Plankway #20 · Wednesday, 14 October 2026

Plankway #20 was the puzzle for Wednesday, 14 October 2026. Its difficulty is medium: a 8 × 8 grid with 14 islands and 2 reefs. The solver needs 22 logical steps; the hardest technique it needs is Avoid isolation.

Wednesday is the first day on an 8 × 8 grid, and the first that needs the network rule: avoid isolation. A reef played a part in 5 steps, hiding a neighbour that would otherwise be in line.

Techniques used

• Every bridge counts: 11 steps
• Bridge limits: 8 steps
• Full island: 1 step
• At least one: 1 step
• Avoid isolation: 1 step

The key step

The hardest step in this puzzle is step 19. This is the board just before it; in blue what the step adds, in red what it rules out.

Step by step

Every step follows from what is already certain. Columns are lettered, rows numbered. Read along to where you got stuck, then try to continue on your own.

1. Every bridge counts: The 4 in C3 needs every bridge it can still get: a double bridge between the 4 in C3 and the 4 in E3 and a double bridge between the 4 in C3 and the 5 in C6. A reef plays a part here.
2. Every bridge counts: The 4 in H1 needs every bridge it can still get: a double bridge between the 4 in H1 and the 4 in H3 and a double bridge between the 2 in D1 and the 4 in H1.
3. Every bridge counts: The 2 in E8 needs every bridge it can still get: a double bridge between the 5 in E6 and the 2 in E8. A reef plays a part here.
4. Bridge limits: The 1 in G6 does not have the room for more: at most one bridge between the 5 in E6 and the 1 in G6.
5. Every bridge counts: The 1 in G6 needs every bridge it can still get: one bridge between the 5 in E6 and the 1 in G6.
6. Bridge limits: The 1 in H7 does not have the room for more: at most one bridge between the 4 in H3 and the 1 in H7.
7. Every bridge counts: The 1 in H7 needs every bridge it can still get: one bridge between the 4 in H3 and the 1 in H7.
8. Bridge limits: The 4 in H3 does not have the room for more: at most one bridge between the 4 in E3 and the 4 in H3.
9. Every bridge counts: The 4 in H3 needs every bridge it can still get: one bridge between the 4 in E3 and the 4 in H3.
10. Bridge limits: The 4 in E3 does not have the room for more: at most one bridge between the 4 in E3 and the 5 in E6.
11. Every bridge counts: The 4 in E3 needs every bridge it can still get: one bridge between the 4 in E3 and the 5 in E6.
12. Bridge limits: The 5 in E6 does not have the room for more: at most one bridge between the 5 in C6 and the 5 in E6.
13. Every bridge counts: The 5 in E6 needs every bridge it can still get: one bridge between the 5 in C6 and the 5 in E6.
14. Bridge limits: The 1 in A3 does not have the room for more: at most one bridge between the 1 in A3 and the 2 in A6. A reef plays a part here.
15. Every bridge counts: The 1 in A3 needs every bridge it can still get: one bridge between the 1 in A3 and the 2 in A6. A reef plays a part here.
16. Bridge limits: The 2 in A6 does not have the room for more: at most one bridge between the 2 in A6 and the 5 in C6 and at most one bridge between the 2 in A6 and the 1 in A8.
17. Bridge limits: The 1 in A8 does not have the room for more: at most one bridge between the 1 in A8 and the 2 in C8.
18. At least one: The 5 in C6 cannot get enough from its other neighbours: at least one bridge between the 5 in C6 and the 2 in C8.
19. Avoid isolation: Otherwise a finished group would be cut off from the rest, so: one bridge between the 5 in C6 and the 2 in C8.
20. Every bridge counts: The 5 in C6 needs every bridge it can still get: one bridge between the 2 in A6 and the 5 in C6.
21. Full island: The 2 in A6 already has all its bridges: no bridge between the 2 in A6 and the 1 in A8.
22. Every bridge counts: The 2 in C8 needs every bridge it can still get: one bridge between the 1 in A8 and the 2 in C8. A reef plays a part here.

The solution

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