Cantle · when nothing is forced
When the board stops moving
You have placed everything that placed itself. Now nothing is forced, and it looks like guesswork. It is not. Three shapes of deadlock, on real Hard boards, from the cheapest to the most general.
What a deadlock is
Every cell still has a choice
Each empty cell has at least two colours left, and none has exactly one. Two things about this moment are worth knowing in advance, because neither is obvious.
In 45% of cases the board deadlocks before a single cell has been placed. It does not move from the start, and that is not your mistake.
The empty cells always form one connected patch. They are never scattered, so there is one place to look rather than a whole board.
All three boards below are real Hard boards: eighteen nodes, about forty lines, thirteen cells to fill.
Shape one · cheapest
The triangle
Two joined cells holding the same two colours take that pair away from every cell touching both.
Those two colours will split between them, one each, whatever happens. For anyone touching both, they are already gone.
How to find it
- Find two empty cells joined by a line.
- Check whether the same two colours are left in both. To see that, look at their coloured neighbours: which two of the four are still free.
- Find a cell that touches both.
Why it holds
The two are joined, so they cannot match. Between them they have exactly two colours. So one takes the first and the other the second — the order is unknown, but both colours get used. For a cell touching both, neither is available.
What it gives you
The shared neighbour loses two colours at once. If it had three left, it is now settled outright, and you can place it without thinking.
Cell 0 had three colours: ●, ■, ◆. Two of them are spoken for by a pair that is not even placed yet, and that is enough to leave one.
And then there is nothing left to do. The moment 0 became ◆, the ordinary rule finished the board on its own: all twelve remaining cells followed. One pair found, and the puzzle is over.
90% of hard boards, together with the first rule
Shape two · wider
The witness
The triangle needs the pair to be the same. Often it is nearly the same — two joined cells drawing on the same three colours. That alone says nothing. But if a witness stands nearby, it becomes enough.
If a cell touching both has only two colours left, and both come from that same three, then one of the pair is certainly the third — the one the witness does not have.
How to find it
- Find a cell with exactly two colours left. That is the witness. On the board below it is 10, holding ■ and ◆.
- Among its neighbours find two that are joined to each other: here 4 and 14.
- Check that those two draw on only three colours between them. Here each holds ●■◆ — three, two of which are the witness's own.
Why it holds
- Could both of the pair land inside the witness's pair, taking ■ and ◆?
- No. They are joined, so they would take one each — and between them they would strip the witness of both its colours. Nothing would be left for it, and that cannot happen on a solvable board.
- So at least one of the two lies outside the witness's pair.
- Outside ■ and ◆ the only colour available to them is ●. So one of them is ●.
What it gives you
You do not know which of the two — and you do not need to. It is enough that ● is taken by one of them: any other cell touching both can no longer be that colour. One colour struck out.
Why 8 loses it and not the witness 10
This is the easiest place to get lost, because three different roles are in play and all three are just cells on a board.
| Role | Which | What happens to it |
|---|---|---|
| Pair | 4 and 14 | One of them is ●. Which one stays unknown, and that bothers nobody. |
| Witness | 10 | Loses nothing. This is the cell that made the conclusion possible, not the cell being narrowed. |
| Target | 8 | Loses ● — and is left with one colour. |
The whole answer is one line: look at which of them has ● at all.
The witness 10 has ■ and ◆ left. ● is not in it — there is nothing to take away.
The target 8 has ● and ■ left. ● is in it, so it is the one that loses it. And since it had exactly two colours, one remains: 8 = ■.
And then
Nothing more to do. The moment 8 became ■, the ordinary rule finished the board by itself — all twelve remaining cells followed.
“Can I put ● on 14 instead?”
The most natural question here, and the answer shows why the rule is built the way it is.
The rule says: one of the pair is ●. It deliberately does not say which. So at that moment putting ● on 14 looks exactly as legal as putting it on 4 — and none of these three rules forbids it.
On this board 14 cannot be ●: exhaustive search confirms it. But here is the thing — no simple rule can see that. Force 14 to ● and “only one left” hits no contradiction at all. There is one, but it lies deeper than the eye reaches.
So the right move is not to decide this at all.
Do the thing the rule gives you for certain: 8 = ■. And immediately after, the ordinary rule lays out the whole board in one wave — including 4 = ● and 14 = ◆.
You do not guess which of the pair is sky. You make one move you are sure of, and a moment later the board says it for you.
95% of hard boards, together with the ones above
The law · both grow from it
Two colours left
A cell with exactly two colours left does not allow any two of its neighbours to take both.
If they did, it would be left with nothing — and an empty cell does not happen on a solvable board.
This is the statement behind both shapes above. The triangle is its case where two joined cells are themselves two-coloured and identical. The witness is its case where the shared neighbour is the two-coloured one. You have been using this law all along, in its two ready-made faces.
The difference is the one thing that makes it feel strange: it strikes out not a colour but a combination. You learn not what a cell is, but what two cells cannot be at the same time. At first that seems weaker, which is why it goes unnoticed. It is stronger, because bans combine with each other and colours do not.
Step 1 · what to find
A cell with exactly two colours left. Call them X and Y. Then any two of its neighbours, A and B. Whether they are joined to each other does not matter; what matters is that both touch this cell.
Step 2 · what to write down
Not a colour, but one sentence: “A and B cannot take X and Y between them”.
Two arrangements drop out: A=X together with B=Y, and A=Y together with B=X. Each of those colours is still possible in each cell on its own — it is the pair that is impossible. That is what throws people: you have written something down and nothing on the board has changed.
Step 3 · turning it into a move
- Take some colour of cell A — call it C. The question is always the same: can A be C?
- If it can, what would B be then? Run through B's colours. Out go the ones equal to C (if A and B are joined), and the ones banned together with C.
- If nothing is left for B, then A cannot be C. Strike it out.
What it gives you
One colour disappears from a cell that neither shortcut could reach. Usually that is enough for the first rule to start working again — and the board fills in from there.
Where neither the triangle nor the witness helps
Roughly one hard board in twenty runs into a place where no shortcut gives a move. Here is such a deadlock — a real Hard board too. Nowhere on it is there a pair with the same two colours, and no witness. There are nine two-coloured cells, but each of them alone says nothing.
Question: can 16 be ●? It had ▲●■ left.
Take its neighbour 7, which has ■ and ◆ left. If 16 really is ●, then 7 is either ■ or ◆. Both options close — but by different cells:
16 = ● and 7 = ■ is banned by cell 5: it holds exactly ● and ■, and both of these cells are its neighbours.
16 = ● and 7 = ◆ is banned by cell 1: it holds exactly ● and ◆, and it touches both as well.
No partner is left — so 16 is not ●. Two options closed by two different two-coloured cells, one each. No single motif would have done it, which is why neither shortcut helps here.
And what did that give? Nothing yet — and that is how it should be
The honest answer to the most important question: cell 16 held ▲●■ and now holds ▲■. Nothing was placed. The board looks the same, and the other twelve cells are as open as they were.
Here is the main difference between the law and the two shortcuts, and the thing I should have said first: the triangle and the witness give you a move, the law only gives you a narrowing. It almost never places a colour outright. You make one strike, then another, and somewhere along the way a cell is left with one colour. Only then does the first rule switch on.
The reward does not come move by move, it comes all at once at the end. Until then it feels like standing still. You are not — you are narrowing, and that is exactly why this rule is hard to love.
And honestly about the cost
To carry such a board to the end on bans alone, you have to hold more in your head than is comfortable. That is bookkeeping, not insight.
Which is why the triangle and the witness matter so much. They close nineteen boards in twenty, and you can play on them. The law is worth knowing not for daily counting, but to understand why those two work — and to know, on the twentieth board, that a way out exists even when you cannot see it.
100% of boards, on all five tiers
The order
What to do, in order
- Place everything forced, while anything moves: three colours among the neighbours means the fourth is the only one left.
- Stuck? Look for a triangle: two joined cells with the same pair. It is the cheapest thing there is, and on nine boards in ten this is where it ends.
- None? Look for a witness: two joined cells drawing on three colours, and a shared neighbour holding two of that same three.
- Not that either? Then the law. Walk the cells with exactly two colours; each forbids its neighbours, pairwise, from taking both. Strike out any colour left without a compatible partner.
- Go back to step 1. One strike almost always starts a chain that runs to the end.
Honestly about the limits
What this does not promise
It cannot be wrong
This follows from reasoning, not from counting: a two-coloured cell whose two neighbours took both colours is left with none. That does not exist on a solvable board, so the ban is always fair.
Completeness, though, is measured, not proved
Two rules were enough for 50,000 game boards, 1000 random planar graphs and 2840 arbitrary ones. Not once did they fail. But that is a check, not a proof: to claim “always” you need mathematics, not statistics.
On an ambiguous board it stops rather than invents
450 boards were made deliberately ambiguous. The method never struck out a colour that appears in some answer, and never named one answer as the right one. It simply stops — which is more honest than a guess.
All three boards on this page come from the game's own generator, and every deduction was checked against the puzzle's stored solution. The numbers: 50,000 game boards across five tiers · 1000 random planar graphs up to 50 nodes · 2840 arbitrary graphs · 450 deliberately ambiguous boards · two independent implementations · zero wrong eliminations anywhere.
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