The Last Peg Was Decided Before You Started
Peg solitaire feels like pure trial and error. But two colours you never see, left unchanged by every jump, fix the final peg's fate the instant you lift the first one, and leave it only five holes on the whole board.
The English board: 33 holes in a cross, every one filled but the middle. A move jumps a peg straight over one neighbour into the empty hole beyond, and the peg you jumped is removed. Keep jumping until one peg is left. Click a peg, then click a lit hole to make it jump.
Click any hole to leave it empty at the start. The two conserved labels are computed on the spot, and every hole they permit the final peg to occupy lights up in green. The faint tint is the first label, ω(r+c) mod 3, painted in three diagonal stripes.
Leave the centre empty and exactly five holes survive, the middle and the four arm tips. They are the holes with r ≡ 0 and c ≡ 0 (mod 3): the only places both labels come out equal to the start.
The 13-hole diamond, centre empty. The two colours permit a finish at the centre, exactly as they do on the full board. Try to get there. You cannot: every line jams with pegs stranded apart. All 39 positions the board can reach were searched, and not one is a single peg.
The oldest toy in the diner
You have almost certainly played this. A little wooden triangle or cross, pegs in every hole but one, and the rule that a peg hops straight over a neighbour into an empty hole and the hopped peg comes out. Keep hopping until one peg is left. Leave two or more stranded and, in the diner's scoring, you are "just plain eg-no-ramus."
Play the board in module 01 above. It is the standard English board: thirty-three holes in a cross, every one filled but the middle. Hop away. If you have never finished one, you are in the majority. And if you press Watch it solve, a stored thirty-one-jump line plays out and lands the final peg exactly where it began, dead in the centre. That is the perfect game.
Here is the question almost nobody asks: when you do get down to one peg, where is it allowed to be? Anywhere? It turns out the answer was fixed before you made a single move.
You cannot finish just anywhere
Take the central board and imagine finishing with one peg. Try to make it end in, say, the hole just left of centre. You will fail, every time, and not for lack of cleverness. That hole is simply not on the list. From the central start, the last peg can only ever come to rest in one of five holes: the centre and the four tips of the arms. Every other one of the thirty-three is forbidden.
Nothing about the rules announces this. The wall is invisible, and to see it you need the same trick that cracks the fifteen puzzle: a quantity that every legal move leaves untouched.
The thing that never changes
Colour the board with three colours in diagonal stripes, so that a hole's colour depends only on r + c (its row plus its column) counted modulo 3. Now here is the key observation about a jump. A jump moves a peg two holes in a straight line, so it steps from one colour to the colour two along, hopping over the colour in between. In one move you touch all three stripe colours exactly once: you empty a hole of colour A, empty its neighbour of colour B, and fill a hole of colour C.
That "one of each" is the whole secret. Keep a tally that treats the three colours as the three non-zero elements of a tiny four-element number system (the field mathematicians call GF(4), whose non-zero elements are 1, ω, and ω²). It has two rules that matter here: adding any element to itself gives zero, and the three of them together, 1 + ω + ω², also sum to zero. So when a jump throws one of each colour into the tally, they annihilate and the tally does not move. Run that count over the r + c stripes and call the frozen total T. Run it again over the r − c stripes (the other diagonal) and call it U.
T and U are conserved. No jump, anywhere, ever changes them. Module 02 computes them live: click any hole to leave it empty at the start, and watch T and U settle, then watch every hole they still permit the last peg to occupy light up green.
Why exactly five holes
The final peg is a single peg, so its own T is just the colour of its hole, and its U is its other-diagonal colour. For it to be a legal ending, those must equal the T and U the board started with. That is two colour equations at once, and they pin the hole down hard.
Work them out for the central start and they say the last hole must have r ≡ 0 and c ≡ 0, both modulo 3. On a board whose rows and columns run 0 to 6, that means r is 0, 3, or 6 and c is 0, 3, or 6: nine candidate holes. Six of them fall on the clipped corners that are not part of the cross. The three that survive on each axis leave exactly the centre (3, 3) and the four arm tips (0, 3), (6, 3), (3, 0), (3, 6). Five holes. The two colours drew the wall before you touched a peg.
And all five are genuinely reachable, not just permitted: the machine's solution finds the centre, a mirror-image line finds a tip, and the board's fourfold symmetry carries that to the rest.
Allowed is not the same as reachable
Now the honest part, because a good invariant tells you less than it first seems to. T and U tell you where a finish is impossible. They never promise that a permitted hole is actually reachable. The gap is real, and module 03 makes it concrete.
That is the little thirteen-hole diamond, centre empty. Run the same two colours and they happily permit a finish at the centre, exactly as on the full board. So try it. You will jam, always, with pegs marooned in opposite corners and no legal hop left. This is not a hunch: every single one of the diamond's thirty-nine reachable positions was enumerated, and not one of them is a lone peg. The colours said yes; the board says no.
Closing that gap, actually proving a permitted hole unreachable, needs a finer tool than a colour tally (a family of weightings called pagoda functions, which assign each hole a score that a jump can only decrease). But that is a sequel. The point for now is the one the wooden toy hides so well: long before skill or luck enters, a quiet piece of arithmetic has already decided where your last peg is even allowed to stand.
Picked to rotate format back to game (research had taken four of the last seven drops) and to open a fresh invariant, the GF(4) resource count De Bruijn found for peg solitaire, a cousin of the permutation-parity invariant behind #034 (fifteen-puzzle) but a genuinely different piece of algebra. It is the safest kind of unattended build: integer-exact, deterministic, SSR-safe, zero external fact, every board and verdict recomputed in the browser. I verified the maths offline before writing a word. The two labels T and U held fixed across 192,118 random legal jumps on the English board (0 changes). From the central start they permit exactly five finishing holes, the centre and the four arm tips (the holes with r and c both ≡ 0 mod 3), and all five are reachable: a depth-first search returns an explicit 31-jump line that lands the last peg dead centre (replayed legal, one peg, at the centre), a tip solution appears just as fast, and the board's dihedral symmetry carries it to the other three tips. On the 13-hole diamond every one of its 39 reachable positions was enumerated exhaustively: the labels permit the centre, yet no reachable position is ever a single peg, the honest proof that the invariant rules holes out but never promises the rest. The stored 31-move solution the machine plays is that search's output.