When the cartesian product of two directed cycles is hypo-Hamiltonian

We show that the Cartesian product Za x Zb of two directed cycles is hypo-Hamiltonian (Hamiltonian) if and only if there is a pair of relatively prime positive integers m and n with ma + nb = ab 1 (ma + nb = ab). The result for hypo-Hamiltonian is new; that for Hamiltonian is known. These are special cases of the fact that there is a simple circuit of length p in Za x Zb if and only if there is a pair of relatively prime non-negative integers m and n with ma + nb = p -s ab. Curran and Witte [ I ] characterized the Hamiltonian paths in the Cartesian product of two directed cycles. They utilized ideas that originated with Rankin [3, 41 and Housman [2], and, by embedding the Cartesian product on a torus in the natural way, they were able to apply the theory of torus knots. Curran has also observed (oral communication) that this method yields an easy proof that the Cartesian product Z,, x Z,, of two directed cycles is Hamiltonian if and only if there is a pair of relatively prime positive integers m and n with ma + nb = ah. [Other (equivalent) characterizations had been given by Trotter and Erdos [6, Theorem 11 and as a special case of an old theorem of Rankin [3, Theorem 41. ) Here, we use the same method to determine when Z,, x Z,, is hypo-Hamiltonian. (A digraph D is hypo-Humiltonian if it is not Hamiltonian but every vertex-deleted subdigraph D\{u} is Hamiltonian.) The reader will need to be familiar with Section 4 of [I]. Thomassen [ 5 ] gives references on hypo-Hamiltonian graphs and digraphs, and shows Z,, x Z, a i is hypoHamiltonian. Journal of Graph Theory, Vol. 7 (1983) 441-443 @ 1983 by John Wiley & Sons, Inc CCC 0364-9024/83/040441-03$01 30 442 JOURNAL OF GRAPH THEORY Theorem 1. Z,, x ZÃˆ is hypo-Hamiltonian if and only if there is a pair of relatively prime positive integers m and n with ma + nb = ab 1. Proof. (+). Let C be a Hamiltonian circuit in Z,, x Zh\{(O, O)}, and let (m, n) E Z x Z be the knot class of C considered as a knot on the torus. Then ma vertices travel by (1, 0) in C, exactly nb vertices travel by (0, I), and there are ah 1 vertices in C. So ma + nb = ab 1. Also, since (m, n) is the knot class of a circuit (namely C) , we have gcd(m, n) = 1 . +) . Clearly a and b are relatively prime, so the Abelian group Z,, x Zi, is generated by ( 1 , 1). (This also implies Z,, x Zi, is not Hamiltonian [6 , Theorem 1 ; or 5 , Theorem 51.) Let H be the spanning subdigraph of Z,, x Z,,\{(O, O)} in which a vertex 41 , 1) (1, 0) travels by (0, 1) if 0 s; d < nb, whereas it travels by (1, 0) if nb < d < ab. [Note nb(I , 1) (1, 0) = (0, 0) in Z,, x Z,,; in terms of Definition 5.2 of [I], we have H = H(nb)\{(0, O)}. ] Then H is a disjoint union of simple circuits (because both the indegree and the outdegree of each vertex are 1) and has a b 1 vertices. So, letting (r, s) = knot(H), we have ru + sb = a b 1. Therefore ( r , s) = (m, n). Hence gcd[knot(H)] = I , which implies that H consists of a single circuit. So H is a Hamiltonian circuit in Z,, x ZÃˆ,\{(O O)}. Because Z', x Z,, is vertex-transitive, there is a Hamiltonian circuit in Z,, x Zh\{u} for any vertex v. I When a and b are relatively prime, there is a unique pair of positive integers m and n with ma + nb = ah 1 , namely, m = a (mod b) and n = b ' (mod a). In practice, we calculate m and n using Euclid's algorithm. For example, 7. , x Z7 is not hypo-Hamiltonian, because m = n = 2 . The above can be generalized as follows: Theorem 2. Let G be a finite additive Abelian group with generating set {x, y}. Put a = ord(x) and b = G : (x)\ . Let e be the unique integer which simultaneously satisfies 0 s; e < a and by = ex. Then, for any p , there is a simple circuit of length p in the Cayley digraph Cay(x, y : G) if and only if there is a pair of relatively prime integers m and n with 0 "s bn < p s: \GI and ma + n(b e) = p. Proof. (+). Let C be a simple circuit of length p , and put (tn, n) = knot(C). The conclusions are immediate. (6). We have ma 3s ne 3 0 and ma + n(b e) c IGI, so it is a consequence of [ l , Theorem 7.2 and the proof of Lemma 7.101 that there is a subdigraph H of Cay(x, y :G) with knot(H) = (m, n). Since gcd(m, n) = 1 , it follows that there is a unique simple circuit C in H , and knot(C) = (m, n). Then the length of C is mu + n(b e) = p . I Specialization to the case where Cay(x, y : G) = Z,, x Zh yields: there is a simple circuit of length p in Z,, x Z,, if and only if there is a pair W H E N PRODUCT OF CYCLES IS HYPO-HAMILTONIAN 443 of relatively prime non-negative integers t n and n with mu + nb = p ss ah.