Combinatorics of topmost discs of multi-peg Tower of Hanoi problem

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1 Combinatorics of tomost discs of multi-eg Tower of Hanoi roblem Sandi Klavžar Deartment of Mathematics, PEF, Unversity of Maribor Koroška cesta 160, 000 Maribor, Slovenia Uroš Milutinović Deartment of Mathematics, PEF, University of Maribor Koroška cesta 160, 000 Maribor, Slovenia Ciril Petr Institute of Information Sciences Prešernova 17, 000 Maribor, Slovenia Abstract Combinatorial roerties of the multi-eg Tower of Hanoi roblem on n discs and egs are studied. To-mas are introduced as mas which reflect tomost discs of regular states. We study these mas from several oints of view. We also count the number of edges in grahs of the multi-eg Tower of Hanoi roblem and in this way obtain some combinatorial identities. 1 Introduction The Tower of Hanoi roblem osed in 1884 [1] is by now very well understood. The classical roblem consists of finding the minimum number of moves necessary to transfer a tower of n discs from one eg to another. Several variants and generalizations of the roblem have been roosed, cf., for instance, [5, 7]. Paers [, 3, 9] nicely survey the toic and/or give the corresonding large bibliograhy. When the classical roblem with three egs is generalized to more egs, the roblem becomes notoriously difficult although Hinz [4] susects that the roblem might be solvable by closer examining the grahs associated to the roblem. Suorted by the Ministry of Science and Technology of Slovenia under the grant J

2 In this aer we introduce a artial descrition of regular states of the roblem by considering only the tomost disc of every eg. Formally this is done via so-called to-mas which assign to every regular state a vector whose comonents are labels of tomost discs. For a fixed number of egs this descrition is olynomial in the number of discs, in contrast to the usual descrition of regular states which is exonential in the number of discs. In the rest of this section we introduce the concets and notations needed later. In the next section we describe several roerties of tomas, for instance we comute sizes of their images and classify their unique reimages. In Section 3 we then aly our considerations to grahs of multi-eg Tower of Hanoi roblem and to obtain certain combinatorial identities related to the Stirling numbers of the second kind. The multi-eg Tower of Hanoi roblem consists of 3 egs numbered 0, 1,... 1 and n 1 discs of different sizes. Discs will be numbered 1,,..., n and we assume that they are ordered by size, disc 1 being the smallest one. Initially all discs lie on eg 0 in small-on-large ordering. The objective is to transfer all the discs to eg 1 in the minimum number of legal moves. A legal move is a transfer of the tomost disc from one eg to another eg such that no disc can be moved onto a smaller one. As usual, a state is regular if no larger disc is laced on a smaller one. A regular state can be uniquely described with an n-tule r ( Z ) n = {0, 1,..., 1} n. More recisely, we set r = (r 1, r,...,r n ), where r d denotes the eg on which the disc d is laced. An examle of a regular state on = 5 egs with n = 6 discs is shown in Figure 1. The corresonding 6-tule r is also given (as well as a 5-tule s to be defined in the next section). Figure 1: A regular state A regular state is erfect if all the discs lie on the same eg. In Figure an examle of a erfect state is shown with all discs on eg 3.

3 Figure : A erfect state We call a regular state a sread state, if all the discs are on different egs, see Figure 3. Clearly, for n > there are no sread states. Figure 3: A sread state A regular state is an almost sread state, if for some k 1, discs 1,...,k are on a common eg, while discs k + 1,...,n are each on a rivate eg, cf. Figure 4. Note that a erfect state is an almost sread state with k = n and a sread state is an almost sread state with k = 1. Let r = (r 1, r,...,r n 1, r n ) be a regular state. Then r is a erfect state if and only if all the comonents r i are equal. The state r is a sread state if and only if r i r j for any i j. Finally, r is an almost sread state if and only if there exists k 1 such that all r i are equal for i k and r i r j r 1 for i, j > k, i j. Finally, as usual, let (n) m = n(n 1) (n m + 1) and let S(n, k) denote the Stirling numbers of the second kind. 3

4 Figure 4: An almost sread state To-mas Information about tomost discs suffices to know all the ossible legal moves from a given regular state. From this reason we introduce maings T n, : ( Z ) n ( Z n+1 ) as T n, (r 1, r,...,r n ) = (s 0, s 1,..., s 1 ), where { 0; rk i for 1 k n, s i = min {k ; r k = i}; otherwise. 1 k n Hence, the comonent s i of s is the tomost disc on eg i, if eg i is nonemty; otherwise s i = 0. We will briefly refer to these maings as to-mas. As usual, R(T n, ) denotes the image of T n, and R(T n, ) its size. Observe that in general a to-ma need not be surjective. For instance, the 4-tules (, 3, 4, 5) and (, 1, 4, 4) are not in the image of some to-ma. In the first case the smallest disc is not resent and in the second case the disc 4 is suosed to be simultaneously on two discs. We begin with the lemma which is useful for comuting R(T n, ). Lemma.1 Let s = (s 0, s 1,..., s 1 ) ( Z n+1 ). Then s R(T n, ) if and only if the following two conditions are fulfilled: (i) i {0,..., 1} : s i = 1, (ii) i, j {0,..., 1}: s i = s j (i = j s i = 0). Proof. The conditions are clearly necessary. Indeed, the smallest disc must be one of the to discs and no disc can be tomost simultaneously on two egs. Assume now that s = (s 0, s 1,..., s 1 ) fulfills the two conditions. We need to show that there exists r = (r 1, r,...,r n ) ( Z ) n with T n, (r) = s. 4

5 We define the comonents r i as follows. For any i with s i 0 we set r si = i. By condition (ii) such i is uniquely determined. For all the other discs i (i.e. for all those which do not aear as the tomost discs) we set r i = r 1. (Note that by condition (i) the comonent r 1 has already been defined.) It is now easy to verify that r ( Z ) n reresents a regular state and that T n, (r) = s. Proosition. For any n 1 and 3 we have 1 ( ) 1 R(T n, ) = (n 1) k 1. k k=0 Proof. Let s R(T n, ). Then, by condition (i) of Lemma.1 we have s i = 1 for some i. There are ossibilities for that. After fixing this i, there may be k emty egs for any k with 0 k 1. For a fixed k there are ( ) 1 k ossible selections of k emty egs. The remaining 1 k egs are nonemty, and by the condition (ii) of Lemma.1, which asserts that the tomost discs must be different, we infer that there are ( )( ) ( ) n 1 (n 1) 1) (n 1) ( 1 k 1) = (n 1) k 1 ossibilities to select the 1 k to discs. Let be fixed. Then, as R(T n, ) ( Z n+1 ), the size of R(T n, ) is a olynomial function in n. This observation can be made more recise as follows: Corollary.3 Let 3. Then R(T n, ) = n 1 + O(n ). Proof. Result follows by noting that the leading term from Proosition. is obtained for k = 0. Recall that the number of regular states is n, i.e. the number of regular states is exonential in the number of discs. Therefore Corollary.3 seems to be interesting from the algorithmic oint of view, because the exonential number of regular states is a source of difficulties in studying the grahs of the Tower of Hanoi roblem. Let s R(T n, ). In order to determine T n, 1 (s) we introduce the following function. For a ositive integer d and a -tule s = (s 0, s 1,...,s 1 ) let h(d, s) = {i {0,..., 1} ; 0 < s i < d}. Now we have: 5

6 Lemma.4 Let s R(T n, ). Then T 1 n, (s) = 1; h(n + 1, s) = n, n h(i, s); otherwise. i=1 i s 0,...,s 1 Proof. Suose that h(n + 1, s) = n. Then every disc in (any element of) T 1 n, (s) is tomost. It follows that the reimage of s is unique, i.e. T 1 n, (s) = 1. Assume now that at least one disc in T 1 n, (s), say i, is not tomost. Then h(i, s) is nonzero and reresents the number of egs on which we can lace the disc i in a regular state corresonding to s. Therefore T 1 n, (s) is at least i h(i, s), where i runs over all discs that are not tomost. On the other hand, a disc i can only be laced on a eg j if i < s j. If there are several discs that can be laced on the same eg, then they must be on this eg sorted by their sizes, i.e. there is only one ossibility to do that. Therefore, T 1 n, (s) is at most i h(i, s). The case when the reimage of s is unique is characterized in the next theorem. Theorem.5 Let T n, (r) = s. Then T n, 1 (s) = {r} if and only if r is an almost sread state. Proof. Suose first that r is a erfect state, a sread state, or an almost sread state. Then, using Lemma.4, it is easy to verify that T 1 n, (s) = 1. Conversely, let T 1 n, (s) = 1. By Lemma.4 we then either have h(n + 1, s) = n or i h(i, s) = 1. In the first case r is a sread state and thus an almost sread state. Assume now that i h(i, s) = 1. It follows that the index set of this roduct is nonemty and for any such i we have h(i, s) = 1. Let i be such number with h(i, s) = 1. Clearly, i > 1. Moreover, Lemma.1, imlies that there exists a eg j such that s j = 1. Therefore, i < s j and as h(i, s) = 1 disc i must lie on eg s j. Hence, all the discs i with h(i, s) = 1 must be on the same disc (i.e. on the disc s j ). Now, if the number of such discs i is n 1, then r is a erfect state which is an almost sread state. Otherwise, for any other disc k we have k = s t for some t. In other words, all the other discs are tomost. They clearly lie on egs different from s j, and we can conclude that we have an almost sread state. 6

7 3 To-mas and combinatorial identities We have seen in the revious section that the image of a to-ma is of olynomial size in the number of discs. Therefore it is natural to ask which information can be deduced from the image of T n, itself. Here we show how to comute the number of legal moves from a regular state r using only information of T n, (r). Proosition 3.1 Let T n, (r) = s. Then the number of legal moves from r is ( h(n + 1, s) 1 ) 1 h(n + 1, s). Proof. Observe first that the number of nonemty egs is h(n + 1, s), and so the number of emty egs is h(n+1, s). Any tomost disc of a eg can be moved to any emty eg. Therefore, there are h(n+1, s)( h(n+1, s)) legal moves of this kind. It remains to consider the legal moves in which a tomost disc is moved to a nonemty eg. Recall that the h(n + 1, s) tomost discs are different. Thus, the smallest one can be moved to any other of h(n+1, s) 1 nonemty egs, the second smallest can be moved to h(n + 1, s), and so on. Therefore, there are h(n+1,s) 1 i=1 i = 1 (h(n + 1, s) 1)h(n + 1, s) different moves from a nonemty onto a nonemty eg. Summing the above exressions we get the result. The number of tomost discs that can be moved to a eg i is equal to h(s i, s). Moreover, if s j = 0, i.e. if eg j is emty, then by definition we have h(s j, s) = 0. Therefore, the number of legal moves in which a tomost disc is moved to an nonemty eg is equal to i h(s i, s) and so we also have (cf. the above roof) 1 h(s i, s) = 1 (h(n + 1, s) 1)h(n + 1, s). i=0 By Theorem 3.1 we thus only need to know the number of nonemty egs h(n + 1, s) in order to comute the number of legal moves. In order to obtain some additional results, we shall consider grahs of the multi-eg tower of Hanoi roblem. They are defined as follows. The grah G n = (V n, En ) of n discs and egs has regular states as vertices, two vertices being adjacent if one state is obtained from the other by a legal move. 7

8 Theorem 3. For any n 1 and 3 we have: (k( 1 ) 1 ) n 1 k S(n, k)() k = ( 1) i+1 ( ) n i 1. k=1 Proof. We will rove the theorem by counting the number of edges in G n in two ways. Note first that the number of regular states where exactly k egs are nonemty is equal to S(n, k)() k. Therefore, using Proosition 3.1, we have E n = 1 k=1 i=0 (k( 1 ) 1 k ) S(n, k)() k. Consider now the set of states in which the largest disc is fixed on some eg. We infer that the corresonding vertices induce a subgrah of G n isomorhic to G n 1. There are such subgrahs and they form a artition of Vn. Two vertices belonging to two such subgrahs are adjacent if and only if they differ exactly in osition of the largest disc. Since all the remaining discs excet the largest one lie on discs (i.e. on the egs that are not involved in the move of the largest disc), there are V n 1 edges connecting two such subgrahs. It follows that the number of edges of G n can be exressed recursively as E n = En 1 + ( ) V n 1. Thus, E n is equal to: ( ) E n 1 + V n 1 = ( ( ) ) ( ) E n + V n + V n 1 = ( ( ( ) ) ( ) ) ( ) E n 3 + V n 3 + V n + V n 1 = ( ( ( ( ) ) ( ) ) ) ( )... E 1 + V 1 + V V n 1 = ( ) ( ) ( ) ( ) ((...( + ( ) 1 ) + ( ) ) +...) + ( ) n 1 = }{{} n 1 n 1 i=0 ( ) i ( ) (n 1) i. } {{ } n 1 8

9 Combining the above exressions the result follows. Using standard methods the right-hand side exression of Theorem 3. can be summed u. In this way we obtain: Corollary 3.3 For any n 1 and 3 we have: k=1 (k( 1 ) 1 k ) S(n, k)() k = ( ) [ n ( ) n ]. In the roof of Theorem 3. we observed that the number of regular states where exactly k egs are nonemty is S(n, k)() k. On the other hand the number of all regular states is n. Thus in assing we get the following well-known identity, cf. [6, 8]: Corollary 3.4 n = S(n, k)() k = k=1 n S(n, k)() k. In this formula the choice of the uer bound for summation makes no difference, since S(n, k) = 0, if n < k, and () k = 0, if < k. k=1 References [1] N. Claus (= E. Lucas), La Tour d Hanoi, Jeu de calcul, Science et Nature 1(8) (1884) [] A.M. Hinz, The Tower of Hanoi, Enseign. Math. 35 (1989) [3] A.M. Hinz, Shortest aths between regular states of the Tower of Hanoi, Inform. Sci. 63 (199) [4] A.M. Hinz, The Tower of Hanoi, Proceedings of the International Congress in Algebra and Combinatorics 1997, Hong Kong (China), to aear. [5] C.S. Klein and S. Minsker, The suer towers of Hanoi roblem: large rings on small rings, Discrete Math. 114 (1993) [6] J.H. van Lint and R.M. Wilson, A Course in Combinatorics, Cambridge University Press, Cambridge,

10 [7] D.G. Pool, The towers and triangles of Professor Claus (or, Pascal knows Hanoi), Math. Magazine 67 (1994) [8] J. Riordan, Combinatorial Identities, John Wiley & Sons, New York, [9] P.K. Stockmeyer, The Tower of Hanoi: A historical survey and bibliograhy, manuscrit, Setember

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