By R. G. Stanton, R. C. Mullin (auth.), Kevin L. McAvaney (eds.)

**Read Online or Download Combinatorial Mathematics VIII: Proceedings of the Eighth Australian Conference on Combinatorial Mathematics Held at Deakin University, Geelong, Australia, August 25–29, 1980 PDF**

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**Extra info for Combinatorial Mathematics VIII: Proceedings of the Eighth Australian Conference on Combinatorial Mathematics Held at Deakin University, Geelong, Australia, August 25–29, 1980**

**Sample text**

If j >0, we let K 2 = ((s,t) e K It >j }. tonian path p in K 2 from x to y. By Lemma 2, there exists a hamil- We may then extend p to get a hamiltonian path in K from x to y. In case 3, if i = m-2 and j = n-2, then y = (m-l, n-l), and so we can apply Lemma 2 to get a hamiltonian path in K from x to y. Hence we may assume either 28 i

Hamiltonian path in K from x to y. This completes Lemma 2. By induc- path Pl in K 3 from (0,1) to (I,i) and a Let p = (0,0)plP2(l,0). Then p is a the Proof of Lemma I. If mn is odd, then there exists a hamiltonian path from any corner- vertex x to any vertex in K with the same colour as x. Proof. Again, by symmetry, we may take x = (0,0) which is coloured blue. y = (i,j) be any blue vertex in K. We have the following two cases Let : Case I. i >0. ,(O,n-l)) there exists a hamiltonian (O,n-l)Pl. and K 2 = K -K I.

We shall prove by in- duction on n that there exists a hamiltonian path from x to any red vertex y in K. If n =2, then y = ( l , 0 ) or (0,1). (i,I)(I,0) or (0,0)(I,0)(1,I)(0,I) We may then choose the path p to be (0,0)(0,I) respectively. Hence Lemma 1 holds. the result holds for n