Stostone

Shade some cells. Each region holds exactly one stone — one connected group — and stones from different regions may not touch. A number gives the size of its region's stone. Then drop the stones: each falls straight down, rigid, and they must come to rest exactly filling the bottom half. Click a cell to cycle stone → blank → empty; right-click cycles the other way. The right-hand board is the same stones, dropped.

the board

dropped

stone — shaded, one per region blank — marked with a dot the line — everything below it must end up full, everything above it empty

The falling rule is not a physics

Rule 5 is written as a simulation: drop the stones, check where they land. It never has to be run. Falling moves cells straight down, so it cannot change how many stone cells a column holds — and the bottom half ending up full means every column ends with H/2 of them. So every column starts with H/2 of them, and a rule about a final configuration turns out to be a local count on the given one. That is the col rung, and it has a strange shape: on an empty board it settles nothing at all, because no column is finished either way yet. Give it one cell from anywhere else and it writes about a sixth of the grid — and taking it out of the solver costs 350,027 assumptions across the 77 shipped boards, against 4,917 for taking out the geometric half. The half of rule 5 that cannot open a board is 71 times the more expensive half to do without.

The rest of rule 5 is one invariant. A cell at row r with b blanks under it in its column has to land on row H−1−b, so it wants to travel H−1−b−r. A stone is rigid, so all of its cells must want to travel the same distance:

blanks-below is the same for every cell of a stone, and that common value is how far the stone falls. No stone moves, no order of falling has to be chosen, nothing is simulated. Two engines in this repository check it both ways and agree on every grid up to 4×5 and 6×2 — every one of the 220.

The count is necessary and not sufficient

The counting half looks like it might be the whole rule, and on short boards it is: on a two-row board the two are literally the same condition, and a single column packs whenever it is balanced. It stops being the whole rule at six rows. Here is the smallest witness — a 4×3 grid where every column carries 2 of 4, neither stone has a vertical hole, and the arch still hangs one row too high:

Counting the region-free grids — no regions, no numbers, just the shadings an H×W board admits — puts numbers on the gap. count is exactly C(H, H/2)^W; the real rule keeps 77.8% of them at 4×2, 49.4% at 4×4 and 7.0% at 6×6. The 4×n counts are 6, 28, 134, 640, 3058, 14612, 69822, 333640, 1594282, 7618204 — not in OEIS, but they satisfy a(n) = 6a(n−1) − 5a(n−2) − 4a(n−3) for every term computed.

What the middle reading buys

Between the two sits a rule you can see: the invariant applied to two cells of one stone in one column says that a stone never has a vertical hole. That is what the drop rung propagates first and what the fit rung uses to throw candidate stones away. It is vacuous below six rows — with only two stone cells per column a stone cannot straddle a gap — and from six rows on it starts to bite: at 6×6 it throws away 16.4 million of the 64 million balanced grids, and the real rule then throws away 43.1 million more. As a rule of the puzzle it is worth less than that sounds. Swap exact for convex and 10 of the 44 shipped 6×6 boards stop having a unique answer — and exactly the same 10 fall if you swap it for plain counting instead. At 8×8, 31 of the 33 lose uniqueness under either, with a median of 37 answers where there had been one.

Five rungs

col is the counting half of rule 5. clue is the numbers, the one-stone-per-region rule and the stones-do-not-touch rule. fit lists every way a region's stone could still be drawn and keeps what they agree on. drop is the invariant. probe is singleton consistency on top.