A NON-HAMILTONIAN GRAPH. W.T. Tutte. (received June 30, 1959)

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1 A NON-HAMILTONIAN GRAPH W.T. Tutte (received June 30, 1959) There is a remarkable cubic graph of 28 vertices, discovered by Prof. H.S.M. Coxeter, which has no circuit of fewer than 7 edges and in which all oriented arcs made up of three edges are equivalent under the automorphism group. The structure of the graph is as follows. There are three disjoint heptagons Ai^i^^S^^S^b^J^l) > B(bjb5b2b > b3b-7b^b^) and C(c^c/ c^^cyc^cocj). We denote the seven remaining vertices by d^, d,... > dy. For each suffix i the three edges incident with dj_ join it to a^, b^ and ci. In drawing a diagram of the graph it seems best to show only the heptagons A, B and C, leaving the rest of the figure to the imagination. The purpose of this note is to establish another property of the graph, that it has no Hamiltonian circuit. A Hamiltonian circuit of a graph is a simple closed curve made up of edges and passing through all the vertices. Now if the given graph G has a Hamiltonian circuit H passing through aj^ and a 2 a 3> then the circuit passes also through b2d2 and C2d (since it does not contain the edge a2d2). We abbreviate this proposition as (a 1 a 2, a2a 3 ) > (^2^2* c 2 d 2)> and we shall make extensive use of implications of this kind. We note the existence of an automorphism U of G which maps X onto x^]_, where xisa, b, cord, 1 ^ i 4 7, and x 8 = x 1# Assume G has a Hamiltonian circuit H. Clearly H does not include all the edges of A. Let s denote the maximum number of consecutive edges of A in H. Then s ^ 6. Moreover Can. Math. Bull, vol.3, no.l, Jan. I960. 1

2 s ^ 2 since each vertex of A is incident with two edges of H. We split up the proof into 5 cases corresponding to the values of s from 2 to 6» Case I: s = 6. Because of the automorphism U we can suppose H contains the edges a x a 2, a 2 a 3, a 3 a 4, a 4 a5, &$&(} and a^ay of A. Since it does not contain a^a7 it must include also a^d^ and ayd^. We note the following implications. (a 1 a 2, a z a 3 ) ^ ( b 2 d 2> c 2 d 2 ) (a 2 a 3, a 3 a 4 ) * ( b 3 d 3> c 3 d 3) (a 3 a 4, a 4 a 5 ) > ( b 4d 4, c 4 d 4 ) (a 4 a 5, a 5 a 6 ) $> ( b 5 d 5> c 5 d 5) (a 5 a 6? a 6 a? ) > ( b ^, c 6 d 6 ) The part of H so far determined is shown by thick lines in Figure 1. We note that it is invariant under an automorphism V of G mapping x^ onto xg_. But H must include one of the two edges b}b 4 and b 4 b7. As these are equivalent under V we may suppose H includes b}b 4. We can now extend our knowledge of H with the following implications. (b x b 4, b 4 d 4 ) ^ (b 3 b 7) h 3 d 3 ) > (b 2 b 6? b d 2 ) > (b 1 b 4, bjb 5 ) > (a^dy, hydj) < ^ (c 5 c 7, c 5 d 5 ) > ( Cl c 3, cjdi) > (b 3 b 7, (b 2 b 6, (bibg, (ajdj, (c 2 c 7, (c 1 c 3, (c 4 c 6, b 7 d 7 ) b ô d 6) b 5 d 5 ) c l d l) c 5 c? ) c 3 d 3 ) c 6 d 6) H is now completely determined. separate circuits It consists of two Hi - (a 1 a 2 a3a 4 a 5 a 6 a 7 d 7 b 7 b i Q3C3C 1 d 1 a 1 ) and H 2 = ( b l b 5 d 5 c 5 c 7 c 2 d 2 b 2 b 6 d 6 c 6 c 4 d 4 b 4 b l)» So H is not Hamiltonian, contrary to assumption, 2

3 > p CFQ* H" OQ O^ O 3

4 Case II: s = 5. Because of the automorphism U we can assume H includes all the edges of A except a 7 a^ and a^a2«but H must include one of these two edges since it has two edges incident with a^. The case s = 5 is thus impossible. Case III: s = 4. We may suppose H contains aja2, a2a 3, a 3 a 4, a 4 a5> a^dj and a^d^. We have the following implications. ( a l a 2> a 2 a 3 ) ~^ < b 2 d 2' c 2 d 2> ( a 2 a 3> a 3 a 4) > ( b 3 d 3> c 3 d 3) (a 3 a 4, a 4 a 5 ) > (h^, c 4 d 4 ) (a 1 a 2, &i d \) > (a^a 7, a 7 d? ) (a 4 a 5, a 5 d 5 ) -> (a 6 a?, a 6 d 6 ) The part of H so far determined is shown thickly in Figure 2. It is invariant under an automorphism W of G mapping x^ into x 6-i' w'here XQ = x 7 and x _ ^ = x^. But H must include one of the two edges b 3 bk and b 3 b 7, which are equivalent under W, We may therefore suppose H contains boby. We continue with the following implications. (b 3 b 7, b 3 d 3 ) > ( b 2 b 6> b 6 d 6^ (b 2 b 6, b 2 d 2 ) -> (b^, b 5 d 5 ) ' ( b 6 d 6> a 6 d 6) ^ ( c l c 6» C 4 C 6) (b^d 5, a 5^5) > ( C 3 C 5> C 5 C 7^ (c 3 c 5, c 3 d 3 ) > ( c l c 6' c l d l) {dl l d l, c 1 d 1 ) => (b x b 4, b 1 b 5 ) (b x b 4, b 4 d 4 ) > (b 3 b 7, b? d? ) (c 4 c è, c 4 d 4 ) > (c 2 c 7, c 2 d 2 ) H is now completely determined as the pair of circuits and (a 1 a z a 3 a 4 a 5 d 5 b 5 b 1 b 4 d 4 c 4 c 6 c 1 d 1 a 1 ) < b 2 d 2 c 2 c 7 c 5 c 3 d 3 b 3 b 7 d 7 a 7 a 6 d 6 b 6 b 2)» contrary to its definition as Hamiltonian. Before going on to the remaining cases we note the existence of an automorphism X defined by the following table. 4

5 X maps each element of the first row into the corresponding element of the second row. a l a 2 a 3 a 4 a 5 a 6 a 7 b l b 5 b 2 b 6 b 3 b 7 b 4 c l c 6 c 4 c 2 c 7 c 5 c 3 d l d 2 d 3 d 4 d 5 d 6 d 7 a 1 a 2 a 3 d 3 c 3 c 1 d 1 d 7 c 7 c 2 c 4 d 4 b 4 b 7 a 6 d 6 b 6 b 2 b 5 d 5 a 5 a 7 d 2 a 4 b 3 c 5 c 6 b 1 Case IV: s = 2. We can suppose H contains the edges a^a 2, a 2 a 3, a^d^ and a3d 3/ But then H contains 4 consecutive edges of the heptagon a i a 2 a 3^3 c 3 c i^i a p which is equivalent to A under X» So this case can be reduced to the preceding cases. Case V: s = 3. We can suppose H contains the edges a 4 a5, z^a^, a 4 d 4 and a 7 d 7. We have the implications a^a-^, (a 4 a 5, a 4 d 4 ) > ( a 2 a 3> a 3 d 3^ ' (a^a 7, a 7 d 7 ) - > (a^a 2, a l d l^9 So H contains a^2, a a3, a^d^ and a3d 3. Accordingly Case V reduces to earlier cases in the same way as Case IV. University of Toronto 5

Graph homomorphisms. Peter J. Cameron. Combinatorics Study Group Notes, September 2006

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