On a question of Leiss regarding the Hanoi Tower problem

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1 Theoretical Computer Sciece 369 (006) O a questio of Leiss regardig the Haoi Tower problem D. Azriel a,, D. Bered a,b a Departmet of Mathematics, Be-Gurio Uiversity of the Negev, Beer Sheva 84105, Israel b Departmet of Computer Sciece, Be-Gurio Uiversity of the Negev, Beer Sheva 84105, Israel Received 16 December 004; received i revised form 18 September 006; accepted 4 September 006 Commuicated by Aviezri Fraekel Abstract The Tower of Haoi problem is geeralized i such a way that the pegs are located at the vertices of a directed graph G, ad moves of disks may be made oly alog edges of G. Leiss obtaied a complete characterizatio of graphs i which arbitrarily may disks ca be moved from the source vertex S to the destiatio vertex D. Here we cosider graphs which do ot satisfy this characterizatio; hece, there is a boud o the umber of disks which ca be hadled. Deote by g the maximal such umber as G varies over all such graphs with vertices ad S, D vary over the vertices. Aswerig a questio of Leiss [Fiite Haoi problems: How may discs ca be hadled? Cogr. Numer. 44 (1984) 1 9], we prove that g grows sub-expoetially fast. Moreover, there exists a costat C such that g C 1/ log for each. O the other had, for each ε>0 there exists a costat C ε > 0 such that g C ε (1/ ε)log for each. 006 Published by Elsevier B.V. Keywords: Haoi Tower problem; Graphs; Digraph; Recurrece equatio 1. Itroductio ad the mai result Cosider the followig geeralizatio of the Tower of Haoi problem: A Haoi graph is a simple, directed, fiite graph G = (V, E) with two distiguished vertices, deoted by S (source) ad D (destiatio), S = D, such that for each vertex v V \{S,D} there is a path from S to v ad a path from v to D. At each vertex of G there is a peg, which we shall idetify with the vertex itself. The source iitially cotais m disks, o two of which are of equal size, such that smaller disks rest o top of larger oes. The task is to move all disks from S to D. To this ed we may use the other vertices of the graph as auxiliary vertices. The trasfer is subject to the followig rules: (1) Each move cosists of takig the topmost disk from a peg ad placig it o top of all disks residig o some other peg. (Thus, pegs behave as stacks.) () A disk may be moved from a peg v to aother peg w oly if there is a edge from v to w, i.e., (v, w) E. (3) At o time may a disk be placed upo a smaller oe. The Haoi Tower problem HAN(G, m), for a Haoi graph G ad m 0, is to trasfer m disks from S to D, subject to the above rules. Correspodig author. Tel.: addresses: azrield@math.bgu.ac.il (D. Azriel), bered@math.bgu.ac.il (D. Bered) /$ - see frot matter 006 Published by Elsevier B.V. doi: /j.tcs

2 378 D. Azriel, D. Bered / Theoretical Computer Sciece 369 (006) The Tower of Haoi problem was composed by Lucas [10] over a 100 years ago. The problem has a simple recursive solutio ad is used i may texts as a example for recursive programmig. A equally simple iterative solutio is preseted i [3]. A thorough discussio o space ad time complexity of various algorithms for solvig the Tower of Haoi problem ca be foud i [4]. Various properties of istaces of the problem were studied. I [7], aalogies betwee Pascal s triagle, the Sierpiński gasket ad the Tower of Haoi are discussed. Properties of the solutios are also discussed, as i [1], where it is show that, with a direct approach codig, a strig which represets a optimal solutio is square free. A commo geeralizatio of the problem is to allow more tha 3 pegs ad put restrictios o the legitimate moves of the disks. This is discussed i [6,8,9,14,15]. The case of 4 pegs istead of the classical 3 ad the correspodece with graphs is discussed thoroughly i [16]. The special cases of 3 pegs with restrictios o the allowed moves are thoroughly discussed i [13]. The correspodece of the solutios i those cases with sequeces ad morphisms is discussed i []. Aother directio was cocered with various geeralizatios, such as havig ay iitial ad fial cofiguratios [5], ad assigig colors to disks [1]. Problem HAN(G, m) is solvable if the task may be accomplished. A Haoi graph G is solvable if HAN(G, m) is such for all m 1. The requiremet that for each vertex v V there is a path from S to v ad a path from v to D is desiged to get rid of iessetial vertices. I fact, if there is o path from S to v or from v to D, the o solutio of HAN(G, m) may use the peg v. Leiss [8] obtaied the followig characterizatio of solvable graphs. (Note that, i his formulatio of the problem, V may cotai iessetial vertices, ad hece the formulatio of the theorem is differet accordigly.) Theorem A (Leiss [8]). Let G = (V, E) be a Haoi graph, ad let G + = (V, E + ) be its trasitive closure: E + = {(v, w) : v = w, there is a path from v to w}. The G is solvable if ad oly if there exist three distict vertices v 1,v,v 3 V such that (v i,v j ) E + for i, j such that 1 i, j 3, i = j. A usolvable graph G has a maximal m for which HAN(G, m) is solvable. Deote that m by M(G). It is easy to see that M(G) may assume arbitrarily large values whe G varies over all usolvable graphs. However, there is clearly a maximal such m for graphs with vertices. Deote this maximum by g, i.e., g = max V (G) = M(G). Example 1.1. Oe may check that g = 1, g 3 =, g 4 = 4, but it takes some time already to see that g 5 = 5. Leiss [8] was iterested i the asymptotic behavior of g. The proof of Theorem A may be show to yield a upper boud, which however exceeds. O the other had, he showed i [9] that g grows super-polyomially. More precisely, he showed that g Ω(log ), ad posed. Questio 1.. Does g grow expoetially fast? I this paper we prove. Theorem 1.3. There exists a costat C such that g C 1/lg for each. O the other had, for each ε>0 there exists a costat C ε > 0 such that g C ε (1/ ε) lg for each. Here we used the otatio lg = log. I Sectio we fid a family Γ of graphs which are (amog) the best withi the family of usolvable graphs; more accurately, g = max G Γ: V = M(G). I Sectios 3 ad 4 we prove the first part of Theorem 1.3 ad i Sectio 5 the secod.. The best usolvable graphs Defiitio.1. A usolvable Haoi graph G = (V, E) is a ladder graph if E(G) is maximal with respect to G beig usolvable (i.e., by addig ay edge to G, oe makes it solvable).

3 D. Azriel, D. Bered / Theoretical Computer Sciece 369 (006) S D Fig. 1. A (graph whose trasitive closure is a) ladder graph. Obviously, ay usolvable graph is a dag ad ca be tured ito a ladder graph by addig edges if ecessary. The umber of disks that ca be moved from S to D does ot decrease by this additio, ad hece for each there exists a ladder graph o vertices that realizes g, so that we may restrict our attetio to ladder graphs. (For the lower boud o g we shall actually use other graphs.) It is kow that the trasitive closure of a dag is a dag as well, ad hece, if G is a ladder graph, the the set of edges, E, coicides with its trasitive closure, E +. Deote the set of strogly coected compoets of a ladder graph G = (V, E) by V. Defie a orderig o V by A B if either A = B or there are vertices u A ad v B such that (u, v) E. It is kow that i a dag the relatio is a partial order o the set of strogly coected compoets. For a ladder graph, this is a total order. Write V ={A 1,...,A r }, where A 1 A A r. Corollary.. Let G = (V, E) be a ladder graph ad A 1,...,A r its strogly coected compoets. The each strogly coected compoet is of size 1 or ad E ={(u, v) : u = v, u A i,v A j,i j}. From here o, we deote V ={v 1,...,v }, ad agree that S = v 1, D = v, ad if A i ={v k } or A i ={v k 1,v k } the A i+1 ={v k+1 } or A i+1 ={v k+1,v k+ }. Fig. 1 depicts a typical ladder graph. To avoid over-cogestio, we have draw oly the edges betwee vertices that belog to the same strogly coected compoet ad betwee vertices belogig to cosecutive A i s, but there are actually edges from each vertex to each vertex to the right of it. (That is, we refer to the trasitive closure of the graph from Fig. 1.) Obviously, the decompositio ito equivalece classes has the property that there are o two cosecutive A i s of size 1. Thus, the umber l of ladder graphs o vertices satisfies the recurrece l = l + l 3, so that l grows approximately as C It is a amusig fact that the umber l of ladder-like graphs, where we drop the restrictio that o two cosecutive A i s are of size 1, satisfies the recurrece l = l 1 + l, which yields the Fiboacci sequece. Also the umber l of ladder graphs with exactly strogly coected compoets satisfies the latter recurrece relatio. However, we shall make o use of these facts. 3. The sequece (g ) Lemma 3.1. The sequece (g ) = is strictly icreasig. Proof. Let G = (V, E) be a ladder graph with vertices with M(G) = g. We shall show that there exists a ladder graph with + 1 vertices such that we ca trasfer (at least) g + 1 disks from S to D. Cosider the partitio A 1,...,A r of V ito strogly coected compoets. Add a ew vertex v to G with icomig edges from, say, the (oe or two) vertices i A 1, ad outgoig edges to all vertices i V \ A 1. Obviously, the resultig graph is a ladder graph as well (or ca be trasformed ito oe by addig the edge (v,v) if A 1 ={v} or the edge (v, v ) if A ={v}). Start with g + 1 disks at S. Trasfer the topmost disk from S to v. Next, trasfer the other g disks from S to D without usig v (which is possible sice M(G) = g ). Fially, move the smallest disk from v to D. Altogether, we have trasferred g + 1 disks. The followig lemma will be useful for estimatig the umbers g.

4 380 D. Azriel, D. Bered / Theoretical Computer Sciece 369 (006) Lemma 3.. Let G be a ladder graph with vertices. If HAN(G, m) is solvable, the it has a solutio i which the largest disk moves oly oce. Proof. Suppose a solutio of HAN(G, m) is give, i which the first move of disk m is from S to v k. If, at the stage whe this move is carried out, D is empty, the we ca chage the give solutio as follows. Replace the first move of disk m by a move from S to D, ad cotiue with all moves of the origial solutio, omittig all moves of disk m.ithe other case, amely if, whe disk m is moved from S to v k,pegd is ot empty, the the edge (D, v 1 ) must belog to E, ad at that stage peg v 1 has to be empty ad D cotais a sigle disk. Replace the move (S, v k ) with the sequece (D, v 1 ), (S, D), (v 1,D), ad cotiue as i the origial solutio, omittig all further moves of the largest disk. Notice that, up to the first move of disk m, the two sequeces yield the same cofiguratios of disks o G. From that poit o, the oly differece betwee the cofiguratios might be the locatio of disk m. Sice disk m is the largest, moves of other disks ca be performed without ay obstructio. I view of the lemma, we may restrict ourselves to solutios of HAN(G, m) i which the largest disk moves but oce. The followig lemma is straightforward. Lemma 3.3. Let v V \{S,D}, ad let A i be the strogly coected compoet cotaiig v. Put 1 = i j=1 A j ad = r j=i A j. Give ay solutio of HAN(G, m), the umber of disks residig at v at the stage whe the largest disk moves from S to D does ot exceed g mi(1, ). Proof. The subgraph G iduced by {u V : (u, v) E} {v}, with S = S ad D = v, is clearly a ladder graph with 1 vertices. Hece M(G ) g 1. Similarly, we see that o more tha g disks ca be trasferred from v to D. Lemma 3.4. We have g = 1, ad for 3: /+1 g i + 1, 0(mod ), g ( 1)/ g i + 3g (+1)/ + 1, 1(mod ). Proof. Let G be a ladder graph with vertices such that HAN(G, g ) is solvable. By Lemma 3.3, the umber of disks residig at ay v i at the stage whe the largest disk is moved from S to D is at most mi(g i+1,g + i ). Hece, the total umber of disks, g, satisfies g 1 i= /+1 g i + 1, 0(mod ), mi(g i+1,g + i ) + 1 = (+1)/ g i + g (+1)/ + 1, 1(mod ). It is possible to further improve this upper boud, i.e., if A k ={v i 1,v i } the it is possible to trasfer at most g i disks from S to each of the vertices of A k. However, the umber of disks residig at either v i 1 or v i is at most g i 1 + g i. We do ot discuss such improvemets sice the upper boud we shall obtai is ot sigificatly chaged by them. 4. The upper boud for (g ) Let (a ) = be the sequece defied by the iitial coditio a = 1 ad, for 3, by the recurrece relatio /+1 a i + 1, 0(mod ), a = ( 1)/ a i + 3a (+1)/ + 1, 1(mod ).

5 D. Azriel, D. Bered / Theoretical Computer Sciece 369 (006) Lemma 4.1. There exists a costat C such that a C 1/lg for. Proof. The recurrece relatio readily gives a +1 a + a (+)/, 3. (1) Put α = l 1. Let N 0 be the miimal umber for which e 3α(/e+3) α l ( + 1). Take C so that the required iequality holds for all <N 0. Assume iductively that a k Ck α l k for each 0 k for some N 0.Bythe iductio hypothesis ad (1): a +1 a + a (+)/ C α + 3 α l(+3)/ l + C. Thus it suffices to prove that C α + 3 α l((+3)/) l + C C( + 1) α l(+1), N 0. () The right-had side may be bouded from below usig Beroulli s iequality: ( ) ( + 1) α l(+1) α l α l ( + 1) 1 +. (3) We boud from above the left-had side of (), + 3 α l((+3)/) = ( + 3)α l(+3) α l ( + 3) α l α l(+3) ( + 3)α l(+3), ad we have ( + 3) α l(+3) = α l e F (), where ( F () = α l )[ ( l + l )]. Boud F () by usig the iequalities l(1 + 3/) 3/ ad 0 l / 1/e for 1: F () 3α [ l + 3 ] [ 3α e + 3 ] [ ] 3α e + 3. Altogether we have + 3 α l((+3)/) α l e 3α(/e+3) α l = C 0. (4) Replace the right-had side of () by that of (3) ad the secod term o the left-had side by the right-had side of (4). We get the stroger iequality α l α l ( ) l α l ( + 1) + C 0 α 1 +, N 0, which holds sice it is equivalet to C 0 α l ( + 1), N 0. Clearly, g a for, ad the first part of Theorem 1.3 follows: g a C 1/lg.

6 38 D. Azriel, D. Bered / Theoretical Computer Sciece 369 (006) The lower boud for (g ) Let be a positive iteger ad cosider the trasitive path digraph H = (V, E), where V ={v 1,...,v } ad E ={(v i,v j ) : i<j}. Mark S = v 1, D = v. Deote h = M(H ). Obviously, h g for each. I this sectio we fid a lower boud for the sequece (h ) =, thus completig the proof of Theorem 1.3. Lemma 5.1. The iequality h h 1 + h / holds for each. Proof. The followig algorithm yields the required iequality. First, trasfer h / disks from S to v /. Next, trasfer h 1 disks o the graph H \{v / }. Fially, trasfer the h / disks residig o v / to D. A result of Mahler [11] ca be used to derive a lower boud for the sequece (h ) i the followig maer. A biary partitio of a o-egative iteger is a represetatio of i the form = with 0, 1, Deote by b the umber of biary partitios of. It is easy to see that { b 1, 1(mod ), b = b 1 + b /, 0(mod ). I [11] Mahler deduced a asymptotic formula for lg b from his aalysis of the fuctios satisfyig a certai class of fuctioal equatios. He showed that lg b (lg ) /. It is easily checked that h b for all 5, ad thus a lower boud for (g ) ca be obtaied. However, we have preferred to keep the paper self-cotaied ad use oly elemetary tools. The followig lemma is used to obtai the lower boud for (g ). Lemma 5.. For each ε>0 there exists a costat C ε > 0 such that h C ε (1/ ε) lg for all. Proof. Let ε>0. Obviously, the lemma is true for ε 1, so we may assume ε<1. Put α = (1 ε)/(l). Let N 0 be the miimal umber for which 1 α l / l( + 1) C/ α l, where C is a costat to be determied. Take C ε so that the required iequality holds for all <N 0. We proceed by iductio o. Assume that h k C ε k α l k for each 0 k for some N 0. By the iductio hypothesis ad Lemma 5.1, h +1 h + h (+1)/ C ε α + 1 α l((+1)/) l + C ε, ad thus it suffices to prove that α + 1 α l((+1)/) l + ( + 1) α l(+1), N 0. (5) The secod term o the left-had side of (5) may be bouded below by + 1 α l((+1)/) ( ) α l(/) α l α l = α l. (6) Usig the iequality l(1 + 1/) 1/, we estimate the right-had side of (5), [ ] ( + 1) α l(+1) α l e α l (+1)/ α l l( + 1) 1 + C (7) for a suitable C. To prove (5) it suffices, by (6) ad (7), to show that [ ] [ ] α l 1 + α l 1 α l l( + 1) 1 + C, N α l 0. (8)

7 This is equivalet to the iequality 1 α l l( + 1) C α l, N 0, which is ideed correct. The secod part of Theorem 1.3 follows: g h C ε (1/ ε) lg. Ackowledgmets D. Azriel, D. Bered / Theoretical Computer Sciece 369 (006) We wish to thak J.P. Allouche, G. Derfel, E.L. Leiss ad P.K. Stockmeyer for helpful discussios, ideas ad commets related to the subject matter. We also thak the referees for their commets o the first ad secod drafts of the paper. Refereces [1] J.P. Allouche, D. Astooria, J. Radall, J. Shallit, Morphisms, squarefree strigs, ad the Tower of Haoi puzzle, Amer. Math. Mothly 101 (1994) [] J.P. Allouche, A. Sapir, Restricted Towers of Haoi ad morphisms, i: C. De Felice, A. Restivo (Eds.), Developmets i Laguage Theory, Lecture Notes i Computer Sciece, Vol. 357, Spriger, Berli, 005, pp [3] P. Buema, L. Levy, The Towers of Haoi Problem, Iform. Process. Lett. 10 (4,5) (1980) [4] P. Cull, E.F. Ecklud Jr., Towers of Haoi ad aalysis of algorithms, Amer. Math. Mothly 9 (6) (1985) [5] M.C. Er, The complexity of the geeralised cyclic Towers of Haoi, J. Algorithms 6 (1985) [6] J.S. Frame, Solutio to advaced problem 3918, Amer. Math. Mothly 48 (1941) [7] A.M. Hiz, Pascal s triagle ad the Tower of Haoi, Amer. Math. Mothly 99 (199) [8] E.L. Leiss, Solvig the Towers of Haoi o graphs, J. Combi. Iform. System Sci. 8 (1983) [9] E.L. Leiss, Fiite Haoi problems: how may discs ca be hadled?, Cogr. Numer. 44 (1984) 1 9. [10] É. Lucas, Récréatios Mathématiques, Vol. III, Gauthier-Villars, Paris, [11] K. Mahler, O a special fuctioal equatio, J. Lodo Math. Soc. 15 (58) (1940) [1] S. Misker, The Towers of Haoi raibow problem: colorig the rigs, J. Algorithms 10 (1989) [13] A. Sapir, The Towers of Haoi with forbidde moves, Comput. J. 47 (1) (004) 0 4. [14] B.M. Stewart, Advaced problem 3918, Amer. Math. Mothly 46 (1939) 363. [15] B.M. Stewart, Solutio to advaced problem 3918, Amer. Math. Mothly 48 (1941) [16] P.K. Stockmeyer, Variatios o the four-post Tower of Haoi puzzle, Cogr. Numer. 10 (1994) 3 1.