Plankway #17 · Sunday, 11 October 2026

Plankway #17 was the puzzle for Sunday, 11 October 2026. Its difficulty is expert: a 10 × 10 grid with 22 islands and 2 reefs. The solver needs 39 logical steps; the hardest technique it needs is Try and see.

Sunday puzzles are the hardest of the week: a 10 × 10 grid, up to five reefs, and aimed at try and see. A reef played a part in 3 steps, hiding a neighbour that would otherwise be in line.

Techniques used

• Every bridge counts: 14 steps
• Bridge limits: 10 steps
• Full island: 6 steps
• Crossing bridges: 1 step
• At least one: 7 steps
• Try and see: 1 step

The key step

The hardest step in this puzzle is step 31. 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 2 in J10 needs every bridge it can still get: a double bridge between the 4 in H10 and the 2 in J10. A reef plays a part here.
2. Every bridge counts: The 2 in D9 needs every bridge it can still get: a double bridge between the 5 in D7 and the 2 in D9.
3. Bridge limits: The 1 in A4 does not have the room for more: at most one bridge between the 1 in A4 and the 3 in A7.
4. Every bridge counts: The 1 in A4 needs every bridge it can still get: one bridge between the 1 in A4 and the 3 in A7.
5. Bridge limits: The 1 in B1 does not have the room for more: at most one bridge between the 1 in B1 and the 3 in D1 and at most one bridge between the 1 in B1 and the 3 in B5.
6. Every bridge counts: The 3 in B5 needs every bridge it can still get: a double bridge between the 3 in B5 and the 6 in D5 and one bridge between the 1 in B1 and the 3 in B5.
7. Full island: The 1 in B1 already has all its bridges: no bridge between the 1 in B1 and the 3 in D1.
8. Bridge limits: The 1 in E8 does not have the room for more: at most one bridge between the 1 in E8 and the 4 in E10.
9. Every bridge counts: The 1 in E8 needs every bridge it can still get: one bridge between the 1 in E8 and the 4 in E10.
10. Bridge limits: The 1 in F5 does not have the room for more: at most one bridge between the 6 in D5 and the 1 in F5 and at most one bridge between the 1 in F2 and the 1 in F5.
11. Bridge limits: The 1 in I1 does not have the room for more: at most one bridge between the 3 in G1 and the 1 in I1.
12. Every bridge counts: The 3 in G1 needs every bridge it can still get: a double bridge between the 3 in D1 and the 3 in G1 and one bridge between the 3 in G1 and the 1 in I1.
13. Bridge limits: The 3 in D1 does not have the room for more: at most one bridge between the 3 in D1 and the 4 in D3.
14. Every bridge counts: The 3 in D1 needs every bridge it can still get: one bridge between the 3 in D1 and the 4 in D3.
15. Bridge limits: The 1 in F2 does not have the room for more: at most one bridge between the 1 in F2 and the 2 in J2.
16. At least one: The 2 in J2 cannot get enough from its other neighbours: at least one bridge between the 2 in J2 and the 3 in J7.
17. At least one: The 3 in J7 cannot get enough from its other neighbours: at least one bridge between the 5 in H7 and the 3 in J7. A reef plays a part here.
18. At least one: The 4 in D3 cannot get enough from its other neighbours: at least one bridge between the 4 in D3 and the 2 in H3 and at least one bridge between the 4 in D3 and the 6 in D5.
19. Crossing bridges: A certain bridge blocks the connections that would cross it: no bridge between the 1 in F2 and the 1 in F5.
20. Every bridge counts: The 1 in F5 needs every bridge it can still get: one bridge between the 6 in D5 and the 1 in F5.
21. Every bridge counts: The 1 in F2 needs every bridge it can still get: one bridge between the 1 in F2 and the 2 in J2.
22. Full island: The 2 in J2 already has all its bridges: one bridge between the 2 in J2 and the 3 in J7.
23. Every bridge counts: The 3 in J7 needs every bridge it can still get: a double bridge between the 5 in H7 and the 3 in J7. A reef plays a part here.
24. Bridge limits: The 2 in H3 does not have the room for more: at most one bridge between the 2 in H3 and the 5 in H7.
25. At least one: The 6 in D5 cannot get enough from its other neighbours: at least one bridge between the 6 in D5 and the 5 in D7.
26. At least one: The 4 in E10 cannot get enough from its other neighbours: at least one bridge between the 4 in E10 and the 4 in H10 and at least one bridge between the 3 in A10 and the 4 in E10.
27. Bridge limits: The 4 in H10 does not have the room for more: at most one bridge between the 5 in H7 and the 4 in H10.
28. At least one: The 5 in H7 cannot get enough from its other neighbours: at least one bridge between the 5 in D7 and the 5 in H7.
29. Bridge limits: The 5 in D7 does not have the room for more: at most one bridge between the 3 in A7 and the 5 in D7.
30. At least one: The 3 in A7 cannot get enough from its other neighbours: at least one bridge between the 3 in A7 and the 3 in A10.
31. Try and see: Assuming the opposite leads to a contradiction a few steps later, so: one bridge between the 5 in H7 and the 4 in H10.
32. Full island: The 4 in H10 already has all its bridges: one bridge between the 4 in E10 and the 4 in H10.
33. Every bridge counts: The 4 in E10 needs every bridge it can still get: a double bridge between the 3 in A10 and the 4 in E10.
34. Full island: The 3 in A10 already has all its bridges: one bridge between the 3 in A7 and the 3 in A10.
35. Every bridge counts: The 3 in A7 needs every bridge it can still get: one bridge between the 3 in A7 and the 5 in D7.
36. Full island: The 5 in D7 already has all its bridges: one bridge between the 5 in D7 and the 5 in H7 and one bridge between the 6 in D5 and the 5 in D7.
37. Every bridge counts: The 5 in H7 needs every bridge it can still get: one bridge between the 2 in H3 and the 5 in H7.
38. Full island: The 2 in H3 already has all its bridges: one bridge between the 4 in D3 and the 2 in H3.
39. Every bridge counts: The 4 in D3 needs every bridge it can still get: a double bridge between the 4 in D3 and the 6 in D5.

The solution

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