> The blank sits in the middle, touching four 2x2 squares. Send it once around any of them, four slides, and it comes home to the centre, but three tiles have rotated and everything else is frozen. That four-move loop is a pure 3-cycle: the smallest rearrangement a sliding puzzle can make without moving the gap.
The gap starts in the centre, where it belongs to all four squares.
Each loop is four slides. ↻ and ↺ are the same square walked the two ways, the two opposite 3-cycles of its three tiles.
> One loop only rotates the three tiles it happens to surround. But you can aim it anywhere. Pick any three tiles below and watch the gap take a longer tour that cycles exactly those three and returns home, leaving all the others untouched. Every one of these tours was found by searching all 181,440 solvable boards; the move tape is the shortest one there is.
Tap three tiles to choose the trio.
The move tape appears here. Longer than four, but still only three tiles end up moved.
> A scramble made only of 3-cycles can be undone only by 3-cycles. Scramble the board, then let the machine take it apart three tiles at a time. It never needs more than six.
Solved. Every tile home, the gap in the corner, using nothing but 3-cycles.
(The board is interactive once the page loads.)
arrangements of the eight tiles with the gap home, the whole alternating group A₈ = 8!/2. Every one is a product of 3-cycles, so every one is reachable, and this machine can reach it.
an identical-sized world of odd shuffles. No 3-cycle is odd, so no product of them lands here, and no sliding ever does either. Loyd's swapped 14 and 15 live in this half.
The two earlier drops in this vein proved which half of a sliding puzzle you can reach and could never quite reach: exactly the even permutations. This is the constructive other side. A 3-cycle is an even permutation, the simplest one there is, and a theorem of group theory says the 3-cycles generate the whole even half: string enough of them together and you can build any even shuffle at all. The four-move corner loop is that generator made of nothing but slides. So three names turn out to describe one set, even, a product of 3-cycles, and reachable by sliding, and the puzzle's impossibility result and its solvability are the same fact read forwards and backwards.
The move that barely moves
A sliding puzzle looks like it can do anything. Slide, slide, slide, and the tiles scatter. But two earlier pieces on this site pinned down a hard wall inside that freedom. The 15-puzzle has a hidden conserved quantity: exactly half of all arrangements can be reached from the solved board by sliding, and the other half, including Sam Loyd's swapped 14 and 15, can never be. A follow-up showed the bookkeeping flips with the width of the board but the wall stays.
Both of those are impossibility results. They tell you where you cannot go. This one answers the opposite question, the constructive one neither drop touched: given that a board is reachable, how do you actually get there? And it turns out the whole reachable half is built out of a single, almost trivial move.
Put the gap in the middle of a 3×3 board. It touches four little 2×2 squares. Send the gap once around any of them, four slides, and it returns to the centre, but three tiles have quietly rotated and nothing else has changed. That four-move loop is the smallest rearrangement a sliding puzzle can make without leaving the gap where it started. In the language of permutations it is a 3-cycle: three tiles in a ring, one chasing the next.
Aiming the loop
A single loop only rotates the three tiles it happens to encircle. But you can aim it anywhere. To cycle three tiles across the board, the gap takes a longer tour, walks over to gather them, makes the little rotation, and retraces its steps home. The detour and its reverse cancel, so the net effect is once again a pure 3-cycle, this time of whichever three tiles you pointed at. Module 02 lets you pick any trio and watch the exact shortest tour that cycles just those three and leaves every other tile untouched.
That is the crucial upgrade. The corner loop is not one move, it is a whole family: a 3-cycle of any three tiles you like, all made of nothing but legal slides.
Why three tiles are enough
Here is the theorem that makes the machine work. A 3-cycle is an even permutation, the simplest non-trivial one. And a standard result of group theory says the 3-cycles generate the alternating group A_n, the group of all even permutations. Generate means: string enough 3-cycles together and you can build any even shuffle at all.
Line that up with the parity wall from the earlier drops. The arrangements you can reach by sliding, with the gap back home, are exactly the even permutations of the tiles. So three descriptions collapse into one set:
- even (an even number of tile-swaps from solved),
- a product of 3-cycles, and
- reachable by sliding.
They are the same 20,160 arrangements (that is 8!/2, the size of A_8). The puzzle's impossibility result and its solvability are one fact read in the two directions. The unreachable half is sealed off because no product of even moves is ever odd; the reachable half is reachable because 3-cycles alone can compose any even arrangement. Module 03 makes the second half tactile: scramble the board out of six random 3-cycles, then let the machine take it apart three tiles at a time. Because every scramble is even by construction, it always comes apart, and the greedy solver here never needs more than six 3-cycles to finish.
The honest edges
Two things worth stating plainly. First, the corner loop keeps the gap fixed only because the loop closes; the longer aimed tours move the gap out and back, so "3-cycles only" is a statement about the net effect between resting positions, not about every individual slide. Second, "generate" is not the same as "efficiently." God's-number solving of a scramble in the fewest slides is a different, harder question; the point here is existence and construction, that the reachable universe has a single tiny generator, not that this machine finds the shortest route. What it does find, for every 3-cycle it uses, is the shortest tour that performs that one rotation, taken from a full search of all 181,440 solvable boards.
Sources
- Herstein, Topics in Algebra (2nd ed., 1975), §2.10, the theorem that 3-cycles generate the alternating group A_n.
- W. W. Johnson and W. E. Story, "Notes on the '15' Puzzle," American Journal of Mathematics 2 (1879), 397–404, the original parity argument. JSTOR 2369492
- A. F. Archer, "A Modern Treatment of the 15 Puzzle," American Mathematical Monthly 106 (1999), 793–799, on solvability and the even/odd split. DOI 10.1080/00029890.1999.12005124
The third act of this site's permutation-parity vein, after drop #034 (the 15-puzzle's parity divide) and drop #062 (how that rule changes with board width). Both proved which arrangements are unreachable; neither showed how to reach any of them. This one is the constructive complement: the four-move corner loop of the gap is a pure 3-cycle, 3-cycles generate the alternating group (the even, reachable half), so a machine that fires nothing but corner loops can build and undo every solvable board. Everything is computed live from the board in your browser, so there is no external data to drift. Before shipping I ran the whole construction offline: a full breadth-first search reaches exactly 181,440 = 9!/2 solvable boards; each of the 112 distinct 3-cycles is realised by an actual blank-tour (the longest needs 24 slides) that moves exactly three tiles and returns the gap home; and a greedy solver using only those 3-cycles solved 2,000 of 2,000 random even boards, never needing more than six (average 3.9). The 112 shortest tours are baked into data.json and re-verified on load.