Doppelblock

The clue doesn't say which cells it's about

In Kakuro the scope of a sum is drawn on the paper. Here it is not. A Doppelblock clue is a statement about the cells between the two blocks — and where those two cells sit is exactly what you are trying to work out. The constraint's variable list is itself a variable.

The first thing to do with such a clue is therefore not arithmetic but geometry: ask what the printed number says about the distance between the blocks, before looking at a single digit. A window of length L holds L distinct values out of 1 … n−2, so its sum lies between L(L+1)/2 and the sum of the L largest — and that interval is exact, every value in it is reachable. Read backwards, the clue names the gaps that are still possible. Here is that rule running live on the board above:

A clue of 0 pins the two blocks adjacent. A clue of n−2 choose everything — 21 on an 8×8 — pins them to the two ends. Everything in between leaves two or three gaps open.

The grid is a Latin square with two symbols blurred together

Each value 1 … n−2 sits once per row and once per column, so it occupies a permutation matrix. What is left over — the blocks — is a 2-regular bipartite graph, and by König's edge-colouring theorem that always splits into two more permutation matrices. So a Doppelblock answer is exactly a Latin square of order n with two of its n symbols painted the same colour.

That is not just a pretty remark; it is the generator. There is no search here and no rejection: peel one random perfect matching off the still-unassigned graph per symbol (König says one always exists), then paint two symbols black. Every draw is a legal answer.

It is also a machine check. The map from Latin squares to grids is onto, and the fibre over a grid has size 2^k where k is the number of cycles in the block pattern — each cycle can be split between the two painted symbols two ways. So the independent engine, run on a board with no clues at all, must satisfy Σ 2^cycles = L(n) exactly:

Five rules, and only one of them changes what gets solved

This ladder is not five different rules. It is one constraint at five consistency strengths: line sees only the permutation part, scope reads the clue as geometry, bounds adds bounds consistency on the sum, gac is full domain consistency by support enumeration, and probe is singleton consistency on top. Adding them one at a time, on unfiltered boards with all 2n sums printed and nothing printed inside the grid:

Share of cells each level pins. line pins nothing at any size: with no cell printed in the grid, the permutation structure on its own has literally no purchase. Every bit of this puzzle is in the sums.

Now take each rule away from the full probe ladder instead. In every other puzzle in this series that table has interesting numbers in it. Here it does not, and that is the finding:

Removing line, scope, bounds or even gac from the full ladder leaves the fixpoint bit-for-bit identical on every board. Only removing probe costs anything. Four of the five rules are pure accelerators — they change the bill, not the answer.

The sums that pin the geometry hardest make the worst puzzles

Sort boards by how vague their clues are about distance — the mean number of window lengths each printed sum leaves open. The sharp boards are the ones full of 0s and maxima, where scope has the most to say. They are also the boards that are almost never unique:

A clue of 0 tells you exactly where the blocks are and nothing whatsoever about any digit. Geometric sharpness and arithmetic content are in tension, and the puzzle needs the second one.

What it costs to print nothing in the grid

A published Doppelblock prints only its 2n sums. Draw an answer at random and that is almost never enough:

Share of random answers whose 2n sums alone pin them. Probing and uniqueness agreed on all 1,000 boards — every board the probe ladder finishes has exactly one answer, and every board with exactly one answer the probe ladder finishes.