THE NEW VARIATION ON TOWERS OF HANOI 汉诺塔新变形的研究

THE NEW VARIATION ON TOWERS OF HANOI 汉诺塔新变形的研究 Team members: Hong Xuechen, Wang Zhiqi Tutor: Wang Kai School: Jiangsu Tianyi High School, Jiangsu, China

THE NEW VARIATION ON TOWERS OF HANOI 汉诺塔新变形的研究 Abstract French mathematician Edouard Lucas introduced Towers of Hanoi puzzle in 1883. Nine years later, W.R.Ball used a recursive algorithm to solve it perfectly. It has been more than one hundred years since the classical Towers of Hanoi was proposed, but modern mathematicians still put great emphasis on this problem and have suggested different variations on the classical Towers of Hanoi. We have seen a lot of variations on Towers of Hanoi. These different variations sparkle our interest and urge us to propose another new variation on the classical Towers of Hanoi puzzle. Many of these variations reduce the strict rules of the classical puzzle to yield more possible moving methods. Thus, we become more curious about this topic and have also been thinking to reduce the restrictions on the classical Towers of Hanoi. We introduce a new variation: changing the maximum number of disks that can be placed on one disk in this paper. After putting relentless effort into the research on our new variation, we can find out a method of moving disks that requires fewer steps and makes the recreational mathematic puzzle more interesting. By using mathematical induction, we prove the recursion formulas and the formulas for general terms of our new variation as following (suppose that f(n) represents the minimum steps of moving disks to complete this variation on Towers of Hanoi which has n disks): n = 1, f(1) = 1 n is a positive even integer recursion formula: f(n + ) = f(n) + formula for general term: f(n) = n+ n is a positive odd integer except 1 recursion formula: f(n + ) = f(n) + 1 formula for general term: f(n) = n+1 + n 3 1 Furthermore, we offer general optimal patterns of moving disks and a new algorithm to obtain the formulas for general terms. Keywords: Towers of Hanoi; variation; recursion formulas; formulas for general terms; optimality; mathematical induction; patterns of moving disks

1 Introduction 1.1 The classical puzzle and different variations Towers of Hanoi puzzle originally derives from a mysterious legend in India. This legend delineates an Indian temple in Kashi Vishwanath 1 which contained a hall with three time-worn posts surrounded by 64 golden disks. Towers of Hanoi, a recreational math puzzle, was proposed by the French mathematician Edouard Lucas. The initial configuration consists of three posts A, B, C and n disks of different sizes which are placed on post A in increasing order of size from top to bottom (i.e. the smallest on the top, the biggest on the bottom). The challenge is to move all the disks from post A to post C with the help of intermediate post B. At the same time the following four rules must be obeyed: 1) Only one disk can be moved at a time. ) Moving one disk from a post to another one is defined as one step. 3) A bigger disk can never be placed on the top of a smaller one. 4) Final configuration: n disks should be placed in the same sequence as that in the initial configuration. In 189, W.R.Ball utilized an optimal recursive algorithm and concluded that the minimum number of steps is n 1 [1]. Many variations of this puzzle have been proposed, in which the set of allowable moves has been extended or restricted. For example, Jeremy Barbay changed the insertion and removal point in each tower to invent Bouncing Tower []. Jonathan Chappelon, Urban Larsson and Akihiro Matsuura studied extension of the one-player setting to two players of Hanoi Towers, invoking classical winning conditions in combinatorial game theory [3].The Apprentices' Towers of Hanoi proposed by Cory B.H.Ball and Robert A.Beeler allows placing a larger disk on the top of a smaller one as long as all the other disks on the post are in increasing order of their diameter [4]. In addition, John McCarthy from the University of Stanford proposed the Towers of Stanford puzzle in which a smaller disk can be placed on the top of a bigger one provided that the disk on the bottom of that stack is the largest one on the post [5]. Furthermore, Lee Badger and Marc Williams from Weber State University offered the solution n n + 1 for the minimum steps to complete the Towers of Stanford with n disks [6]. Additionally, The super towers of Hanoi problem by Carole S.Klein and Steven Minsker stipulated that the distribution of disks was completely random in the initial configuration and the movement of disks should obey the first three rules of the classical Towers of Hanoi [7]. 1. Our research and process We try to claim a new variation and can get interesting results. The classical Towers of Hanoi puzzle has strict restrictions. We reduce the restrictions on the method of moving steps in this variation, which yields more feasible approaches. Therefore, we need to consider the optimality of more possibilities to find out the minimum steps and have a deeper analysis on finding out the optimal method of moving disks. Furthermore, it is essential for us to prove it, which is the core in this passage. Additionally, we find out the algorithm of the new variation is similar to the original one and there is a special relationship between the optimal patterns of moving disks in the classical puzzle and our new variation. The following is our research process: Firstly, we redefine the maximum number of disks that can be put on a disk. Secondly, we prove the optimality of a possibility and then apply mathematical induction to obtain the recursion 1 The name of an Indian temple. François Édouard Anatole Lucas was a French mathematician. He proposed Towers of Hanoi puzzle in 1883.

formulas and formulas for general terms. Thirdly, we state the optimal patterns of moving disks and discover the relationship of optimal moves between our variation and the classical puzzle. The new variation We change part of the rules and consider another variation on the Towers of Hanoi puzzle. The initial configuration is the same as classical Towers of Hanoi. Rules 1) and ) in the classical Towers of Hanoi puzzle should also be obeyed in this new variation. However, we redefine rule 3) and 4): 1) Only one disk can be moved at a time. ) Moving one disk from a post to another one is defined as one step. 3) n disks are marked as disks 1,, 3 n from the smallest disk to the biggest one. At most i disks can be put on the top of the disk which is marked as i ( 1 i n and i is an integer ) in any intermediate configurations. So in some configurations, a larger disk can be placed on a smaller one. 4) Final configuration: n disks are all placed on post C and their sequence only needs to obey the new rule 3). 3 Proof: mathematical induction 3.1 Proof for optimality In this passage we define the optimal way as the approach which requires the fewest steps. Additionally, function f(n) in terms of n stands for the minimum number of steps transporting n disks from post A to post C. From the statistics in the Appendix, we discover that for a certain f(n + ), we deal with the move of two disks at a time. Additionally, disk n + 1 is always on the bottom of post C. Disk n + is always adjacent to disk n + 1 and only can be placed on the top of disk n + 1 in the optimal final configuration (a vertical line stands for a post in all following figures): the first n disks disk n+ disk n+1 Figure 3-1 We first prove the optimality of the final configuration above: According to the new rule 3), for a certain f(n + ), only disk n + 1 and disk n + can be placed on the bottom of post C in the final configuration. Since disk n + 1 and disk n + are placed on the bottom of the post A in the initial configuration, we must make sure all of the first n disks are moved to post B and guarantee that post C is completely empty in order to move disk n + 1 or disk n + to the bottom of post C. As a result, the following three intermediate situations (X,Y and Z) are possible after moving disk n + 1 (the sequence of the first n disks is certain, optimal and the same in situation X and Y):

X disk n+1 disk n+ the first n disks A B Figure 3- C Y disk n+ the first n disks disk n+1 A B Figure 3-3 C Z part 1 disk n+1 disk n+ part A B (Part 1 and part are combinations of some disks, part 1+ part = the first n disks) Figure 3-4 C It is clear to see that disk n + 1 and disk n + can be moved only after all of the other n disks are removed from post A. So in order to attain situation Z, we need to first extract disk n + 1 from post A and then plug it back into the middle of the first n disks. However, we only need to place disk n + 1 on the top of the first n disks. Therefore, situation X is better than situation Z. Then, we need to compare the optimality of situation X and Y. These two situations can attain the following two intermediate situations *X and *Y respectively:

*X the first n disks disk n+1 disk n+ A B Figure 3-5 C *Y the first n disks disk n+ disk n+1 A B Figure 3-6 C Both situations *X and *Y need to move the first n disks from post B to post C to attain different final configurations which meet the requirements. Since the optimal way to move the first n disks is certain, the two situations need the same number of steps to attain the viable final configuration. We need to move two more steps from X to *X: move disk n + from post A to post C and then move disk n + 1 from post B to post C. However, we only need to move one more step from Y to *Y: move disk n + from post A to post C. As a result, situation Y is better than situation X. We obtain the conclusion that situation Y is the optimal one. 3. Proof for even terms Then we use mathematical induction to prove the recursion formula for even terms: f() = According to the Figure 3-3, 3-6 and 3-1, we can find the relationship between f(n + ) and f(n). For a certain even integer n, the minimum steps of moving n + disks from post A to post C can be represented by the sum of the following three parts: 1) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from post A to B ) the steps of moving the two disks which are marked as n + 1 and n + from post A to the bottom of post C 3) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from B to C The initial configurations of the first n disks in the above part 1) and f(n) are the same: disks 1,, 3 n are

placed on post A in increasing order of size from the top to the bottom. Since the optimal move for n disks in the same sequence is certain, the methods of moving disks in part 1) and f(n) are completely identical and they require the same number of steps. In addition, the steps required in the above part ) is. Then we need to figure out the minimum number of steps required in part 3). It is obvious that the steps of part 1) and 3) are continuous manipulations when we only pay attention to the position change of the first n disks. As a result, the initial permutation of disks in part 3) is the same as the ultimate one in part 1). Since the optimal move is certain for n disks, moving the first n disks back into the permutation in the beginning of part 1) is also the optimal way. Therefore, part 3) and part 1) are exactly inverse manipulations, which indicates the number of steps in part 3) equals the number of steps required to move n disks in both part 1) and f(n). From the induction above we can conclude that f(n + ) = f(n) + where n is a positive even integer. According to the proven recursion formula f(n + ) = f(n) + for even integers, we can estimate the function: f(n) = n+ We also apply mathematical induction to prove this general formula for positive even integers. Proof: 1. n = f() = + = This case satisfies the general formula for positive even integers.. We suppose that the case n = k (k can be any positive even integer except ) satisfies the general formula above. Then f(k) = k+. 3. n = k + According to the case n = k and the recursion formula, we can obtain that f(k + ) = f(k) + = ( k+ ) + = (k+)+. It satisfies the general formula f(n) = n+ Finally, we can elicit the conclusion that the recursion formula is f(n + ) = f(n) + and the general formula is f(n) = n+ (n is a positive even integer). 3.3 Proof for odd terms After excluding f(1) = 1, we utilize mathematical induction to prove the recursion formula for odd terms in a similar way. f(3) = 4 According to the Figure 3-3, 3-6 and 3-1, we can find the relationship between f(n + ) and f(n). For a certain odd integer n, the minimum steps of moving n + disks from post A to post C can also be represented by the sum of the following three parts: 1) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from post A to B

) the steps of moving the two disks which are marked as n + 1 and n + from post A to the bottom of post C 3) the steps of moving the first n disks (except the two disks which are marked as n + 1 and n + ) from B to C The same procedure in Chapter 3. (even integer terms) can be easily adapted to obtain that the number of steps in part 1) equals f(n) and part ) needs two steps to complete. Then we compare part 1) with part 3). We first neglect the existence of the smallest three disks among the first n disks (disks 1, and 3) since their positions are more changeable. The same method in Chapter 3. (inverse manipulations) can be utilized to obtain that the number of steps to move the n-3 disks in the middle are identical in part 1) and part 3). Then we focus on disks 1, and 3 which are ignored in the previous discussion. The first time to move these three disks from post A requires four steps according to the optimal situation when n = 3 that has been listed as f(3) = 4 in the Appendix. By listing all the possible methods of moving three disks, there are four different viable final configurations of the smallest three disks (from top to bottom) after switching their position for the first time: a) 1 3 b) 1 3 c) 3 1 d) 1 3 At least 5, 4, 4, 6 steps are required to attain a), b), c) and d) configurations respectively. We first consider the position of disk 1. If disk 1 is on the top, it will require more than three steps to move all the three disks to another post. Otherwise disk 1 will be on the bottom among the three disks, which contradicts the new rule. Therefore, placing disk 1 in the middle of disk and disk 3 is the most optimal one. As a result, situation c) and d) are more optimal beginnings for further moves. Since three steps are required to move the first three disks from one post to another each time in both permutations (3 1 ) and ( 1 3), we claim that we need at least three steps every time to move all the three disks from one post to another except the first time. Compared to d), situation c) is more optimal since it can be obtained more easily. (According to the case n=3 in the Appendix) Then, we can compare the steps of moving disks 1, and 3 in part 1) and 3). Since we need to move n disks in both part 1) and part 3) and the optimal method of moving is certain (for odd terms, the optimal method of moving is to move two disks at a time except the smallest three disks), we need to switch the position of disks 1, and 3 for the same number of times. Here we define the number of times to switch the position of the smallest three disks in both part 1) and 3) as m. From the poof above, we know that the first time to move the smallest three disks from post A in the initial configuration needs four steps and every following move of the three disks requires three steps. Hence, we can calculate the total steps of moving disks 1, and 3 in part 1) and 3). The number of steps in part 1) equals 4 + 3(m 1) = 3m + 1 And the number of steps in part 3) equals 3m As a result, the total steps of moving three smallest disks in part 1) require one more step than those in part 3). Since we have proved that the number of steps required to move the n-3 disks in the middle are identical in part 1) and part 3), we can conclude that the total number of steps in part 1) equals the total number of steps in part 3) plus1 step. As we claimed before in this Chapter, the methods of moving disks in part 1) and f(n) are the same. Therefore, the total number of steps in part 3) is f(n) 1. Finally, we can draw out our conclusion that f(n + ) = f(n) + 1 According to the proven recursion formula, we can also estimate the formula for general odd terms except n = 1: f(n) = n+1 + n 3 1 (n is a positive odd integer and n 3)

We exploit mathematical induction to prove the general formula for odd terms except n=1. Proof: 1. n = 3 f(3) = 3+1 + 3 3 1 = 4 This case satisfies the general formula for positive odd integers.. We suppose that the case n = k (k can be any positive odd integer except 1 and 3) satisfies the general formula above. Then f(n) = k+1 + k 3 1. 3. n = k + According to the case n = k and the recursion formula, we can obtain that f(k + ) = f(k) + 1 = ( k+1 + k 3 1) + 1 = (k+1)+ + (k 3)+ 1 It satisfies the general formula f(n) = n+1 + n 3 1. Ultimately, we concluded that the recursion formula is f(n + ) = f(n) + 1 and the general formula is f(n) = n+1 + n 3 1 (n is a positive odd integer except 1). 3.4 Results of Proof Here we demonstrate our results of proof: Theorem: n = 1, f(1) = 1 n is a positive even integer recursion formula: f(n + ) = f(n) + formula for general term: f(n) = n+ n is a positive odd integer except 1 recursion formula: f(n + ) = f(n) + 1 formula for general term: f(n) = n+1 + n 3 1 We compare the results in our new variation with those in classical Towers of Hanoi puzzle. The two results are similar in some ways: both are related to the exponent of. There is a special relationship between the optimal patterns of moving disks for classical puzzle and for our new variation. This relationship will be stated in Chapter 4. However, unlike the results on classical Towers of Hanoi, our results are different for even and odd positive integers since the optimal methods of moving are a bit different in even and odd cases. In addition, the numerical value in our results for every term is smaller than that in the classical case, which indicates the new variation needs fewer steps. These differences are rational because the restrictions in our new variation are fewer.

4 The optimal patterns of moving disks 4.1 Definition of dividing group We define a new term called group. A group can contain j continuous adjacent disks in the initial configuration ( j n and j is a positive integer). It is not necessary that every group contains the same number of disks. The permutation of the disks in one group can be changed in intermediate positions, but it is on the basis of obeying the four rules in the new variation. Additionally, the disks in one group cannot be separated in intermediate positions and they are viewed as a whole. 4. Pattern for even terms We offer a general pattern for even terms in which n disks can be moved from post A to post C in the optimal way: In the initial configuration for even number of disks, we divide every two adjacent disks as a group from the top to the bottom and we mark them as groups 1,, 3 n (group 1 consists of disk 1 and disk, group n consists of disk n 1 and disk n). Here we prove a lemma. Lemma: The n groups in even cases should be placed in increasing order of size (from group 1 on the top to group n on the bottom) in any intermediate and final configurations. Proof: It is obvious that group n is on the bottom, otherwise the configuration will oppose to the rule 3)3 in the new variation. Since the disks in one group can't be separated, group n can only be placed above group n. Then we consider group n 4. It can be placed either in the middle of group n and group n or above group n. When group n 4 is middle of group n n and group, there can't be any groups in the middle of group n 4 and group n since the largest disk in the other groups is disk n 6 and the maximum number of disks on this disk is only n 6. As a result, the other groups are all above group n 4. In this case, we know that there are at least n 4 disks above disk n 5 in group n 4, which contradict the rule 3) in the new variation. Consequently, group n 4 can only be placed above group n n 6. In the same manner, group must be placed above group n 4 n 8, group result, we have proved the sequence of the groups. must be placed above group n 6 group 1 must be placed above group. As a Since the sequence of the groups is certain and requirements of the groups are the same as those in the classical Towers of Hanoi puzzle, the groups must obey the rule of the classical puzzle. According to the minimum steps of 3 The rule 3) was stated in Chapter.

moving n disks in the classical Towers of Hanoi puzzle 4, the minimum steps of moving n groups equals n 1 Since two steps are required to move the disks in a group when one group is moved once, the minimum steps of moving groups equals ( n 1) = n+ = n+ The result is the same as the formula for even terms from Theorem in Chapter 3.4.We also know the optimal pattern of moving even number of disks: move n groups in the same way as the optimal one in the classical Towers of Hanoi puzzle after defining every two adjacent disks in the initial configuration from top to bottom as a group. Furthermore, we find out there is a relationship between our algorithm and that in the classical Towers of Hanoi puzzle, whose optimal pattern is also related to ours: the results of our new variation can be calculated by utilizing W.R.Ball's recursive algorithm; the optimal pattern in the classical puzzle can be applied to move the disks in our optimal way in our the new variation. 4.3 Pattern for odd terms We also offer a general pattern to move odd number of disks in the fewest steps (n 5): In the initial configuration for odd number of disks, we divide every two adjacent disks except the three disks on the top (disks 1, and 3) as a group from the top to the bottom and we mark them as groups 1,, 3 n 3 (group 1 consists of disk 4 and disk 5, group n-3 consists of disk n 1 and disk n). According to the lemma in Chapter 4., the first n 3 groups can only be placed in increasing sequence of size in any configuration. Like Chapter 4., we can also employ the recursive algorithm presented by W.R.Ball from the classical puzzle's solution to calculate the fewest number of steps to move these n 3 groups which is n 3 1 The number of steps required is also two when we move one group once. Hence, the minimum steps of moving these n-3 disks can be represented as ( n 3 1) = n 3 Now we consider the moves of the smallest three disks (disks 1, and 3). According to the optimal way to move the three disks in Chapter 3.3, the first time to move these three disks require four steps and we require three steps to move the three disks every time except the first time. From the statistics listed in the Appendix, the numbers of times to move all the three smallest disks from one post to another for cases n = 5, 7, 9, 11 are, 4, 8, 16 respectively. Thus, we can conclude that the number of times to move the three disks is for odd integers (n 5). The total number of steps to move disks 1, and 3 is n 3 4 The result from W.R.Ball in 189 was stated in Chapter 1.1.

4 + 3( n 3 1) = 3 n 3 + 1 Hence, the minimum number of steps to move odd number of disks is ( n 3 ) + (3 n 3 + 1) = 5 n 3 1 The general formula for odd integers is also the same as that in Theorem (Chapter 3.4) since 5 n 3 1 = 4 n 3 + n 3 1 = n+1 + n 3 1 = n+1 + n 3 1 Additionally, we find out the optimal pattern of moving disks for odd terms: define every two adjacent disks as a group from top to bottom except disks 1, and 3 and move these n 3 groups in the same way as the optimal pattern in the classical Towers of Hanoi. Our algorithm and optimal pattern for odd terms are also related to W.R.Ball's algorithm and his optimal pattern: the algorithm of the new variation for odd terms is based on the original recursive algorithm presented by W.R.Ball and the optimal pattern of moving the n 3 as that in the classical Towers of Hanoi puzzle. groups is the same 5 Conclusion This new variation on Towers of Hanoi is originally proposed by us. We have also found out the minimum number of steps and proved it. Meanwhile, we offer general optimal patterns of moving disks. Compared to the results in the classical Towers of Hanoi puzzle, the steps required in our variation are fewer. In the future, we may do further research to look for applications. References: [1] W.R.Ball. Mathematical Recreations and Essays. McMillan, London, 189. [] Jeremy Barbay. Bouncing Towers move faster than Hanoi Towers, but still require exponential time. Computer Science, 016. [3] J.Chappelon, U.Larsson, A.Matsuura. Two-Player Tower of Hanoi, Computer Science, 015. [4] B.H.Ball, R.A.Beeler. The Apprentices' Tower of Hanoi. Journal of Mathematical Science, 016. [5] J.McCarthy. The Tower of Stanford, Problems and Solutions, American Mathematical Monthly, Vol.109, Number 7 (August-September 00), 664. [6] L.Badger, M.Williams.The Tower of Stanford, Problems and Solutions, American Mathematical Monthly, Vol.111, Number 4 (April 004), 364-365. [7] C.S.Klein, S.Minsker. The super towers of Hanoi problem. Discrete Mathematics. 114 (1993), 83-95.

Appendix: The patterns of the first eleven terms We search for general patterns about the new variation by listing the final configuration, the optimal way to attain the final configuration and the minimum numbers of steps for n = 1,, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively since we can find the patterns after 11 terms. (For convenience, we use i K J to denote every step. It means disk number i is moved from post K to post J, 1 i n and i is an integer, K and J can be post A, B or C): n = 1(1step) 1 A C Final configuration ( from top to bottom):1 n = ( steps ) 1 A C A C Final configuration (from top to bottom): 1 n = 3(4steps) 1 A B A C 1 B C 3 A C Final configuration (from top to bottom): 3 1 n = 4 ( 6 steps ) 1 A B A B 3 A C 4 A C B C 1 B C Final configuration (from top to bottom): 1 4 3 n = 5 ( 9 steps ) 1 A C A B 1 C B 3 A B 4 A C 5 A C 3 B C 1 B C B C Final configuration (from top to bottom): 1 3 5 4 n = 6 ( 14 steps ) 1 A C A C 3 A B 4 A B C B 1 C B 5 A C 6 A C 1 B A B A 4 B C 3 B C A C 1 A C Final configuration ( from top to bottom ): 1 3 4 6 5 n = 7 ( 19 steps ) 1 A B A C 1 B C 3 A C 4 A B 5 A B 3 C B 1 C B C B 6 A C 7 A C B A 1 B A 3 B A 5 B C 4 B C 3 A C 1 A C A C Final configuration ( from top to bottom ): 1 3 4 5 7 6 n = 8 ( 30 steps ) 1 A B A B 3 A C 4 A C A C 1 A C 5 A B 6 A B 1 C A C A 4 C B 3 C B A B 1 A B 7 A C 8 A C 1 B C B C 3 B A 4 B A C A 1 C A 6 B C 5 B C 1 A B A B 4 A C 3 A C B C 1 B C Final configuration ( from top to bottom ): 1 3 4 5 6 8 7 n = 9 ( 39 steps ) 1 A C A B 1 C B 3 A B 4 A C 5 A C

3 B C 1 B C B C 6 A B 7 A B C A 1 C A 3 C A 5 C B 4 C B 3 A B 1 A B A B 8 A C 9 A C B C 1 B C 3 B C 4 B A 5 B A 3 C A 1 C A C A 7 B C 6 B C A B 1 A B 3 A B 5 A C 4 A C 3 B C 1 B C B C Final configuration ( from top to bottom ): 1 3 4 5 6 7 9 8 n = 10 ( 6 steps ) 1 A C A C 3 A B 4 A B C B 1 C B 5 A C 6 A C 1 B A B A 4 B C 3 B C A C 1 A C 7 A B 8 A B 1 C B C B 3 C A 4 C A B A 1 B A 6 C B 5 C B 1 A C A C 4 A B 3 A B C B 1 C B 9 A C 10 A C 1 B A B A 3 B C 4 B C A C 1 A C 5 B A 6 B A 1 C B C B 4 C A 3 C A B A 1 B A 8 B C 7 B C 1 A C A C 3 A B 4 A B C B 1 C B 6 A C 5 A C 1 B A B A 4 B C 3 B C A C 1 A C Final configuration ( from top to bottom ): 1 3 4 5 6 7 8 10 9 n = 11 ( 79 steps ) 1 A B A C 1 B C 3 A C 4 A B 5 A B 3 C B 1 C B C B 6 A C 7 A C B A 1 B A 3 B A 5 B C 4 B C 3 A C 1 A C A C 8 A B 9 A B C B 1 C B 3 C B 4 C A 5 C A 3 B A 1 B A B A 7 C B 6 C B A C 1 A C 3 A C 5 A B 4 A B 3 C B 1 C B C B 10 A C 11 A C B A 1 B A 3 B A 4 B C 5 B C 3 A C 1 A C A C 6 B A 7 B A C B 1 C B 3 C B 5 C A 4 C A 3 B A 1 B A B A 9 B C 8 B C A C 1 A C 3 A C 4 A B 5 A B 3 C B 1 C B C B 7 A C 6 A C B A 1 B A 3 B A 5 B C 4 B C 3 A C 1 A C A C Final configuration ( from top to bottom ): 1 3 4 5 6 7 8 9 11 10 From the statistics above, we can obtain the following table: n 1 3 4 5 6 7 8 9 10 11 Steps 1 4 6 9 14 19 30 39 6 79 Table 1 According to the table, we can find out that there are different patterns for even and odd terms except n=1.