Peg Solitaire

Choose a board

Jump a peg over its neighbour into an empty hole and take the jumped peg off. Leave one peg standing.

Peg Solitaire

This collection was written by Claude Opus 5 for TedPonderings.com — three boards, one rule, and a game that has been quietly defeating people since the court of Louis XIV. What follows is how to play, and why it is far more mathematical than it looks.

Why we file this under solitaire, not puzzles

TedPonderings.com classifies this as a solitaire game, and that is not a loose use of the word. It is the original one. The peg game was being played in France in the seventeenth century — there is a 1697 engraving of the Princesse de Soubise at the board — and it carried the name solitaire from the start, meaning simply a game for one player sitting alone.

Card patience did not exist yet. It surfaces in northern Europe roughly a century later, in the 1780s, and only afterwards borrowed the older game's name. So when an American says "solitaire" and pictures a deck of cards, that is the newcomer wearing a borrowed coat. The pegs had the word first, by about a hundred years, and we see no reason to demote the original to a mere puzzle on the strength of a latecomer's popularity.

How to play

  1. Every hole starts filled but one.
  2. Click a peg. Legal destinations are ringed in brass.
  3. Click a ringed hole to jump. The peg you jumped over is removed.
  4. A peg jumps over exactly one adjacent peg into the empty hole directly beyond it.
  5. You win by reducing the board to a single peg. Landing that peg in the finishing hole is a clean finish.

On the two cross boards, jumps run up, down, left and right only. On the triangular golf-tee board, jumps also run along both diagonals — six directions in all. Undo is unlimited; the Z key works too.

The arithmetic of a game

Every jump removes exactly one peg, so the length of a game is fixed before you start. If the board holds n holes and you begin with one empty:

pegs at start = n − 1 jumps to finish = n − 2 English 33 holes → 32 pegs → 31 jumps European 37 holes → 36 pegs → 35 jumps Golf tee 15 holes → 14 pegs → 13 jumps

So you cannot win faster by jumping more. What varies is whether you reach one peg at all. Counted the traditional way, where a single peg making a chain of jumps counts as one move, the English centre-to-centre game has been solved in 18 moves (Ernest Bergholt, 1912), and John Beasley proved in 1964 that 18 is the minimum.

Pagoda functions: proving a position is hopeless

Assign a real number ƒ to every hole. Call ƒ a pagoda function if, for every legal jump from hole a over hole b into hole c:

ƒ(a) + ƒ(b) ≥ ƒ(c) for all jumps a → c over b Define the board's score: S = Σ ƒ(h) over occupied holes h Then every jump changes S by: ΔS = ƒ(c) − ƒ(a) − ƒ(b) ≤ 0

The score can never rise. So if your target position scores higher than the position in front of you, no sequence of jumps will ever get you there — no search required. This is Conway's resource count, and it is the standard way to prove a peg solitaire problem impossible.

The most celebrated application is Conway's Soldiers. Pegs fill the half-plane below a line; how far above it can one peg be driven? Weight each hole by φ raised to its taxicab distance from the target, where φ is the reciprocal golden ratio:

φ = (√5 − 1) / 2 ≈ 0.6180339887… φ² + φ = 1 ⟹ ƒ(a) + ƒ(b) = φ^(d+2) + φ^(d+1) = φ^d = ƒ(c) Σ over the whole half-plane = 1, attained only in the limit

The infinite sum converges to exactly the weight of the fifth row — and never exceeds it. An army of any finite size can reach row four and never row five. The golden ratio, uninvited, decides the answer.

The rule of three

Label each hole on a square-grid board by its coordinates modulo 3:

ℓ₁(r, c) = (r + c) mod 3 ℓ₂(r, c) = (r − c) mod 3

Any three holes in a line — a, b, c — carry three different labels under each rule. So a single jump removes one peg from two classes and adds one to the third: all three class counts change by exactly ±1, and therefore all three parities flip together. The differences

n₀ − n₁ (mod 2) n₁ − n₂ (mod 2)

are invariant up to the known number of jumps. Run that condition over the English board starting from the centre and only five holes survive it:

English, vacancy at centre → the only possible survivors are the centre itself, and the tip of each of the four arms. All 28 other holes are ruled out before you touch a peg.

Why the European board starts off-centre

Run the same test on the 37-hole European board with the centre hole empty and something startling happens: no hole at all passes. Not the centre, not the arms, not one of the thirty-seven. The European centre-vacancy game cannot be reduced to a single peg by any sequence of jumps whatsoever. The four filled corners that make the board look friendlier are exactly what break it.

That is why this board opens with the hole directly above centre and asks you to finish directly below it. That pairing does pass the invariant, and a 35-jump solution was verified before the board was published here.

The triangle

The golf-tee board has its own invariant. Label hole i in row r:

ℓ(r, i) = (r + i) mod 3 The surviving peg always lands on a hole with the SAME label as the hole you started empty.

Start at the apex, ℓ = 0, and the game can only ever end on a hole labelled 0 — which on a 15-hole triangle means the apex, either bottom corner, or the middle of the bottom row. All four are genuinely reachable. Pleasantly, every one of the fifteen starting holes admits a one-peg finish, so there is no wasted opening.

How hard is it, really?

For these three boards, exhaustively. But peg solitaire generalised to arbitrary boards is NP-complete (Uehara and Iwata, 1990) — deciding whether a given starting position can be reduced to one peg is as hard as any problem in NP. The English board yields to a computer in seconds only because 2³³ is a small number by modern standards. Widen the board a little and the game stops being a pastime and starts being a research problem.

A practical strategy

  • Work inward. Pegs stranded on the rim have few jumps available and become dead weight.
  • Avoid leaving isolated pegs with no neighbour — a peg with no adjacent peg can never move or be removed.
  • Think in packages: the classic three-peg and six-peg blocks that clear themselves and return the jumping peg to its start. Chaining these is how the 18-move solution is built.
  • Keep the centre populated as long as you can. It has the most jumps running through it.

Written by Claude Opus 5 for TedPonderings.com.
Solvability of each board was checked by exhaustive search before release.

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