TECHNIQUE 1 OF 4

Maximum capacity: find routes that must be used

Start with islands that need many bridges but have few visible neighbours. The two-bridges-per-route limit can force an exact connection or just a minimum of one.

The rule

An island with exactly two usable neighbours can receive at most four bridges. If its clue is 4, both routes must be double. If its clue is 3, each route must carry at least one bridge, but the clue alone does not decide which becomes double. Count only nearest visible neighbours in the four straight directions.

A common mistake

For an island marked 3, drawing two doubles exceeds its clue. Drawing a single on both routes records a minimum, not a complete solution for that island.

TRY IT 1 · bridges-capacity-1

Bridge networkB–A: 0 or 1 or 2 bridges; B–D: 0 or 1 or 2 bridges; C–B: 0 or 1 or 2 bridges; D–F: 0 or 1 or 2 bridges; E–G: 0 or 1 or 2 bridges; E–A: 0 or 1 or 2 bridges; F–I: 0 or 1 or 2 bridges; G–H: 0 or 1 or 2 bridgesA2B4C1D4E2F4G3H2I2

What can you justify?

What range can the displayed capacity deduction establish for route C–B?

Written explanation

Island C needs 1 individual bridges. Its routes currently allow 0 to 2 in total. Holding the other routes within their limits restricts C–B to 1. A domain “1 or 2” proves a connection but does not yet choose single or double.

Play the complete starting board →

This question begins at the displayed checkpoint. The full practice link starts at the board’s opening position.

TRY IT 2 · bridges-capacity-2

Bridge networkA–C: 0 or 1 or 2 bridges; A–F: 0 or 1 or 2 bridges; B–G: 0 or 1 or 2 bridges; B–A: 0 or 1 or 2 bridges; C–E: 0 or 1 or 2 bridges; D–C: 0 or 1 or 2 bridges; F–E: 0 or 1 or 2 bridges; G–D: 0 or 1 or 2 bridges; G–H: 0 or 1 or 2 bridgesA4B3C5D1E4F2G3H2

What can you justify?

What range can the displayed capacity deduction establish for route B–G?

Written explanation

Island B needs 3 individual bridges. Its routes currently allow 0 to 4 in total. Holding the other routes within their limits restricts B–G to 1 or 2; B–A to 1 or 2. A domain “1 or 2” proves a connection but does not yet choose single or double.

Play the complete starting board →

This question begins at the displayed checkpoint. The full practice link starts at the board’s opening position.

At least one does not identify the double

Island A has clue 3 and exactly two usable routes, A–B and A–C. Neither route has a placed bridge; each neighbour can accept up to two.

The tempting move

Declare A–B double and A–C single solely because A needs three bridges.

Why it fails

A–B single and A–C double satisfies the same local information. The total forces both routes to be used, but it does not distinguish their exact counts.

What is actually justified

Record a minimum of one bridge on each route. Use another constraint to decide where the third bridge belongs.

Transfer the idea

An island marked 5 has exactly three available neighbours, each able to supply up to two bridges. Must every route be used? Are any exact counts fixed by this information alone?

Check your reasoning

Every route needs at least one bridge: omitting a route leaves only four bridges of capacity. The counts must be 2, 2 and 1 in some order, but this information alone does not identify the single route.

Playfield Arcade · Updated 3 October 2026 · AI-assisted writing and code; examples checked against explicit constraints. Our method · Report a correction