PlankwayPlankway #23 · Saturday, 17 October 2026
Plankway #23 was the puzzle for Saturday, 17 October 2026. Its difficulty is medium: a 9 × 9 grid with 20 islands and 2 reefs. The solver needs 39 logical steps; the hardest technique it needs is Avoid isolation.
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
- Bridge limits: 11 steps
- Every bridge counts: 12 steps
- Full island: 6 steps
- Crossing bridges: 3 steps
- At least one: 6 steps
- Avoid isolation: 1 step
The key step
The hardest step in this puzzle is step 27. 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.
- 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 2 in B3.
- 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 H2 and at most one bridge between the 1 in F2 and the 2 in F4.
- Bridge limits: The 1 in H8 does not have the room for more: at most one bridge between the 2 in D8 and the 1 in H8 and at most one bridge between the 2 in H6 and the 1 in H8.
- Bridge limits: The 1 in B8 does not have the room for more: at most one bridge between the 5 in B6 and the 1 in B8. A reef plays a part here.
- Every bridge counts: The 5 in B6 needs every bridge it can still get: a double bridge between the 5 in B6 and the 5 in D6, one bridge between the 5 in B6 and the 1 in B8 and a double bridge between the 2 in B3 and the 5 in B6.
- Full island: The 2 in B3 already has all its bridges: no bridge between the 1 in B1 and the 2 in B3. A reef plays a part here.
- Every bridge counts: The 1 in B1 needs every bridge it can still get: one bridge between the 1 in B1 and the 3 in D1.
- Bridge limits: The 1 in G3 does not have the room for more: at most one bridge between the 3 in G1 and the 1 in G3 and at most one bridge between the 1 in G3 and the 2 in I3. A reef plays a part here.
- Bridge limits: The 1 in F7 does not have the room for more: at most one bridge between the 2 in F4 and the 1 in F7 and at most one bridge between the 1 in F7 and the 3 in I7.
- Bridge limits: The 1 in I9 does not have the room for more: at most one bridge between the 3 in I7 and the 1 in I9.
- Every bridge counts: The 1 in I9 needs every bridge it can still get: one bridge between the 3 in I7 and the 1 in I9.
- At least one: The 2 in D8 cannot get enough from its other neighbours: at least one bridge between the 5 in D6 and the 2 in D8. A reef plays a part here.
- At least one: The 2 in H2 cannot get enough from its other neighbours: at least one bridge between the 2 in H2 and the 4 in H4.
- Crossing bridges: A certain bridge blocks the connections that would cross it: no bridge between the 1 in G3 and the 2 in I3.
- Every bridge counts: The 1 in G3 needs every bridge it can still get: one bridge between the 3 in G1 and the 1 in G3. A reef plays a part here.
- Crossing bridges: A certain bridge blocks the connections that would cross it: no bridge between the 1 in F2 and the 2 in H2.
- 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 F4.
- Every bridge counts: The 2 in H2 needs every bridge it can still get: a double bridge between the 2 in H2 and the 4 in H4.
- Bridge limits: The 2 in F4 does not have the room for more: at most one bridge between the 2 in F4 and the 4 in H4.
- At least one: The 4 in H4 cannot get enough from its other neighbours: at least one bridge between the 4 in H4 and the 2 in H6.
- Bridge limits: The 2 in H6 does not have the room for more: at most one bridge between the 5 in D6 and the 2 in H6.
- At least one: The 3 in I7 cannot get enough from its other neighbours: at least one bridge between the 2 in I3 and the 3 in I7.
- Bridge limits: The 2 in I3 does not have the room for more: at most one bridge between the 2 in I1 and the 2 in I3.
- At least one: The 2 in I1 cannot get enough from its other neighbours: at least one bridge between the 3 in G1 and the 2 in I1.
- Bridge limits: The 3 in G1 does not have the room for more: at most one bridge between the 3 in D1 and the 3 in G1.
- At least one: The 3 in D1 cannot get enough from its other neighbours: at least one bridge between the 3 in D1 and the 5 in D6.
- Avoid isolation: Otherwise a finished group would be cut off from the rest, so: one bridge between the 4 in H4 and the 2 in H6.
- Every bridge counts: The 4 in H4 needs every bridge it can still get: one bridge between the 2 in F4 and the 4 in H4.
- Full island: The 2 in F4 already has all its bridges: no bridge between the 2 in F4 and the 1 in F7.
- Every bridge counts: The 1 in F7 needs every bridge it can still get: one bridge between the 1 in F7 and the 3 in I7.
- Full island: The 3 in I7 already has all its bridges: one bridge between the 2 in I3 and the 3 in I7.
- Every bridge counts: The 2 in I3 needs every bridge it can still get: one bridge between the 2 in I1 and the 2 in I3.
- Full island: The 2 in I1 already has all its bridges: one bridge between the 3 in G1 and the 2 in I1.
- Every bridge counts: The 3 in G1 needs every bridge it can still get: one bridge between the 3 in D1 and the 3 in G1.
- Full island: The 3 in D1 already has all its bridges: one bridge between the 3 in D1 and the 5 in D6.
- Crossing bridges: A certain bridge blocks the connections that would cross it: no bridge between the 2 in H6 and the 1 in H8.
- Every bridge counts: The 2 in H6 needs every bridge it can still get: one bridge between the 5 in D6 and the 2 in H6.
- Full island: The 5 in D6 already has all its bridges: one bridge between the 5 in D6 and the 2 in D8.
- Every bridge counts: The 2 in D8 needs every bridge it can still get: one bridge between the 2 in D8 and the 1 in H8. A reef plays a part here.