Amidakuji and Games Steven T. Dougherty Jennifer Franko Vasquez Department of Mathematics University of Scranton Scranton, PA 18510 USA October 10, 2011

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Japanese Ladders

Combinatorial mathematics often lends itself to constructing interesting puzzles. We shall describe here some games that arise from permutations. Consider a collection of people standing in a row. You want to rearrange the people to a specified order but the only tool at your disposal is switching people two at time, but only if they are standing next to each other. Take the added constraint that you must do it in a minimal number of steps and you have the first game we shall describe. There is a natural visual description of this game which makes the game fun to play and easy to learn. Permutations are the base of our games. They are at the heart of combinatorial mathematics and also appear in many areas of recreational mathematics. For example, every line of a Sudoku puzzle is a permutation of the numbers from 1 to 9. In this paper, we describe a very visual way of understanding permutations using a traditional Japanese technique and then we will describe three interesting combinatorial games that arise in this setting. These games, while easy to learn to play, can be quite challenging and fun. The first game was given as an example of a mathematical game in [1]. Moreover, they are related to some deep mathematics like the Braid group [2] and Markov chains [3]. The Japanese technique is called Amidakuji (Japanese ladders). It is used in Japan to construct a bijective map from a set to itself. For example, if five people wanted to split up five chores they could use a Japanese ladder to assign the chores to each person. Figure 1 shows an example of a visual representations of this. Each number falls down the ladder on the post where it is placed. When two numbers hit a rung, they must be switched.

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This means that a rung between the i-th and (i + 1)-st post represents the transposition (i, i + 1). Follow each number down the ladder and you can see where each element goes. Then the permutation can be determined by the location where each number ends. 1

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Figure 1: The paths on a simple Japanese ladder

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Game 1

Lets begin by trying a game. Consider the following Figure 2: 1

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4 2 3 1 Figure 2: Game One

The goal is to have 1, 2, 3, and 4 fall down the posts of this ladder and end up in the locations indicated at the bottom. To do this, you place horizontal rungs between the posts such that when meeting a rung, you must cross it. To insure that there is no ambiguity, no two horizontal rungs may touch. Here, 1 starts on the first post then crosses to the second and finally ends on the third. The reader is invited to follow the path of 2 and 3. Going back to Figure 2, we see that 1 needs to get to the fourth post so we can begin by placing rungs as in Figure 3. This ladder does not take 2,3, and 4 to their desired places. We must alter it without affecting the path that 1 takes. To fix 2 we can simply add the rung as in Figure 4. 2

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1 Figure 3: Following 1 1

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2 1 Figure 4: Following 1 and 2

This is not the only way, but it is a way that works. To complete this game, you need to add one more rung. You can do it with more, but we wish to accomplish this with a minimal number of rungs. The reader is now invited to complete this game.

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Mathematics of Game 1

Every permutation on the set {1, 2, . . . , n} can be represented by a Japanese ladder. However, the representation of the ladder is not unique. In fact, there are infinitely many ladders representing each permutation π. It can be shown that there is a unique minimal number of rungs for a given permutation π as follows. Each ladder has a nonnegative number of rungs. Let Aπ = {i ∈ N | there is a ladder representing π with i rungs}. Since Aπ is a nonempty subset of N, there is a least element. Let µ(Aπ ) be this least element. Any ladder representing π with µ(Aπ ) rungs is said to be optimal. Later we 3

give and algorithm to find an optimal ladder. Notice that for a given permutation, the optimal ladder is not necessarily unique. As an example, Figure 5 shows two different ladders representing (1, 3) both of which are optimal. 1

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3 2 1 3 2 1 Figure 5: Equivalent forms of (1, 3)

It is well known that a permutation can be written only in evenly many transpositions or oddly many transpositions. It is immediate then that a ladder that represents an even permutation has evenly many rungs and a ladder that represents an odd permutation has oddly many rungs. It is easy to see graphically that the inverse of a permutation has a ladder that is the ladder of the original permutation turned upside down.

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Game 1 Details

The first game is a purely combinatorial game. Essentially the game is to construct a minimal ladder for a given permutation. We shall now describe what this means. Essentially we shall give the numbers at the bottom of a Japanese ladder, assuming the top are in their natural order, and will ask the player to construct a ladder that represents the described permutation. The procedure for producing the optimal number is not always the most natural or intuitive technique. For example, to construct a ladder representing (1, 3)(2, 4), one may simply construct ladders for (1, 3) and a ladder for (2, 4) and multiply them together. However, this would produce a ladder with at least 6 rungs and it is possible to construct a ladder representing (1, 3)(2, 4) with only four rungs. This ladder is given in figure 6. An inversion is a pair (j, k) such that π −1 (j) > π −1 (k) and j < k. The number of inversions is the number of rungs in an optimal ladder. We are now prepared to describe how to always find an optimal ladder for a given permutation. ! 1 2 ... n Algorithm 1. Given a permutation π ∈ Sn , write π = . Define π(1) π(2) . . . π(n) K = |{(i, j)|i < j, π(i) > π(j)}|. Notice that K is the number of inversions in π. To find the 4

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3 4 1 2 3 4 1 2 Figure 6: Two ladders representing the permutation π = (1, 3)(2, 4)

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π(1) π(2) . . . π(n) . Now, rearrange the top row so that it is in 1 2 ... n order keeping π(i) in the same column as i. Finally define K 0 = |{(i, j)|π(j) < π(i), j > i}|. K = K 0 and gives an explicit algorithm for constructing an optimal ladder for π. inverse of π, consider

This algorithm can be! seen as “combing” the permutation. Given π ∈ Sn , consider π(1) π(2) . . . π(n) and using using transpositions of the form (k, k + 1), move the 1 2 ... n columns so that the top row becomes 1, 2, . . . , n. ! 1 2 3 4 5 Example 1. Given π = (15324) ∈ S5 . Write π = . Then there are 8 5 4 2 1 3 inversions and an optimal ladder can be constructed as follows: (3,4)

(2,3)

(1,2)

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(4,5)

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(4,5)

54213 −−→ 54123 −−→ 51423 −−→ 15423 −−→ 15243 −−→ 12543 −−→ 12534 −−→ 12354 −−→ 12345 Writing the rungs as transpositions, this can be seen as the ladder in Figure 7. 1

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4 3 5 2 1 Figure 7: Example of the algoritm

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4 1 2 Figure 8: Game 1

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Summary of Game 1 : A final state is given at the bottom of n rungs. To win the game you must construct a Japanese ladder that gives that final state and consists of the minimal number of rungs. We now invite the reader to try the game in Figure 8. While the algorithm constructs a solution it is not always the most natural way for a person to play. We have found that people playing the game use a variety of strategies. Some like to draw rungs for the path of 1 and then continue in order to the path for n. In this technique, the player needs to be sure that they do not interfere with previous paths and they do not add extraneous rungs. Some players first identify the element that moves the farthest and fixes that path first. These seem to be the most popular techniques.

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Game 2

This game can be generalized by replacing the rungs with arrows so that there is a positive and negative direction. Then the elements in their final state can be either positive or negative since passing an arrow in the negative direction multiplies the element by −1. For example, we might be given the game in Figure 9. The reader is invited to place arrows between the posts to the have elements reach their final state with the desired signs. A solution to the game in Figure 9 with the fewest arrows will have at least one arrow that goes from left to right and at least one arrow that goes from right to left. The second game is a variation on the first where the final state consists of ±i where i ∈ {1, 2, . . . , n}. That is, the permutation is an enhanced permutation. An enhanced 6

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4 −1 −3 Figure 9: Game 2

permutation is a map π from {1, 2, . . . , n} to {±1, ±2, . . . , ±n} where if π(i) = ±π(j) then i = j. Namely, the function is a permutation on the set {1, 2, . . . , n} with the image may be multiplied by a −1. The number of possible enhanced permutation is n! 2n , since there are n! permutations on n letters and then there are 2 choices of sign for each of the n letters. But not all of these are realizable as ladders. In fact only half are, namely there are n! 2n−1 enhanced permutations that are realizable. As before, consider the set of all possible enhanced ladders representing an enhanced permutation. Each element of this set has a non-negative number of rungs. Therefore, there is a least element. That is for any enhanced permutation there is a minimal number of rungs necessary to build the ladder. This is not necessarily the same as the minimal number of rungs for the underlying permutation. For example, consider the enhanced permutation: 1 → −1, 2 → −2. The underlying permutation is the identity which has a ladder representation with 0 rungs. However, no assignment of arrows to 0 rungs can build the minimal enhanced ladder. The minimal has two rungs as given in Figure 10. 1

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Figure 10: Two Rungs are Needed

Summary of Game 2 : A final state of an enhanced permutation is given at the bottom of n rungs. To win the game you must construct an enhanced Japanese ladder that gives 7

that final state and consists of the minimal number of rungs. We now invite the reader to try the game in Figure 11. 1

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4 1 −2 Figure 11: Game 2

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Notice that each time a transposition is made one element is multiplied by a −1. This means that that if the signs of the elements in the final state are multiplied together then it must be = (−1)number of rungs . This gives the following very important theorem. This theorem distinguishes Game 2 from Game 1. It also give the necessary information to construct a realizable final state. Theorem 3.1. The only allowable final states for an enhanced permutation π for Game 2 have n Y (Sign(π(i)) = (−1)par(π) , i=1

where par(π) is the parity of the underlying permutation of π. Notice that 1 and −1 are not inverses of each other and so this is not a zero sum game. As a final note, we mention that one can construct a third game by adding elements of a specified group to the rungs. Then when crossing a rung in the positive direction with an α on it, the element is multiplied by α and when crossing the rung in a negative direction it is multiplied by α−1 .

References [1] Dougherty, S.T. and Vasquez, J.F., Ladder Games, MAA Focus, June/July, 2001, page 25. 8

[2] Allocca, M.P., Dougherty, S.T., and Vasquez, J.F., Japanese Ladders and the Braid Group, in preparation. [3] Les Lange and James W. Miller, A Random Ladder Game: Permutations, Eigenvalues, and Convergence of Markov Chains, Mathematics Magazine, Vol. 23, No. 5, 1992.

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