The smallest army capable of moving forward eight rows, when diagonal jumps are allowed [P2]. These pages were created and are maintained by George Bell (gibell@comcast.net) Last Modified February 28th, 2015 Copyright © 2006–2015 by George I. Bell ... a good maths problem to do in your head when you don't want to think about something else because you can make it as complicated as you need to fill your brain by making the board as big as you want and the moves as complicated as you want. Mark Haddon, The Curious Incident of the Dog in the Night-Time [B3]

# Introduction

The peg solitaire army problem is a one-player game played on an infinite board in which every square below a certain horizontal line is occupied by a peg (or "man"). It is also known by as "Conway's Soldiers". Play proceeds by jumping a man horizontally or vertically over another man into an unoccupied space, removing the jumped man. Surprisingly, no matter how the game is played, it is impossible for any man to advance more than four rows beyond the initial line.

The armies in the above diagram are capable of reaching their final goal, marked by an "X". It is relatively easy to discover the sequence of jumps that accomplishes this for the first three cases, but the right-most army to level 4 is harder to verify. The ghost men marked "B" represent an alternative to the men marked "A".

The above armies are also of the smallest possible size capable of reaching this target hole. Alternate armies reaching level 4 are shown below: a minimum size army, (20 men, left), and a symmetrical 21-man army (right). Already we can see there is a trade-off between army symmetry and the minimum size army capable of reaching a certain level.

# Types of Army

In the standard version of the game, jumps can only be made along rows and columns. However we can consider various generalizations with other possible jumps, such as diagonal jumps. We can also play the game on other lattice types other than the square lattice. Or, we could play the game in more than 2 dimensions.

The variations we will consider are:

1. Conway's army: Peg solitaire under the usual rules, allowing jumps only along rows and columns. First considered by John Conway in 1961. Analysis of the problem in Winning Ways for your Mathematical Plays, Vol 4 (2004 edition).
2. Skew army: Start from the usual position on a square lattice board, but allow only diagonal jumps.
3. Pablito's army: Peg solitaire on a hexagonal grid on a triangular board. Assemble an army on the lower part of a triangular board, with the top n rows empty. The goal is to jump pegs and finish with the last peg in the topmost hole. The problem was named and introduced as a Macalester College Problem of the Week.
4. Hexagonal Army: Like the previous, only we use an infinite board. We do not force the initial peg distribution to lie on a triangular board, with the finishing hole the top one. This version is equivalent to Conway's army with the addition of diagonal jumps in one direction only. This problem was considered by John Duncan and Donald Hayes [P1].
5. Diagonal army: Conway's army with diagonal jumps in both directions allowed. Considered by Aigner (1997) and Eriksen et. al. (2000).

# Pagoda Functions

The tables below show the powers of σ given to each peg location, where σ ~ 0.618 satisfies σ2 + σ = 1, σ = (sqrt(5)-1)/2. The finishing hole is the one labeled "0", carrying a value of σ0 = 1. The pattern extends indefinitely below and on the sides. The sum of all the values below a certain row must be greater than 1 for the problem to be solvable. Below we show the pattern of exponents and the highest level that can be reached.

The following identities are useful in order to calculate the sum below a certain row. They are only valid for a σ that satisfies σ2 + σ = 1.

1. Conway's:
5 4 3 2 1 0 1 2 3 4 5
6 5 4 3 2 1 2 3 4 5 6
7 6 5 4 3 2 3 4 5 6 7
8 7 6 5 4 3 4 5 6 7 8
9 8 7 6 5 4 5 6 7 8 9

Sum of row n and beyond:
Sn1 = σn-5
Highest level that can be reached: 4 (see table below)

2. Skew:
5 4 3 2 1 0 1 2 3 4 5
5 4 3 2 1 1 1 2 3 4 5
5 4 3 2 2 2 2 2 3 4 5
5 4 3 3 3 3 3 3 3 4 5
5 4 4 4 4 4 4 4 4 4 5

Sum of row n and beyond:
Sn2 = σn-4(2σ3 + nσ2+1) = σn-3((n-1)σ+3)
Highest level that can be reached: 6 (see table below)

Again this pagoda function is particularly simple, and all upward jumps lose nothing. Any horizontal jump loses an amount equal to the hole jumped over, and any downward jump loses twice this amount. The best solutions involve almost exclusively upward jumps.

3. Pablito's:
0
1 1
2 2 2
3 3 3 3
4 4 4 4 4

Sum of row n and beyond:
Sn3 = σn-4(nσ2 + 1) = σn-3((n+1)σ+1)
Highest level that can be reached: 6 (see table below)

This version has a particularly simple pagoda function. Note that any upward jump loses nothing in regard to this pagoda function. Any horizontal jump loses an amount equal to the hole jumped over, and any downward jump loses twice this amount. The best solutions involve almost exclusively upward jumps.

4. Hexagonal Grid:
5 4 3 2 1 0 1 2 3 4 5
5 4 3 2 1 1 2 3 4 5
6 5 4 3 2 2 2 3 4 5 6
6 5 4 3 3 3 3 4 5 6
7 6 5 4 4 4 4 4 5 6 7

Sum of row n and beyond:
Sn4 = σn-4(nσ2 + 2σ + 1) = σn-3((n+1)σ+3)
Highest level that can be reached: 7 (see table below)

5. Fully Diagonal:
5 4 3 2 1 0 1 2 3 4 5
5 4 3 2 1 1 1 2 3 4 5
5 4 3 2 2 2 2 2 3 4 5
5 4 3 3 3 3 3 3 3 4 5
5 4 4 4 4 4 4 4 4 4 5

Sum of row n and beyond:
Sn5 = σn-5(2nσ3 + 2σ2 + 1) = σn-5((4n-2)σ + 3-2n)
Highest level that can be reached: 8 (see table below)

The table below shows the value of the sums Sni for various values of n and i, showing the highest level that can be reached.

n Sn1 Sn2 Sn3 Sn4 Sn5
1 6.8541 7.8541 5.8541 11.0902 15.3262
2 4.2361 5.8541 4.6180 7.8541 11.4721
3 2.6180 4.2361 3.4721 5.4721 8.3262
4 1.6180 3.0000 2.5279 3.7639 5.9098
5 1.0000 2.0902 1.7984 2.5623 4.1246
6 0.6180 1.4377 1.2574 1.7295 2.8409
7 0.3820 0.9787 0.8673 1.1591 1.9361
8 0.2361 0.6606 0.5917 0.7721 1.3081
9 0.1459 0.4427 0.4001 0.5116 0.8773

# Minimum Size Armies

The following table summarizes the minimum size of an army capable of advancing n levels (n rows forward). The lower bound comes from pagoda function arguments and upper bounds from best known solutions. Click on an army size to see a diagram of an army advancing to the given level.

n Minimal Army Size
Conway's
A014225
Skew Pablito's Hexagonal grid
A014227
Fully diagonal
A125730
1 2 2 2 2 2
2 4 3 3 3 3
3 8 5 5 5 5
4 19 ≤ 20 9 ≤ 9 9 ≤ 9 9 ≤ 9 7 ≤ 8
5 Impossible 18 ≤ 19 18 ≤ 19 16 ≤ 17 13 ≤ 13
6 43 ≤ 46 51 ≤ 53 35 ≤ 36 23 ≤ 23
7 Impossible 140 ≤ Size ≤ 145, 149
156 Duncan & Hayes (1991)
45 ≤ 46
8 Impossible 122 ≤ 123
≥ 9 All Impossible
Notation: Lower Bound ≤ Optimal Value ≤ Upper bound
The lower bound comes from a pagoda function.
The upper bound is the best known solution.

As an example of an army in action, the following diagram shows a 19-man Pablito's army reaching 5 steps forward. This diagram, and all armies in action, can be viewed by clicking on the appropriate entry in the table above.

The table below shows minimal armies divided into "regiments". For example, all the red men form a regiment which is capable of reaching the top hole colored red. The red and blue men are usually mirror images of one another. The final hole to be reached is marked by an X. Sometimes it is the case that one regiment must move before the others. Note that the army diagrams presented above do not necessarily proceed forward by regiment. In general, there are many ways that a specific army can move forward to reach its final target hole.

n Minimum Army Size
Skew Pablito's Hexagonal grid Fully diagonal
4 9
5 19
6 46
7 Impossible
8 Impossible
≥ 9 All Impossible
Notation: Lower Bound ≤ Optimal Value ≤ Upper bound
The lower bound comes from a pagoda function.
The upper bound is the best known solution.

The minimum army size needed to reach level n can be found in The On-Line Encyclopedia of Integer Sequences as A014225. The same question for a hexagonal lattice generates A014227.

# Peg Solitaire Army References

### Books

 B1. John D. Beasley, The Ins & Outs of Peg Solitaire, Oxford Univ. Press, 1985 (the paperback Edition 1992, contains an additional page: Recent Developments) ISBN 0-19-286145-X (paperback). B2. John H. Conway, Elwyn R. Berlekamp, Richard K. Guy, Winning Ways for Your Mathematical Plays, Volume 4, AK Peters, 2004 (second edition). B3. Mark Haddon, The Curious Incident of the Dog in the Night-Time, Vintage, 2004.

### Papers

 P1. John Duncan and Donald Hayes, Triangular Solitaire, Journal of Recreational Mathematics, 1991, Vol 23, p. 26-37. P2. George I. Bell, Dan Hirschberg and Pablo Guerrero, The minimum size required of a solitaire army, INTEGERS Electronic Journal of Combinatorial Number Theory, Vol 7, G7, 2007. P3. George I. Bell, The shortest game of Chinese Checkers and related problems, INTEGERS Electronic Journal of Combinatorial Number Theory, Vol 9, G1, 2007. This paper considers what happens when the rules are changed so that a man is not removed by a jump. In addition, a "step move" is allowed, where one peg moves into an empty space in any direction. These are the jumping rules in the game of Chinese Checkers. Clearly, any level can now be reached, so the interesting question is how quickly can one get an army from one place to another (transfer a set of men in the smallest number of moves).

### Web References

 W1. Eric Weisstein has a page on mathworld.wolfrom.com. W2. Mathematical Mysteries: The Solitaire Advance, Plus Online Maths Mag. Issue 12, Sept. 2000. W3. "Pablito's Solitaire" was the Macalester College Problem of the Week (#860, Spring 1998 and #981, Spring 2003). W4. Home page of Pabl(it)o Guerrero García, contains some preprints of Pablo's papers. W5. The On-Line Encyclopedia of Integer Sequences, see sequences A014225, A014227, and A125730. W6. Stephen Hartke has studied the problem in reverse, i.e. starting with one man and moving him down to form an army. In this paper he and Pieter Blue consider the puzzle where an infinite number of jumps are allowed. W7. Simon Tatham has also considered the puzzle where the number of jumps can be infinite. He shows how Conway's Army can reach level 5 in this infinite setting.