TECHNIQUE 4 OF 4

Connectedness: do not finish a separate island group

A board can satisfy every number and still fail. Look for a route connecting each group to the rest of the board before spending its last available bridges.

The rule

All islands must be reachable from one another through placed bridges. A group cannot use up all of its capacity internally while other islands remain outside. If a group has only one possible route to the rest of the board, that route must carry at least one bridge; its exact count may need another deduction.

A common mistake

Connecting two islands marked 1 to each other may satisfy both clues, but it isolates them if the puzzle contains other islands. Checking numbers alone misses this failure.

TRY IT 1 · bridges-connectivity-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?

Consider connectivity alone: removing B–A splits the graph of remaining possible routes. Which range does this connectivity test establish for that route, without using additional island-count deductions?

Written explanation

Removing B–A would split the graph of all still-possible routes. Every island must belong to one network, so B–A must carry at least one bridge. This does not by itself decide between one and two.

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-connectivity-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?

Consider connectivity alone: removing G–H splits the graph of remaining possible routes. Which range does this connectivity test establish for that route, without using additional island-count deductions?

Written explanation

Removing G–H would split the graph of all still-possible routes. Every island must belong to one network, so G–H must carry at least one bridge. This does not by itself decide between one and two.

Play the complete starting board →

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

Matching every clue can still leave two networks

A separate four-island puzzle has A at R1C1, B at R1C5, C at R5C1 and D at R5C5. Every island's clue is 2. The proposed answer has double connections A–B and C–D and no vertical connections.

The tempting move

Declare the puzzle solved because every island has two bridges and nothing crosses.

Why it fails

A and B are disconnected from C and D. There is no route through placed bridges from the top pair to the bottom pair.

What is actually justified

The proposal is invalid. One repair is a single bridge on each side of the rectangle: A–B, B–D, D–C and C–A. Every clue still totals two, and the four islands are connected.

Transfer the idea

An island marked 1 can connect only to another 1 or to an island in the rest of a larger puzzle. Can it use the other 1? What if those were the only two islands in the puzzle?

Check your reasoning

In the larger puzzle, connecting the two 1s would exhaust both and leave a separate component, so that route is impossible. If they are the only two islands, their single connection forms the whole network and is valid. The rest of the board changes the conclusion.

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