Peg Solitaire

Peg Solitaire is a game that consists of a board with 33 holes arranged in the pattern given in the pictures below. At the start, every hole except the center is filled with a peg. The player then starts jumping pegs. Any peg that is jumped over is removed, just as in checkers. Vertical and horizontal jumps are allowed, but diagonal jumps are forbidden. The goal is to reach a position where only one peg remains, and that peg is in the center hole.

\begin{aligned}• & • & • \\ • & • & • \\ • & • & • & • & • & • & • \\ • & • & • & ∘ & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & Start \end{aligned}  \begin{aligned}∘ & ∘ & ∘ \\ ∘ & ∘ & ∘ \\ ∘ & ∘ & ∘ & ∘ & ∘ & ∘ & ∘ \\ ∘ & ∘ & ∘ & • & ∘ & ∘ & ∘ \\ ∘ & ∘ & ∘ & ∘ & ∘ & ∘ & ∘ \\ ∘ & ∘ & ∘ \\ ∘ & ∘ & ∘ \\ & & Finish \end{aligned}

You can vary the problem by choosing some hole other than the center hole. In fact, you may pick any of the 33 holes, leave it empty at the start, and finish with one peg in that hole. We shall use the notation below to describe the spaces in the board.

                                         c1  d1  e1
                                         c2  d2  e2
                                  a3  b3  c3  d3  e3  f3  g3
                                  a4  b4  c4  d4  e4  f4  g4
                                  a5  b5  c5  d5  e5  f5  g5
                                         c6  d6  e6
                                         c7  d7  e7

Coordinates

Block Moves

Before tackling a big problem, a good strategy is to start with several smaller problems. Some examples of smaller problems are given below. In each case, you need to reduce the given configuration to one peg. These configurations are called block moves or packages.

\begin{matrix}• \\ • \\ ∘ & • & • \\ 3–removal 1 \end{matrix}  \begin{matrix}• & ∘ & • & • \\ • & • \\ • & • \\ 6–removal 1 \end{matrix}  \begin{matrix}∘ \\ • \\ • & • & • \\ • & • & • \\ 6–removal 2 \end{matrix}

\begin{matrix}• \\ • & • & • \\ ∘ \\ 3–removal 2 \end{matrix} \begin{matrix}• & • \\ • & • & • \\ • & • & ∘ \\ ∘ \\ 6–removal 3 \end{matrix}  • • • ; • ; ∘ • • ; • ; L–removal

In the 3–removals, you should try to remove the line of 3 pegs and retain the other peg. Note that the second 3–removal is very similar to the first. In the 6–removals, try to remove the rectangular block of 6 pegs and retain the other peg. In the L–removal, remove the L–shaped collection of pegs.

The problem described on the first page is the d4–reversal. This means we start with pegs in every hole except d4, and we end with one peg in d4 and holes elsewhere. The diagram below shows one solution of this problem. The first block is 3–removal 1, the second block is 6–removal 1, etc.

6 •

• 1
5  ∘ 2 4

3

Instead of the d4–reversal, you can pick any other space on the board and do the corresponding reversal problem. At first glance, you might think that this means that there are 33 different one–peg reversal problems. But by rotation and reflection, these can be reduced to 7 essentially distinct problems. Besides the d4–reversal, the problems are the c1, c2, c3, d1, d2, and d3–reversals. Most of these problems can be solved using block moves, and I leave them to you as exercises. The one exception is the d1-reversal.

Homework

1. Do the c1–c2–reversal problem. In other words, start with holes only at c1 and c2, and end with pegs only at c1 and c2.

2. Do the c1–d1–reversal problem.

3. Do the c1–c4–reversal problem.

4. Do the c4–e4–reversal problem.

\begin{aligned}× & • & • \\ × & • & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & Problem 1 \end{aligned}  \begin{aligned}× & × & • \\ • & • & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & Problem 2 \end{aligned}  \begin{aligned}× & • & • \\ • & • & • \\ • & • & • & • & • & • & • \\ • & • & × & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & Problem 3 \end{aligned}  \begin{aligned}• & • & • \\ • & • & • \\ • & • & • & • & • & • & • \\ • & • & × & • & × & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & Problem 4 \end{aligned}

The four problems mentioned above are examples of two–peg reversal problems. There are 81 such problems, and all but 4 of them are possible. The impossible ones are d1–d2, d2–d4, d2–d6, and d2–b4.

\begin{aligned}• & × & • \\ • & × & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & d1–d2 \end{aligned}  \begin{aligned}• & • & • \\ • & × & • \\ • & • & • & • & • & • & • \\ • & • & • & × & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & d2–d4 \end{aligned}  \begin{aligned}• & • & • \\ • & × & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & × & • \\ • & • & • \\ & & d2–d6 \end{aligned}  \begin{aligned}• & • & • \\ • & × & • \\ • & • & • & • & • & • & • \\ • & × & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & d2–b4 \end{aligned}

                          Impossible Problems

Other problems

Here are some other simple problems that can be done on a Peg Solitaire board. The symbol • represents a peg at the beginning of the problem; the symbol × represents a peg present at the end of the problem.

• •_{• • ×• • •} • • Cross •_{• • •}_{• • × • •}_{• • •} • Diamond

×

• •  • •
• • • • •
•

Crossbow

×_{•  •}_{• •  •  • •}_{•  •  •}_{• • •} • Longbow

Here are three problems known collectively as “the lecturer and her audience.” All three problems start from the position in the first picture below. In the first problem, the lecturer remains a motionless gongoozler while everyone departs except for three sleeping students in the back row c7/d7/e7. The second problem ends with two sleeping students in the corners a5/g5. In the third problem, the lecturer remains motionless until the end, when, disgusted with the course of events, she collects the three remaining students and marches out the back door.

\begin{aligned}& & ∗ \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & & Start \end{aligned}  \begin{matrix}∗ \\ • & • & • \\ Problem 1 \end{matrix}  \begin{aligned}& ∗ \\ • & & • \\ & Problem 2 \end{aligned}  \begin{matrix}∗ \\ • \\ • \\ • \\ Problem 3 \end{matrix}

Homework

5. Do the Cross problem.

6. Do the Diamond problem.

7. Do the Crossbow problem.

8. Do the Longbow problem.

9. Do the first Lecturer problem.

Peg Algebra

Let r, s, t and u denote four spaces on a peg board. Suppose we start with pegs in s and t. We can jump from t to r. Let us think of st as equal to r. Alternatively, we can jump from s to u, so st = u

\begin{matrix}∘ & • & • & ∘ & ⟶ & • & ∘ & ∘ & ∘ \\ r & s & t & u & & r & s & t & u \end{matrix}  \begin{matrix}∘ & • & • & ∘ & ⟶ & ∘ & ∘ & ∘ & • \\ r & s & t & u & & r & s & t & u \end{matrix}

Since st = r and st = u, this means that in our algebra, r = u. In other words,

places three apart in a line are equal.

Starting from the equations st = u and tu = s, we multiply to get st^{2}u = su. Cancelling out the factors of su, we get t^{2} = 1. In other words,

two pegs in the same place cancel.

Combining these first two rules, we see that ru = 1; i.e.

two pegs three apart cancel.

Since r = st and r^{2} = 1, we get rst = 1. Thus

any three pegs in a line cancel.

With the last rule, it is easy to see that the value of the whole board is 1. Thus any position is equal to its complement.

What do these strange algebraic rules tell us about Peg Solitaire? They give us an easy way to show that certain problems are impossible. For example, is it possible to start with a hole in the center (d4) and end with a peg at d3? No, because the two positions do not have the same algebraic values. On the other hand, it is possible to end with one peg in d1. (Try it. If you need help, go back and look at the solution we gave earlier to the d4–reversal problem.) It is also possible to end with one peg in any of the positions a4, g4, or d7; just take the d1 solution and use rotation.

Problems

10. Suppose we start with a peg in every hole except c4, and we jump pegs until only one peg is on the board. According to our algebraic rules, there are four different possible locations for the last peg. Find these four locations, and solve each of the corresponding problems.

11. Start with pegs in every hole except d4. End with pegs in c4 and e4.

12. Start with pegs in every hole except d4. End with pegs in c3, e3, c5, and e5.

\begin{aligned}• & • & • \\ • & • & • \\ • & • & • & • & • & • & • \\ • & • & ×• & ×• & • & • \\ • & • & • & • & • & • & • \\ • & • & • \\ • & • & • \\ & Problem 11 \end{aligned}  \begin{aligned}• & • & • \\ • & • & • \\ • & • & ×• & • & ×• & • & • \\ • & • & • & • & • & • \\ • & • & ×• & • & ×• & • & • \\ • & • & • \\ • & • & • \\ & & Problem 12 \end{aligned}

Further Reading

Much more information about Peg Solitaire can be found in the following two books.

1. The Ins and Outs of Peg Solitaire by John Beasley. This is the definitive reference for Peg Solitaire. It is published by Oxford University Press. A paperback edition is due in August 1992; the cost is about $9.

2. Winning Ways for Your Mathematical Plays by E.R. Berlekamp, J. H. Conway,

and R.K. Guy. This two–volume work, published by Academic Press, covers a number of topics in recreational mathematics. Chapter 23 of the second volume is devoted to Peg Solitaire.