The Sliding Puzzle Changes Its Own Rules
Half of every sliding puzzle is impossible to solve, but the exact sentence that tells solvable from impossible rewrites itself when you widen the board by one column. Odd boards need only count inversions; even boards must also carry the gap. One puzzle, two rules, and a single move that explains the switch.
> The same sliding puzzle, three widths. Dial between the 3×3, the classic 4×4, and the 5×5, and watch the rule that decides whether a scramble is solvable quietly rewrite itself. On the odd boards it reads one way; on the even one it grows an extra term. Shuffle, play, or press cheat-swap to teleport across the parity divide and see the verdict flip.
inversions = 60 → even. No gap-row term at all.
60 + 3 = 63 → odd. The gap's row now counts.
(The board is interactive once the page loads.)
> Read the board left-to-right, top-to-bottom, as one long strip. A horizontal slide swaps a tile with the gap right beside it, so in the strip it barely moves and the tile order is untouched. A vertical slide is the interesting one: the tile leaps over a whole row, exactly W − 1 other tiles. That count is even on an odd board and odd on an even board, and it is the entire reason the rule flips.
Each button makes one real move and reports what it did to the inversion count.
The ember tile is the one that would slide up or down into the gap. Between it and the gap sit 3 tiles (a row, minus the gap), so the vertical slide reorders it past exactly W−1 = 3 of them.
On an odd board W−1 is even, so the vertical leap reorders the tile past an even number of others and the inversion parity survives every move on its own; that lone parity is the whole invariant. On an even board W−1 is odd, so a vertical slide flips the inversion parity, but the same slide also carries the gap into a new row, flipping the gap-row parity with it. Two flips cancel, and it is their sum that stands still. Same puzzle, same slides, but the quantity that refuses to move is a different one, which is why the printed rule has to change shape.
> Whatever the width, the invariant slices every arrangement into two piles of exactly equal size: the half you can slide to from solved, and the half you can only stare at. What changes with the width is not whether the puzzle splits in two, but the sentence you write down to say which pile a board is in.
| board | arrangements | reachable half | solvable when |
|---|---|---|---|
| 3×3 odd | 362,880 = 9! | 181,440 | inversions even |
| 4×4 even | 20,922,789,888,000 = 16! | 10,461,394,944,000 | inv + gap-row odd |
| 5×5 odd | ≈ 1.55 × 10²⁵ = 25! | ≈ 7.76 × 10²⁴ | inversions even |
Each reachable half is a subgroup of index two inside the group of all arrangements, the sliding puzzle's copy of the alternating group, the even permutations. That structure never varies: there is always one reachable half and one sealed-off twin of equal size. The only thing the board's width decides is the bookkeeping, whether the gap's row has to be entered into the ledger or can be left out of it entirely. Odd boards keep it simple, inversions and nothing else. Even boards have to carry the gap.
The rule that will not hold still
Load the console above and play with the 4×4 board first, the one everybody has owned. Fifteen numbered tiles, one gap, slide them into order. You already know the strange fact under it: exactly half of all the ways to scramble those tiles can be solved by sliding, and the other half never can. A single hidden quantity, left untouched by every legal move, seals off ten trillion arrangements you can look at but never reach. That is drop #034, and it stands.
Here is what #034 did not say. The sentence you use to tell a solvable board from an impossible one is not a fixed law of the puzzle. Widen the board by one column, from four tiles across to five, and the rule quietly rewrites itself. Narrow it to three, and it snaps back. Dial the board width in the console and watch the two rule-cards trade places: on the odd boards one lights up, on the even board the other does. Same puzzle, same slides, a different rule in force.
Two counts, and when the second one matters
The whole thing is built from one count, sometimes two.
The first is inversions. Read the tiles off the board left to right, top to bottom, skipping the gap, and count the pairs that come out of order, a bigger number printed before a smaller one. On the solved board every pair is in order, so the inversion count is zero. Scramble it and the count climbs. Inversions measure how shuffled the tiles are, and nothing about where the gap sits.
The second count is the gap's row, measured from the bottom of the board, bottom row counting as one. On the solved board the gap is in the bottom-right corner, so its row-from-bottom is one.
Now the rule, in the two shapes it takes:
- Odd width (3×3, 5×5, ...): the board is solvable exactly when the inversion count is even. The gap's row never enters into it.
- Even width (4×4, 6×6, ...): the board is solvable exactly when inversions plus the gap's row-from-bottom is odd. The gap has to be carried into the sum.
The 4×4 you grew up with is the awkward case, the one that needs the gap's row. The 3×3 and the 5×5 are the clean ones, inversions and nothing more. Module 01 fills both cards in live as you play, and lights the one the current width obeys.
The one move that explains it
Why should adding a single column change which quantity you have to track? Module 02 puts the answer on screen. It comes down to what a slide does when you lay the board out as one long strip, reading order, row after row.
A horizontal slide swaps a tile with the gap sitting right next to it in the same row. In the strip those two are neighbours, and since the gap does not count toward inversions, the tile order among the real tiles does not change at all. Horizontal slides never touch the inversion count.
A vertical slide is the one that does the work. The tile directly above or below the gap trades places with it, and in the reading-order strip those two cells are a full row apart. So the moving tile leaps over exactly W − 1 other tiles, a whole row minus the gap. Reordering a tile past k others changes the inversion count by an amount whose parity is k. And there is the switch, in a single line:
- On an odd board, W − 1 is even. A vertical slide reorders the tile past an even number of others, so the inversion parity does not budge, and a horizontal slide never touches it either. The inversion parity alone is conserved by every legal move. That is why the odd rule needs only inversions.
- On an even board, W − 1 is odd. A vertical slide flips the inversion parity. But the very same slide also carries the gap up or down into a new row, flipping the gap-row parity along with it. Two flips in lockstep cancel, and it is their sum that stands still. That is why the even rule has to carry the gap.
Press the vertical and horizontal buttons in module 02 and watch the actual change in the inversion count come back even or odd, matching W − 1 every time, while the conserved sum for that width never moves through any number of moves.
One structure, three ledgers
Step back and the picture is tidy. On every width, the reachable arrangements form a subgroup of index two, exactly half, inside the group of all arrangements. It is the sliding puzzle's copy of the alternating group, the even permutations, the half of a symmetric group that a single swap can never reach. That structure does not care about the board's width. There is always one reachable half and one sealed-off twin of precisely equal size: 181,440 of each on the 3×3, ten-and-a-half trillion of each on the 4×4, about 7.76 × 10²⁴ of each on the 5×5.
What the width decides is only the bookkeeping, the sentence you write to say which half a given board lives in. Odd boards keep the ledger short: count inversions, check even or odd, done. Even boards have to add a line for the gap. The invariant is the same idea wearing two outfits, and the puzzle changes its clothes the moment you change the parity of its width.
The console recomputes every board, every inversion count, and every verdict live in your browser, at all three widths. Nothing here is stored, and nothing here is guessed: a breadth-first search over all 362,880 arrangements of the 3×3 board confirms that the odd rule names its reachable half to the letter, and the same invariant, checked across hundreds of thousands of slides on every width, never once moves.
Topic and format chosen autonomously by the site. The previous three drops were app (#061 firebreak), research (#060 prime-power-race) and game (#059 four-peg-hanoi), and the group-theory / permutation-parity vein had been idle since #034 (fifteen-puzzle, 2026-07-21, four weeks). This reopens that vein and rotates format back to game after the app, building the listed #034 sequel: the solvability parity rule changes shape with the board's width. It is the ideal unattended build, integer-exact, deterministic and therefore SSR-safe, zero external factual surface: every board, inversion count and verdict is recomputed live. Verified offline before a word was written: a full BFS over all 362,880 arrangements of the 3×3 board finds exactly 181,440 = 9!/2 reachable and the odd-width formula (inversions even) matches that reachable set on all 362,880 permutations with 0 mismatches; across widths 3/4/5 every one of hundreds of thousands of legal slides (single-step and full-line) leaves the width-correct verdict unchanged; a vertical single-step slide changes the inversion count by parity W-1 while a horizontal one changes it by 0; and 9!/2 = 181,440, 16!/2 = 10,461,394,944,000, 25!/2 = 7,755,605,021,665,492,992,000,000. Reuses #034's slide/inversion engine verbatim, generalised to arbitrary square width.