By B. Bollobás (Eds.)

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G, is a hamilton-connected graph. * Research performed while the author was visiting the UniversitC de Paris VI and was partially supported by a grant from the Wisconsin Alumni Research Foundation and a grant from the National Science Foundation. A. 6 that G, has a hamilton cycle for n 2 3. As was pointed out by Balinski and Russakoff [l], the fact that G, has'a hamilton cycle is an immediate consequence of a well-known algorithm for generating the permutations in S,. 7 [4]. For n 2 2 the connectivity of G, is A,.

Since W does not contain m - 4 k vertices of degree at least m - k - 1 in W, each endvertex xi of P is adjacent to at least two vertices of A. Therefore we can select yl, y2 E A, y1 # y2, such that xi is adjacent to yi, i = 1 , 2 . 0 Armed with our lemmas, we shall show now that G does contain a Hamiltonian cycle. Let P be the x 1 - x 2 path whose existence is guaranteed by Lemma 7. Omit V ( P ) from G together with every edge joining vertices in W and add a new vertex x' to the remainder. Join x' to y t and y2.

Lemma 3. The set R consists of independent vertices. Proof. Let P be a longest path in G [ R ]and suppose p = 1P13 2. Let a , and a2 be the endvertices of P. Then each a, ( i = 1 , 2 ) is joined to at most p - 1 vertices of R and so at least 2 ( m - k - p + 1) edges join { u l ,a2} to L. I f x1 and x2 are vertices of L at distance d on L and 1c d =sp , then either Y , is not adjacent to a , or x2 is not adjacent to a2, since otherwise there is a cycle longer than L, as shown in Fig. 3 . Hence at most two edges join any set of p + 1 consecutive vertices of L to the set { a , , a2}.