Determining the Hamilton-Connectedness of Certain Vertex-Transitive Graphs Ming Jiang Frank Ruskey ? Department of Computer Science , University of Victoria Victoria , B.C. , V8W 3P6 , Canada Abstract Faber and Moore have proposed a class of vertex-transitive digraphs as a model of directed inconnection networks . These networks have attractive degree versus diameter properties . We show that these digraphs are Hamiltonian and provide necessary and sufficient conditions for the existence of a Hamilton path between any two vertices . Particular cases of these digraphs are the right rotation Cayley digraphs Cay -LRB- Xn : Sn -RRB- where Xn is the set of right rotations -LRB- 1 2 ? ? ? i ? 1 i -RRB- , for i = 2 ; 3 ; : : : ; n. Another class of vertex-transitive graphs that have been proposed as models of inconnection networks are the arrangement graphs introduced by Day and Tripathi . They have the same vertex set as the Faber-Moore digraphs . We prove that arrangement graphs are Hamilton-connected . AMS Classification : 05C45 , 05C20 . 1 Introduction Mathematical interest in the Hamiltonicity of vertex-transitive graphs is well established , particularly for Cayley graphs . See , for example , the survey of Witte and Gallian -LSB- WiGa -RSB- . There is also potential computer engineering interest since vertextransitive graphs have been proposed as a natural model for interconnection networks and Hamiltonicity is an important property in relation to sorting algorithms on these networks . See , for example , Akers and Krishnamurthy -LSB- AkKr -RSB- and Nigham , Sahni and Krishnamurthy -LSB- NiSaKr -RSB- . ? Research supported in part by the Natural Sciences and Engineering Research Council of Canada under Grant A3379 . The purpose of this paper is to investigate the Hamiltonicity properties of two classes of vertex-transitive -LRB- di -RRB- graphs that have been proposed as models of interconnection networks . In both classes the vertex set consists of all k-permutations of an n-set . These graphs were first defined in the papers of Faber and Moore -LSB- FaMo -RSB- and Day and Tripathi -LSB- DaTr -RSB- . In both cases , we determine simple necessary and sufficient conditions for the existence of a Hamilton path between any two specified vertices . Let P -LRB- n ; k -RRB- denote the set of k-permutations of -LSB- n -RSB- = f1 ; 2 ; : : : ; ng . For 2 < = i < = k , we define the right rotation functions ri : P -LRB- n ; k -RRB- ! P -LRB- n ; k -RRB- as follows . ri -LRB- x1x2 ? ? ? xi ? 1xixi +1 ? ? ? xk -RRB- = xix1x2 ? ? ? xi ? 1xi +1 ? ? ? xk : For m 2 -LSB- n -RSB- we also define the partial functions qm : P -LRB- n ; k -RRB- ! P -LRB- n ; k -RRB- as follows . If m 2 fx1 ; x2 ; : : : ; xkg then qm is undefined and otherwise qm -LRB- x1x2 ? ? ? xk ? 1xk -RRB- = mx1x2 ? ? ? xk ? 1 : The Faber-Moore digraphs G -LRB- n ; k -RRB- are defined as digraphs whose vertex set is P -LRB- n ; k -RRB- . Each vertex x = x1x2 ? ? ? xk has n ? 1 outgoing arcs with endpoints r2 -LRB- x -RRB- ; r3 -LRB- x -RRB- ; : : : ; rk -LRB- x -RRB- and qm1 -LRB- x -RRB- ; qm2 -LRB- x -RRB- ; : : : ; qmn?k -LRB- x -RRB- ; where fm1 ; m2 ; : : : ; mn?kg = -LSB- n -RSB- n fx1 ; x2 ; : : : ; xkg . The graphs G -LRB- n ; k -RRB- are vertextransitive and have diameter k -LRB- see Faber and Moore -LSB- FaMo -RSB- -RRB- . They are clearly n ? 1 regular and have order -LRB- n -RRB- k , where -LRB- n -RRB- k = n -LRB- n ? 1 -RRB- ? ? ? -LRB- n ? k + 1 -RRB- is the \ falling factorial '' . We regard the arcs of G -LRB- n ; k -RRB- as being labeled either ri or qm , for some i or m . If k = n then the only adjacencies in G -LRB- n ; k -RRB- are those given by the right rotations . In this case the graph G -LRB- n ; n -RRB- is denoted Rn and is called a right rotation digraph . The graph Rn is isomorphic to the Cayley graph Cay -LRB- X : Sn -RRB- where X consists of the right rotations -LRB- 1 2 ? ? ? i -RRB- for i = 2 ; 3 ; : : : ; n . The arrangement graphs A -LRB- n ; k -RRB- are defined as graphs whose vertex set is P -LRB- n ; k -RRB- . Two vertices are adjacent if they differ in exactly one position . Clearly , these graphs are k -LRB- n ? k -RRB- regular and have order -LRB- n -RRB- k . It is proven by Day and Tripathi -LSB- DaTr -RSB- that these graphs are vertex and edge-transitive and have diameter b32kc . Below are a few definitions and conventions that we will use in the rest of the paper . We denote the complete graph on n vertices by Kn . In order to simplify the discussion , we regard both K1 and K2 as being Hamiltonian and having Hamiltonian paths . A graph is Hamilton-connected if either -LRB- a -RRB- it is not bipartite and there is a Hamilton path between every pair of distinct vertices , or -LRB- b -RRB- it is bipartite and there is a Hamilton path between every pair of vertices that are in different partite sets . If ss is a permutation , then ss -LRB- i -RRB- denotes the i-th element of the permutation -LRB- e.g. , if ss = 3146 , then ss -LRB- 3 -RRB- = 4 -RRB- . 2 Right rotation Cayley digraphs We begin by proving a necessary condition for the existence of Hamilton paths in G -LRB- n ; k -RRB- . Lemma 1 Let N = -LRB- n -RRB- k . If there is a Hamilton path ss1 ; ss2 ; ? ? ? ; ssN in G -LRB- n ; k -RRB- from ss1 = 12 ? ? ? k to ssN = x1x2 ? ? ? xk , then x1 = 2 . Proof : Consider the directed multigraph Cn ; k formed by condensing , for i = 1 ; 2 ; : : : ; k , into a single supervertex si all those -LRB- n ? 1 -RRB- k ? 1 permutations ss for which ss -LRB- i -RRB- = 2 . The -LRB- n ? 1 -RRB- k permutations that do not contain a 2 are condensed into the supervertex s0 . A Hamilton path H starting at 12 ? ? ? k in G -LRB- n ; k -RRB- becomes a walk W in Cn ; k starting at s2 . Let d ? -LRB- si -RRB- denote the number of arcs in W of the form sj ! si for some j , and d + -LRB- si -RRB- the number of the form si ! sj for some j . Since H is a Hamilton path , P d ? -LRB- si -RRB- = P d + -LRB- si -RRB- = -LRB- n -RRB- k ? 1 . Since each permutation occurs exactly once and the path starts at 12 ? ? ? k , it must be the case that d ? -LRB- s2 -RRB- = -LRB- n ? 1 -RRB- k ? 1 ? 1 , d ? -LRB- si -RRB- = -LRB- n ? 1 -RRB- k ? 1 for all i < > ; 2 , and d ? -LRB- s0 -RRB- = -LRB- n ? 1 -RRB- k. Observe that the only arcs entering s2 are those from s1 -LRB- since the only way that 2 can be brought into the second position is by applying r2 to a permutation with 2 in the first position -RRB- . Thus , d + -LRB- s1 -RRB- = -LRB- n ? 1 -RRB- k ? 1 ? 1 . But this means that the walk W must end at s1 . 2 Note that the preceding proof implies that any Hamilton cycle in G -LRB- n ; k -RRB- must use arc labeled r2 at least -LRB- n ? 1 -RRB- k ? 1 times . To prove a sufficient condition we will proceed by induction , but the inductive hypothesis must be strengthened . Definition 1 A -LRB- j ; n -RRB- - pair in a path in Rn is a pair of successive permutations ss and ss0 on the path such that the arc ss ! ss0 is labeled rn and ss -LRB- n -RRB- = ss0 -LRB- 1 -RRB- = j. Lemma 2 Let n > = 3 . For any permutation x2x3 ? ? ? xn of -LSB- n -RSB- nf2g , there is a Hamilton path H in Rn starting at 12 ? ? ? n and ending at 2x2x3 ? ? ? xn . Furthermore , the path H has a -LRB- j ; n -RRB- - pair for each j = 1 ; 2 ; : : : ; n. Proof : Our proof is by induction on n. For n = 3 , there are two paths . 123 213 321 132 312 231 and 123 312 231 321 132 213 . The -LRB- j ; 3 -RRB- - pairs are indicated in the lists above by underlining j. Suppose that the lemma holds for n ? 1 ; we will show that it holds for n . There are two cases to consider , depending upon whether xn = n or not . -LSB- Case I -RSB- If xn = n then we form a list of all -LRB- n ? 1 -RRB- ! permutations with an n in the final position . The list starts at 12 ? ? ? n and ends at 2x2x3 ? ? ? xn . Such a list exists by the inductive assumption . Furthermore , the list has -LRB- j ; n ? 1 -RRB- - pairs for each j = 1 ; 2 ; : : : ; n ? 1 . Let ss , ss0 be a -LRB- j ; n ? 1 -RRB- - pair where , say ss = y2 ? ? ? yn ? 1jn and so ss0 = jy2 ? ? ? yn ? 1n . The arc ss ! ss0 is replaced with the following list . ss = y2 ? ? ? yn ? 1jn rn o = ny2 ? ? ? yn ? 1j ... ... o = y2 ? ? ? yn ? 1nj rn ss0 = jy2 ? ? ? yn ? 1n The list from o to o contains all -LRB- n ? 1 -RRB- ! permutations that end in j . It exists by the inductive assumption . We do the above replacement for each j = 1 ; 2 ; : : : ; n ? 1 and thereby obtain a Hamilton path H in Rn from 12 ? ? ? n to 2x2 ? ? ? xn . Note that ss , o form an -LRB- n ; n -RRB- - pair and that o , ss0 form a -LRB- j ; n -RRB- - pair , and thus H has a -LRB- j ; n -RRB- - pair for each j = 1 ; 2 ; : : : ; n. -LSB- Case II -RSB- If xn < > n , then there is some 1 < = i < n such that xi = n. Inductively , there are lists of all permutations ending with n or xn that can be joined into a single path of length 2 -LRB- n ? 1 -RRB- ! in Rn as shown below . = 12 ? ? ? -LRB- n ? 1 -RRB- n ... ... = 2x2 ? ? ? xi ? 1xi +1 ? ? ? xnn rn = n2x2 ? ? ? xi ? 1xi +1 ? ? ? xn ... ... = 2x2x3 ? ? ? xn Note that the above list contains the -LRB- n ; n -RRB- - pair ; . Lists of permutations ending with -LSB- n -RSB- nfn ; xng can be spliced in as they were in Case I ; we do the splicing only in the sublist starting at and ending at . Each such splicing , say of those permutations ending xk , creates both a -LRB- xn ; n -RRB- - pair and a -LRB- xk ; n -RRB- - pair . Thus , after all splicings , we obtain a -LRB- j ; n -RRB- - pair for each j = 1 ; 2 ; : : : ; n. 2 Theorem 1 For n > = 2 , there is a Hamilton path in Rn starting at 12 ? ? ? n and ending at x1x2 ? ? ? xn if and only if x1 = 2 . Proof : This theorem follows from Lemmas 1 and 2 , together with the observation that 12 , 21 is the only Hamilton path in G2 . 2 By setting the ending vertex to be 2134 ? ? ? n in the preceding theorem , we obtain the following corollary . Corollary 1 The graph Rn is Hamiltonian for all n > = 3 . 3 General Faber-Moore digraphs Recall that the vertices in Rn are permutations of -LSB- n -RSB- . Given an n-set A = fa1 ; a2 ; : : : ; ang , we denote by Rn -LRB- A -RRB- the graph , isomorphic to Rn , but in which the vertices are permutations of A under the bijection which maps i to ai . By Theorem 1 of Section 2 , we have the following corollary . Corollary 2 Let A = fa1 ; a2 ; : : : ; ang . There is a Hamilton path in Rn -LRB- A -RRB- starting at a1a2 ? ? ? an and ending at x1x2 ? ? ? xn if and only if x1 = a2 . There are two subsections in this section . In the first subsection , we study kcombinations from -LSB- n -RSB- . In the second subsection , we study the Hamiltonicity of G -LRB- n ; k -RRB- . 3.1 The combination graph Definition 2 By GC -LRB- n ; k -RRB- we denote the graph whose vertices are the k-subsets of -LSB- n -RSB- . Two vertices are adjacent if their intersection has size k ? 1 . It is known that GC -LRB- n ; k -RRB- is Hamiltonian -LRB- e.g. , -LSB- ReNiDe -RSB- -RRB- ; we will show the stronger result that it is Hamilton-connected . First we prove an easy lemma . Lemma 3 The complete graph Kn is Hamilton-connected . Proof : Assume the vertices of Kn are labeled 1 ; 2 ; : : : ; n . The proof follows immediately from the observation that any permutation of -LSB- n -RSB- is a Hamilton path in Kn . 2 Theorem 2 The graph GC -LRB- n ; k -RRB- is Hamilton-connected . Proof : If k = or k = n , then GC -LRB- n ; k -RRB- consists of a single vertex , which is Hamilton-connected by definition . If k = 1 or k = n ? 1 , then GC -LRB- n ; k -RRB- is isomorphic to Kn , which is Hamilton connected by Lemma 3 . Thus we may assume that 2 < = k < = n ? 2 . Our proof is by induction on n. Consider the subgraphs Gk ? 1 and Gk of GC -LRB- n ; k -RRB- induced by those vertices that do and do not contain n , respectively . Clearly , Gk ? 1 is isomorphic to GC -LRB- n ? 1 ; k ? 1 -RRB- and Gk is isomorphic to GC -LRB- n ? 1 ; k -RRB- . Thus we may assume inductively that Gk ? 1 and Gk are Hamilton-connected . Let u and v be arbitrary vertices of GC -LRB- n ; k -RRB- . We need to show that there is a Hamilton path from u to v . There are three cases to consider , depending upon whether u and v are in Gk ? 1 or Gk . -LSB- Case I : Both u and v are in Gk . -RSB- Inductively , there is a Hamilton path Hk from u to v in Gk . Let 1 , 2 be two successive vertices on Hk . Then there are elements x ; y ; z , with 1 < = x ; y ; z < = n ? 1 , such that 1 and 2 have the form 1 = fa1 ; a2 ; : : : ; ak ? 2 ; x ; yg and 2 = fa1 ; a2 ; : : : ; ak ? 2 ; x ; zg : Also inductively , there is a Hamilton path Hk ? 1 in Gk ? 1 from 1 = fa1 ; a2 ; ? ? ? ; ak ? 2 ; y ; ng to 2 = fa1 ; a2 ; ? ? ? ; ak ? 2 ; z ; ng : Since -LRB- 1 ; 1 -RRB- and -LRB- 2 ; 2 -RRB- are edges of GC -LRB- n ; k -RRB- , we can insert Hk ? 1 between 1 and 2 , thus obtaining a Hamilton path in GC -LRB- n ; k -RRB- from u to v. -LSB- Case II : Both u and v are in Gk ? 1 . -RSB- The proof here is similar to Case I and is omitted . -LSB- Case III : One of u , v is in G1 and the other in G2 . -RSB- Without loss of generality , we may assume that u is in Gk and v is in Gk ? 1 . Let , where < > u , be a k-combination in -LSB- n ? 1 -RSB- . According to the inductive assumption , there is a Hamilton path Hk in Gk starting at u and ending at . Since k > = 2 , there is an element a in such that = nfag -LSB- fng satisfies < > v. Again , inductively , there is a Hamilton path Hk ? 1 in Gk ? 1 starting at and ending at v . Since and are adjacent in GC -LRB- n ; k -RRB- , Hk followed by Hk ? 1 is a Hamilton path in GC -LRB- n ; k -RRB- from u to v. 2 3.2 Hamilton paths in G -LRB- n ; k -RRB- A necessary condition for the existence of Hamilton paths in G -LRB- n ; k -RRB- was given earlier as Lemma 1 . The proof of the corresponding sufficient condition is split into two lemmas , one for k > 2 and the other for k = 2 . The reason that k = 2 must be handled separately is that the proof for k > 2 appeals to Lemma 2 -LRB- with n = k -RRB- , which requires n > 2 . If k = 1 then G -LRB- n ; k -RRB- is isomorphic to Kn . Lemma 4 Let 3 < = k < = n and x2x3 ? ? ? xk be a -LRB- k ? 1 -RRB- - permutation of -LSB- n -RSB- n f2g . Then there is a Hamilton path in G -LRB- n ; k -RRB- that starts at 12 ? ? ? k and ends at 2x2x3 ? ? ? xk . Proof : According to Lemma 2 , the lemma holds if n = k . If k < n , there are two cases to consider , depending on whether all xi 2 -LSB- k -RSB- -LRB- i = 2 ; : : : ; k -RRB- or not . -LSB- Case I :] If all xi 2 -LSB- k -RSB- , then according to Lemma 2 , there is a Hamilton path Hk in Rk starting at 12 ? ? ? k and ending at 2x2x3 ? ? ? xk . By Theorem 1 there is a Hamilton path H = C1 ; C2 ; : : : in GC -LRB- n ; k -RRB- , where C1 = f1 ; 2 ; : : : ; kg . Our construction first splices the permutations of C2 into those of C1 -LRB- as given by the path Hk -RRB- , then the permutations of C3 into those of C2 , and so on . Let Ci = fa1 ; a2 ; ? ? ? ; ak ? 1 ; akg ; and Ci +1 = fa1 ; a2 ; ? ? ? ; ak ? 1 ; ak +1 g be two successive vertices in H. Also , let ss ; ss0 be a -LRB- ak ; k -RRB- - pair in the Hamilton path H -LRB- Ci -RRB- in Rk -LRB- Ci -RRB- where ss = a1a2 ? ? ? ak ? 1ak and ss0 = aka1a2 ? ? ? ak ? 1 . According to Lemma 2 , there exists a Hamilton path H -LRB- Ci +1 -RRB- in Rk -LRB- Ci +1 -RRB- starting at o = ak +1 a1a2 ? ? ? ak ? 1 and ending at o = a1a2 ? ? ? ak ? 1ak +1 . Since there are arcs ss ! o and ss0 ! o , we can replace arc ss ! ss0 by inserting the permutations of Ci +1 between ss and ss0 and obtain the list shown below . ss = a1a2 ? ? ? ak ? 1ak qak +1 o = ak +1 a1 ? ? ? ak ? 2ak ? 1 ... ... o = a1a2 ? ? ? ak ? 1ak +1 qak ss0 = aka1 ? ? ? ak ? 2ak ? 1 If we apply the above procedure to all the vertices -LRB- k-combinations -RRB- in H , then we obtain a Hamilton path in G -LRB- n ; k -RRB- starting at 12 ? ? ? k and ending at 2x2x3 ? ? ? xk . -LSB- Case II :] There is at least one xi 62 -LSB- k -RSB- . In the graph GC -LRB- n ; k -RRB- , we denote the subgraph induced by vertices -LRB- combinations -RRB- that contain a 2 by GC2 and the subgraph induced by the remaining vertices by GC2 . Obviously , GC2 is isomorphic to GC -LRB- n ? 1 ; k ? 1 -RRB- and GC2 isomorphic to GC -LRB- n ? 1 ; k -RRB- . Similarly , in the graph G -LRB- n ; k -RRB- , we denote the subgraph induced by vertices -LRB- permutations -RRB- that contain a 2 by G2 . Let N = ? n ? 1 k ? 1 ? . According to Theorem 1 , there is a Hamilton path H = C1 ; C2 ; : : : ; CN in GC2 , where C1 = f1 ; 2 ; : : : ; kg and CN = f2 ; x2 ; x3 ; : : : ; xkg . This Hamilton path is characterized by two sequences of integers y2 ; y3 ; : : : ; yN and z1 ; z2 ; : : : ; zN where Ci +1 = Ci -LSB- fyi +1 g n fzig . By Corollary 2 , for 1 < i < N , there is a listing of the permutations of Ci starting from any permutation of the form yi2 ? ? ? and ending at any permutation of the form 2 ? ? ? zi . By then applying qyi +1 we get a permutation of Ci +1 of the form yi +12 ? ? ? , and thus obtain a path through those vertices of G2 whose underlying sets are in Ci or Ci +1 . Also by Corollary 2 , there is a listing of the permutations of C1 that starts at 12 ? ? ? k and ends at a permutation of the form 2 ? ? ? z1 , and a listing of the permutations of CN that starts at any permutation of the form yN2 ? ? ? and ends at 2x2x3 ? ? ? xk . By applying the above procedure successively for i = 1 ; 2 ; : : : ; N we obtain a spanning path in G2 from 12 ? ? ? k to 2x1x2 ? ? ? xk . It remains to splice in the permutations that do not contain a 2 . By Theorem 2 there is a Hamilton path H in GC2 starting at f1 ; 3 ; 4 ; ? ? ? ; k + 1g . Since C1 is adjacent to f1 ; 3 ; 4 ; ? ? ? ; k + 1g in GC -LRB- n ; k -RRB- , we can successively splice permutations of all k-combinations in H into the previously constructed path in the same way as described in Case I . The final result is the required Hamilton path in G -LRB- n ; k -RRB- . 2 Lemma 5 For any n > = 2 and x 2 -LSB- n -RSB- n f2g , there is a Hamilton path in G -LRB- n ; 2 -RRB- starting at 12 and ending at 2x . Furthermore , the Hamilton path contains at least one pair of twins , which are successive vertices of the form pq ; qp . Proof : We proceed by induction on n . If n = 2 , the Hamilton path is 12 ; 21 . Clearly , 12 and 21 are twins . The following cycle C in G -LRB- n ; 2 -RRB- contains all permutations of P -LRB- n ; 2 -RRB- that contain an n. C = 1n ; n1 ; 2n ; n2 ; : : : ; pn ; np ; : : : ; -LRB- n ? 1 -RRB- n ; n -LRB- n ? 1 -RRB- If x < > n , then inductively we may assume the existence of a Hamilton path in G -LRB- n ; 2 -RRB- starting at 12 and ending at 2x which contains at least one pair of twins , say pq and qp . Using the arcs pq ! np and pn ! qp we can splice in the cycle C to obtain a Hamilton path in G -LRB- n ; 2 -RRB- . If x = n , then inductively we may assume the existence of a Hamilton path H in G -LRB- n ? 1 ; 2 -RRB- starting at 12 and ending at 21 . Using the arc 21 ! n2 we can splice in the cycle C to obtain a Hamilton path in G -LRB- n ; 2 -RRB- that ends at 2n . 2 Theorem 3 For n > = 2 , there is a Hamilton path in G -LRB- n ; k -RRB- starting at 12 ? ? ? k and ending at x1x2 ? ? ? xk if and only if x1 = 2 . Proof : This theorem follows from Lemmas 1 , 4 and 5 . 2 Theorem 4 For n > = 2 , there is a Hamilton path in G -LRB- n ; k -RRB- starting at x1x2 ? ? ? xk and ending at y1y2 ? ? ? yk if and only if x2 = y1 . Proof : This theorem follows from the previous theorem and the fact that G -LRB- n ; k -RRB- is vertex-transitive . 2 By setting the ending vertex be any vertex that is adjacent to the starting vertex in the previous theorem , we obtain the following corollary . Corollary 3 The digraph G -LRB- n ; k -RRB- is Hamiltonian for all 1 < = k < = n. 4 Arrangement graphs are Hamilton-connected We now turn our attention to arrangement graphs . Let u and v be two adjacent vertices in A -LRB- n ; k -RRB- . By d -LRB- u ; v -RRB- we denote the integer in v that is different than any integer in u. For example d -LRB- 4163 ; 4563 -RRB- = 5 and d -LRB- 4563 ; 4163 -RRB- = 1 . The following result was proven by Tcheunte -LSB- Tc -RSB- . Note that A -LRB- n ; n ? 1 -RRB- is bipartite and is isomorphic to the so-called \ star graph '' -LRB- -LSB- AkKr -RSB- -RRB- . If k < n ? 1 , then A -LRB- n ; k -RRB- is not bipartite . Theorem 5 -LRB- Tcheunte -RRB- The graph A -LRB- n ; n ? 1 -RRB- is Hamilton-connected . Theorem 6 Let x1x2 ? ? ? xk 2 P -LRB- n ; k -RRB- be distinct from 12 ? ? ? k and 1 < k < n ? 1 . Then there is a Hamilton path ss1 ; ss2 ; : : : ; ssN -LRB- where N = -LRB- n -RRB- k -RRB- in A -LRB- n ; k -RRB- from 12 ? ? ? k to x1x2 ? ? ? xk such that d -LRB- ss1 ; ss2 -RRB- = k + 1 if k < n ? 2 . Proof : The graphs A -LRB- n ; k -RRB- possess many symmetries , one of which results by consistently permuting the positions of the elements of the k-permutations . In particular , for any permutation o of -LSB- k -RSB- , there is a Hamilton path from 12 ? ? ? k to x1x2 ? ? ? xk if and only if there is a Hamilton path from o -LRB- 1 -RRB- o -LRB- 2 -RRB- ? ? ? o -LRB- k -RRB- to xo -LRB- 1 -RRB- xo -LRB- 2 -RRB- ? ? ? xo -LRB- k -RRB- . Note that this symmetry preserves the property of d -LRB- ss1 ; ss2 -RRB- = k + 1 . Another symmetry is obtained by permuting the elements of the k-permutations . In particular , if oe is a permutation of -LSB- n -RSB- that fixes the elements of -LSB- n -RSB- n -LSB- k -RSB- , then there is a Hamilton path from 12 ? ? ? k to x1x2 ? ? ? xk if and only if there is a Hamilton path from oe -LRB- 1 -RRB- oe -LRB- 2 -RRB- ? ? ? oe -LRB- k -RRB- to oe -LRB- x1 -RRB- oe -LRB- x2 -RRB- ? ? ? oe -LRB- xk -RRB- . This symmetry also preserves the property of d -LRB- ss1 ; ss2 -RRB- = k + 1 since oe -LRB- k + 1 -RRB- = k + 1 . As a result of the preceding discussion , without loss of generality , we may limit our attention to Hamilton paths that start at 12 ? ? ? k and end at x1x2 ? ? ? xk where x1 < > 1 . For example , to show the existence of a Hamilton path from 1234 to 1245 , it suffices to find a Hamilton path from 1234 to 2534 . This follows by taking oe = -LRB- 1 3 -RRB- -LRB- 2 4 -RRB- to get a Hamilton path from 3412 to 4512 , and then also taking o = -LRB- 1 3 -RRB- -LRB- 2 4 -RRB- to get a Hamilton path from 1234 to 1245 . We first consider the case of k = 2 . If n = 4 and k = 2 then the following lists satisfy the conditions of the theorem . 12 12 12 12 12 12 12 12 12 13 32 13 13 13 13 13 13 13 14 34 14 14 14 14 14 14 14 24 24 24 24 24 24 34 34 34 23 14 23 21 21 21 32 31 31 21 13 21 23 23 23 31 32 32 31 43 31 43 43 43 41 42 42 34 23 34 41 42 41 42 43 43 32 21 32 42 41 42 43 41 41 42 31 42 32 31 32 23 21 21 41 41 43 31 34 34 24 24 23 43 42 41 34 32 31 21 23 24 If k = 2 and n > 4 then let p1 ; p2 ; : : : ; pn = 1 ; 2 ; : : : ; x1 ? 1 ; x1 + 1 ; : : : ; n ; x1 . Each pi gets expanded into a set Pi of n ? 1 permutations in P -LRB- n ; 2 -RRB- . The permutations of Pi have the form pij , where j 2 -LSB- n -RSB- n fpig . We now need to order the permutations in the sets Pi for i = 1 ; 2 ; : : : ; n so that the result is a Hamilton path in A -LRB- n ; 2 -RRB- . Let f -LRB- pi -RRB- , s -LRB- pi -RRB- , and l -LRB- pi -RRB- denote the second integer in first , second and last permutations of the ordering of Pi , respectively . For example , if n = 5 and pi = 3 , then one ordering of Pi is 34 ; 31 ; 35 ; 32 , for which f -LRB- pi -RRB- = 4 , s -LRB- pi -RRB- = 1 , and l -LRB- pi -RRB- = 2 . A Hamilton path results if and only if the following constraints are satisfied : f -LRB- p1 -RRB- = 2 , s -LRB- p1 -RRB- = 3 , f -LRB- pi +1 -RRB- = l -LRB- pi -RRB- for i = 1 ; 2 ; : : : ; n ? 1 , and f -LRB- p1 -RRB- = 2 , l -LRB- pn -RRB- = x2 . Since n ? 1 > 3 these constraints can easily be met . It is instructive to note that the construction does not work if n = 4 and x1x2 = 42 . We now show that if 2 < = k < = n ? 3 and A -LRB- n ; k -RRB- satisfies the conditions of the theorem , then A -LRB- n ; k + 1 -RRB- satisfies the conditions of the theorem . Given a Hamilton path H = ss1 ; ss2 ; : : : ; ssN from 12 ? ? ? k to x = x1x2 ? ? ? xk in A -LRB- n ; k -RRB- , we will show how to construct a Hamilton path from 12 ? ? ? k -LRB- k +1 -RRB- to x1x2 ? ? ? xkxk +1 in A -LRB- n ; k + 1 -RRB- , for any xk +1 2 -LSB- n -RSB- nfx1 ; x2 ; : : : ; xkg . We do this by expanding -LRB- as before -RRB- each vertex z = z1z2 ? ? ? zk into the the n ? k vertices zj = z1z2 ? ? ? zkj , where j 2 -LSB- n -RSB- n fz1 ; z2 ; : : : ; zkg . The f ; s ; l notations are extended in the natural way ; i.e. , f -LRB- ss -RRB- is the -LRB- k +1 -RRB- - st element in the first permutation of the expansion of ss . Define f -LRB- ssN +1 -RRB- to be xk +1 and ~ ss to be the set of integers in ss -LRB- e.g. , if ss = 4163 , then ~ ss = f4 ; 1 ; 6 ; 3g -RRB- . The expansion is done , one vertex at a time , starting at x and ending at 12 ? ? ? k , by executing the three steps shown below for i = N ; N ? 1 ; : : : ; 2 , in that order . -LRB- 1 -RRB- l -LRB- ssi -RRB- f -LRB- ssi +1 -RRB- -LRB- 2 -RRB- f -LRB- ssi -RRB- some element of -LSB- n -RSB- n -LRB- ~ ssi -LSB- fl -LRB- ssi -RRB- ; d -LRB- ssi ; ssi ? 1 -RRB- g -RRB- -LRB- 3 -RRB- Fill in the remaining elements of the expansion from -LSB- n -RSB- n -LRB- ~ ssi -LSB- ff -LRB- ssi -RRB- ; l -LRB- ssi -RRB- g -RRB- Now expand ss1 to satisfy f -LRB- ss1 -RRB- = k + 1 and l -LRB- ss1 -RRB- = f -LRB- ss2 -RRB- . This is possible since d -LRB- ss1 ; ss2 -RRB- = k + 1 and thus f -LRB- ss2 -RRB- < > k + 1 . Now if k < n ? 3 then we also want to have s -LRB- ss1 -RRB- = k +2 . This can be done by swapping two permutations in the expansion of ss1 unless f -LRB- ss2 -RRB- = l -LRB- ss1 -RRB- = k + 2 ; in this case we must also swap two permutations of the expansion of ss2 . These swaps are always possible if k < n ? 3 , but not if k = n ? 3 . 2 Corollary 4 For 1 < = k < = n ? 1 , the graph A -LRB- n ; k -RRB- is Hamilton-connected . Proof : If k = 1 then A -LRB- n ; 1 -RRB- is isomorphic to the complete graph Kn which is Hamilton-connected by Lemma 3 . For 2 < = k < = n ? 2 the result follows from Theorem 6 . For k = n ? 1 the result follows from Theorem 5 . 2 5 Final Remarks We considered the digraph Rn where the elements of Sn are generated by right rotations . Similarly , we could consider the Cayley digraph Ln = Cay -LRB- fl2 ; l3 ; : : : ; lng : Sn -RRB- where the generators are the left rotations li = -LRB- i i ? 1 ? ? ? 2 1 -RRB- and obtain the theorem stated below . The graph Rn and Ln are not isomorphic in general but the exactly the same methods of proof work for Ln . Theorem 7 There is a Hamilton path in Ln starting at 12 ? ? ? n and ending at x1x2 ? ? ? xn if and only if x2 = 1 . It would be interesting to have efficient algorithms for generating Hamilton paths and cycles in A -LRB- n ; k -RRB- and G -LRB- n ; k -RRB- . The ideal algorithms would have running times O -LRB- -LRB- n -RRB- k -RRB- and use space O -LRB- n -RRB- . We have recently learned that Day and Tripathi -LSB- DaTr2 -RSB- have shown that A -LRB- n ; k -RRB- is Hamiltonian . Their proof is by an induction from -LRB- n ? 1 ; k ? 1 -RRB- to -LRB- n ; k -RRB- and is longer and more complicated than our proof . Our proof can be modified to produce a very simple proof of Hamiltonicity . 6 Acknowledgement We thank Mike Dinneen for suggesting the problem of the Hamiltonicity of the FaberMoore digraphs and he and Gary MacGillivray for helpful comments on an earlier draft . References -LSB- AkKr -RSB- S.B. Akers and B. Krishnamurthy , \ A Group-Theoretic Model for Symmetric Interconnection Networks '' , IEEE Trans . Computers , 38 -LRB- 1989 -RRB- 555-565 . -LSB- AkDuFe -RSB- S. G. Akl , J. Duprat and A. G. Ferreira , \ Building Hamiltonian Circuits and Paths in Star Graphs '' , Technical Report No . 90-22 , Laboratorie di l'Informatique du Parrallelisme , Ecole Normale Superieure de Lyon , Lyon , France , October 1990 . -LSB- CoFi -RSB- F. Comellas and M. A. Fiol , \ Vertex symmetric digraphs with small diameter '' , manuscript , July 1990 . -LSB- DaTr -RSB- K. Day and A. Tripathi , \ Arrangement graphs : A class of generalized star graphs '' , Information Processing Letters , 42 -LRB- 1992 -RRB- 235-241 . -LSB- DaTr2 -RSB- K. Day and A. Tripathi , \ Embedding of Cycles in Arrangement graphs '' , TR 91-58 , Computer Science Dept. , U. Minnesota , November 1991 . -LSB- FaMo -RSB- V. Faber and J.W. Moore , \ High-Degree Low-Diameter Interconnection Networks with Vertex Symmetry : the Directed Case '' , Technical Report LA-UR - 88-1051 , Los Alamos National Laboratory , Los Alamos , New Mexico , 1988 . -LSB- NiSaKr -RSB- M. Nigam , S. Sahni , and B. Krishnamurthy , Embedding Hamiltonians and Hypercubes in Star Interconnection Graphs , Proc . International Conference on Parallel Processing , -LRB- 1990 -RRB- 1-8 . -LSB- ReNiDe -RSB- E.M. Reingold , J. Nievergelt and N. Deo , Combinatorial Algorithms , PrenticeHall , Inc. , Englewood Cliffs , New Jersey , 1977 . -LSB- Tc -RSB- M. Tchuente , \ Generation of permutations by graphical exchanges '' , ARS Combinatoria , 14 -LRB- 1982 -RRB- 115-122 . -LSB- WiGa -RSB- D. Witte and J.A. Gallian , \ A survey -LCB- Hamiltonian cycles in Cayley graphs '' , Discrete Math. , 44 -LRB- 1984 -RRB- 293-304 .

O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O 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O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O 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O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O 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