Twin Towers of Hanoi

TWIN TO WERS OF HANOI ZORAN ˇ SUNI ´ C De dic ate d to Antonio Machi on t he o c c asio n of his r etir ement Abstract. In the Twin T o we rs of Hanoi v ersion of the well kno wn T o wers of Hanoi Problem there are tw o coupled sets of p egs. In eac h mov e, one chooses a pair of p egs in one of the sets and p erforms the only possi ble legal transfer of a disk b etw een the c hosen p egs (the smallest disk from one of the pegs i s mov ed to the other p eg), but also, sim ultaneously , b etw een the corresponding pair of pegs in the coupled set (th us the same sequence of mo ves is al wa ys used in both sets). W e pro vide upp er and low er b ounds on the length of the optimal solutions to problems of the following t ype. Given an initial and a final p osition of n disks in eac h of the coupled sets, what is the smal l est n umber of mo ve s needed to simu ltaneously obt ain the final position from the i ni tial one in eac h set? Our analysis is based on the use of a gr oup, called Hanoi T o wers group, of rooted ternary tree automorphisms, whic h models the original problem i n suc h a w ay that the confi gurations on n di s ks are th e vertices at lev el n of the tree and the action of the generators of the group represen ts the three possi ble mov es b etw een the three pegs. The t win version of the problem is analyzed b y considering the act ion of Hanoi T o wers gr oup on pairs of ve rtices. 1. To wers of H anoi and Twin Towers of Hanoi W e first describ e the well known Hanoi T ow er s Problem on n disks a nd 3 p egs . The n disks hav e different size. Allow ed p o sitions (whic h we call c o nfigurations ) of the disk s on the p eg s ar e those in whic h no disk is on top of a smaller disk. An example o f a configur ation on 4 disks is provided in Figure 1). In a single mo ve, the top disk from one of the p egs can b e transfer red to the top po sition on another p eg as long as the newly obta ined pos ition of the disks is allowed (it is a configuration). t t a 02 * * i i a 01 5 5 i i a 12 5 5 1 4 2 3 0 1 2 Figure 1. A configura tion o n four disks This material is based upon work supp orted by the National Science F oundation. 1 2 ZORAN ˇ SUNI ´ C Lab el the three p egs by 0, 1 and 2. A t any moment, regar dless o f the cur rent configuratio n, there are exactly three po ssible mov es, denoted by a 01 , a 02 , and a 12 . The mo ve a ij transfers the smalle st disk from p egs i and j betw een these t wo peg s. More pre cisely , if the smallest disk on p egs i and j is on i the mo ve a ij transfers it to j , and if it is on j the mov e transfers it to i . F or instance, the mov e a 01 applied to the configura tion in Figure 1 transfers disk 2 from p eg 1 to pe g 0, a 02 transfers disk 1 from peg 2 to peg 0, and a 12 transfers disk 1 from peg 2 to peg 1. W e do not need to sp ecify the direction of the transfer , since it is uniquely determined by the disks (by their size) that are curr ently on p egs i and j . In the exceptional case when there are no disks on e ither p eg i or j , the mov e a ij leav es suc h a c onfiguration unch anged. In the classical T ow ers of Hano i Pro blem on n disk s all disk s are initially on one of the pegs and the g oal is to trans fer all o f them to another (prescrib ed) p eg in the smallest p o s sible num b er o f mov es . It is w ell known that the optimal solution is unique and consis ts of 2 n − 1 mov es. One may p ose a more general pr oblem such as, g iven some initial and final configur a tions on n disks, what is the smallest nu mber of mov es needed to obtain the final configur ation from the initial one. It turns out that this problem a lways has a solution (regardless of the chosen initial and final configurations) and that the optimal solution is either unique or there are exactly tw o solutions. The latter happens for a relatively small n umber of ch oices of initial a nd final co nfigurations. F o r a survey on topics and r esults rela ted to Hanoi T ow ers Problem se e [Hin89] and for an optimal solution (repre s ented/obtained b y a finite automa ton) for a ny pair of co nfigurations s ee [Rom06]. Note that, in this setting, none o f the instances of the gener al problem is mo r e difficult (in terms of the optimal n umber of mov es) than the cla ssical problem. In the Tw in T ow ers of Hanoi version tw o sets o f three pegs labeled b y 0, 1 and 2 ar e coupled up. W e often r e fer to the t wo sets as the top and the b ottom set. A coupled co nfiguration on n disks is a pair of configurations on n disk s, one in each set (see, for instance, the coupled configuratio n on 4 disks in Figure 2). A mov e a ij applied to a coupled configura tion consists of application of the mov e a ij to eac h configuration in the coupled pair. F or instance, the mov e a 01 applied to the coupled configura tion in Figur e 2 transfers disk 1 in the top set to p eg 1 and, simult aneous ly , disk 1 in the bottom set to p eg 0. The mo ve a 02 applied to the same coupled co nfiguration, transfers disk 1 in the top set and disk 2 in the bottom set to peg 2 (in their sets), and a 12 changes nothing in the top set a nd transfers disk 1 in the bottom set to p eg 2 . In the setting of Twin T ow ers w e pos e three pr oblems. Problem 1 (Twin T ow e r s Switch) . Given the initial coupled co nfiguration in which all disks in the top set are on peg 0 and all dis ks in the b ottom set are on p eg 2, how many mo ves a re needed to o btain the final coupled configura tio n in whic h all disks in the top set are o n p eg 2, and all disks in the bottom set a re on peg 0? Note that the Twin T ow ers Switch Problem asks for simultaneous solution of t wo instances of the classical Hanoi T o wers Pr oblem (a ll disks are, simultaneously , using the same sequence of mo ves, transferr ed from peg 0 to p eg 2 in the top set, and from peg 2 to p eg 0 in the b ottom set). Problem 2 (Small Disk Shift) . Given the initial co upled config ur ation in Figure 2, how many mo ves a re needed to o btain the final coupled configura tio n in whic h all TWIN TOWERS OF HANOI 3 1 2 3 4 0 1 2 4 1 3 2 Figure 2. Initial po sition for the Small Disk Shift Pro ble m disks are in the same p ositions as in the initial one, except the smallest disk in each set is mo ved one peg to the right (disk 1 in the to p configura tion to peg 1, and disk 1 in the bottom c o nfiguration to peg 2)? Problem 3 (General Pr oblem) . Given any initial coupled configuration and any final co upled configuration wha t is the smallest n umber of mo ves needed to o btain the final configuration from the initial one? W e provide an upp er bound for the Twin T ow ers Switch, exact a ns wer for the Small Disk Shift, a nd lo wer a nd upp er b ounds for the General Pr oblem restr ic ted to basic coupled configuratio ns (defined below). Theorem TTS (Twin T ow er s Switch) . The smal lest numb er of moves ne e de d to solve the Twin T owers S witch Pr oblem on n disks is no gr e ater than a ( n ) , wher e a ( n ) = ( 1 , n = 1 , 4 3 · 2 n − ( − 1) n 3 , n ≥ 2 . Remark. The sequence a ( n ) satisfies the Jacobshtal linear recurs ion a ( n ) = a ( n − 1 ) + 2 a ( n − 2) , for n ≥ 4 , with initial condition a (1) = 1, a (2) = 5, and a (3) = 11. Conjecture TTS. The smal le st n umb er of moves ne e de d to solve the Twin T owers Switch pr oblem on n disks is exactly a ( n ) . Note that the Twin T ow ers Switch, req uir ing no more than r oughly 4 3 2 n mov es is not considerably more difficult than the classical problem of moving a single to wer, which require s roughly 2 n mov es . I n fact, there are more difficult problems that can be posed in the context o f coupled sets (recall that there are no problems that ar e more difficult than the classical problem when only one se t of disks is co nsidered). F or instance, the next result implies that the Small Disk Shift Pr oblem requires more mo ves than the Twin T owers Switch P r oblem. 4 ZORAN ˇ SUNI ´ C Theorem SDS (Small Disk Shift) . The s m al lest numb er of moves d ( n ) ne e de d to solve the Smal l Disk Shift Pr oblem on n disks is e qual to d ( n ) =      2 , n = 1 , 6 , n = 2 , 2 · 2 n , n ≥ 3 . In order to state our result on the Gener al Problem, we need the no tion of compatible coupled configurations . An initial coupled co nfiguration I on n disks is c omp atible to the fina l coupled co nfiguration F on n disks if F ca n b e obtained fr om I in a finite n umber of mov es. A coupled config uration is called b asic if the smalles t disks in its top configura tion a nd the smallest disk in its b ottom configur ation are not on co rresp onding pegs (it is not the cas e that b oth are on p eg 0, b o th on peg 1, or bo th o n peg 2). Note that, based o n the bra nching structure of Hanoi T ow ers group described by Grig orch uk and the author in [G ˇ S07], D’Angeli and Donno show in [DD07] that Hanoi T o wers group a cts dis tance 2 -transitively on the levels of the roo ted terna ry tree. This provides a c hara c terization of the pairs of compatible coupled configu- rations. In particular, their result implies that a ll basic coupled configur a tions ar e compatible. W e quote their re sult in more detail (Theorem 1), after we sufficient ly develop the necessar y terminology . Along the wa y we provide a different pro o f (w e need it for our upp er b ound estimate on the Gener al Problem). Note that an in ter- esting consequence of the result of D’Angeli and Donno is that the Hanoi T ow er s group induces an infinite sequence of finite Gel ′ fand pairs (see [DD07] for details). Theorem GP (General pro ble m for bas ic config urations) . The numb er of moves ne e de d to obtain one b asic c ouple d c onfigu r ation on n disks fr om another is no gr e ater than 11 3 × 2 n = 3 . 66 × 2 n . Note that the coupled co nfig urations in Theorem SDS are ba sic. Thus, The- orem SDS implies that fo r at least one pa ir of basic coupled config urations the smallest num b er of moves that is needed to obtain one from the other is exactly 2 × 2 n . Obtaining go o d upp er b ound seems to b e a difficult task, since one nee ds to solve all instances of the problem in optimal or nea rly optimal w ay . Low er b ounds seem a bit ea sier to o btain since they may be derived from lower b ounds from some sp ecific, well chosen, instances . The low er bound (2 × 2 n ) and the upper b ound (3 . 66 × 2 n ) provided here differ by less than a factor of tw o. All results men tione d so far will b e reca st in the following sectio ns in the natural setting of group actions on roo ted trees. The r eason is that this setting provides a conv e nient language and to ols to prov e our results. Ac knowledgmen t. T he author is thankful to T ullio Ceccherini-Silbers tein and Alfredo Donno for their help, useful remarks, and corrections. 2. En coding by w ords and tree a utomorphisms W e start by an encoding of the original Hanoi T ow er s Proble m o n three pe g s, as originally presen ted in [G ˇ S06] (and further elab orated in [G ˇ S07, G ˇ S08]), by a group of ro oted ternary tree automorphisms. TWIN TOWERS OF HANOI 5 Lab el the disks by 1 , 2 , . . . , n ac c o rding to their s iz e (smallest to larg est). The configuratio ns can b e enco ded b y words over the finite a lpha be t X = { 0 , 1 , 2 } . The letters in this alphab et represent the p egs. The w ord x 1 x 2 . . . x n represents the unique configuration on n dis ks in which, for i = 1 , . . . , n , the dis k i is on p eg x i . F or example, the w ord 212 0 represents the configur ation in Figure 1. Note that there are exactly 3 n configuratio ns o n n disks. The mov es a ij are enco ded as the transformations o f the s e t o f a ll finite words X ∗ ov e r X defined b y a 01 (2 . . . 20 u ) = 2 . . . 21 u, a 02 (1 . . . 10 u ) = 1 . . . 12 u, a 12 (0 . . . 01 u ) = 0 . . . 02 u, a 01 (2 . . . 21 u ) = 2 . . . 20 u, a 02 (1 . . . 12 u ) = 1 . . . 10 u, a 12 (0 . . . 02 u ) = 0 . . . 01 u, a 01 (2 . . . 2) = 2 . . . 2 , a 02 (1 . . . 1) = 1 . . . 1 , a 12 (0 . . . 0) = 0 . . . 0 , for any word u in X ∗ . Thus, a ij changes the fir st occur rence of i or j to the other of these tw o symbols. The p oint of, say , a 01 “ignoring ” initia l prefixe s of the form 2 ℓ is that such prefixes repr esent small disks on p eg 2, and a 01 should ig no re such disks, since it is supp osed to tr ansfer a disk be tw een p eg 0 a nd p eg 1. The firs t o ccurrence of 0 or 1 represents the smallest disk on o ne of these tw o p egs a nd changing this oc c ur rence of the sym b ol 0 or 1 to the other one in the code o f the given configura tio n transfers the corresp onding disk to the other peg. Note that if a ij is applied to (a co de of ) a configuration that has no o cc ur rences o f i or j it leav es such a c o nfiguration unchanged. This corres p o nds to the situatio n in which there are no disks on p egs i ad j a nd the mov e a ij has no effect on such a configuration since there are no disks to be moved. In order to work with more compact notation, set a 01 = a, a 02 = b, a 12 = c. In this notation, the mo ves a , b and c act on the set of all finite words X ∗ by a (2 . . . 20 u ) = 2 . . . 21 u, b (1 . . . 10 u ) = 1 . . . 12 u, c (0 . . . 01 u ) = 0 . . . 02 u, a (2 . . . 21 u ) = 2 . . . 20 u, b (1 . . . 12 u ) = 1 . . . 10 u, c (0 . . . 02 u ) = 0 . . . 01 u, (1) a (2 . . . 2) = 2 . . . 2 , b (1 . . . 1) = 1 . . . 1 , c (0 . . . 0) = 0 . . . 0 . Hanoi graph on n disks, denoted by Γ n , is the gr a ph on 3 n vertices representing the configura tions o n n disks . Two vertices u and v are connec ted b y an edg e lab eled b y s ∈ { a, b, c } if the configur ations represented by u and v can b e obtained from ea ch other by a pplication o f the mov e s (note that each of the mov es is an inv olutio n). The Hanoi graph on 3 disks is depicted in Figure 3. Graphs very similar to the graphs we just defined hav e a lr eady a pp ea red in the literature in connection to Hanoi T ow ers Problem (see, for instance, [Hin89]). The difference is that the edges are usually not labe led and there are no loops at the corner s. The set of all words X ∗ has the str ucture of a ro oted ternar y tree in which the ro ot is the empty word, level n of the tree consis ts of the 3 n words of length n ov e r X , and each v ertex (eac h word) u has three children, u 0, u 1 a nd u 2. The transformatio ns a , b and c a ct on the tree X ∗ as tree a utomorphisms (in particular, they preserve the roo t and the levels of the tree). Thus, a , b and c gener ate a gr oup of automo rphisms of the ro oted terna ry tree X ∗ . The g roup H = h a, b, c i , called Hanoi T owers group, w as defined in [G ˇ S06]. The Hano i g raph Γ n is the Schreier graph, with resp ect to the generating set { a, b, c } , o f the action of H on the w ords of length n in X ∗ (Sc hreier gr aph o f the action on lev el n in the tree). 6 ZORAN ˇ SUNI ´ C   • a        c 5 5 5 5 5 5 5 b 111   • c        b 011   • a 5 5 5 5 5 5 5 211   • b        a 5 5 5 5 5 5 5 021   • c        b 5 5 5 5 5 5 5 201   • a        c 221   • b 121   • a 101   • c 5 5 5 5 5 5 5 001   • c        b 5 5 5 5 5 5 5 220   • b        a 5 5 5 5 5 5 5 002   • b        a 120   • c 5 5 5 5 5 5 5 020   • a        c 202   • b 5 5 5 5 5 5 5 102   • a        c 5 5 5 5 5 5 5 100   • b        a 5 5 5 5 5 5 5 010   • c        b 5 5 5 5 5 5 5 212   • a        c 5 5 5 5 5 5 5 122   • b c 000   • a 200   • c 210   • b 110   • a 112   • c 012   • b 022   • a 222 Figure 3. Γ 3 , the Hanoi graph on 3 disks A sequence of mo ves is a word ov er S = { a , b, c } . The order in whic h mov es are applied is from right to left as in the following calcula tion caba (022 0 ) = ca b (1220) = ca (102 0) = c (0020) = 0010 . The structure of the Hanoi g raphs is fairly well understo o d. In particular, for n ≥ 0, the Hano i g raph Γ n +1 is o btained fr om the Ha noi graph Γ n as follows [G ˇ S07]. Three copies of Γ n are constructed by app ending the la bel 0, 1, and 2, resp ectively , to every vertex la b el in Γ n . Then the tw o lo ops la b e led by c at the vertices 0 n 1 and 0 n 2 are deleted and replaced b y an edge b etw een 0 n 1 and 0 n 2 labeled b y c , the tw o lo o ps la b eled b y b at the vertices 1 n 0 and 1 n 2 are deleted a nd replaced by an edg e b e tw een 1 n 0 and 1 n 2 lab eled by b , and the t wo loo ps lab eled by a at the vertices 2 n 0 and 2 n 1 are deleted and replaced by an edge betw een 2 n 0 a nd 2 n 1 lab eled b y a . Indeed, this “r ewiring” o n the next level (level n + 1) needs to b e done as indicated since c (0 n 1) = 0 n 2, b (1 n 0) = 1 n 2 and a (2 n 0) = 2 n 1. In general, the gra phs for even and o dd n hav e the form pr ovided in Figure 4 and Figure 5. These figures suffice for our purp oses, since only the region near the path from 0 n to 2 n (near the bo ttom) and near the path from 0 n to 1 n (near the left side) play significant role in our consider ations. The follo wing lemma , pro viding a non-recurs ive, optimal solution to the classical Hanoi T ow e r s Problem is part of the folklore (it has b een proved and expressed in many disguises and our setting may b e cons idered one of them). TWIN TOWERS OF HANOI 7   • c    a 7 7 7 b   • a    b   • c 7 7 7   • b    c 7 7 7   • a    b 7 7 7   • c   a   • b   • c   • a c     • a    b 7 7 7   • b    c   • a 7 7 7   • c    a 7 7 7   • b    c 7 7 7   • a    b   • c   • a   • b   • b    c 7 7 7   • c    a   • b 7 7 7   • a    b 7 7 7   • c    a 7 7 7   • b    c   • a   • b   • c 7 7 7 a b   • c    a 7 7 7   • a    b 7 7 7   • b    c 7 7 7   • c    a 7 7 7   • a    b   • c 7 7 7   • b    c   • a 7 7 7   • c    a   • b 7 7 7   • a    b   • c 7 7 7   • b    c 7 7 7   • a    b 7 7 7   • c    a 7 7 7   • b    c 7 7 7   • a    b 7 7 7   • c    a 7 7 7   • b    c 7 7 7   • a    b 7 7 7   • a c   • b   • c   • a   • b   • c   • a   • b   • c   • a   • b   • c c   • a   • b   • c   • a Figure 4. Hanoi graph on even num ber of disks Lemma 1. The dia meter of the Hanoi T owers gr aph Γ n is 2 n − 1 . It is achieve d as the distanc e b etwe en any two of the c onfigur ations 0 n , 1 n , and 2 n . The unique se quenc e of moves of length 2 n − 1 b etwe en any t wo of t hese c onfigu ra tions is given in the fol lowing t able. even n o dd n from \ to 0 n 1 n 2 n 0 n 1 n 2 n 0 n × ( cab ) m ( n ) ( cba ) m ( n ) × a ( cba ) m ( n ) b ( cab ) m ( n ) 1 n ( bac ) m ( n ) × ( bca ) m ( n ) a ( bca ) m ( n ) × c ( bac ) m ( n ) 2 n ( abc ) m ( n ) ( acb ) m ( n ) × b ( acb ) m ( n ) c ( abc ) m ( n ) × wher e m ( n ) = 1 3 (2 n − 1) , for even n , and m ( n ) = 1 3 (2 n − 2) , for o dd n . Our goa l is to provide some understanding o f the co upled Hanoi g raph CΓ n on n disk s . The vertices of this gra ph are the 3 2 n pairs of words  u T u B  of leng th n ov e r X (representing the top a nd the bottom configuration on n disks in a co upled configuratio n). Two vertices in CΓ n are co nnected by an edge labe led by s in { a, b, c } if the coupled configur ations repres ent ed by these vertices ca n be obtained from each o ther by applica tio n of the move s . The coupled Hanoi gr aph on 1 disk is depicted in Figur e 6. The coupled Hano i graph CΓ n is the Schreier graph, with resp ect to the gener a ting set { a, b, c } , of the action of H on the pairs of words of length n in X ∗ defined by s  u T u B  =  s ( u T ) s ( u B )  , for s in { a, b, c } . 8 ZORAN ˇ SUNI ´ C   • a    c 7 7 7 b   • c    b   • a 7 7 7   • b    a 7 7 7   • c    b 7 7 7   • a   c   • b   • a   • c c     • b    a 7 7 7   • a    c   • b 7 7 7   • c    b 7 7 7   • a    c 7 7 7   • b    a   • c   • b   • a   • a    c 7 7 7   • c    b   • a 7 7 7   • b    a 7 7 7   • c    b 7 7 7   • a    c   • b   • a   • c 7 7 7 b c   • c    b 7 7 7   • b    a 7 7 7   • a    c 7 7 7   • b    a 7 7 7   • b    a   • c 7 7 7   • a    c   • b 7 7 7   • c    b   • a 7 7 7   • a    c   • b 7 7 7   • a    c 7 7 7   • b    a 7 7 7   • c    b 7 7 7   • a    c 7 7 7   • b    a 7 7 7   • c    b 7 7 7   • c    b 7 7 7   • a    c 7 7 7   • b c   • a   • c   • b   • a   • c   • b   • a   • c   • b   • a   • c b   • a   • c   • b   • a Figure 5. Hanoi graph on o dd num b er of disks ( 0 1 ) ( 2 0 ) ( 1 2 )   • a c D D D D D D D D D D D b Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q   • c z z z z z z z z z z z b a D D D D D D D D D D D   • b m m m m m m m m m m m m m m m m m m a z z z z z z z z z z z c   • a         c 2 2 2 2 2 2 2 2 b ( 1 1 )   •   •   •   • c b   • a ( 1 0 ) ( 0 2 ) ( 2 1 ) ( 0 0 ) ( 2 2 ) Figure 6. CΓ 1 , the coupled Hanoi graph on 1 disk 3. Twin Towers Switch In this section we provide an upp er bo und on the n umber of mov es needed to solve the Twin T owers Switch Problem. In the lang ua ge of coupled Hanoi graphs the same result is expressed as follows. Theorem TTS ′ . The distanc e b etwe en the c ouple d c onfigur ations  000 . . . 0 222 . . . 2  and  222 . . . 2 000 . . . 0  TWIN TOWERS OF HANOI 9 in the c ouple d Hanoi gr aph CΓ n (on n disks) is no gr e ater than a ( n ) = ( 1 , n = 1 , 4 3 · 2 n − ( − 1) n 3 , n ≥ 2 . Pr o of. Let n = 2 . W e have, by us ing (1), ababa (00) = abab (1 0) = aba (12) = ab (02) = a (2 2) = 22 . Since ababa is a pa lindrome, it has o rder 2 (as a g r oup element) and, therefor e, ababa (22) = 00. Thus the distance betw een the initial a nd the final coupled co n- figurations is no greater than 5 (it can be shown that it is a ctually 5). Assume that n is even and n ≥ 4. Consider the s e q uence of a ( n ) = 4 3 · 2 n − 1 3 mov es ababa ( cacaba ba ) 1 3 (2 n − 1 − 2) . Notice the pattern   • b    c 7 7 7   • a    b 7 7 7   • c    a 7 7 7   • a   • b   • c   • a   • b   • c   • that rep eats along the b o ttom edge in Figure 4, indicating that the r e s ult o f the action of cacabab a and ( cba ) 2 on the leftmost vertex in the pattern is the s ame and it is equal to the rightmost vertex in the pattern (which is the leftmost vertex in the next occur rence of the pattern). Therefore, by (1) a nd Lemma 1, ababa ( cacaba ba ) 1 3 (2 n − 1 − 2) (000 . . . 0) = ababa ( cba ) 1 3 (2 n − 4) (000 . . . 0) = = ababaabc ( cba ) 1 3 (2 n − 1) (000 . . . 0) = abac (22 2 . . . 2) = = aba (122 . . . 2) = ab (0 2 2 . . . 2) = = a (222 . . . 2) = 2 22 . . . 2 . Since ababa ( cacab aba ) 1 3 (2 n − 1 − 2) is a palindrome it has order 2. Thus ababa ( cacaba ba ) 1 3 (2 n − 1 − 2) (222 . . . 2) = 000 . . . 0 and the distance b etw ee n the initial and the final co upled configuratio ns is no greater than a ( n ). Let n = 1. The distance b etw een the coupled co nfigurations  0 2  and  2 0  is 1 (see Figure 6). Assume that n is odd a nd n ≥ 3. Consider the sequence of a ( n ) = 4 3 · 2 n + 1 3 mov es aca ( cbcbcaca ) 1 3 (2 n − 1 − 1) . Notice the pattern   • a    c 7 7 7   • b    a 7 7 7   • c    b 7 7 7   • b   • a   • c   • b   • a   • c   • that rep eats along the b o ttom edge in Figure 5, indicating that the r e s ult o f the action of cbcbcaca and ( cab ) 2 on the leftmost vertex in the pattern is the s a me a nd it is equal to the rightmost vertex in the pattern (which is the leftmost vertex in the next occur rence of the pattern). Therefore, by (1) a nd Lemma 1, aca ( cbcbcaca ) 1 3 (2 n − 1 − 1) (000 . . . 0) = aca ( cab ) 1 3 (2 n − 2) (000 . . . 0) = = acabb ( cab ) 1 3 (2 n − 2) (000 . . . 0) = acab (222 . . . 2) = = aca (0 22 . . . 2) = ac (122 . . . 2 ) = = a (222 . . . 2) = 2 22 . . . 2 . 10 ZORAN ˇ SUNI ´ C Since aca ( cbcbcaca ) 1 3 (2 n − 1 − 1) is a palindrome it has order 2. Thus aca ( cbcbcaca ) 1 3 (2 n − 1 − 1) (222 . . . 2) = 000 . . . 0 and the distance b etw ee n the initial and the final co upled configuratio ns is no greater than a ( n ).  Remark. There are s e veral s o lutions of le ngth a ( n ), for n ≥ 2. F or instance, another solution, for odd n , is cac ( ababacac ) 1 3 (2 n − 1 − 1) and, for ev en n , is cbcbc ( acacbcb c ) 1 3 (2 n − 1 − 2) . W e rephras e Co njectur e TTS as follo ws. Conjecture TTS ′ . The distanc e b etwe en the c ouple d c onfigur ations  000 . . . 0 222 . . . 2  and  222 . . . 2 000 . . . 0  in the c ouple d Hanoi gr aph CΓ n (on n disks) is e qual to a ( n ) . 4. Smal l Disk Shift F or the considera tions tha t follow, the concept of parity will b e useful. Definition 2. F or a configura tion u = x 1 . . . x n in X ∗ and x ∈ X , let p x ( u ) b e the parity of the num b er of a pp ea rances of the letter x in u . F or a co upled configura tio n U =  u T u B  , let p x ( U ) be the parit y of the sum of the parities p x ( u T ) and p x ( u B ). Call any of the config urations 0 n , 1 n , 2 n a c orner c onfigur ation . Call a coupled configuratio n a c orner c ouple d c onfigur ation if at least one of the configur a tions in it is a co rner configuration. Application of a ij to an y non corner configuration changes the parities o f b oth i and j . Therefo re, application o f a ij to a no n cor ner coupled configuration do es not change any par ities. Theorem SDS ′ (Small Disk Shift) . The distanc e b etwe en the c ouple d c onfi gur a- tions  000 . . . 0 100 . . . 0  and  100 . . . 0 200 . . . 0  in the c ouple d Hanoi gr aph CΓ n (on n disks) is d ( n ) =      2 , n = 1 , 6 , n = 2 , 2 · 2 n , n ≥ 3 . Pr o of of Theo rem SDS ′ : upp er b oun d. Assume that n is even and n ≥ 4. Consider the sequence of 2 · 2 n mov es bab ( abc ) 1 3 (2 n − 4) a ( cba ) 1 3 (2 n − 1) c. TWIN TOWERS OF HANOI 11 A simple and convincing wa y to verify that this sequence of mov es accomplishes the goal would b e to tra c e the action in Fig ure 4 . A mo re for mal a pproach, using (1) and Lemma 1, gives bab ( abc ) 1 3 (2 n − 4) a ( cba ) 1 3 (2 n − 1) c (000 . . . 0) = bab ( abc ) 1 3 (2 n − 4) a ( cba ) 1 3 (2 n − 1) (000 . . . 0) = = bab ( abc ) 1 3 (2 n − 4) a (222 . . . 2) = bab ( abc ) 1 3 (2 n − 4) (222 . . . 2) = = babcba ( abc ) 1 3 (2 n − 1) (222 . . . 2) = babcba (0 00 . . . 0) = = ba b cb (100 . . . 0) = babc (120 . . . 0) = = bab (220 . . . 0) = ba (0 20 . . . 0) = = b (1 20 . . . 0) = 100 . . . 0 , and bab ( abc ) 1 3 (2 n − 4) a ( cba ) 1 3 (2 n − 1) c (100 . . . 0) = bab ( abc ) 1 3 (2 n − 4) a ( cba ) 1 3 (2 n − 1) (200 . . . 0) = = bab ( abc ) 1 3 (2 n − 4) acba (200 . . . 0) = bab ( abc ) 1 3 (2 n − 4) acb (210 . . . 0 ) = = ba b ( abc ) 1 3 (2 n − 4) ac (010 . . . 0 ) = bab ( abc ) 1 3 (2 n − 4) a (020 . . . 0) = = bab ( abc ) 1 3 (2 n − 4) (120 . . . 0) = bab (120 . . . 0 ) = = ba (100 . . . 0) = b (00 0 . . . 0) = = 200 . . . 0 , where, in the transition betw een the first a nd second row, we used the fact that 1 3 (2 n − 1) is o dd. Assume that n is o dd and n ≥ 3. Consider the sequence of 2 · 2 n mov es bab ( abc ) 1 3 (2 n − 5) bcba ( cba ) 1 3 (2 n − 2) . A simple and convincing w ay to v erify that this sequence o f mov es ac c omplishes the go al would b e to trace the action in Figure 5. A more formal approach, using (1) and Lemma 1, gives bab ( abc ) 1 3 (2 n − 5) bcba ( cba ) 1 3 (2 n − 2) (000 . . . 0) = bab ( abc ) 1 3 (2 n − 5) bcb (111 . . . 1) = = bab ( abc ) 1 3 (2 n − 5) aabc (111 . . . 1) = baba ( bca ) 1 3 (2 n − 5) abc (111 . . . 1 ) = = baba ( bca ) 1 3 (2 n − 5) ab (211 . . . 1 ) = baba ( bca ) 1 3 (2 n − 5) a (011 . . . 1) = = ba b a ( bca ) 1 3 (2 n − 5) (111 . . . 1) = babcba a bca ( bca ) 1 3 (2 n − 5) (111 . . . 1) = = babcbaa ( bca ) 1 3 (2 n − 2) (111 . . . 1) = babcba (0 00 . . . 0) = = babcb (100 . . . 0) = babc (1 20 . . . 0) = = ba b (220 . . . 0 ) = b a (020 . . . 0) = = b (120 . . . 0) = 100 . . . 0 , and bab ( abc ) 1 3 (2 n − 5) bcba ( cba ) 1 3 (2 n − 2) (100 . . . 0) = bab ( abc ) 1 3 (2 n − 5) b (100 . . . 0 ) = = bab ( abc ) 1 3 (2 n − 5) (120 . . . 0) = bab (120 . . . 0 ) = = ba (1 00 . . . 0) = b (000 . . . 0 ) = 200 0 . . . 0 . When n = 1, a solution of length 2 is giv en by the sequence of mov es ba and, for n = 2, a solution of length 6 is given by the s e quence of mo ves bcacba .  Remark. Note that the ab ov e sequences o f mov es of length 2 · 2 n are not unique. F or instance, for even n , n ≥ 4, one co uld use caba ( bac ) 1 3 (2 n − 4) b ( cab ) 1 3 (2 n − 1) . 12 ZORAN ˇ SUNI ´ C Pr o of of Theo rem SDS ′ : lower b ound. Since the 0 -parities for the initial and final coupled configurations are p 0  000 . . . 0 100 . . . 0  = 1 and p 0  100 . . . 0 200 . . . 0  = 0 , somewhere on the w ay fro m the initial to the final co upled configur ation the 0 -parity changes. This parity cannot b e changed at the cor ner coupled configurations  0 n v  and  v 0 n  , where v is no t a corner config uration. Since the 0 -parity must b e changed, any sequence of mov es that starts a t the initial coupled configuration  000 ... 0 100 ... 0  and accomplishes this change in volv es a corner a -lo op of a corner b -lo op application in either the top or in the b ottom configuratio n. The 4 pos sibilities are given as cases T op a , T op b , Bot a and Bot b (standing for top configuration inv olved in a corner a - lo op, top configura tion inv olved in a corner b - lo op, etc.) in T able 1, where, in each case initial 0-parity change / / even n o dd n T op a :  00 ... 0 10 ... 0  / /  2 n ∗  a / /  2 n ∗  / /  10 ... 0 ∗  , 2 · 2 n − 2 2 · 2 n − 1 T op b :  00 ... 0 10 ... 0  / /  1 n ∗  b / /  1 n ∗  / /  10 ... 0 ∗  , 2 · 2 n − 1 2 · 2 n − 2 Bot a :  00 ... 0 10 ... 0  / /  ∗ 2 n  a / /  ∗ 2 n  / /  ∗ 20 ... 0  , 2 · 2 n − 2 2 · 2 n − 2 Bot b :  00 ... 0 10 ... 0  / /  ∗ 1 n  b / /  ∗ 1 n  / /  ∗ 20 ... 0  , 2 · 2 n − 2 2 · 2 n − 2 T able 1. Changing the 0-par ity case, ∗ denotes s ome configuration different from the one with whic h it is coupled. The last tw o columns provide the num b er o f steps in the unique shortest path of the given form, for even and o dd num ber o f disk s. Note that the a b ov e co nsiderations alr eady show that d ( n ) ≥ 2 · 2 n − 2 and that the largest disk has to be moved in at least one coupled set of disks. F urther, any element g in H fo r which g  000 ... 0 100 ... 0  =  100 ... 0 200 ... 0  m ust act o n the first le tter as the p ermutation (01 2), which is an ev en permutation. Therefore, the length of g m ust be ev en. T o complete the pro of, all we ne e d to show is that none of the shortest paths (sequences of mov es) of length 2 · 2 n − 2 implicitly mentioned in T able 1 solves the Small Disk Shift Pro ble m. F or the unique s ho rtest path g o f length 2 · 2 n − 2 in Case T op a , e ven n , such that for the top configuration we hav e g (000 . . . 0 ) = 100 . . . 0, tracing the a ction in Figure 4 for the bottom co nfiguration, w e obtain ( bca ) 1 3 (2 n − 1) ( cba ) 1 3 (2 n − 1) (100 . . . 0) = 2 0 1 . . . 1 6 = 200 . . . 0 . F or the unique shortest path g of length 2 · 2 n − 2 in Case Bot a , ev en n , suc h tha t for the b ottom configuration we hav e g (10 0 . . . 0) = 20 0 . . . 0, tracing the a ction in Figure 4 for the top c onfiguration, we o btain ( cbc )( abc ) 1 3 (2 n − 4) ( acb ) 1 3 (2 n − 1) (000 . . . 0) = 101 . . . 1 6 = 10 0 . . . 0 . TWIN TOWERS OF HANOI 13 F or the unique shor test path g o f le ngth 2 · 2 n − 2 in Ca s e Bot b , even n , such that for the b ottom configuration we hav e g (10 0 . . . 0) = 20 0 . . . 0, tracing the a ction in Figure 4 for the top c onfiguration, we o btain ac ( bac ) 1 3 (2 n − 4) ( bca ) 1 3 (2 n − 1) c (000 . . . 0) = 1 02 . . . 2 6 = 100 . . . 0 . F or the unique shor test path g of le ng th 2 · 2 n − 2 in Case T op b , o dd n , such that for the top configuration we hav e g (000 . . . 0 ) = 100 . . . 0, tracing the a ction in Figure 5 for the bottom co nfiguration, w e obtain ( bca ) 1 3 (2 n − 2) ba ( cba ) 1 3 (2 n − 2) (100 . . . 0) = 222 . . . 2 6 = 20 0 . . . 0 . F or the unique shortest path g o f length 2 · 2 n − 2 in Case Bot a , o dd n , such that for the b ottom configuration we hav e g (10 0 . . . 0) = 20 0 . . . 0, tracing the a ction in Figure 5 for the top c onfiguration, we o btain ( acb ) 1 3 (2 n − 2) a ( bca ) 1 3 (2 n − 2) c (000 . . . 0) = 1 11 . . . 1 6 = 100 . . . 0 . Finally , fo r the unique shortest path g o f length 2 · 2 n − 2 in Ca se Bot b , o dd n , such that for the b o tto m configur ation w e ha ve g (1 00 . . . 0) = 20 0 . . . 0, tracing the action in Figure 5 for the top c o nfiguration, we obta in c ( bca ) 1 3 (2 n − 2) b ( acb ) 1 3 (2 n − 2) (000 . . . 0) = 122 . . . 2 6 = 10 0 . . . 0 .  5. General P roblem In this section w e describ e the compatible coupled co nfigurations (r ecov ering the result o f D’Angeli a nd Do nno fro m [DD07]) and then provide an upp er b ound on the distance betw een any compatible co upled configurations. In order to ac c omplish the goa ls of this section, w e need a bit more infor mation on the Hanoi T ow ers group H . In particular, w e rely o n the s e lf- s imilarity of the action of H on the tr e e X ∗ . More o n s elf-similar actions in ge ne r al can b e found in [Nek05]. F or our purpos es the following obser v ations suffice. The ac tion of a , b and c on X ∗ given b y (1 ) can b e rewritten in a recursive form as follows. F or any word u over X , a (0 u ) = 1 u, b (0 u ) = 2 u, c (0 u ) = 0 c ( u ) , a (1 u ) = 0 u, b (1 u ) = 1 b ( u ) , c (1 u ) = 2 u, (2) a (2 u ) = 2 a ( u ) , b (2 u ) = 0 u, c (2 u ) = 1 u. This implies that, for any sequence g of mov es, there ex is t a p ermutation π g of X and three sequences of mov es g 0 , g 1 and g 2 such that, for every w ord u o ver X , (3) g (0 u ) = π g (0) g 0 ( u ) , g (1 u ) = π g (1) g 1 ( u ) , g (2 u ) = π g (2) g 2 ( u ) . The p e rmutation π ( g ) is ca lled the r o ot p ermutation and it indicates the action of g on the first level of the tree (just b elow the ro o t), while g 0 , g 1 and g 2 are called the se ctions of g a nd indicate the action of g b elow the vertices o n the first level. When (3) holds, w e write g = π g ( g 0 , g 1 , g 2 ) and call the expression o n the rig ht a de c omp osition o f g . Note tha t (3 ) may be correct for many different sequences of mov es g 0 (or g 1 or g 2 ), but all these se q uences represent the same elemen t of the group H . Decompo s itions o f the generators a , b and c are giv en b y (4) a = (01) (1 , 1 , a ) , b = (02) (1 , b, 1) , c = (12) ( c, 1 , 1) , 14 ZORAN ˇ SUNI ´ C where 1 denotes the e mpty sequence of mov es (the trivia l automorphism of the tree). Two decomp ositions may b e multiplied b y using the formula (see [Nek0 5] or [G ˇ S07]) (5) g h = π g ( g 0 , g 1 , g 2 ) π h ( h 0 , h 1 , h 2 ) = π g π h ( g h (0) h 0 , g h (1) h 1 , g h (2) h 2 ) . The decompositions of the generators a , b and c given in (4) and the decomp ositio n pro duct formula (5) are s ufficient to ca lculate a deco mpo sition fo r any sequence of mov es . W e refer to such calculations a s decompo sition calculations. Theorem 1 (D’Angeli and Donno [DD07]) . Two c ouple d c onfigur ations U =  u T u B  and V =  v T v B  on n disks ar e c omp atible if and only if the length of the longest c ommon pr efi x of u T and u B is the same as the length of the longest c ommon pr efix of v T and v B . Remark. Note that Theorem 1 implies that the n + 1 sets CΓ n, 0 , CΓ n, 1 , . . . , CΓ n,n , where CΓ n,i consists o f the coupled config ur ations  u T u B  such that the length of the longest co mmon pre fix of u T and u B is i , ar e the connected comp onents of the coupled Hanoi g raph CΓ n . The largest of these sets is CΓ n, 0 . It consists of 6 · 9 n − 1 vertices, whic h are the basic coupled co nfigurations (defined in the intro duction). More generally , the set CΓ n,i has 3 i · 6 · 9 n − 1 − i vertices, for i = 0 , . . . , n − 1, and CΓ n,n has 3 n vertices (moreover, CΓ n,n is canonically isomorphic to Γ n through the isomorphism u ↔  u u  ). Since every tree a utomorphism preserves prefixes, the connected comp onents of the coupled Hanoi graph m ust b e subsets of the sets CΓ n,i . Thus, o nly the other direction (s howing that ea ch of the sets CΓ n,i is connected) is interesting and needs to be proved. Consider the s ubg roup A = h cba, acb , bac i ≤ H (in tro duced in [GN ˇ S06] and called Ap ollonian group, beca us e its limit space is the Ap ollonian g asket). It is known that this subgro up ha s index 4 in H and H / A = C 2 × C 2 (where C 2 is cyclic of order 2). A seq uence of mo ves g b elongs to A if and only if the parities of the nu mber of o ccurrences o f the mov es a , b and c in g a re a ll odd or all even. The elements 1 , a , b, c for m a tra nsversal for A in H . The Schreier graph of the subgr oup A in H is g iven in Figure 7. The vertices are denoted by the coset representativ es (for instance, the v ertex b is the coset bA ).   • a a c b > > > > > > > >   • b a   • 1 c b           • c Figure 7. The Schreier g r aph of A in H Lemma 3 . The Ap ol lonian sub gr oup acts tr ansitively on every level of t he tr e e X ∗ . Pr o of. The claim follows from the fact that H acts transitively on every level o f the tree and tha t, for every generator s in { a, b, c } , there is a lo o p lab eled by s in the Hanoi graph Γ n . Indeed, if g ( u ) = v , for some se q uence of mo ves g , and g is in, sa y , the coset aA , then g ′ g ( u ) = v a nd g ′ g is in A , w he r e g ′ = h − 1 ah and h is any sequence of mov es TWIN TOWERS OF HANOI 15 from v to the vertex 2 n (note that g ′ ∈ aA and g ′ ( v ) = h − 1 ah ( v ) = h − 1 a (2 n ) = h − 1 (2 n ) = v ).  Remark. A small mo dification o f the ab ov e a rgument (using the corner loo ps to mo dify the pa rity o f the num b er o f o c currences o f any ge ner ator) shows that the commutator subgroup H ′ also acts transitively on every lev el of the tree . The fact that H ′ acts trans itively was pro ved in a different wa y by D’Angeli and Donno and used in their pro of of Theorem 1 . W e provide a different proo f o f Theorem 1, based on the transitivity of the action of A , enabling us to pr ovide go o d es timates in the General Problem for basic coupled configurations. Lemma 4. The set CΓ n, 0 of b asic c ouple d c onfigura tions on n disks is c onne cte d. Pr o of. Let  u T u B  and  v T v B  be coupled configurations in CΓ n, 0 . Since H acts transitively on every level of the tr e e , there exists a s equence of mov es h such that h ( u T ) = v T . Let h ( u B ) = v ′ B . Without loss of gene r ality , assume that the top configuratio n v T starts by 2, while the b ottom co nfiguration v B starts b y 0. The co nfiguration v ′ B may start by either 0 or 1. If it starts by 1, a s ingle application of the sequence of 3 mo ves cab = (01) ( a, cb , 1) , do es not affect v T (note the triv ia l se ction a t 2), and c hanges the first letter in the bo ttom configuration to 0. Thus, we may a ssume that both v ′ B and v B start b y 0. W e ar e in terested in sequences of mov es g that do not affect an y configura tio ns that star t b y 2 (and th us do not affect v T ) and k eep the fir st letter in the b ottom configuratio n equal to 0. In other w o rds, we ar e in terested in sequences of mov es that decompos e as g = ( g 0 , ∗ , 1 ) , where ∗ represents the section at 1, in which we are no t in terested. Three s uch sequences ar e (this can b e verified by direct decompo sition ca lcula- tions) cabcab = ( cba, ∗ , 1) bacacaba = ( acb , ∗ , 1 ) , bcbcacac = ( bac , ∗ , 1) . Since h cba, acb , bac i = A , these thr e e decompositio ns imply that, for every sequence of mo ves g 0 in A , ther e exists a sequence of mo ves g in H , and in fact in A , suc h that g = ( g 0 , ∗ , 1 ) . Let v ′ B = 0 v ′ and v B = 0 v . Since A a c ts transitiv ely on each level of the tree, there exists g 0 in A such that g 0 ( v ′ ) = v . Therefor e , there exists g in A such that g ( v ′ B ) = v B and g ( v T ) = v T , completing the pro of that CΓ n, 0 is connected.  The rest of the pro of of Theo rem 1 follows, essent ially , the sa me steps as the original pro of of D’Angeli and Donno and, b eing short, is included for completeness . Indeed, once it is known that the larges t sets CΓ n, 0 are connected, it is s ufficient to observe that H is a self-replicating group. 16 ZORAN ˇ SUNI ´ C Lemma 5. Hanoi T owers gr oup H is a self-r eplic ating gr oup of tr e e automor- phisms, i.e. , for every wor d u over X and every se quenc e of moves g in H , t her e exists a se quenc e of moves h in H s uch t hat, for every wor d w over X , h ( uw ) = ug ( w ) . Pr o of. Let w b e any word over X . Since a (2 w ) = 2 a ( w ) , cbc = 2 b ( w ) , bcb (2 w ) = 2 c ( w ) , it is clear that, for every sequence of mov es g , there exists a sequence of mov es h such that h (2 w ) = 2 g ( w ). By symmetry , for every letter x in X and every sequence of mo ves g , there e xists a sequence of mo ves h such that h ( xw ) = xg ( w ) and the claim easily extends to w ords ov er X (and not just letters).  Pr o of of The or em 1. Let u and u ′ be words of length i and  uw T uw B  and  u ′ w ′ T u ′ w ′ B  be t wo coupled configuratio ns in CΓ n,i . Since H acts transitively on the levels of the tree, there exists a sequence of moves h ′ in H suc h that h ′  uw T uw B  =  u ′ w ′′ T u ′ w ′′ B  , for some w ′′ T and w ′′ B (in fact, o ne ma y easily find suc h h ′ for which w ′′ T = w T and w ′′ B = w B , but this do es not matter). Since CΓ n − i, 0 is connected, there exists a sequence o f mov es g suc h that g  w ′′ T w ′′ B  =  w ′ T w ′ B  . By the self-replicating pro pe r ty of H , there exists a sequence of mo ves h in H such that hh ′  uw T uw B  = h  u ′ w ′′ T u ′ w ′′ B  =  u ′ g ( w ′′ T ) u ′ g ( w ′′ B )  =  u ′ w ′ T u ′ w ′ B  .  Theorem GP ′ (General Problem for bas ic co nfigurations) . The diameter D ( n ) of the lar gest c omp onent CΓ n, 0 of the c ouple d Hanoi gr aph CΓ n (on n disks) satisfies, for n ≥ 3 , the ine qualities 2 × 2 n ≤ D ( n ) ≤ 3 . 66 × 2 n . Pr o of. W e follow the pro of of Lemma 4, but k eep trac k o f the lengths of the se- quences of moves in volv ed and, when w e have a c hoice (a nd know how to make it), try to use short sequences. Let U =  u T u B  and V =  v T v B  be coupled configurations in the larg e st comp onent CΓ n, 0 of the coupled Hanoi gr aph. Without loss of genera lity , assume that the top configuratio n v T starts by 2, while the b ottom configur ation v B starts by 0. There ex ists a sequence of mov es h of length a t most 2 n + 2 such that h ( u T ) = v T and h ( u B ) = v ′ B , for some co nfig uration v ′ B that starts b y 0. Indeed, at mos t 2 n − 1 steps are needed to c hange the top configuratio n from u T to v T , and then at most three mo re steps (r e c all that c ab = (01)( a, cb , 1)) a re needed to make sure that the bo ttom configuration starts b y 0. Let v ′ B = 0 v ′ and v B = 0 v . W e claim that there exists a sequence of mov es g 0 in A such that g 0 ( v ′ ) = v and the num b er of mov es in the sequence g 0 is no greater than 2 n − 1. Indeed, if the shortest sequence of moves g s betw een v ′ and v happ ens to be in A we may set g 0 = g s (note that v ′ and v ar e vertices in the Hanoi graph Γ n − 1 of diameter 2 n − 1 − 1 ). If g s happ ens to b e, say , in the coset aA , we ma y set g 0 = g (2) ag (1) , where g (1) is the shortest sequence of moves from v ′ to 2 n − 1 and g (2) is the shortest seq ue nc e of mov es fro m 2 n − 1 to v . The length of the sequence g 0 = g (2) ag (1) is no g reater than 2 (2 n − 1 − 1) + 1 = 2 n − 1. Since TWIN TOWERS OF HANOI 17 the sequence of mo ves g − 1 s g (2) g (1) represents a closed path in the gra ph Γ n − 1 that do es not go throug h any of the corner lo ops and s ince all cycles in Γ n − 1 other than the three co rner lo ops ar e lab eled by elements in A , the sequence g − 1 s g (2) g (1) is in A . Therefore g 0 A = g (2) ag (1) A = ag (2) g (1) A = ag s A = aaA = A, which is what we needed. Direct decomp osition calculatio ns give bab ( cba ) 2 bab = ( acaba, ∗ , 1) abc ( acb ) 2 cba = ( bab cb, ∗ , 1) , cb ( cba ) 2 bc = ( cbca c, ∗ , 1 ) , and therefore, for any k ≥ 0, bab ( cba ) 2 k +2 bab = ( a ( cab ) k +1 a, ∗ , 1 ) , abc ( acb ) 2 k +2 cba = ( b ( abc ) k +1 b, ∗ , 1 ) , (6) cb ( cba ) 2 k +2 bc = ( c ( bca ) k +1 c, ∗ , 1 ) . This calculatio n justifies the en tries in the top thr ee rows of T able 2. In this case f f 0 ℓ ( f ) ℓ ( f 0 ) ratio a − ← − a bab ( cba ) 2 k +2 bab a ( cab ) k +1 a 6 k + 12 3 k + 5 2 . 4 b − ← − b abc ( acb ) 2 k +2 cba b ( abc ) k +1 b 6 k + 12 3 k + 5 2 . 4 c − ← − c cb ( cba ) 2 k +2 bc c ( bca ) k +1 c 6 k + 10 3 k + 5 2 cabcab cba 6 3 2 c − ← − a cabcb ( cba ) 2 k +2 bab cb ( cab ) k +1 a 6 k + 14 3 k + 6 2 . 34 bacacaba acb 8 3 2 . 67 a − ← − b bacacac ( acb ) 2 k +2 cba ac ( abc ) k +1 b 6 k + 16 3 k + 6 2 . 67 bcbcacac bac 8 3 2 . 67 b − ← − c bcbcacbcb ( cba ) 2 k +1 bc ba ( bca ) k +1 c 6 k + 14 3 k + 6 2 . 34 bcbcacbcab baba 10 4 2 . 5 b − ← − a bcbcacbcb ( cba ) 2 k +2 bab bab ( ca b ) k +1 a 6 k + 18 3 k + 7 2 . 58 cabacaba cbcb 8 4 2 c − ← − b cabacac ( acb ) 2 k +2 cba cbc ( abc ) k +1 b 6 k + 16 3 k + 7 2 . 29 babcbabc acac 8 4 2 a − ← − c bab ( cba ) 2 k +3 bc aca ( bca ) k +1 c 6 k + 14 3 k + 7 2 T able 2. Sequences of mov es fixing v T and moving v ′ B table, f is a sequence of mo ves a nd f 0 is the cor resp onding section a t 0. The first letter of any word is fixed by f and the section a t 2 is trivial. In other words, f decomp oses as f = ( f 0 , ∗ , 1 ) . 18 ZORAN ˇ SUNI ´ C The lengths of the sequences f and f 0 , as written, are ℓ ( f ) and ℓ ( f 0 ), and the ratio in the la st column is the r atio ℓ ( f ) / ℓ ( f 0 ) (in the rows that dep end on k , the r atio is the maxim um p ossible ratio, taken for k ≥ 0 and ro unded up). The entries in the re maining rows in T able 2 are easy to verify . F or insta nce, for case c − ← − a , b y direct decomp osition calculation, (7) cabcab = ( cba, ∗ , 1) and the en try in the next row is obtained simply b y multiplying the equalit y (7) and the first equalit y in (6) cabcb ( cba ) 2 k +2 bab = ( cabc )( abb a ) b ( cba ) 2 k +2 bab ) = ( cabcab )( bab ( cba ) 2 k +2 bab ) = = ( cba, ∗ , 1)( a ( cab ) k +1 a, ∗ , 1 ) = ( cbaa ( cab ) k +1 a, ∗ , 1 ) = = ( cb ( cab ) k +1 a, ∗ , 1 ) . All other ca ses are equally easy to verify (by verifying directly the basic ca se, and then m ultiplying it by a co rresp onding e q uality from (6) to obtain the ca ses depe nding on k ). Consider g 0 as defined ab ov e . There is no occur rence of aa , bb o r cc in this sequence (since w e alwa ys chose the shortest paths a s w e built g 0 ) and it is in A . The sequence g 0 is a pr o duct of factors each of which has the form of o ne of the ent ries in co lumn f 0 in T able 2 or their inv erses. Mo reov er, the decompo sition is s uch that the length of g 0 is the sum of the lengths of the factor s. Indeed, the entries in co lumn f 0 and their in verses are all possible sequences of moves in A without o ccurrenc e o f aa , bb o r cc for which no prop er s uffix is in A . Such sequences corresp ond pr ecisely to pa ths without bac ktracking in the Schreier graph in Fig ure 7 that star t at 1, end a t 1 a nd do not visit the v e rtex 1 except at the very b eginning and at the very end. There are 18 such t yp es of paths, three choices for the first s tep ( a , b or c ) to leav e v ertex 1, three c hoices for the la st step ( a , b or c ) to go back to vertex 1, and tw o choices for the o rientation (order) used to lo op around the three vertices (co s ets) a , b a nd c b efore the return to 1 (positive or nega tive orient ation). The column f 0 in the table only lists the 9 p o ssible paths with negative orientation (and classifies the 9 cas e s by the first and last mov e), since the other 9 are just in verses of the en tries in the table. F o r instance, the notation c − ← − a indicates pa ths (sequences of mo ves) that start by the mov e a and end b y the mov e c . Once g 0 is appr opriately facto red, T able 2 can b e used to define g of length no greater than 2 . 66 ℓ ( g 0 ) ≤ 2 . 66(2 n − 1) such that g  v T v ′ B  =  v T v B  . Thu s, we may a rrive from the initial coupled co nfiguration  u T u B  to the final coupled configuration  v T v B  in no mo r e than (2 n + 2 ) + 2 . 66(2 n − 1 ) ≤ 3 . 66 × 2 n mov es .  It is evident that go o d understanding of the structure of CΓ n, 0 , for a ll n , provides go o d understa nding of CΓ n,i , for all n and i . F or ins tance, the understanding of the graphs CΓ 1 , 0 (6 vertices, dia meter 2) a nd CΓ 2 , 0 (54 vertices, diameter 6) enabled the author to determine the exact v alues o f the diameter of the t wo smallest nontrivial comp onents CΓ n,n − 1 and CΓ n,n − 2 , fo r any num b er of disk s . F or insta nce, the diameter of CΓ n,n − 1 is, for n ≥ 1, equal to 7 6 2 n − 3 + ( − 1) n 6 . TWIN TOWERS OF HANOI 19 The details will appear in a future work. 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