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hashi · 8 min read

How to Play Hashi

Build the architectural infrastructure between numbered logic islands.

Hashi (also called Bridges or Hashiwokakero) is a graph puzzle. Numbered islands sit on a grid, and you draw bridges connecting them so that: every island has exactly the right number of bridges, no two bridges cross, and the whole network is one connected piece.

Despite the visual difference from grid-fill puzzles, Hashi is just as deductive — almost every easy puzzle can be solved by repeatedly finding islands whose required count forces their bridges uniquely.

The Puzzle

A 7×7 grid with nine islands. Each circle's number says how many bridges that island needs in total.

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Starting puzzle: 9 islands, 0 bridges. The numbers are the bridge counts each island needs.

The Three Rules

Rule 1: Bridges run only horizontally or vertically

Bridges connect two islands in the same row or same column. There must not be any other island between them along that line — bridges always go to the nearest island in a given direction.

Rule 2: At most 2 bridges between any pair

Two islands can be linked by 0, 1, or 2 bridges. Never 3 or more. The two bridges in a double-link are drawn as two parallel lines.

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A double bridge is shown as two parallel lines.

Rule 3: Bridges may not cross

Once you draw a bridge across the board, no other bridge may cross over it. Plan ahead — sometimes a tempting bridge is impossible because it would require a perpendicular crossing later.

Rule 4: Every island's count must be matched exactly

The number on the island is the sum of all bridges connected to it (1 from each single bridge, 2 from each double bridge). Not less, not more.

Rule 5: All islands form one connected network

After all bridges are drawn, you must be able to walk from any island to any other by following bridges. No isolated sub-groups.

The Forced-Bridge Technique

For an island with R bridges still needed and N reachable neighbors:

  • Maximum total bridges available = 2N (since each pair allows up to 2 bridges).
  • Minimum bridges per neighbor = max(0, R − 2(N − 1)).

When the minimum is 1 or 2, you have a forced bridge. The cleanest case: an island with only one accessible neighbor must send all its bridges to that neighbor.

Walkthrough

Step 1 — Island 7 (top, req 2) has only one accessible neighbor

Look at the “2” at row 1, column 3 (the topmost island). Same row: nothing else. Same column: there's a 4 directly below and beyond it a 3 even lower — but the 4 sits between, so the 2 can only reach the 4. With one neighbor and a need of 2 bridges, the only option is a double bridge.

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The 2 at the top can only reach the 4 below it.
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Step 1: double bridge between the 2 and the 4 below it.

Step 2 — Island 4 (req 2) is in the same boat

The 2 on the left (row 3, column 2) has a single accessible neighbor: the 3 directly below it at row 7, column 2. Force a double bridge.

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Step 2: another double bridge — the 2 to the 3 below.

Step 3 — Island 3 needs one more bridge

The 3 at row 7, column 2 used up 2 bridges with island 4. It needs one more. Its only other accessible neighbor is the 3 at row 7, column 6 (same row). One bridge.

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Step 3: a single bridge along the bottom row.

Step 4 — Island 0 needs two more

The 3 at the bottom-right (row 7, column 6) just received 1 bridge from step 3. It still needs 2 more. Same row is exhausted. Same column: the 3 directly above it. So the only option is a double bridge upward.

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Step 4: a double bridge straight up the right side.

Step 5 — Island 1 needs one more

The 3 at row 6, column 6 has used 2 of 3 bridges. The remaining accessible neighbor is the 2 at row 2, column 6 (long vertical bridge). One single bridge.

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Step 5: the long single bridge running up the right column.

Step 6 — Island 6 needs one more

The 2 at row 2, column 6 used 1 bridge. One left. The only other accessible island is the 4 to its left (row 2, column 3).

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Step 6: a single horizontal bridge across the top half.

Step 7 — Island 8 needs one more

The 4 at row 2, column 3 has used 2 (down to island 7) + 1 (right to island 6) = 3. It needs one more. Its remaining accessible neighbor is the 3 at row 6, column 3 (same column).

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Step 7: a single vertical bridge connecting the two columns.

Step 8 — Island 5 fills up the last connection

The 3 at row 6, column 3 has used 1 bridge. It needs 2 more. Its only remaining accessible neighbor is the 2 to its right at row 6, column 5 (the 4 between them is gone — wait, there is no island between). So a double bridge to the right.

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Solved! Every island's count is exactly satisfied. No crossings. One connected network.

Tips for Beginners

  • Start with islands that have one accessible neighbor. They're forced.
  • Then look at islands with high required counts relative to their neighbors. An island needing 8 bridges with 4 neighbors must have 2 bridges to each.
  • Subtract as you go. Once you draw a bridge, reduce the “remaining” count for both islands. The cascade continues from those updated counts.
  • Don't isolate. A common late-game mistake is finishing a sub-group of islands without leaving room to connect to the rest. If a planned bridge would close off a small group, look for an alternative.
  • Plan around crossings. Two perpendicular bridges can't both exist. If two pairs of islands both want a bridge that would cross, one of them must use a different connection.

Ready to try one yourself? Hit the button below to play your first Hashi.