Plankway #16 · Saturday, 10 October 2026

Plankway #16 was the puzzle for Saturday, 10 October 2026. Its difficulty is expert: a 9 × 9 grid with 20 islands and 2 reefs. The solver needs 33 logical steps; the hardest technique it needs is Try and see.

Saturday puzzles are 9 × 9 and usually need the only way out, sometimes 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: 13 steps
• Bridge limits: 9 steps
• Full island: 6 steps
• At least one: 4 steps
• Try and see: 1 step

The key step

The hardest step in this puzzle is step 26. 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 F2 needs every bridge it can still get: a double bridge between the 3 in D2 and the 2 in F2. A reef plays a part here.
2. Every bridge counts: The 2 in A9 needs every bridge it can still get: a double bridge between the 3 in A7 and the 2 in A9.
3. Bridge limits: The 3 in D2 does not have the room for more: at most one bridge between the 3 in B2 and the 3 in D2 and at most one bridge between the 3 in D2 and the 2 in D5.
4. Every bridge counts: The 3 in B2 needs every bridge it can still get: one bridge between the 3 in B2 and the 3 in D2 and a double bridge between the 3 in B2 and the 4 in B4.
5. Full island: The 3 in D2 already has all its bridges: no bridge between the 3 in D2 and the 2 in D5.
6. Bridge limits: The 1 in I1 does not have the room for more: at most one bridge between the 1 in I1 and the 2 in I6 and at most one bridge between the 3 in G1 and the 1 in I1.
7. Bridge limits: The 1 in F5 does not have the room for more: at most one bridge between the 2 in D5 and the 1 in F5. A reef plays a part here.
8. Every bridge counts: The 1 in F5 needs every bridge it can still get: one bridge between the 2 in D5 and the 1 in F5. A reef plays a part here.
9. Bridge limits: The 2 in D5 does not have the room for more: at most one bridge between the 2 in D5 and the 3 in D7.
10. Every bridge counts: The 2 in D5 needs every bridge it can still get: one bridge between the 2 in D5 and the 3 in D7.
11. Bridge limits: The 3 in A7 does not have the room for more: at most one bridge between the 3 in A3 and the 3 in A7 and at most one bridge between the 3 in A7 and the 3 in D7.
12. Every bridge counts: The 3 in A3 needs every bridge it can still get: a double bridge between the 3 in A1 and the 3 in A3 and one bridge between the 3 in A3 and the 3 in A7.
13. Full island: The 3 in A7 already has all its bridges: no bridge between the 3 in A7 and the 3 in D7.
14. Every bridge counts: The 3 in D7 needs every bridge it can still get: a double bridge between the 3 in D7 and the 3 in G7.
15. Bridge limits: The 3 in G7 does not have the room for more: at most one bridge between the 4 in G4 and the 3 in G7.
16. Every bridge counts: The 3 in G7 needs every bridge it can still get: one bridge between the 4 in G4 and the 3 in G7.
17. Bridge limits: The 3 in A1 does not have the room for more: at most one bridge between the 3 in A1 and the 2 in C1.
18. Every bridge counts: The 3 in A1 needs every bridge it can still get: one bridge between the 3 in A1 and the 2 in C1.
19. Bridge limits: The 2 in C1 does not have the room for more: at most one bridge between the 2 in C1 and the 3 in G1.
20. Every bridge counts: The 2 in C1 needs every bridge it can still get: one bridge between the 2 in C1 and the 3 in G1.
21. At least one: The 3 in F8 cannot get enough from its other neighbours: at least one bridge between the 3 in F8 and the 3 in I8 and at least one bridge between the 2 in B8 and the 3 in F8. A reef plays a part here.
22. Bridge limits: The 2 in B8 does not have the room for more: at most one bridge between the 4 in B4 and the 2 in B8.
23. At least one: The 4 in B4 cannot get enough from its other neighbours: at least one bridge between the 4 in B4 and the 4 in G4.
24. At least one: The 3 in I8 cannot get enough from its other neighbours: at least one bridge between the 2 in I6 and the 3 in I8.
25. At least one: The 4 in G4 cannot get enough from its other neighbours: at least one bridge between the 3 in G1 and the 4 in G4.
26. Try and see: Assuming the opposite leads to a contradiction a few steps later, so: a double bridge between the 3 in F8 and the 3 in I8. A reef plays a part here.
27. Full island: The 3 in F8 already has all its bridges: one bridge between the 2 in B8 and the 3 in F8. A reef plays a part here.
28. Full island: The 3 in I8 already has all its bridges: one bridge between the 2 in I6 and the 3 in I8.
29. Every bridge counts: The 2 in B8 needs every bridge it can still get: one bridge between the 4 in B4 and the 2 in B8.
30. Full island: The 4 in B4 already has all its bridges: one bridge between the 4 in B4 and the 4 in G4.
31. Every bridge counts: The 4 in G4 needs every bridge it can still get: a double bridge between the 3 in G1 and the 4 in G4.
32. Full island: The 3 in G1 already has all its bridges: no bridge between the 3 in G1 and the 1 in I1.
33. Every bridge counts: The 2 in I6 needs every bridge it can still get: one bridge between the 1 in I1 and the 2 in I6.

The solution

Play today’s puzzle