example when n = 4 α I β (12) (12)(34) (23) η λ η λ Single Court starting with (12) β α

Example when n = 4 α I β (12) (12)(34) (23) η λ η λ β α Reverse Bob starts with (12)(34)

Euler Characteristic The Cayley graphs we have examined so far are not always planar graphs. We can embed a graph that we are interested in in a planar way (no edges crossing except at vertices) on a surface with a certain number of holes. If a connected graph with n vertices and e edges has a planar embedding in a plane or on a sphere then n e + f = 2 where f is the number of faces (minimal [having no chords] cycles) of the graph. Example: n = 4, e = 6, f = 4 (don't forget the outside face) 4 6 + 4 = 2

Euler Characteristic The quantity n e + f is called the Euler characteristic of the graph and is related to the type of surface in which the graph may have a planar embedding. The result here is: Theorem: Suppose that a graph with n vertices, e edges and f faces has a simple imbedding in a closed surface S, then (1) If S is a sphere with g handles then n-e+f = 2 2g (2) If S is a sphere with c cross-caps then n-e+f = 2 - c. A simple imbedding is one in which each face (including the infinite face) is topologically a disk. The surfaces of type (1) are orientable while those of type (2) are nonorientable.

Embedding Graphs Sphere with 1 handle = Torus (Doughnut)

Embedding Graphs Torus Möbius Strip

Projective Plane Embedding Graphs Cross Cap

Embedding Rapaport's Graph The Rapaport graph has n! vertices, 3n!/2 edges and its faces are the n!/4 4-cycles, n!/12 12-cycles and ½(n-1)! 2n-cycles for n > 4. Thus the Euler characteristic is n! - 3n!/2 + (n!/4 + n!/12 + n!/2n) = n! ( 1 3/2 + 1/4 + 1/12 + 1/2n) = n!(-1/6 + 1/2n) n! (3 n)/6n = (n-1)! (3 n)/6 = 2 [(n-1)(n-2)(n-3) 2 (n-4)! + 12]/6. For n = 4 we have 24 vertices, 36 edges, 6 + 4 + 3 faces, so n e + f = 24 36 + 13 = 1 = 2 1 and so, it may be embedded in a projective plane.