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Solving techniques

What to try when the grid stops giving things away.

The order to try things

New to the puzzle? Start with the rules — this page assumes them.

Bounds: the extremes give themselves away

A cage target near the top or bottom of what is arithmetically possible usually has one answer, and finding those first costs nothing. In a 6×6 the digits are 1 to 6, so:

Compare that with 1− across two cells, which is any of 1-2, 2-3, 3-4, 4-5 or 5-6. Same size of cage, five times the work. Spend your attention on the first kind.

Two-cell cages can never repeat

A two-cell cage is always two neighbouring cells, so it always lies inside one row or one column — and rule one forbids a repeat there. That is worth holding onto, because it is the one place where "a cage may repeat a number" never applies.

It also prunes: 12× across two cells is 2 and 6 or 3 and 4, never "two of something". Whereas a three-cell cage bent into an L has the option, and often needs it.

Factor the products

Multiplication cages are the most informative clues on the board, because the numbers 1 to 6 factor in very few ways. Three cells in a 6×6:

A large product in a small cage is close to being told the answer. A cage of 180× in a 6×6 has exactly one multiset of numbers, and the only question left is which cell holds which.

The reverse is useful too. A cage whose product is odd contains no even numbers at all, which in a 6×6 leaves only 1, 3 and 5 — often a sharper constraint than the target itself.

Let the shape decide whether a repeat is legal

This is the deduction most players never make, and it is the one that separates a hard grid from an impossible-feeling one.

Three cells in a straight line — all in one row, or all in one column — cannot repeat anything. Three cells bent into an L cover two rows and two columns, so exactly one pair of them can hold the same number, provided that pair shares neither.

Now put that together with the factoring above. 180× needs 5, 6 and 6. A straight three-cell cage therefore cannot be 180× at all — so if you see that target, the cage is bent, and the two 6s sit at the ends of the L, never in the corner.

The same logic rules things out. 8× could in principle be 2, 2 and 2, but three equal numbers need three different rows and three different columns, and an L-shaped cage only spans two of each. So 8× is 1, 2 and 4 wherever it appears in a three-cell cage.

Close a line with its sum

Every row and every column holds each digit once, so every line adds up to the same number: 21 in a 6×6, 45 in a 9×9, 10 in a 4×4. That total is a clue the puzzle never prints but always obeys.

Say a row contains a two-cell 11+ cage, a single-cell 3, and a three-cell cage whose target you have not worked out. The 11+ contributes 11 and the given contributes 3, so the remaining three cells must add to 21 − 11 − 3, which is 7. Three different numbers from 1 to 6 adding to 7, none of them 3, 5 or 6, and you are nearly done without touching that cage’s own clue.

The technique survives cages that poke out of the line. If a cage is mostly inside a row with one cell hanging below it, everything inside still has to add with the rest to 21 — so the cell hanging out is the difference between the cage total and what the row has room for. Killer sudoku players call these innies and outies, and they work here for the same reason.

Pigeonhole: where else could it go?

Instead of asking what a cell can hold, ask where a number can live. Take a row and a number — say 6 — and cross off every cell that cannot take it: cells in columns that already have a 6, and cells in cages where 6 would break the target. If one cell survives, it holds the 6, even if that cell had four candidates a moment ago.

This is usually what breaks a grid that feels stuck. Cell-by-cell reasoning runs out long before number-by-number reasoning does.

Two habits worth keeping

Try one

Play today’s puzzle — a fresh grid every day, free, in the browser. Or read the rules first if you have not met a cage grid before: how to play.

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