Plankway #25 · Monday, 19 October 2026

Plankway #25 was the puzzle for Monday, 19 October 2026. Its difficulty is easy: a 7 × 7 grid with 10 islands and 2 reefs. The solver needs 12 logical steps; the hardest technique it needs is Every bridge counts.

Monday puzzles are the gentlest of the week: a 7 × 7 grid where counting is enough. A reef played a part in 4 steps, hiding a neighbour that would otherwise be in line.

Techniques used

• Every bridge counts: 8 steps
• Bridge limits: 4 steps

The key step

The hardest step in this puzzle is step 1. 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 A1 needs every bridge it can still get: a double bridge between the 2 in A1 and the 4 in A4.
2. Every bridge counts: The 4 in A4 needs every bridge it can still get: a double bridge between the 4 in A4 and the 6 in C4. A reef plays a part here.
3. Every bridge counts: The 2 in E2 needs every bridge it can still get: a double bridge between the 3 in C2 and the 2 in E2. A reef plays a part here.
4. Every bridge counts: The 2 in E4 needs every bridge it can still get: a double bridge between the 6 in C4 and the 2 in E4. A reef plays a part here.
5. Bridge limits: The 3 in C2 does not have the room for more: at most one bridge between the 3 in C2 and the 6 in C4.
6. Every bridge counts: The 3 in C2 needs every bridge it can still get: one bridge between the 3 in C2 and the 6 in C4.
7. Bridge limits: The 6 in C4 does not have the room for more: at most one bridge between the 6 in C4 and the 4 in C6.
8. Every bridge counts: The 6 in C4 needs every bridge it can still get: one bridge between the 6 in C4 and the 4 in C6.
9. Bridge limits: The 1 in A6 does not have the room for more: at most one bridge between the 1 in A6 and the 4 in C6. A reef plays a part here.
10. Every bridge counts: The 4 in C6 needs every bridge it can still get: a double bridge between the 4 in C6 and the 3 in F6 and one bridge between the 1 in A6 and the 4 in C6.
11. Bridge limits: The 3 in F6 does not have the room for more: at most one bridge between the 1 in F3 and the 3 in F6.
12. Every bridge counts: The 3 in F6 needs every bridge it can still get: one bridge between the 1 in F3 and the 3 in F6.

The solution

Play today’s puzzle