- Email: [email protected]
Open Problems for Hanoi and Sierpiński Graphs
Available online at www.sciencedirect.com
Electronic Notes in Discrete Mathematics 63 (2017) 23–31
www.elsevier.com/locate/endm
Open Problems for Hanoi and Sierpi´nski Graphs Andreas M. Hinz 1 Mathematical Institute, LMU M¨ unchen Munich, Germany and Institute of Mathematics, Physics and Mechanics Ljubljana, Slovenia
Abstract We list some unsolved problems concerning two meanwhile famous classes of graphs, namely the Hanoi graphs and the Sierpi´ nski graphs. Whereas many graph parameters have been determined and metric properties are well-understood for the latter, the former give rise to notoriously difficult questions, in particular when it comes to distance-related tasks. Keywords: Hanoi graphs, Sierpi´ nski graphs, graph parameters, graph distance.
1
The Frame-Stewart Conjecture
Hanoi graphs were introduced as a mathematical model for the famous Tower of Hanoi puzzle; see [4, Chapter 0] for the historical background and [4, Chapters 2 and 5] for all statements in this note which are not directly referred to A.M.Hinz, c 2016 1 Email: [email protected] https://doi.org/10.1016/j.endm.2017.10.058 1571-0653/© 2017 Elsevier B.V. All rights reserved.
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A.M. Hinz / Electronic Notes in Discrete Mathematics 63 (2017) 23–31
the list of references. These graphs are defined for base p ∈ N3 (number of pegs) and exponent n ∈ N0 (number of discs) as follows. 2 Definition 1.1 Let [p]0 := {0, . . . , p − 1}, [n] := {1, . . . , n}. Then 3 V (Hpn ) = [p]n0 (= {sn . . . s1 | ∀ d ∈ [n] : sd ∈ [p]0 }) , E Hpn = {sis, sjs} | i, j ∈ [p]0 , i = j, s ∈ ([p]0 \ {i, j})d−1 , d ∈ [n] . Here sd is the peg where disc d lies in the state of the game represented by s = sn . . . s1 and edge {sis, sjs} stands for a move of disc d from peg i to peg j or vice versa. The original puzzle was for p = 3 pegs and was later extended to p = 4 (The Reve’s puzzle) and then to general p which led to increasing difficulty. Whereas almost everything is known for p = 3 (see [4, Chapter 2]), the case of more pegs resulted in many open questions like the notorious Frame-Stewart conjecture which relates to the number of moves needed to get from perfect state 4 0n to perfect state 1n , i.e. it refers to shortest 0n , 1n -paths in Hpn (see [4, Chapter 5]). To make things a little easier, Sierpi´ nski graphs were considered, which are defined as follows. Definition 1.2 Let p ∈ N2 and n ∈ N0 . Then V (Spn ) = [p]n0 ,
d−1 sij , sjid−1 | i, j ∈ [p]0 , i = j, d ∈ [n] . E Spn = They can be interpreted as state graphs for the Switching Tower of Hanoi, where disc d changes position with the subtower consisting of all smaller discs from [d − 1] (including the case d = 1!). Note that S2n ∼ = P2n , the path graph on 2n vertices, such that this case is included just for comparison and because it is connected to another ancient puzzle, namely the Chinese Rings (cf. [4, Chapter 1]). Let us also mention another related class of graphs, namely Sierpi´ nski triangle graphs; see [5, Section 0.2.1]. We will not discuss these graphs here. The difficulty of the problems addressed in the present note as applied to them will lie between that of the Sierpi´ nski graphs proper and of the Hanoi graphs. 2
We write Nk for the set of natural numbers greater than or equal to k. The set N1 is denoted by N, because it is the most natural set of natural numbers. 3 We write sn . . . s1 for elements s of [p]n0 . 4 i.e. with all discs on one and the same peg Exponents in sn . . . s1 are interpreted as repetitions of entries, e.g., 0n = 0 . . . 0 ∈ [p]n0 .
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All graphs defined here are connected, 5 leading to the canonical distance functions d on the vertex sets, respectively, given by the length of a shortest path between vertices. This can be seen by recognizing that there are alternative recursive definitions: for fixed p and starting from the trivial graphs Hp0 = Sp0 with just one vertex (no disc, i.e. n = 0), one gets the graph for n + 1 by taking p copies of the graph for n and linking pairs of these copies by a certain number of edges, namely (p − 2)n for Hanoi graphs, but only 1 for Sierpi´ nski graphs; this makes all the difference! In fact, H3n ∼ = S3n and Hp1 ∼ = Kp ∼ = Sp1 , but Hpn is not isomorphic to Spn for p ∈ N4 and n ∈ N2 anymore. The historical background and basic properties of Sierpi´ nski graphs can be found in [4, Chapters 1 and 4], more advanced ones are collected in the recent survey [5]. Metric properties can be obtained from the algorithm in [3] which decides, for a given pair of vertices (s, t), about the shortest s, tpath(s) and whether they employ one or two moves of the largest (moving) disc, called LDMs. In Hanoi graphs, there may be up to p − 1 LDMs necessary in a shortest path and we were forced to use computational methods in [1] to approach their patterns. Computer power was also used to attack other problems in connection with distances in Hanoi graphs. For instance, whereas the diameter of Spn is always 2n − 1 (independent of p), and it is easy to see that diam(Hpn ) ≤ 2n − 1, the essentially best analytical result we have for Hanoi graphs is contained in the following theorem (F denoting the Fibonacci sequence). Theorem 1.3 ∀ p ∈ N4 ∀ n ∈ N0 : diam(Hpn ) ≤ Fn+3 − 2 . Proof. Clearly, diam(Hp0 ) = 0 and diam(Hp1 ) = 1, so we may assume that n ∈ N2 . Let is, jt ∈ [p]n0 . If i = j, then d(is, jt) = d(s, t) ≤ diam(Hpn−1 ) . Let i = j. Choose ∈ [p]0 \ {i, j, sn−1 } and k ∈ [p]0 \ {i, j, }. (Here we need p ≥ 4.) Then d(is, jt) ≤ d(is, isn−1 n−2 ) + d(isn−1 n−2 , ikn−2 ) + d(ikn−2 , jkn−2 ) + d(jkn−2 , jt) ≤ diam(Hpn−2 ) + 2 + diam(Hpn−1 ) . The statement of the theorem now follows by solving the recurrence x0 = 0, x1 = 1, ∀ n ∈ N2 : xn = xn−1 + xn−2 + 2 . 5
This is why we excluded p = 2 for Hanoi graphs.
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Although it is possible to improve the upper bound by somewhat more sophisticated arguments, the numerical values remain unsatisfactory. For instance, from Theorem 1.3 we get diam(H415 ) ≤ 2582 << 215 − 1, but our computer experiments (see below) yield diam(H415 ) = 130. This is, by the way, the first (with respect to n) example of Korf ’s phenomenon, namely that diam(H4n ) > F S4n , where the Frame-Stewart numbers are defined by Definition 1.4 F Sp0 = 0, F S3n = 2n − 1, n m F Sp+1 = min 2F Sp+1 + F Spn−m | m ∈ [n]0 . The Frame-Stewart conjecture is that d(0n , 1n ) = F Spn in Hpn . It is clearly true for p = 3 and has been proved in 2014 by Thierry Bousch for p = 4 [2], but remains the greatest open problem for Hanoi graphs when p > 4. For p = 5 our computations confirmed it up to and including n = 20, for p = 6 to n = 16, and for p = 7 to n = 21. The latter effort was taken in connection with [4, Conjecture 5.41] which we confirmed for p ∈ [7] \ [2] and n ∈ p2 0 . We now present a couple of other challenges. In the sequel we will assume, if not otherwise stated, that p ∈ N4 and n ∈ N2 . Note that Hp0 = Sp0 is trivial and Hp1 ∼ = Kp ∼ = Sp1 are the complete graphs, for which (almost) all is known.
2
Eccentricities
For a study of eccentricities in Sierpi´ nski graphs, see [7], where all the corresponding answers to the subsequent problems can be found. Eccentricities in Hanoi graphs are being calculated on a large, but, of course, limited scale in our project pr87mo at SuperMUC (Garching, Germany). (i) Calculate ε(0n ) in Hpn . (ii) Determine the average eccentricity ε(Hpn ) = p−n
s∈V
(iii) Find a general formula for ε(s), s ∈ V
ε(s).
(Hpn )
(Hpn ).
(iv) Construct an efficient algorithm to produce shortest path(s) from s to 0n in Hpn . (v) Give formulas for the radius of Hpn and find the center C(Hpn ).
3
Largest Disc Moves
The problems in this section have been solved for Sierpi´ nski graphs in [3]. For Hanoi graphs one can find some partial analytical solutions and the results of
A.M. Hinz / Electronic Notes in Discrete Mathematics 63 (2017) 23–31
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computer experiments in [1]. (i) How many LDMs are included in a shortest path from s to t in Hpn ? (ii) Construct an algorithm which determines the LDMs included in a(ny) shortest path from s to t in Hpn .
4
Diameters
For Sierpi´ nski graphs we always have d(0n , 1n ) = 2n − 1 6 and since this is also an upper bound for eccentricities, diam(Spn ) = 2n − 1. In Hanoi graphs the situation is much more intricate. For instance, for p = 4, Korf’s phenomenon in its weaker form, namely ε(0n ) > F S(H4n ), has been confirmed by computer experiments for n ∈ {15, 20, 21, 22, 23, 24, 25}, but the behavior of ε(0 n)−F S(H4n) is rather eratic. The Korf-Felner conjecture is that this function is positive for n ≥ 20. Diameters of H4n have been computed by R. Korf so far for n ≤ 17 with the result that diam(H4n ) = F S(H4n ) in that interval with the exception that diam(H415 ) = F S(H415 ) + 1 (= ε(015 )). In particular, we know of no instance where ε(0n ) < diam(Hpn ), i.e. that 0n does not belong to the periphery of Hpn . Therefore, most of the following is open. (i) For which p and n are inequalities in d(0n , 1n ) ≤ ε(0n ) ≤ diam(Hpn ) strict? (ii) Find diam(Hpn ) and the periphery P (Hpn ).
(iii) Determine the average distance d(Hpn ) = p−2n
d(s, t).
(s,t)∈V (Hpn )2
(iv) Construct an efficient algorithm to produce shortest path(s) from s to t in Hpn . (v) Determine the average distance of s ∈ V (Hpn ), i.e. d(s) = p−n d(s) where
d(s) = d(s, t) . t∈V (Hpn )
(vi) Find the proximity of Hpn and the median M (Hpn ). To get an impression how difficult problem (iii) may be, have a look at the corresponding formula for Sierpi´ nski graphs: d(Spn ) = 6
(p − 1)(2p4 + 6p3 − 17p2 + 26p − 16) n 2 p(2p − 1)(p3 + 4p2 − 4p + 8)
This is why vertices of the form k n are called extreme vertices in Sierpi´ nski graphs.
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p2 + 3p − 6 p−2 + p−n p (2p − 1)(p2 − 7p + 8) n λp,+ p(p − 1)γp,+ − 2 2(p − 7p + 8)(p3 + 4p2 − 4p + 8) p2 n λp,− p(p − 1)γp,− − 2 , 2(p − 7p + 8)(p3 + 4p2 − 4p + 8) p2 −
where αp = p4 − 12p3 + 56p2 − 104p + 68 , 1 1√ αp , λp,± = p2 − p + 1 ± 2 2 √ γp,± = (p2 + 3p − 2) ∓ (p4 + p3 − 30p2 + 58p − 36) αp . Whereas (iv) has been solved in [3] for Sierpi´ nski graphs, problems (v) and (vi) are still open also for them, even for p = 3. For p = 2 we know that S2n ∼ = P2n , so we start by looking at path graphs P1+k , k ∈ N. Let the vertex set be V = [k + 1]0 and vertices being neighbors iff they are neighbors on the number line. An easy calculation shows that ∀ v ∈ V : d(v) = k(k+1) − v(k − v), so that d(v) = k2 − v(k−v) . For odd k 2 k+1 k+1 the minimum prox(P1+k ) = 4 is achieved for v ∈ C(P1+k ) = k−1 . , k+1 2 2 n n n−2 n n Putting k = 2 − 1, we obtain prox(S2 ) = 2 . Moreover, M (S2 ) = C(S2 ) = {01n−1 , 10n−1 }.
5
Domination
All Sierpi´ nski graphs possess perfect codes and therefore their domination n +p n +1 number is known to be γ(Spn ) = pp+1 , if n is even, and γ(Spn ) = pp+1 , if n 2 n is odd. Similarly, we find that γ(Hp ) = p. But otherwise Hp do not contain perfect codes. (i) Determine the domination number of Hpn for n ∈ N3 .
6
Metric Dimension
Due to the complete mastery of metric properties of Sierpi´ nski graphs, their metric dimension can be deduced to be μ(Spn ) = p − 1 for n ∈ N with any p − 1 extreme vertices forming a basis. Also μ(Hp2 ) = p − 1 with every p − 1 out of the perfect states k 2 , k ∈ [p]0 , forming a basis. This can be seen as follows. By [4, Theorem 5.34] there are just 2 equisets (symmetry classes) of vertices, represented by 00 and 01 (cf. [4,
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Figure 5.6] for p = 4). The distance representation of 00 with respect to all perfect states consists of one 0 and otherwise 3s, so that the position of the 2-tower can be reconstructed. For 01 the distance representation contains one 1, one 3 and otherwise 2s; the position of the 1 shows the peg where disc 2 is lying, the position of the 3 gives the peg of disc 1. It is also clear from this, that p − 1 distances suffice to reconstruct the pth. Finally, to see that a set U with |U | = p − 2 can not be resolving for p ≥ 4, one has to investigate all 13 equisets of tasks (cf. [4, Theorem 5.35]), i.e. pairs of vertices, and show that for each such pair there are two different vertices with equal distance representation. For instance, if U = {02 , 12 }, then 22 and 32 have the same distance representation with respect to U , namely dU (22 ) = (3, 3) = dU (32 ). Again, a look at [4, Figure 5.6] facilitates the job for all pairs; the pegs with labels larger than 3 do not play a role. However, for Hp3 not even the whole set of perfect states U = {03 , . . . , (p − 1)3 } is resolving! This is so because (cf. [4, Figure 5.10]) dU (100) = (5, 3, 4, . . . , 4) = dU (101) . So we are left with (i) Determine the metric dimension of Hpn for n ∈ N3 and find a basis in these graphs.
7
Genera and Crossing Numbers
Hanoi graphs and Sierpi´ nski graphs share the same domain of planarity, namely p ≤ 3 or n = 0 or (p = 4 and n ∈ {1, 2}). For n = 1, i.e. the complete graphs Kp , the genera are known (Ringel and Youngs, 1968) and Guy and others made the conjecture about the crossing number to be p−2 p−3 1 p p − 1 cr(Kp ) = . 4 2 2 2 2 This conjecture has so far been confirmed for p ≤ 12. The only other provable value for our non-planar graphs is cr(S43 ) = 12. (i) Determine g(Hpn ) and g(Spn ) for p = 4 and n ∈ N3 and for p ∈ N5 and n ∈ N2 . (ii) Determine cr(H43 ) (known to be ≤ 72) and cr(Hpn ) and cr(Spn ) for p = 4 and n ∈ N4 , for p ∈ N5 and n ∈ N2 and for p ∈ N13 and n = 1.
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8
Further Questions
There are, of course, many more properties of Hanoi graphs and Sierpi´ nski graphs to be explored. For instance, colorings have been considered in [6], where the classical coloring numbers—vertex, edge and total—were determined for both classes with the result that they behave quite similarly. It would be interesting to extend these findings to some of the meaningful types of colorings and so-called labellings (cf. [5, Section 4.2]). Other challenges in connection with the Tower of Hanoi and with graphs deriving from variations of the puzzle (see [4, Chapters 3 and 6 to 8]) can be found in [4, Chapter 9]. (Some already solved!) For variants of Sierpi´ nski graphs, see [5]. Let us finally mention that computational methods also led to the solution of some very old tasks asked in English magazines by Henry Ernest Dudeney, the most outstanding being d(020 , 120 ) = 111 = F S520 in H520 ; see [8, Proposition 1]. There remains Dudeney’s Last Question, namely the value of d(035 , 135 ) in H535 . To verify computationally that it is equal to F S535 = 351 is out of the reach of current computing devices, so an analytical solution is needed! Acknowledgements. I thank Ciril Petr (Maribor) for his work on our project pr87mo at SuperMUC. We are both supported by the Slovenian Research Agency (research core funding No. J1-7110) which also provided funds for my travel to the International Conference on Current Trends in Graph Theory and Computation held in September 2016 in New Delhi. I am grateful to Deepa Sinha (South Asian University, New Delhi) and Purnima Gupta (University of Delhi) for their generous hospitality during my extended visit and to Shariefuddin Pirzada (Srinagar) for his support. The cheerful volunteers made my stay even more enjoyable.
References [1] Aumann, S., K. A. M. G¨ otz, A. M. Hinz, and C. Petr, The number of moves of the largest disc in shortest paths on Hanoi graphs, Electron. J. Combin. 21 (2014), P4.38. [2] Bousch, T., La quatri`eme tour de Hano¨ı, Bull. Belg. Math. Soc. Simon Stevin 21 (2014), 895–912. [3] Hinz, A. M., and C. Holz auf der Heide, An efficient algorithm to determine all shortest paths in Sierpi´ nski graphs, Discrete Appl. Math. 177 (2014), 111–120.
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[4] Hinz, A. M., S. Klavˇzar, U. Milutinovi´c, and C. Petr, “The Tower of Hanoi— Myths and Maths,” Springer, Basel, 2013. [5] Hinz, A. M., S. Klavˇzar, and S. S. Zemljiˇc, A survey and classification of Sierpin ´ski-type graphs, Discrete Appl. Math. 217 (2017), 565–600. [6] Hinz, A. M., and D. Parisse, Coloring Hanoi and Sierpi´ nski graphs, Discrete Math. 312 (2012), 1521–1535. [7] Hinz, A. M., and D. Parisse, The Average Eccentricity of Sierpi´ nski Graphs, Graphs Combin. 28 (2012), 671–686. [8] Hinz, A. M., and C. Petr, Computational Solution of an Old Tower of Hanoi Problem, Electron. Notes Discrete Math. 53 (2016), 445–458.
Electronic Notes in Discrete Mathematics 63 (2017) 23–31
www.elsevier.com/locate/endm
Open Problems for Hanoi and Sierpi´nski Graphs Andreas M. Hinz 1 Mathematical Institute, LMU M¨ unchen Munich, Germany and Institute of Mathematics, Physics and Mechanics Ljubljana, Slovenia
Abstract We list some unsolved problems concerning two meanwhile famous classes of graphs, namely the Hanoi graphs and the Sierpi´ nski graphs. Whereas many graph parameters have been determined and metric properties are well-understood for the latter, the former give rise to notoriously difficult questions, in particular when it comes to distance-related tasks. Keywords: Hanoi graphs, Sierpi´ nski graphs, graph parameters, graph distance.
1
The Frame-Stewart Conjecture
Hanoi graphs were introduced as a mathematical model for the famous Tower of Hanoi puzzle; see [4, Chapter 0] for the historical background and [4, Chapters 2 and 5] for all statements in this note which are not directly referred to A.M.Hinz, c 2016 1 Email: [email protected] https://doi.org/10.1016/j.endm.2017.10.058 1571-0653/© 2017 Elsevier B.V. All rights reserved.
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A.M. Hinz / Electronic Notes in Discrete Mathematics 63 (2017) 23–31
the list of references. These graphs are defined for base p ∈ N3 (number of pegs) and exponent n ∈ N0 (number of discs) as follows. 2 Definition 1.1 Let [p]0 := {0, . . . , p − 1}, [n] := {1, . . . , n}. Then 3 V (Hpn ) = [p]n0 (= {sn . . . s1 | ∀ d ∈ [n] : sd ∈ [p]0 }) , E Hpn = {sis, sjs} | i, j ∈ [p]0 , i = j, s ∈ ([p]0 \ {i, j})d−1 , d ∈ [n] . Here sd is the peg where disc d lies in the state of the game represented by s = sn . . . s1 and edge {sis, sjs} stands for a move of disc d from peg i to peg j or vice versa. The original puzzle was for p = 3 pegs and was later extended to p = 4 (The Reve’s puzzle) and then to general p which led to increasing difficulty. Whereas almost everything is known for p = 3 (see [4, Chapter 2]), the case of more pegs resulted in many open questions like the notorious Frame-Stewart conjecture which relates to the number of moves needed to get from perfect state 4 0n to perfect state 1n , i.e. it refers to shortest 0n , 1n -paths in Hpn (see [4, Chapter 5]). To make things a little easier, Sierpi´ nski graphs were considered, which are defined as follows. Definition 1.2 Let p ∈ N2 and n ∈ N0 . Then V (Spn ) = [p]n0 ,
d−1 sij , sjid−1 | i, j ∈ [p]0 , i = j, d ∈ [n] . E Spn = They can be interpreted as state graphs for the Switching Tower of Hanoi, where disc d changes position with the subtower consisting of all smaller discs from [d − 1] (including the case d = 1!). Note that S2n ∼ = P2n , the path graph on 2n vertices, such that this case is included just for comparison and because it is connected to another ancient puzzle, namely the Chinese Rings (cf. [4, Chapter 1]). Let us also mention another related class of graphs, namely Sierpi´ nski triangle graphs; see [5, Section 0.2.1]. We will not discuss these graphs here. The difficulty of the problems addressed in the present note as applied to them will lie between that of the Sierpi´ nski graphs proper and of the Hanoi graphs. 2
We write Nk for the set of natural numbers greater than or equal to k. The set N1 is denoted by N, because it is the most natural set of natural numbers. 3 We write sn . . . s1 for elements s of [p]n0 . 4 i.e. with all discs on one and the same peg Exponents in sn . . . s1 are interpreted as repetitions of entries, e.g., 0n = 0 . . . 0 ∈ [p]n0 .
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All graphs defined here are connected, 5 leading to the canonical distance functions d on the vertex sets, respectively, given by the length of a shortest path between vertices. This can be seen by recognizing that there are alternative recursive definitions: for fixed p and starting from the trivial graphs Hp0 = Sp0 with just one vertex (no disc, i.e. n = 0), one gets the graph for n + 1 by taking p copies of the graph for n and linking pairs of these copies by a certain number of edges, namely (p − 2)n for Hanoi graphs, but only 1 for Sierpi´ nski graphs; this makes all the difference! In fact, H3n ∼ = S3n and Hp1 ∼ = Kp ∼ = Sp1 , but Hpn is not isomorphic to Spn for p ∈ N4 and n ∈ N2 anymore. The historical background and basic properties of Sierpi´ nski graphs can be found in [4, Chapters 1 and 4], more advanced ones are collected in the recent survey [5]. Metric properties can be obtained from the algorithm in [3] which decides, for a given pair of vertices (s, t), about the shortest s, tpath(s) and whether they employ one or two moves of the largest (moving) disc, called LDMs. In Hanoi graphs, there may be up to p − 1 LDMs necessary in a shortest path and we were forced to use computational methods in [1] to approach their patterns. Computer power was also used to attack other problems in connection with distances in Hanoi graphs. For instance, whereas the diameter of Spn is always 2n − 1 (independent of p), and it is easy to see that diam(Hpn ) ≤ 2n − 1, the essentially best analytical result we have for Hanoi graphs is contained in the following theorem (F denoting the Fibonacci sequence). Theorem 1.3 ∀ p ∈ N4 ∀ n ∈ N0 : diam(Hpn ) ≤ Fn+3 − 2 . Proof. Clearly, diam(Hp0 ) = 0 and diam(Hp1 ) = 1, so we may assume that n ∈ N2 . Let is, jt ∈ [p]n0 . If i = j, then d(is, jt) = d(s, t) ≤ diam(Hpn−1 ) . Let i = j. Choose ∈ [p]0 \ {i, j, sn−1 } and k ∈ [p]0 \ {i, j, }. (Here we need p ≥ 4.) Then d(is, jt) ≤ d(is, isn−1 n−2 ) + d(isn−1 n−2 , ikn−2 ) + d(ikn−2 , jkn−2 ) + d(jkn−2 , jt) ≤ diam(Hpn−2 ) + 2 + diam(Hpn−1 ) . The statement of the theorem now follows by solving the recurrence x0 = 0, x1 = 1, ∀ n ∈ N2 : xn = xn−1 + xn−2 + 2 . 5
This is why we excluded p = 2 for Hanoi graphs.
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Although it is possible to improve the upper bound by somewhat more sophisticated arguments, the numerical values remain unsatisfactory. For instance, from Theorem 1.3 we get diam(H415 ) ≤ 2582 << 215 − 1, but our computer experiments (see below) yield diam(H415 ) = 130. This is, by the way, the first (with respect to n) example of Korf ’s phenomenon, namely that diam(H4n ) > F S4n , where the Frame-Stewart numbers are defined by Definition 1.4 F Sp0 = 0, F S3n = 2n − 1, n m F Sp+1 = min 2F Sp+1 + F Spn−m | m ∈ [n]0 . The Frame-Stewart conjecture is that d(0n , 1n ) = F Spn in Hpn . It is clearly true for p = 3 and has been proved in 2014 by Thierry Bousch for p = 4 [2], but remains the greatest open problem for Hanoi graphs when p > 4. For p = 5 our computations confirmed it up to and including n = 20, for p = 6 to n = 16, and for p = 7 to n = 21. The latter effort was taken in connection with [4, Conjecture 5.41] which we confirmed for p ∈ [7] \ [2] and n ∈ p2 0 . We now present a couple of other challenges. In the sequel we will assume, if not otherwise stated, that p ∈ N4 and n ∈ N2 . Note that Hp0 = Sp0 is trivial and Hp1 ∼ = Kp ∼ = Sp1 are the complete graphs, for which (almost) all is known.
2
Eccentricities
For a study of eccentricities in Sierpi´ nski graphs, see [7], where all the corresponding answers to the subsequent problems can be found. Eccentricities in Hanoi graphs are being calculated on a large, but, of course, limited scale in our project pr87mo at SuperMUC (Garching, Germany). (i) Calculate ε(0n ) in Hpn . (ii) Determine the average eccentricity ε(Hpn ) = p−n
s∈V
(iii) Find a general formula for ε(s), s ∈ V
ε(s).
(Hpn )
(Hpn ).
(iv) Construct an efficient algorithm to produce shortest path(s) from s to 0n in Hpn . (v) Give formulas for the radius of Hpn and find the center C(Hpn ).
3
Largest Disc Moves
The problems in this section have been solved for Sierpi´ nski graphs in [3]. For Hanoi graphs one can find some partial analytical solutions and the results of
A.M. Hinz / Electronic Notes in Discrete Mathematics 63 (2017) 23–31
27
computer experiments in [1]. (i) How many LDMs are included in a shortest path from s to t in Hpn ? (ii) Construct an algorithm which determines the LDMs included in a(ny) shortest path from s to t in Hpn .
4
Diameters
For Sierpi´ nski graphs we always have d(0n , 1n ) = 2n − 1 6 and since this is also an upper bound for eccentricities, diam(Spn ) = 2n − 1. In Hanoi graphs the situation is much more intricate. For instance, for p = 4, Korf’s phenomenon in its weaker form, namely ε(0n ) > F S(H4n ), has been confirmed by computer experiments for n ∈ {15, 20, 21, 22, 23, 24, 25}, but the behavior of ε(0 n)−F S(H4n) is rather eratic. The Korf-Felner conjecture is that this function is positive for n ≥ 20. Diameters of H4n have been computed by R. Korf so far for n ≤ 17 with the result that diam(H4n ) = F S(H4n ) in that interval with the exception that diam(H415 ) = F S(H415 ) + 1 (= ε(015 )). In particular, we know of no instance where ε(0n ) < diam(Hpn ), i.e. that 0n does not belong to the periphery of Hpn . Therefore, most of the following is open. (i) For which p and n are inequalities in d(0n , 1n ) ≤ ε(0n ) ≤ diam(Hpn ) strict? (ii) Find diam(Hpn ) and the periphery P (Hpn ).
(iii) Determine the average distance d(Hpn ) = p−2n
d(s, t).
(s,t)∈V (Hpn )2
(iv) Construct an efficient algorithm to produce shortest path(s) from s to t in Hpn . (v) Determine the average distance of s ∈ V (Hpn ), i.e. d(s) = p−n d(s) where
d(s) = d(s, t) . t∈V (Hpn )
(vi) Find the proximity of Hpn and the median M (Hpn ). To get an impression how difficult problem (iii) may be, have a look at the corresponding formula for Sierpi´ nski graphs: d(Spn ) = 6
(p − 1)(2p4 + 6p3 − 17p2 + 26p − 16) n 2 p(2p − 1)(p3 + 4p2 − 4p + 8)
This is why vertices of the form k n are called extreme vertices in Sierpi´ nski graphs.
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p2 + 3p − 6 p−2 + p−n p (2p − 1)(p2 − 7p + 8) n λp,+ p(p − 1)γp,+ − 2 2(p − 7p + 8)(p3 + 4p2 − 4p + 8) p2 n λp,− p(p − 1)γp,− − 2 , 2(p − 7p + 8)(p3 + 4p2 − 4p + 8) p2 −
where αp = p4 − 12p3 + 56p2 − 104p + 68 , 1 1√ αp , λp,± = p2 − p + 1 ± 2 2 √ γp,± = (p2 + 3p − 2) ∓ (p4 + p3 − 30p2 + 58p − 36) αp . Whereas (iv) has been solved in [3] for Sierpi´ nski graphs, problems (v) and (vi) are still open also for them, even for p = 3. For p = 2 we know that S2n ∼ = P2n , so we start by looking at path graphs P1+k , k ∈ N. Let the vertex set be V = [k + 1]0 and vertices being neighbors iff they are neighbors on the number line. An easy calculation shows that ∀ v ∈ V : d(v) = k(k+1) − v(k − v), so that d(v) = k2 − v(k−v) . For odd k 2 k+1 k+1 the minimum prox(P1+k ) = 4 is achieved for v ∈ C(P1+k ) = k−1 . , k+1 2 2 n n n−2 n n Putting k = 2 − 1, we obtain prox(S2 ) = 2 . Moreover, M (S2 ) = C(S2 ) = {01n−1 , 10n−1 }.
5
Domination
All Sierpi´ nski graphs possess perfect codes and therefore their domination n +p n +1 number is known to be γ(Spn ) = pp+1 , if n is even, and γ(Spn ) = pp+1 , if n 2 n is odd. Similarly, we find that γ(Hp ) = p. But otherwise Hp do not contain perfect codes. (i) Determine the domination number of Hpn for n ∈ N3 .
6
Metric Dimension
Due to the complete mastery of metric properties of Sierpi´ nski graphs, their metric dimension can be deduced to be μ(Spn ) = p − 1 for n ∈ N with any p − 1 extreme vertices forming a basis. Also μ(Hp2 ) = p − 1 with every p − 1 out of the perfect states k 2 , k ∈ [p]0 , forming a basis. This can be seen as follows. By [4, Theorem 5.34] there are just 2 equisets (symmetry classes) of vertices, represented by 00 and 01 (cf. [4,
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Figure 5.6] for p = 4). The distance representation of 00 with respect to all perfect states consists of one 0 and otherwise 3s, so that the position of the 2-tower can be reconstructed. For 01 the distance representation contains one 1, one 3 and otherwise 2s; the position of the 1 shows the peg where disc 2 is lying, the position of the 3 gives the peg of disc 1. It is also clear from this, that p − 1 distances suffice to reconstruct the pth. Finally, to see that a set U with |U | = p − 2 can not be resolving for p ≥ 4, one has to investigate all 13 equisets of tasks (cf. [4, Theorem 5.35]), i.e. pairs of vertices, and show that for each such pair there are two different vertices with equal distance representation. For instance, if U = {02 , 12 }, then 22 and 32 have the same distance representation with respect to U , namely dU (22 ) = (3, 3) = dU (32 ). Again, a look at [4, Figure 5.6] facilitates the job for all pairs; the pegs with labels larger than 3 do not play a role. However, for Hp3 not even the whole set of perfect states U = {03 , . . . , (p − 1)3 } is resolving! This is so because (cf. [4, Figure 5.10]) dU (100) = (5, 3, 4, . . . , 4) = dU (101) . So we are left with (i) Determine the metric dimension of Hpn for n ∈ N3 and find a basis in these graphs.
7
Genera and Crossing Numbers
Hanoi graphs and Sierpi´ nski graphs share the same domain of planarity, namely p ≤ 3 or n = 0 or (p = 4 and n ∈ {1, 2}). For n = 1, i.e. the complete graphs Kp , the genera are known (Ringel and Youngs, 1968) and Guy and others made the conjecture about the crossing number to be p−2 p−3 1 p p − 1 cr(Kp ) = . 4 2 2 2 2 This conjecture has so far been confirmed for p ≤ 12. The only other provable value for our non-planar graphs is cr(S43 ) = 12. (i) Determine g(Hpn ) and g(Spn ) for p = 4 and n ∈ N3 and for p ∈ N5 and n ∈ N2 . (ii) Determine cr(H43 ) (known to be ≤ 72) and cr(Hpn ) and cr(Spn ) for p = 4 and n ∈ N4 , for p ∈ N5 and n ∈ N2 and for p ∈ N13 and n = 1.
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8
Further Questions
There are, of course, many more properties of Hanoi graphs and Sierpi´ nski graphs to be explored. For instance, colorings have been considered in [6], where the classical coloring numbers—vertex, edge and total—were determined for both classes with the result that they behave quite similarly. It would be interesting to extend these findings to some of the meaningful types of colorings and so-called labellings (cf. [5, Section 4.2]). Other challenges in connection with the Tower of Hanoi and with graphs deriving from variations of the puzzle (see [4, Chapters 3 and 6 to 8]) can be found in [4, Chapter 9]. (Some already solved!) For variants of Sierpi´ nski graphs, see [5]. Let us finally mention that computational methods also led to the solution of some very old tasks asked in English magazines by Henry Ernest Dudeney, the most outstanding being d(020 , 120 ) = 111 = F S520 in H520 ; see [8, Proposition 1]. There remains Dudeney’s Last Question, namely the value of d(035 , 135 ) in H535 . To verify computationally that it is equal to F S535 = 351 is out of the reach of current computing devices, so an analytical solution is needed! Acknowledgements. I thank Ciril Petr (Maribor) for his work on our project pr87mo at SuperMUC. We are both supported by the Slovenian Research Agency (research core funding No. J1-7110) which also provided funds for my travel to the International Conference on Current Trends in Graph Theory and Computation held in September 2016 in New Delhi. I am grateful to Deepa Sinha (South Asian University, New Delhi) and Purnima Gupta (University of Delhi) for their generous hospitality during my extended visit and to Shariefuddin Pirzada (Srinagar) for his support. The cheerful volunteers made my stay even more enjoyable.
References [1] Aumann, S., K. A. M. G¨ otz, A. M. Hinz, and C. Petr, The number of moves of the largest disc in shortest paths on Hanoi graphs, Electron. J. Combin. 21 (2014), P4.38. [2] Bousch, T., La quatri`eme tour de Hano¨ı, Bull. Belg. Math. Soc. Simon Stevin 21 (2014), 895–912. [3] Hinz, A. M., and C. Holz auf der Heide, An efficient algorithm to determine all shortest paths in Sierpi´ nski graphs, Discrete Appl. Math. 177 (2014), 111–120.
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[4] Hinz, A. M., S. Klavˇzar, U. Milutinovi´c, and C. Petr, “The Tower of Hanoi— Myths and Maths,” Springer, Basel, 2013. [5] Hinz, A. M., S. Klavˇzar, and S. S. Zemljiˇc, A survey and classification of Sierpin ´ski-type graphs, Discrete Appl. Math. 217 (2017), 565–600. [6] Hinz, A. M., and D. Parisse, Coloring Hanoi and Sierpi´ nski graphs, Discrete Math. 312 (2012), 1521–1535. [7] Hinz, A. M., and D. Parisse, The Average Eccentricity of Sierpi´ nski Graphs, Graphs Combin. 28 (2012), 671–686. [8] Hinz, A. M., and C. Petr, Computational Solution of an Old Tower of Hanoi Problem, Electron. Notes Discrete Math. 53 (2016), 445–458.