When the cartesian product of two directed cycles is hyperhamiltonian

We say a digraph G is hyperhamiltonian if there is a spanning closed walk in G which passes through one vertex exactly twice and all others exactly once. We show the cartesian product Zg x Zb of two directed cycles is hyperhamiltonian if and only if there are positive integers m and n with ma + nb = ab + 1 and gcd(m, n) = 1 or 2, We obtain a similar result for the vertex-deleted subdigraphs of Za x Zb. S. Curran [4, Theorem 4.31 observed that by using the theory of torus knots it is easy to prove that the cartesian product Za X Zb of two directed cycles is hamiltonian if and only if there are posi t ive integers m and n with ma + nb = ab and gcd(m, n) = 1 . Using Curran's ideas, Penn and Witte [2] proved that Za X Zb is hypohamiltonian if and only if there are positive integers m and n with ma + nb = ab 1 and gcd(m, n) = 1 . ( A digraph is said to be hypohamiltonian if it is not hamiltonian but every vertex-deleted subdigraph is hamiltonian.) Motivated by these results we define a digraph to be hyperhamiltonian if there is a spanning closed walk which passes through one vertex exactly twice and all others exactly once and determine when Za x Zb is I hyperhamiltonian and when a vertex-deleted subdigraph of Za X Zb is hyperhamiltonian. We assume the reader is familiar with [2] and with the background on torus knots given in [I , Section 41. Journal of Graph Theory, Vol. 11, No. 1, 21-24 (1987) @ 1987 by John Wiley & Sons, Inc. CCC 0364-9024/87/010021-04$04.00 22 JOURNAL OF GRAPH THEORY Theorem 1. The cartesian product Za X Zb of two directed cycles is hyperhamiltonian if and only if there are posit ive integers m and n with ma + nb = ab + 1 andgcd(m,n) = 1 or 2. Proof. Let C be a hyperhamiltonian closed walk in Zg X Zb. Then C decomposes uniquely into a pair of edge-disjoint circuits C1 and Ca with a common vertex. Let (m, , n l ) and (m2, n2) be the knot classes of C1 and C2, respectively. Then (ml + mJa + (nl + n Jb = ab + 1 . If C l crosses Cz, then the algebraic intersection number is k1 , so  1 = m\n^ n,ma = (ml + m& (n, + n&nz and it follows that gcd(ml + ma,n l + n2) = 1. If Cl does not cross C2, then the intersection number is nzln2 n\m^ = 0. From this and the fact that m , , n , , m2, n2 are non-negative and gcd(ml , n I ) = 1 = gcd(m2, n2), we obtain mi = m2 and n l = n2. Thus 2mla + 2 n l b = ab + 1 and gcd(2ml,2nl) = 2. Now assume there are positive integers m and n such that ma + nb = ab + 1 and gcd(m, n) = 1 or 2. Clearly a and b are relatively prime, so the group Za x Zb is generated by (1, 1). Let H be the spanning subdigraph of Za x Zb in which a vertex d(1 , l ) travels by ( 0 , l ) if 0 5 d < nb and it travels by (1,O) if nb 5 d -^ ab. Since the in-degree and the out-degree of each vertex are equal, H is the union of edge disjoint circuits C l , Ca, . . . , CN. All vertices of H have out-degree 1, except (0,O) which has out-degree 2, so, by renumbering if necessary, we may assume that C l and Ca are the only circuits in the decomposition of H which intersect. We wish to show that Cl and Ca are, in fact, the only circuits in the decomposition of H , so that H is the union of two circuits with a single point of intersection, for then H can obviously be realized as a closed walk with one repeated vertex. To this end, suppose N ? 3. Then, for i 2 3, C, is disjoint from Cl and Ca, so knot(Cl) = knot(C,) = knot(C2). Set (r, s) = N knot(Cl) = xZl knot(C,), Then ra + sb = ab + 1. Since a and b are relatively prime, only one pair of positive integers u and v may satisfy ua + vb = a b + 1 . Thus ( r , s ) = ( m , n ) . Hence gcd(m, n) = gcd(r, s) = N > 2, a contradiction. I We next consider vertex-deleted subdigraphs Za X Zh {v}. Since Zu X Zh is vertex-transitive, the particular vertex which is deleted is unimportant. Theorem 2. Let v â Zo X Z b . Then Za X Zb {v} is hyperhamiltonian if and only if there are positive integers m and n with ma + nb = a b and gcd(m, n) = 1 or 2. Proof. The proof of necessity is similar to that in Theorem 1 . To prove sufficiency, let Ho be a spanning subdigraph of Zn X Zb with knot(//,,) = (m, n ) . (Namely: Put mo = ma/lcm(a, b) and no = nb/lcm(a, b). Let mo cosets of (1, 1) travel by (1,O) in Ho; and let the other no cosets travel by (0, I).) Since mo and fly are positive, there must be some vertex w, such that w travels by (0, I), and w (1,O) travels by (1,O). Replacing Ho by a translate if necessary, we assume w = v. Set v= v (1, O), v, = v + (0, l ) , and vy = v + (1, 1). In Ho, there is an arc from vto v, and from v to v+. Create a new digraph H by removing the vertex v from Ho (and removing the two arcs inciA CARTESIAN PRODUCT IS HYPERHAMILTONIAN 23