Playfield Arcade

FIXED PRACTICE BOARD

Bridges study 11

Solve at your own pace. Ask what follows from the position before applying the suggested move.

Study ID: bridges-11 · Original board: bridges-2026-09-12

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

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Work from the current board

Use the route buttons beneath the diagram. They cycle unknown (0, 1 or 2) → exactly 1 → exactly 2 → 0 → at least 1 (1 or 2) → unknown. Dotted lines are possible routes, not established bridges.

Tab to any control and use Enter or Space.

Interactive help is separate from the written starting-board solution below.

Review the techniques

Rules and proof limits

Join nearest visible islands horizontally or vertically with zero, one or two bridges per route. A bridge cannot pass through an island or cross another bridge between islands. Each island’s number counts individual bridges, and all islands must belong to one connected network.

A visible neighbour is only a possible route. Its remaining capacity and crossings may make it unusable. Correct island totals are insufficient without connectedness. Local capacity deductions need not finish every board; do not call a connection forced simply because it creates a tidy network.

Show the checked route from the starting position (11 steps)

This route starts from the original study position. It does not describe moves you made differently. Use “Explain a next step” for a deduction from your current marks.

  1. 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.
  2. 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.
  3. Island C needs 5 individual bridges. Its routes currently allow 0 to 6 in total. Holding the other routes within their limits restricts A–C to 1 or 2; C–E to 1 or 2; D–C to 1 or 2. A domain “1 or 2” proves a connection but does not yet choose single or double.
  4. Island D needs 1 individual bridges. Its routes currently allow 1 to 4 in total. Holding the other routes within their limits restricts D–C to 1; G–D to 0. A domain “1 or 2” proves a connection but does not yet choose single or double.
  5. Island C needs 5 individual bridges. Its routes currently allow 3 to 5 in total. Holding the other routes within their limits restricts A–C to 2; C–E to 2. A domain “1 or 2” proves a connection but does not yet choose single or double.
  6. Island A needs 4 individual bridges. Its routes currently allow 3 to 6 in total. Holding the other routes within their limits restricts A–F to 0 or 1. A domain “1 or 2” proves a connection but does not yet choose single or double.
  7. Island E needs 4 individual bridges. Its routes currently allow 2 to 4 in total. Holding the other routes within their limits restricts F–E to 2. A domain “1 or 2” proves a connection but does not yet choose single or double.
  8. Island F needs 2 individual bridges. Its routes currently allow 2 to 3 in total. Holding the other routes within their limits restricts A–F to 0. A domain “1 or 2” proves a connection but does not yet choose single or double.
  9. Island A needs 4 individual bridges. Its routes currently allow 3 to 4 in total. Holding the other routes within their limits restricts B–A to 2. A domain “1 or 2” proves a connection but does not yet choose single or double.
  10. Island B needs 3 individual bridges. Its routes currently allow 3 to 4 in total. Holding the other routes within their limits restricts B–G to 1. A domain “1 or 2” proves a connection but does not yet choose single or double.
  11. Island G needs 3 individual bridges. Its routes currently allow 2 to 3 in total. Holding the other routes within their limits restricts G–H to 2. A domain “1 or 2” proves a connection but does not yet choose single or double.
Bridge networkA–C: 2 bridges; A–F: 0 bridges; B–G: 1 bridge; B–A: 2 bridges; C–E: 2 bridges; D–C: 1 bridge; F–E: 2 bridges; G–D: 0 bridges; G–H: 2 bridges××A4B3C5D1E4F2G3H2

Final check

Every island has its exact bridge count, routes do not cross, and a graph traversal reaches every island.

Back to the centreTry the regular game

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