Hamilon Cycle which Exend Tranpoiion Maching in Cayley Graph of S n Frank Rukey Deparmen of Compuer Science Unieriy of Vicoria, P. O. Box 1700 Vicoria, B. C. V8W 2Y2 CANADA frukey@cr.uic.ca Carla Saage y Deparmen of Compuer Science Norh Carolina Sae Unieriy, Box 8206 Raleigh, NC 27612-8206 cd@ccadm.ncu.edu Abrac Le B be a bai of ranpoiion for S n and le Cay(B :S n )behecayley graph of S n wih repec o B. Iwa hown by Kompel'makher and Likoe ha Cay(B :S n ) i hamilonian. We exend hi reul a follow. Noe ha eery ranpoiion b in B induce a perfec maching M b in Cay(B :S n ). We how here when n>4 ha for any b 2 B, here i a Hamilon cycle in Cay(B :S n ) which include eery edge of M b. Tha i, for n>4, for any baib of ranpoiion of S n, and for any b 2 B, i i poible o generae all permuaion of 1 2 ::: n by ranpoiion in B o ha eery oher ranpoiion i b. Keyword. Cayley graph, perfec maching, hamilonian graph, ranpoiion. AMS(MOS) ubec claicaion. 05C25, 05C45. Reearch uppored by he Naural Science and Engineering Reearch Council of Canada under gran A3379. y Reearch uppored by he Naional Science Foundaion Gran No. CCR8906500 1
1 Inroducion For a nie group G wih generaing e X, he Cayley graph of G wih repec o he generaing e X i he graph Cay(X :G) wiherex e G, in which g and gx are oined by an undireced edge for eery g 2 G and x 2 X. We will hink of he edge fg gxg a being labeled x. A compelling queion in graph heory i wheher eery undireced Cayley graph i hamilonian. Alhough here are reul uch a [CuWi] and [KeWi] which how ha he anwer i ye for cerain ubclae of Cayley graph, he general queion remain open. If we require only a Hamilon pah, he queion i ill open and i, in fac, a pecial cae of he more general conecure of Loaz ha eery conneced, undireced, erex raniie graph ha a Hamilon pah [Lo]. If we reric our aenion o he cae when G = S n, he ymmeric group of all permuaion of [n] =f1 2 ::: ng, i i ill an open problem wheher eery Cayley graph of S n i hamilonian. The queion remain open een when we require ha eery generaor x 2 X aify x 2 = id. Whaiknown i ha for eery generaing e X of ranpoiion, hecayley graph of S n i hamilonian. Thi wa r hown by Kompel'makher and Likoe [KoLi]. Slaer howed in [Sl] ha one could alway nd a Hamilon pah in Cay(X :S n ) which ended a a permuaion wih a in poiion k for any k 2 [n]. Tchuene generalized boh of hee reul by howing ha any wo permuaion of dieren pariy are oined by a Hamilon pah in Cay(X :S n ) [Tc]. A an example, he well-known algorihm of Seinhau [S], ohnon [o], and Troer [Tr], for generaing permuaion by adacen ranpoiion, gie a Hamilon cycle hrough he Cayley graph of S n wih generaing e f(12) (23) (34) ::: (n;1 n)g. Howeer, an elemen ofs n of order wo need no be a ranpoiion, o i remain open wheher he Cayley graph of S n on a e of generaor, each of order wo, i hamilonian. Recenly i ha been hown ha he Cayley graph of Coxeer group, generaed by order wo elemen which are geomeric reecion, are hamilonian [CoSlWi]. A relaed reul i ha for A n generaed by he e of 3-cycle f(12n) (13n) ::: (1 n;1 n)g, hecayley graph i hamilonian [GoRo]. 2
In hi paper, we conider S n wih any generaing e of ranpoiion, X. Noe ha each x 2 X dene a perfec maching in Cay(X :S n ), ha i, a e M x of edge of he graph wih he propery haeacherex of Cay(X :S n ) i he end of exacly one edge in M x : M x = ffg gxgg 2 S n g: Knowing ha Cay(X :S n ) i hamilonian by [KoLi], we can ak if M x exend o a Hamilon cycle. Such a cycle correpond o a liing of all permuaion of [n], in which ucceie permuaion dier by a ranpoiion in X, o ha alernae ranpoiion correpond o he elemen x. The graph C = Cay(f(12) (23) (34)g:S 4 )ihown in Figure 1. The ripled line of he gure indicae edge in he maching M (23), and he li of permuaion of Figure 2 i a Hamilon cycle in C ha conain eery edge of M (23). A pecic inance of hi problem aroe iniially in he work of Pruee and Rukey on liing he linear exenion of cerain poe by ranpoiion [PrRu]. Le R be he cla of ranked poe in which eery non-maximal elemen ha a lea wo upper coer. Example of poe in R include he odd fence, crown, he Boolean algebra laice, he laice of ubpace of a nie-dimenional ecor pace oer GF (q), and pariion laice. In [PrRu] i i proen ha he linear exenion of any poe in R can be lied o ha eery exenion dier by a ranpoiion from i predeceor in he li. Their proof required a cyclic liing of all permuaion of [n] by ranpoiion o ha eery oher ranpoiion wa an exchange of he elemen in poiion 1 and 2. Alhough hey were able o how uchaliingwa alway poible, in ome cae he ranpoiion were no of elemen in adacen poiion hee ranpoiion were he only one in he proof ha were nonadacen. In [RuSa] we howed ha i i poible o li permuaion of [n] by adacen ranpoiion o ha eery oher ranpoiion exchange he elemen in poiion 1 and 2. See Figure 3 for an example when n =5. Thi reul i equialen o howing ha in he Cayley 3
2341 2431 2413 2314 2134 QQ Q Q Q 1234 2143 QQ Q Q Q 1243 1423 4123 4132 4213 4312 4231 3124 1324 3142 1342 1432 4321 3214 3241 3421 3412 Figure 1: The graph Cay(f(12) (23) (34)g : S 4 ) wih M (23). 1234 1324 3124 3214 2314 2134 2143 2413 2431 2341 3241 3421 3412 3142 1342 1432 4132 4312 4321 4231 4213 4123 1423 1243 b 1 2 3 4 Figure 2: B = f(12) (23) (34)g and b = (23) (read acro.) 4
12345 21345 23145 32145 31245 13245 13425 31425 34125 43125 43152 34152 31452 13452 13542 31542 31524 13524 13254 31254 32154 23154 23514 32514 32541 23541 25341 52341 53241 35241 35421 53421 54321 45321 45231 54231 52431 25431 25413 52413 54213 45213 42513 24513 24531 42531 42351 24351 23451 32451 32415 23415 24315 42315 43215 34215 34251 43251 43521 34521 34512 43512 45312 54312 53412 35412 35142 53142 53124 35124 35214 53214 52314 25314 25134 52134 51234 15234 15324 51324 51342 15342 15432 51432 54132 45132 41532 14532 14352 41352 41325 14325 14235 41235 42135 24135 21435 12435 12453 21453 24153 42153 41253 14253 14523 41523 45123 54123 51423 15423 15243 51243 52143 25143 21543 12543 12534 21534 21354 12354 b 1 2 3 4 5 Figure 3: B = f(12) (23) (34) (45)g and b = (12) (read acro.) graph Cay(X :S n ) where X = f(12) (23) ::: (n;1 n)g, he perfec maching M (12) exend o a Hamilon cycle. For n = 4 here i a Hamilon pah including eery edge of M (12), bu no Hamilon cycle. A conequence of hi reul, which i a pecial cae of our main heorem below, i ha he linear exenion of he poe in R can, in fac, be lied by adacen ranpoiion. Our maor reul in hi paper i he following heorem. Main Theorem. LeX be a generaing e of ranpoiion for S n,wheren>4. Then for any x 2 X, M x exend o a Hamilon cycle in Cay(X :S n ). A bai for S n i a minimal e of generaor for S n. Wihou lo of generaliy, we may aume ha our generaing e of ranpoiion for S n i a bai, call i B, o ha he ranpoiion can be decribed a a ree T B : he erice of T B are he poiion 1 2 ::: n, where i and are oined by an edge if and only if (i) ia 5
7 @@ @ ; ;; 6 ; ;; b @@ @ 8 3 4 9 1 2 5 8 @@ @ ; ;; 7 ; ;; b 1 @@ @ 9 3 5 10 4 2 6 Figure 4: Excepional combinaion: ar (lef) and are (righ) wih b a indicaed. ranpoiion in B. For b 2 B we refer o he ordered pair ht B bi a a combinaion. A combinaion ht B bi i aid o be ordinary if here are wo edge e 1 e 2 in T B uch ha (a) e 1 6= b, e 2 6= b, and (b) he edge e 1 and e 2 are no adacen. A combinaion ha i no ordinary i excepional. The reaon for diinguihing beween ordinary and excepional combinaion i ha our baic proof echnique i o plice ogeher Hamilon cycle in cerain induced ubgraph. Thi plicing i baed on mall cycle ha don' conain any edge labeled b. IfhT B bi i ordinary hen eery erex of Cay(B :S n ) i on a 4-cycle wih no edge labeled b. Specically, ifc d 6= b are nonadacen edgeoft B hen (cd) 2 = id, o for any erex of Cay(B :S n ), he equence c cd cdc cdcd = i a 4-cycle. Howeer, if ht B bi i excepional, any edgec d 6= b of T B are adacen, o generaor c and d do no commue. In hi cae here will be no 4-cycle in Cay(B :S n ) no conaining b. Inead, (cd) 3 = id, which gie rie o 6-cycle no conaining b. A ar i a ree of n erice in which oneerex ha degree n ; 1 and a are i a ree of n erice in which oneerex ha degree n ; 2andoneerex ha degree 6
2. We refer o a erex of degree one in a ree a a leaf of he ree. See Figure 4. Excepional combinaion are characerized in he following lemma, which we ae wihou proof. Lemma 1 An excepional combinaion ht (i)i for n>4 mu eiher be a ar or be a are in which i i a leaf and i a erex of degree 2 (or ice-era). The proof of he Main Theorem ue a dieren conrucion for each ofhe following hree familie of combinaion: 1. Ordinary combinaion. 2. Excepional combinaion in which n > 4andT i a are. 3. Excepional combinaion in which n > 4andT i a ar. Wihin each family, he conrucion relie induciely only on member of he ame family, o he hree cae can be handled independenly. Secion 2 concern ordinary combinaion. Excepional combinaion are handled in Secion 3. Secion 4 conain exenion and open problem. 2 Ordinary Combinaion An ordinary combinaion ht (i)i i minimal if for eery leaf k 6= i he combinaion ht ; k (i)i i excepional. The following lemma ieailyproen. Lemma 2 There are hree non-iomorphic minimal ordinary combinaion. They are hown in Figure 2, 3, 5. For b 2 B, dene a b-alernaing pah (cycle) o be a pah (cycle) in Cay(B :S n ) in which alernae edge are labeled b. Furhermore, in he cae of a b-alernaing pah, he r and la edge of he pah mu be labeled b. For example, he cycle in Figure 3 i a (12)-alernaing Hamilon cycle in Cay(B :S n )whereb = f(12) (23) ::: (n;1 n)g. 7
12345 21345 21354 12354 13254 31254 31245 13245 14235 41235 41253 14253 12453 21453 21435 12435 13425 31425 31452 13452 14352 41352 43152 34152 34125 43125 42135 24135 23145 32145 32154 23154 25134 52134 52143 25143 24153 42153 45123 54123 54132 45132 41532 14532 14523 41523 42513 24513 24531 42531 43521 34521 34512 43512 45312 54312 51342 15342 13542 31542 35142 53142 53124 35124 31524 13524 15324 51324 52314 25314 25341 52341 53241 35241 32541 23541 23514 32514 35214 53214 51234 15234 12534 21534 21543 12543 15243 51243 54213 45213 45231 54231 52431 25431 25413 52413 51423 15423 15432 51432 53412 35412 35421 53421 54321 45321 42351 24351 23451 32451 34251 43251 43215 34215 32415 23415 24315 42315 41325 14325 3 b 1 2 4 5 Figure 5: B = f(12) (23) (24) (45)g and b = (12) (read acro.) In hi ecion we will how ha when B i a bai of ranpoiion for S n, wih b 2 B and ht B bi i an ordinary combinaion, hen Cay(B :S n ) ha a b-alernaing Hamilon cycle. The proof i by an inducie conrucion and will require a omewha ronger hypohei. If Q i any b-alernaing cycle in Cay(B :S n ), an (i )-inerion pair for Q i a pair of conecuie erice, on Q aifying (1) (i) =(i)= and (2) he edge oining and i no labeled b (i.e., f g 62 M b ). Theorem 1 Le B be a bai of ranpoiion for S n, and le b 2 B be uch ha ht B bi i an ordinary combinaion. Then Cay(B :S n ) ha a b-alernaing Hamilon cycle Q. Furher, Q can be choen o ha for eery i 2 [n], Q ha an (i )- inerion pair, and for each i 2 [n] here i ome 2 [n] for which Q ha wo diinc 8
(i )-inerion pair. Proof. If he ordinary combinaion ht B bi i minimal, hen by Lemma2imu be iomorphic o one of he combinaion in Figure 2, 3, or 5, each hown wih a cycle Q aifying he condiion of he heorem. Oherwie, aume induciely ha he heorem i rue for all ordinary combinaion wih fewer erice han T B.SincehT B bi i no minimal, T B conain a leaf,, no inciden wih he edge labeled b, uchhaht B ; bi i an ordinary combinaion. Le z be he unique erex of T B adacen o. The Cayley graph of S n on he e B nf(z)g ha n componen G 1 G 2 ::: G n, where G k i he ubgraph of Cay(B :S n ) induced by all permuaion wih () =k. Le G 0 denoe he Cayley graph of permuaion of [n] nfg, generaed by he e B nf(z)g. Then he inducion hypohei hold for G 0.EachG k i iomorphic o G 0, o by inducion, G, in paricular, ha a b-alernaing Hamilon cycle Q. Furher, for each i aifying i 6=, 6=, Q ha an (i )-inerion pair and for each i 6= here i ome 6= for which Q ha wo (i )-inerion pair. For k 6=, inerchanging k and in eery permuaion on Q gie a b-alernaing Hamilon cycle, Q k,ing k. Now, o obain he deired cycle Q for Cay(B :S n ), each of he cycle Q k, where k 6=, i pliced ino he cycle Q aa(z k)-inerion pair of Q (z i he unique erex of T B adacen o.) Thi i done a follow (ee Figure 6). Le be he (z k) inerion pair of Q. Le 0 0 be he correponding pair on he cycle Q k. Tha i, 0 i obained from by inerchanging and k, and imilarly for and 0. Then imply delee edge and 0 0 and add edge 0 and 0 correponding o he generaor (z) inb. I remain o how haaferallq k are pliced ino Q o form Q, ha here i ill an (i )-inerion pair for eery i 2 [n] and ha for each i, here i ome for which Q ha wo (i )-inerion pair. Fir conider i 6= z and 6=. The cycle Q ha an (i )-inerion pair and for ome 6= here are wo (i )-inerion pair. Thee pair are ill in he nal cycle 9
Q Q k = = z # # # # k k u u z k k = = 0 0 Figure 6: Splicing cycle Q k ino cycle Q a a (z k) inerion pair in proof of Theorem 1. = = # # # # # # k k l l Q l l z i z i # # # k k Q k k k Q l z i l l l l Figure 7: Coneraion of (i )-inerion pair when i 6= z and 6=. l l = = = = 0 0 10
Q unle ome Q k wa pliced ino Q a a (z k)-inerion pair which wa alo an (i )-inerion pair. Bu hen, for any l 6= k conider he conecuie pair on Q l obained by wapping elemen and l in each of (ee Figure 7). Then i an (i )-inerion pair on Q l. Noe ha (z) = (z) =k. Bu, in plicing Q l ino Q, Q l i pli only a a pair wih elemen in poiion z, o i ill an (i )-inerion pair in Q. Thu, afer plicing, here i no ne lo in inerion pair for i 6= z and 6=. For i 6= z and =, chooe k 6=. InQ here wa an (i k)-inerion pair. Inerchanging elemen and k in each of gie an (i )-inerion pair on Q k. Since i 6= z, hi i no he pair in Q k which wa pli when Q z wa pliced ino Q. Thu each Q k, k 6= conribue an (i )-inerion pair o Q. If i =, he number of ( k)-inerion pair on Q k i (n ; 1)!=2. During he plicing, only one pair i pli for k 6= and only n;1 pair for k =. So,Q conain a( k)-inerion pair for eery k, awell a wo ( )-inerion pair. In he cae ha i = z, plicing Q k ino Q for k 6= can only pli Q k a a (z )-inerion pair for =. So,een afer plicing, Q k will conain (z )-inerion pair for eery 6= k. Chooe any l m, diinc from k and. Then each ofq l and Q m conain a (z k)-inerion pair, een afer plicing. Finally, wemu check fora(z )-inerion pair. The cycle Q ha none and each Q k, k 6= ge pli a a (z )-inerion pair during plicing. inducion, here i ome 6= for which Q Howeer, by conain wo diinc (z )-inerion pair. Correponding o hee, Q conain wo (z )-inerion pair. Thu, een afer plicing, Q conain a (z )-inerion pair. 2 3 Sar and Flare: Excepional Combinaion Le B be a bai of ranpoiion for S n and b 2 B. We conider here he cae where T B i a ar or a are in which b oin he erex of degree 2 wih a leaf (ee Figure 4.) In boh cae, any wo edge in B nfbg are adacen, o he echnique ued for 11
12345 21345 21435 12435 12534 21534 21354 12354 12453 21453 21543 12543 15243 51243 51423 15423 15324 51324 51234 15234 15432 51432 51342 15342 13542 31542 31452 13452 13254 31254 31524 13524 13425 31425 34125 43125 43215 34215 34512 43512 43152 34152 34251 43251 43521 34521 35421 53421 53124 35124 35214 53214 53412 35412 35142 53142 53241 35241 32541 23541 25341 52341 52143 25143 25413 52413 52314 25314 25134 52134 52431 25431 24531 42531 45231 54231 54321 45321 45123 54123 54213 45213 45312 54312 54132 45132 41532 14532 14352 41352 41253 14253 14523 41523 41325 14325 14235 41235 42135 24135 24315 42315 42513 24513 24153 42153 42351 24351 23451 32451 32154 23154 23514 32514 32415 23415 23145 32145 31245 13245 4 b 1 2 3 5 Figure 8: Bai cae for are (read acro.) ordinary combinaion will no work. We focu aenion on are, and hen how ha ar can be handled imilarly. If T B i a are, we can aume ha B = F n = f(12) (23) (34) (35) ::: (3n)g. For n 5 are are iomorphic o ordinary combinaion, unle b = (12). We hownow ha een in hi cae, Cay(F n :S n ) ha a (12)-alernaing Hamilon cycle. Theorem 2 For n 5, Cay(F n :S n ) ha a (12)-alernaing Hamilon cycle H aifying 1. For n odd, here areconecuie permuaion n n on H aifying n (3) = 2 n (n ; 1) = 1 n (n) =n n (3) = n n (n ; 1) = 1 n (n) =2 12
2. For 0 k<(n ; 1)=2 when n i een and for 1 k<(n ; 1)=2 when n i odd, here areconecuie permuaion (k) n and (k) n on H aifying (k) n (3) = 2k +1 (k) n (n) =2k +2 (k) n (3) = 2k +2 (k) n (n) =2k +1 Proof. The heorem i rue when n = 5, a demonraed in Figure 8. Noe ha on he cycle of Figure 8, he required conecuie permuaion 5 and 5 are 34215 and 34512. The required conecuie permuaion (1) 5 and (1) 5 are 52314 and 52413. (Noe ha he order doe no maer a long a he permuaion appear conecuiely.) Aume ha for ome n 5, Cay(F n :S n ) ha a (12)-alernaing Hamilon cycle H aifying condiion (1) and (2) of he heorem. If we appendn + 1 o eery permuaion on H, wehae ab-alernaing cycle in Cay(F n+1 :S n+1 ), call i H n+1, ill aifying (1) and (2). For 1 i n, he ubgraph of Cay(F n+1 :S n+1 ), induced by he elemen ofs n+1 wih i in poiion n+1, i iomorphic o Cay(F n :S n ), and herefore i conain a (12)- alernaing Hamilon cycle H i. Noe ha gien any permuaion wih (n +1)= i, and any ranpoiion of he form (3k) 2 F n,wemay aume ha i followed by an edge labeled (3k) onh i. (Some edge labeled (3k) mu appear on H i ince F n i a bai for S n. Simply arrange he cyclic li of generaor correponding o he edge along H i o begin wih (3k) and apply hem, aring wih permuaion. Thi yield a new H i wih he required propery.) The idea of he conrucion i o plice H 1 ::: H n ino H n+1 in uch away o obain a (12)-alernaing Hamilon cycle H in Cay(F n+1 :S n+1 ) and preere properie (1) and (2) of he heorem. For n odd, we r plice H 1, H 2,andH n ino H n+1 a he pair 0 n 0 n on H n+1 correponding o n n on H (ee Figure 9.) To do hi we ue he fac ha he following compoiion of ranpoiion i he ideniy: 13
A B n 0 = ab 2 1 nn+1 n 0 = ab n 12n+1 cd 2k+1 2k+2 n+1 cd 2k+2 2k+1 n+1 ab 1 n+1 2 n ab n+1 12n cd n+1 2k+1 2k+2 cd 2k+1 n+1 2k+2 ab n n+1 2 1 ab 2 n+1 n 1 6 H n+1 H n H 1 H 2 6 new ((n;1)=2) ((n;1)=2) pair n+1 n+1 6 new (0) H 2k+2 H 2k+1 cd n+1 2k+2 2k+1 cd 2k+2 n+1 2k+1 new (k) n+1 (k) n+1 pair ab n+1 1 n 2 ab 1 n+1 n 2 n+1 (0) n+1 pair B A Figure 9: For n odd, plicing he cycle H i ino H n+1. 14
(3 n)(3 n+1)(3 n;1)(3 n+1)(3 n)(3 n+1)(3 n;1)(3 n+1) = id We know ha 0 n and 0 n appear conecuiely on H n+1 and, a dicued aboe, we may aume wihou lo of generaliy ha: [ 0 n (3 n+1)] and [ 0 n (3 n+1)](3 n;1) appear conecuiely on H n, [ 0 n(3 n+1)(3 n;1)(3 n+1)] and [ 0 n(3 n+1)(3 n;1)(3 n+1)](3 n) appear conecuiely on H 1, and [n 0 (3 n+1)(3 n;1)(3 n+1)(3 n)(3 n+1)] and [ 0 n (3 n+1)(3 n;1)(3 n+1)(3 n)(3 n+1)](3 n;1) appear conecuiely on H 2. In each pair aboe, a well a for he pair 0 n 0 n, delee he edge oining he wo elemen of he pair in heir repecie cycle. Then ue edge correponding o he generaor (3 n+1) o oin ogeher he cycle a hown in Figure 9. Noe from Figure 9 ha hi conrucion proide u wih he required pair (0) n+1 (0) n+1 and ((n;1)=2) n+1 ((n;1)=2) n+1 for he (12)-alernaing cycle H being conruced in Cay(F n+1 :S n+1 ). For 0 k<(n ; 1)=2 when n i een and 1 k<(n ; 1)=2 when n i odd, we plice H 2k+1 and H 2k+2 ino H n+1 a he conecuie paironh n+1 correponding o (k) n n (k) on H, imilar o he mehod aboe, bu uing he ideniy (3 n)(3 n+1)(3 n)(3 n+1)(3 n)(3 n+1) = id (ee Figure 9 and 10.) Noe from Figure 9 ha hi proide for he cycle H he pair (k) n+1 (k) n+1 for 1 k<(n ; 1)=2 and, from Figure 10, when n i een, gie n+1 n+1. 2 The cae of ar can be handled imilarly. For a bai B of ranpoiion of S n, if T B i a ar, we may aume ha B = R n = f(31) (32) (34) (35) :::(3n)g, and ha he diinguihed edge b of B i (32) (ee Figure 4.) In hi cae we hae he following heorem. 15
A H n+1 H 2 H 1 (0) n = ab 1 2 n+1 n (0) = ab 2 1 n+1 ab n+1 12 6 ab 1 n+1 2 ab 2 n+1 1 ab n+1 21 A new n+1 n+1 pair B (k) n = cd 2k+22k+1 n+1 cd n+1 2k+1 2k+2 n (k) = cd 2k+12k+2 n+1 cd 2k+1 n+1 2k+2 H 2k+2 H 2k+1 6 cd n+1 2k+2 2k+1 cd 2k+2 n+1 2k+1 new (k) n+1 (k) n+1 pair B Figure 10: For n een, plicing he H i ino H n+1 16
Theorem 3 For n 5, Cay(R n :S n ) ha a (32)-alernaing Hamilon cycle H aifying 1. For n odd, here areconecuie permuaion n n on H aifying n (3) = 2 n (n ; 1) = 1 n (n) =n n (3) = n n (n ; 1) = 1 n (n) =2 2. For 0 k<(n ; 1)=2 when n i een and for 1 k<(n ; 1)=2 when n i odd, here areconecuie permuaion (k) n and (k) n on H aifying (k) n (3) = 2k +1 (k) n (n) =2k +2 (k) n (3) = 2k +2 (k) n (n) =2k +1 Proof. The heorem i rue when n = 5, a demonraed in Figure 11. Noe ha on he cycle of Figure 11, he required conecuie permuaion 5 and 5 are 34215 and 34512. The required conecuie permuaion (1) 5 and (1) 5 are 25314 and 25413. The remainder of he proof i idenical o he proof of Theorem 2. 2 4 Final Remark There hae been ome oher paper wrien abou nding Hamilon cycle hrough pecied maching in graph, bu no in connecion wih Cayley graph ([Ha],[Wo]). For example, Haggki [Ha] ha hown ha if d(u) +d() V (G) + 1 for all nonadacen erice u and of G, heng ha a Hamilon pah hrough any gien perfec maching. By deleing all odd permuaion from our li we obain liing of he alernaing group A n. In he cae of a ar, where B = f(1 n) (2 n) ::: (n;1 n)g and b =(1n), noe ha ince (1 n)( n) =(1n), our reul proide anoher proof of he reul of Gould and Roh [GoRo] ha he digraph Cay(X :A n ) i hamilonian for n 5, where X = f(1 n) 1 <<ng. 17
12345 13245 13542 15342 15432 14532 54132 51432 51234 52134 52314 53214 23514 25314 25413 24513 24153 21453 41253 42153 42351 43251 43521 45321 35421 34521 34125 31425 31245 32145 32415 34215 34512 35412 35142 31542 51342 53142 53241 52341 52431 54231 54321 53421 53124 51324 31524 35124 35214 32514 32154 31254 31452 34152 34251 32451 32541 35241 25341 23541 23145 21345 21543 25143 15243 12543 12453 14253 14352 13452 43152 41352 41532 45132 45312 43512 53412 54312 54213 52413 52143 51243 51423 54123 14523 15423 15324 13524 13254 12354 12534 15234 25134 21534 21354 23154 23451 24351 24531 25431 45231 42531 42135 41235 21435 24135 24315 23415 43215 42315 42513 45213 45123 41523 41325 43125 13425 14325 14235 12435 5 4 3 1 b 2 Figure 11: Bai cae for ar (read acro.) Tchuene [Tc] howed ha here i a Hamilon pah beween any wo permuaion of oppoie pariy incay(b :S n )forany bai of ranpoiion B. The nex lemma how ha i i no in general he cae ha here i a b-alernaing pah conaining M b beween any wo permuaion of oppoie pariy. Le ht B bi be a combinaion. If he edge b i remoed from T B hen wo ree remain hee ree induce a pariion of [n] ino wo e, ay X and Y. Lemma 3 Le X Y be he pariion of [n] induced byht B bi. Any b-alernaing Hamilon pah in Cay(B :S n ) ha ar a he permuaion and end a he permu- 18
aion 0 mu aify he following condiion. [ (i) = [ 0 (i) i2x i2x Proof. Conider he muligraph M formed from Cay(B :S n )by condening ino a ingle erex, for each k-ube S of [n], hoe permuaion for which f(i) i 2 Xg = S. Thu M ha n k erice and each erex i regular of degree k!(n ; k)!. Eery edge of M i labeled b ince eery ranpoiion oher han b eiher wap wo elemen wih poiion in X or wap wo elemen wih poiion in Y. A b- alernaing pah in in Cay(B :S n ) ha conain eery edge of M b become an Euler our in M. Clearly, hi our ha o ar and end a he ame condened erex. 2 If n = 4 hen here are wo non-iomorphic excepional combinaion ht B bi, namely he ar B = f(12) (13) (14)g wih b = (12), and he pah B = f(12) (23) (34)g again wih b = (12). In hee cae i i no oo dicul o how ha here i no b-alernaing Hamilon cycle. Howeer, here are b-alernaing Hamilon pah conaining M b a hown in Figure 12. Below we li ome queion for furher ineigaion. 1. I here an ecien algorihm o generae he permuaion on a b-alernaing Hamilon cycle? We would like an algorihm whoe oal orage requiremeni O(n) and whoe oal running ime i O(n!). A raighforward implemenaion of our proof lead o algorihm ha require (nn!) ime and (nn!) pace. 2. I he neceary condiion of Lemma 3, ogeher wih he condiion ha and 0 hae oppoie pariy, alo a ucien condiion for he exience of a Hamilon pah from o 0?We conecure ha he condiion i ucien. 3. Gien a maching M in he n-cube Q n, i here a Hamilon cycle in Q n ha include eery edge of M? 19
1234 2134 3124 1324 2314 3214 4213 2413 3412 4312 1342 3142 4132 1432 2431 4231 3241 2341 4321 3421 1423 4123 2143 1243 4 2 1 3 1234 2134 2314 3214 3124 1324 1342 3142 3412 4312 4321 3421 3241 2341 2431 4231 4213 2413 2143 1243 1423 4123 4132 1432 1 2 3 4 Figure 12: (12)-alernaing Hamilon pah. 4. If X i a e of generaor for a group G, andx 2 X i an inoluion (i.e., x 2 = id), hen x induce a perfec maching, M x,in Cay(X :G). A naural queion i wheher here i a x-alernaing pah in Cay(X :G). In general, here i no x-alernaing Hamilon pah in Cay(X : G). For example, if X = f(1 2) (1 2 n)g, wheren 3 i odd, hen he following lile argumen, imilar o he proof of Lemma 3, how ha Cay(X :S n ) ha no (1 2)- alernaing pah. Condene ino ingle uper-erice all hoe permuaion inequialen under he roaion (1 2 n). The reuling muligraph ha (n;1)! erice, each of degree n,andanyhamilon pah in Cay(X :S n ) become a Eulerian cycle in he muligraph. Clearly, here i no Eulerian cycle if n i odd. On he oher hand, here doe appear o be a x-alernaing pah in Cay(X :G) 20
if G i a reecion group and x i one of i generaor, bu we hae no proen hi fully. In [CoSlWi], he Cayley graph of reecion group are hown o be Hamilonian. Reference [CoSlWi] [CuWi] [GoRo] [Ha] [o] [KeWi] [KoLi] [Lo] [PrRu] [RuSa] [Sl]. H. Conway, N.. A. Sloane, and A. R. Wilk,\Gray code for reecion group," Graph and Combinaoric, 5 (1989) 315-325, S.. Curran and D. Wie, \Hamilon pah in careian produc of direced cycle," in Cycle in Graph, Annal of Dicree Mahemaic 27, B. R. Alpach and C. D. Godil, ed., Norh Holland, The Neherland (1985). R.. Gould and R. Roh, \Cayley digraph and (1 n) equencing of he alernaing group A n," Dicree Mahemaic 66 (1987) 91-102. R. Haggki, \On F-Hamilonian Graph," in Graph Theory and Relaed Topic, Academic Pre, New York and London, (1979) 219-231. S. M. ohnon, \Generaion of permuaion by adacen ranpoiion," Mah. Comp. 17 (1963) 282-285. K. Keaing and D. Wie, \On Hamilon cycle in Cayley graph of group wih cyclic commuaor ubgroup," in Cycle in Graph, B. R. Alpach and C. D. Godil, ed., Annal of Dicree Mahemaic 27, Norh Holland, The Neherland (1985). V. L. Kompel'makher and V. A. Likoe, \Sequenial generaion of arrangemen by mean of a bai of ranpoiion," Kiberneica, 3 (1975) 17-21. L. Loaz, Problem 11 in Combinaorial Srucure and heir Applicaion, Gorden and Breach, London 1970. G. Pruee and F. Rukey, \Generaing he linear exenion of cerain poe by ranpoiion," SIAM. Dicree Mah., 4 (1991) 413-422. F. Rukey and C. Saage, \Generaing permuaion by rericed adacen ranpoiion," unpublihed manucrip (1989). P.. Slaer, \Generaing all permuaion by graphical ranpoiion," ARS Combinaoria 5 (1978) 219-225. [S] H. Seinhau, One Hundred Problem in Elemenary Mahemaic, Baic (1964). [Tc] M. Tchuene, \Generaion of permuaion by graphical exchange," ARS Combinaoria 14 (1982) 115-122. 21
[Tr] [Wo] H. F. Troer, \PERM (Algorihm 115)", Communicaion of he ACM 5, No. 8 (1962) 434-435. A.P. Woda, \Hamilonian Cycle Through Maching", Demonraio Mah., 22 (1988) 547-553. 22