← Strategy guide

Tier 3

Simple undercounting

Two colours that fit in only two rows fill those rows completely. Pigeonhole logic, and the first truly hard technique.

Here is the shape of the argument. Take two colours. Work out which rows each one can still occupy. If between them they can only reach two rows, then those two rows are taken: one colour in each, in some order you do not need to determine.

Consequence: every square in those two rows belonging to any other colour is dead.

It generalises. Three colours confined to three rows lock those three rows. Four to four. The counting is the same; only the bookkeeping gets heavier, which is why most solvers stop at three.

Doing it by hand without going mad

Write the row set for each unresolved colour as you go — literally, on paper or in your head as a short list. Then look for pairs whose lists are identical or nested:

red    → rows {2, 5}
blue   → rows {2, 5}
green  → rows {1, 2, 5, 7}

Red and blue together cover exactly two rows, so rows 2 and 5 belong to them. Green loses its squares in rows 2 and 5 and is left with rows 1 and 7 — which may well resolve it outright by region containment.

Columns work identically

Everything above holds with columns substituted for rows. Check both. Boards are frequently built so that the row version finds nothing and the column version cracks it open.

The other direction

Counting rows against colours instead of colours against rows is the rule of crowding. Different positions trigger each one.