An Expert 11×11 Puzzle Solved with Forced Chain in 56 Steps

11×11 · expert · 56 steps

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This 11×11 board rates as expert (difficulty score 2587). Solving it from scratch takes 56 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Line confinement, Paired regions, Forced chain. Every step below is forced: nothing here is a guess.

First, together, two regions have open cells only in rows 10 and 11; with one marker each, those two markers have to fill exactly those two rows, so every other region's open cells there can be cleared.
Next, if E1 were the marker, it would force F5, F6, F7, and 4 more cleared (together, two regions have open cells only in columns F and G), then H3, H4, H7, and 2 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then I5, I6, I7, and 5 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, that one can't be the marker: at F1 it would force G4, G5, G6, and 3 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then E9 (the last open cell in region 9), then H2, H7 and H8 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and row 3 would be left with no open cell for its marker.
After that, marking G1 would force F11 (the last open cell in region 11), then E9 (the last open cell in region 9), then B10 (the last open cell in region 10), and the chain continues, and then row 3 would be left with no open cell for its marker. So G1 can't be the marker there.
Now, if H1 were the marker, it would force F2, F3, F4, and 7 more cleared (together, two regions have open cells only in columns F and G), then E9 (the last open cell in region 9), then I8 (the last open cell in region 8), and the chain continues, and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, marking J1 would force I8 and I9 cleared (together, two regions have open cells only in columns K and I), then H2, H5 and H6 cleared (every open cell left in this region sits in column H, so its marker has to land there), then F2, F3, F4, and 8 more cleared (together, two regions have open cells only in columns F and G), and the chain continues, and then column H would be left with no open cell for its marker. So J1 can't be the marker there.
Following that, marking E2 would force F5, G5, A6, and 10 more cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), then F6, F7, G3, and 2 more cleared (together, two regions have open cells only in columns F and G), then H3, H4, H7, and 2 more cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and then region 7 would be left with no open cell for its marker. So E2 can't be the marker there.
First, if F2 were the marker, it would force G5, G6, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then E9 (the last open cell in region 9), then H3, H4, H7, and 1 more cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, that one can't be the marker: at G2 it would force F11 (the last open cell in region 11), then E9 (the last open cell in region 9), then B10 (the last open cell in region 10), and the chain continues, and region 4 would be left with no open cell for its marker.
Then, marking H2 would force J4, J5, J6, and 5 more cleared (together, two regions have open cells only in columns J and I), then K6, K7, K8, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and then region 7 would be left with no open cell for its marker. So H2 can't be the marker there.
After that, that one can't be the marker: at J2 it would force I8 and I9 cleared (together, two regions have open cells only in columns K and I), then H5 and H6 cleared (every open cell left in this region sits in column H, so its marker has to land there), then F3, F4, F5, and 6 more cleared (together, two regions have open cells only in columns F and G), and the chain continues, and row 5 would be left with no open cell for its marker.
Now, that one can't be the marker: at E3 it would force F5, G5, A6, and 10 more cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), then F6, F7, G6, and 1 more cleared (together, two regions have open cells only in columns F and G), then H4, H7, H8, and 1 more cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and region 7 would be left with no open cell for its marker.
From there, marking F3 would force G5, G6, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then E9 (the last open cell in region 9), then H4, H7 and H8 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and then region 3 would be left with no open cell for its marker. So F3 can't be the marker there.
Following that, that one can't be the marker: at H3 it would force F4, F5, F8, and 4 more cleared (together, two regions have open cells only in columns F and G), then E9 (the last open cell in region 9), then I8 (the last open cell in region 8), and the chain continues, and row 5 would be left with no open cell for its marker.
First, if J3 were the marker, it would force A1, B1, C1, and 5 more cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then D1, C5, D5, and 2 more cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), then K1 (the last open cell in row 1), and the chain continues, and region 11 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, if E4 were the marker, it would force F6, F7, G3, and 3 more cleared (together, two regions have open cells only in columns F and G), then I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then H7, H8 and H9 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, that one can't be the marker: at F4 it would force I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then G6, G7, G8, and 1 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then E9 (the last open cell in region 9), and the chain continues, and region 8 would be left with no open cell for its marker.
After that, marking G4 would force F11 (the last open cell in region 11), then H6 (the last open cell in region 6), then E9 (the last open cell in region 9), and the chain continues, and then region 3 would be left with no open cell for its marker. So G4 can't be the marker there.
Now, if H4 were the marker, it would force F5, F8, F9, and 2 more cleared (together, two regions have open cells only in columns F and G), then E9 (the last open cell in region 9), then I8 (the last open cell in region 8), and the chain continues, and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, that one can't be the marker: at K3 it would force I7, I8 and I9 cleared (every open cell left in this region sits in column I, so its marker has to land there), then J9 cleared (every open cell left in this region sits in column J, so its marker has to land there), then H5 and H6 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and region 5 would be left with no open cell for its marker.
Following that, if B1 were the marker, it would force G10 (the last open cell in row 10), then I3 (the last open cell in row 3), then K2 (the last open cell in row 2), and the chain continues, and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, that one can't be the marker: at I1 it would force A2, B2, C2, and 6 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then K2 (the last open cell in row 2), then D4 (the last open cell in row 4), and the chain continues, and row 6 would be left with no open cell for its marker.
Next, that one can't be the marker: at D2 it would force K1 (the last open cell in row 1), then I4 (the last open cell in row 4), and row 3 would be left with no open cell for its marker.
Then, if D3 were the marker, it would force K1 (the last open cell in row 1), then I2 (the last open cell in row 2), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, if B4 were the marker, it would force G10 (the last open cell in row 10), then I3 (the last open cell in row 3), then K2 (the last open cell in row 2), and the chain continues, and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, that one can't be the marker: at I4 it would force A1, C1, A2, and 4 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then K2 (the last open cell in row 2), then D1 (the last open cell in row 1), and the chain continues, and region 9 would be left with no open cell for its marker.
From there, marking C3 would force I2 (the last open cell in region 3), then D7, D8 and D9 cleared (every open cell left in this region sits in column D, so its marker has to land there), then K1, J5, K5, and 1 more cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then column B would be left with no open cell for its marker. So C3 can't be the marker there.
Following that, that one can't be the marker: at A5 it would force I2 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then K2 (the last open cell in row 2), then D1 (the last open cell in row 1), and row 4 would be left with no open cell for its marker.
First, marking B5 would force G10 (the last open cell in row 10), then I3 (the last open cell in row 3), then K2 (the last open cell in row 2), and the chain continues, and then row 4 would be left with no open cell for its marker. So B5 can't be the marker there.
Next, if C5 were the marker, it would force A6, E6, A7, and 7 more cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), then E9 (the last open cell in column E), then D7 (the last open cell in region 5), and the chain continues, and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, marking E5 would force D1 (the last open cell in column D), then F7, G3, G6, and 1 more cleared (together, two regions have open cells only in columns F and G), then H6 (the last open cell in region 6), and the chain continues, and then region 4 would be left with no open cell for its marker. So E5 can't be the marker there.
After that, if F5 were the marker, it would force E9 (the last open cell in column E), then G3 and G7 cleared (every open cell left in this region sits in column G, so its marker has to land there), then H6 (the last open cell in region 6), and the chain continues, and region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, that one can't be the marker: at G5 it would force F7 (the last open cell in region 6), and region 11 would be left with no open cell for its marker.
From there, marking A1 would force I2 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then K2 (the last open cell in row 2), then D4 (the last open cell in row 4), and the chain continues, and then row 7 would be left with no open cell for its marker. So A1 can't be the marker there.
Following that, that one can't be the marker: at C1 it would force D7, D8 and D9 cleared (every open cell left in this region sits in column D, so its marker has to land there), then I2 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then K2 (the last open cell in row 2), and the chain continues, and row 7 would be left with no open cell for its marker.
First, if C2 were the marker, it would force K1 (the last open cell in row 1), then D4 (the last open cell in row 4), then H5 (the last open cell in row 5), and the chain continues, and row 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, marking K2 would force D1 (the last open cell in row 1), then H5 (the last open cell in row 5), then A3 and B3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), and the chain continues, and then row 7 would be left with no open cell for its marker. So K2 can't be the marker there.
Then, marking A4 would force I2 (the last open cell in row 2), and then row 3 would be left with no open cell for its marker. So A4 can't be the marker there.
After that, that one can't be the marker: at C4 it would force I2 (the last open cell in row 2), and row 3 would be left with no open cell for its marker.
Now, if K1 were the marker, it would force D4 (the last open cell in row 4), then H5 (the last open cell in row 5), then A7, B7, C7, and 8 more cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, place a marker at D1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region.
Following that, every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
First, place a marker at H5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column H, and its region.
Next, together, two regions have open cells only in rows 8 and 9; with one marker each, those two markers have to fill exactly those two rows, so every other region's open cells there can be cleared.
Then, marking B2 would force G10 (the last open cell in row 10), then I3 (the last open cell in region 3), then K4 (the last open cell in region 4), and the chain continues, and then region 8 would be left with no open cell for its marker. So B2 can't be the marker there.
After that, if I2 were the marker, it would force J9 (the last open cell in region 8), then K4 (the last open cell in region 4), and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, place a marker at A2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region.
From there, marking I3 would force K4 (the last open cell in region 4), then J7 (the last open cell in region 7), and then region 8 would be left with no open cell for its marker. So I3 can't be the marker there.
Following that, place a marker at G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region.
First, place a marker at F11. Region has exactly one open cell left. Placing here clears the rest of row 11, column F, and its region.
Next, place a marker at E9. Region has exactly one open cell left. Placing here clears the rest of row 9, column E, and its region.
Then, place a marker at I8. Region has exactly one open cell left. Placing here clears the rest of row 8, column I, and its region. Placing here also clears its 1 touching neighbour.
After that, place a marker at B10. Region has exactly one open cell left. Placing here clears the rest of row 10, column B, and its region.
Now, place a marker at C7. Region has exactly one open cell left. Placing here clears the rest of row 7, column C, and its region.
From there, place a marker at K6. Region has exactly one open cell left. Placing here clears the rest of row 6, column K, and its region.
Finally, place a marker at J4. Region has exactly one open cell left.

The key move here was Forced chain. Once you can spot that, this board falls quickly. ∎