Plankway #24 · Sunday, 18 October 2026

Plankway #24 was the puzzle for Sunday, 18 October 2026. Its difficulty is expert: a 10 × 10 grid with 22 islands and 2 reefs. The solver needs 40 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 6 steps, hiding a neighbour that would otherwise be in line.

Techniques used

• Every bridge counts: 12 steps
• Bridge limits: 10 steps
• Full island: 7 steps
• Crossing bridges: 2 steps
• At least one: 7 steps
• Avoid isolation: 1 step
• Try and see: 1 step

The key step

The hardest step in this puzzle is step 33. 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 J1 needs every bridge it can still get: a double bridge between the 2 in J1 and the 4 in J3. A reef plays a part here.
2. Bridge limits: The 1 in G10 does not have the room for more: at most one bridge between the 3 in C10 and the 1 in G10 and at most one bridge between the 2 in G5 and the 1 in G10.
3. Bridge limits: The 1 in G1 does not have the room for more: at most one bridge between the 4 in E1 and the 1 in G1 and at most one bridge between the 1 in G1 and the 2 in G5. A reef plays a part here.
4. Bridge limits: The 1 in D7 does not have the room for more: at most one bridge between the 2 in A7 and the 1 in D7 and at most one bridge between the 1 in D7 and the 3 in D9.
5. Every bridge counts: The 3 in D9 needs every bridge it can still get: a double bridge between the 3 in D9 and the 3 in H9 and one bridge between the 1 in D7 and the 3 in D9.
6. Full island: The 1 in D7 already has all its bridges: no bridge between the 2 in A7 and the 1 in D7.
7. Crossing bridges: A certain bridge blocks the connections that would cross it: no bridge between the 2 in G5 and the 1 in G10.
8. Every bridge counts: The 1 in G10 needs every bridge it can still get: one bridge between the 3 in C10 and the 1 in G10.
9. Bridge limits: The 3 in H9 does not have the room for more: at most one bridge between the 3 in H9 and the 3 in J9.
10. Every bridge counts: The 3 in J9 needs every bridge it can still get: a double bridge between the 5 in J5 and the 3 in J9 and one bridge between the 3 in H9 and the 3 in J9.
11. Bridge limits: The 1 in I8 does not have the room for more: at most one bridge between the 2 in E8 and the 1 in I8.
12. Every bridge counts: The 1 in I8 needs every bridge it can still get: one bridge between the 2 in E8 and the 1 in I8.
13. Bridge limits: The 2 in E8 does not have the room for more: at most one bridge between the 4 in E3 and the 2 in E8.
14. Every bridge counts: The 2 in E8 needs every bridge it can still get: one bridge between the 4 in E3 and the 2 in E8.
15. Crossing bridges: A certain bridge blocks the connections that would cross it: no bridge between the 3 in C5 and the 2 in G5.
16. At least one: The 3 in C5 cannot get enough from its other neighbours: at least one bridge between the 4 in C3 and the 3 in C5 and at least one bridge between the 3 in C5 and the 3 in C10. A reef plays a part here.
17. Bridge limits: The 3 in C10 does not have the room for more: at most one bridge between the 2 in A10 and the 3 in C10.
18. At least one: The 2 in A10 cannot get enough from its other neighbours: at least one bridge between the 2 in A7 and the 2 in A10.
19. Bridge limits: The 2 in A7 does not have the room for more: at most one bridge between the 2 in A5 and the 2 in A7.
20. At least one: The 5 in J5 cannot get enough from its other neighbours: at least one bridge between the 4 in J3 and the 5 in J5 and at least one bridge between the 2 in G5 and the 5 in J5.
21. Bridge limits: The 4 in J3 does not have the room for more: at most one bridge between the 4 in E3 and the 4 in J3.
22. At least one: The 4 in E1 cannot get enough from its other neighbours: at least one bridge between the 4 in E1 and the 4 in E3 and at least one bridge between the 3 in A1 and the 4 in E1.
23. At least one: The 3 in A1 cannot get enough from its other neighbours: at least one bridge between the 3 in A1 and the 3 in A3.
24. At least one: The 2 in A5 cannot get enough from its other neighbours: at least one bridge between the 3 in A3 and the 2 in A5. A reef plays a part here.
25. Bridge limits: The 3 in A3 does not have the room for more: at most one bridge between the 3 in A3 and the 4 in C3.
26. At least one: The 4 in C3 cannot get enough from its other neighbours: at least one bridge between the 4 in C3 and the 4 in E3.
27. Avoid isolation: Otherwise a finished group would be cut off from the rest, so: one bridge between the 2 in A7 and the 2 in A10.
28. Every bridge counts: The 2 in A10 needs every bridge it can still get: one bridge between the 2 in A10 and the 3 in C10.
29. Full island: The 3 in C10 already has all its bridges: one bridge between the 3 in C5 and the 3 in C10.
30. Every bridge counts: The 3 in C5 needs every bridge it can still get: a double bridge between the 4 in C3 and the 3 in C5. A reef plays a part here.
31. Every bridge counts: The 2 in A7 needs every bridge it can still get: one bridge between the 2 in A5 and the 2 in A7.
32. Full island: The 2 in A5 already has all its bridges: one bridge between the 3 in A3 and the 2 in A5. A reef plays a part here.
33. Try and see: Assuming the opposite leads to a contradiction a few steps later, so: one bridge between the 4 in E3 and the 4 in J3.
34. Full island: The 4 in E3 already has all its bridges: one bridge between the 4 in C3 and the 4 in E3 and one bridge between the 4 in E1 and the 4 in E3.
35. Full island: The 4 in J3 already has all its bridges: one bridge between the 4 in J3 and the 5 in J5.
36. Every bridge counts: The 4 in C3 needs every bridge it can still get: one bridge between the 3 in A3 and the 4 in C3.
37. Full island: The 3 in A3 already has all its bridges: one bridge between the 3 in A1 and the 3 in A3.
38. Every bridge counts: The 5 in J5 needs every bridge it can still get: a double bridge between the 2 in G5 and the 5 in J5.
39. Full island: The 2 in G5 already has all its bridges: no bridge between the 1 in G1 and the 2 in G5.
40. Every bridge counts: The 4 in E1 needs every bridge it can still get: one bridge between the 4 in E1 and the 1 in G1 and a double bridge between the 3 in A1 and the 4 in E1.

The solution

Play today’s puzzle