An Expert 11×11 Puzzle Solved with Forced Chain in 51 Steps

11×11 · expert · 51 steps

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This 11×11 board rates as expert (difficulty score 2328). Solving it from scratch takes 51 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, every open cell left in this region sits in column F, so its marker has to land there; that clears every other open cell in the column.
Next, every remaining candidate in this region touches E10 and G10, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Then, that one can't be the marker: at I1 it would force K5, K10, K11, and 7 more cleared (together, two regions have open cells only in columns K and J), then H11 (the last open cell in region 11), then G4, G6, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), and region 8 would be left with no open cell for its marker.
After that, if K1 were the marker, it would force J8 (the last open cell in region 7), then A11, B11, C11, and 4 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then A9, B9, C9, and 6 more cleared (together, two regions have open cells only in rows 9 and 10), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking H2 would force J7, J9, J10, and 3 more cleared (together, two regions have open cells only in columns J and K), then I11 (the last open cell in region 11), then G4, G6 and G7 cleared (every open cell left in this region sits in column G, so its marker has to land there), and the chain continues, and then row 4 would be left with no open cell for its marker. So H2 can't be the marker there.
From there, if I2 were the marker, it would force K5, K6, K7, and 4 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then J8 (the last open cell in region 7), then H11 (the last open cell in region 11), and the chain continues, and column F would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, that one can't be the marker: at J2 it would force K4 (the last open cell in region 3), and region 7 would be left with no open cell for its marker.
First, marking K2 would force J8 (the last open cell in region 7), then A11, B11, C11, and 4 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then A9, B9, C9, and 6 more cleared (together, two regions have open cells only in rows 9 and 10), and then region 9 would be left with no open cell for its marker. So K2 can't be the marker there.
Next, if A3 were the marker, it would force J4, J5, J6, and 7 more cleared (together, two regions have open cells only in columns J and K), then B11, C11, D11, and 3 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then D4, E4, B5, and 8 more cleared (every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column), 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, that one can't be the marker: at B3 it would force J4, J5, J6, and 7 more cleared (together, two regions have open cells only in columns J and K), then A11, C11, D11, and 3 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then D4, E4, D5, and 6 more cleared (every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 8 would be left with no open cell for its marker.
After that, if D3 were the marker, it would force A1, B1, C1, and 1 more cleared (every open cell left in this region sits in row 1, so its marker has to land there), then K4 (the last open cell in region 3), then J8 (the last open cell in region 7), 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.
Now, if G3 were the marker, it would force J7, J9, J10, and 3 more cleared (together, two regions have open cells only in columns J and K), then A11, B11, C11, and 3 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then H1, H6 and H7 cleared (together, two regions have open cells only in columns I and H), 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.
From there, that one can't be the marker: at H3 it would force J7, J9, J10, and 3 more cleared (together, two regions have open cells only in columns J and K), then I11 (the last open cell in region 11), then G1, G6 and G7 cleared (every open cell left in this region sits in column G, so its marker has to land there), and the chain continues, and column F would be left with no open cell for its marker.
Following that, that one can't be the marker: at K3 it would force J8 (the last open cell in region 7), then A11, B11, C11, and 4 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then A9, B9, C9, and 6 more cleared (together, two regions have open cells only in rows 9 and 10), and region 9 would be left with no open cell for its marker.
First, if H4 were the marker, it would force J7, J8, J9, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then I11 (the last open cell in region 11), and the chain continues, and column C 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 I4 it would force J1 (the last open cell in region 3), then E5, E6, E7, and 3 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and region 8 would be left with no open cell for its marker.
Then, marking J4 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
After that, if K4 were the marker, it would force J8 (the last open cell in region 7), then A11, B11, C11, and 4 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then A9, B9, C9, and 6 more cleared (together, two regions have open cells only in rows 9 and 10), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking C3 would force J1 (the last open cell in region 3), then E2 (the last open cell in region 2), then G4 (the last open cell in row 4), and the chain continues, and then region 8 would be left with no open cell for its marker. So C3 can't be the marker there.
From there, marking B5 would force G4 (the last open cell in row 4), then K6 cleared (every open cell left in this region sits in row 6, so its marker has to land there), then J7, J8, J9, and 7 more cleared (together, two regions have open cells only in columns J and I), and the chain continues, and then region 8 would be left with no open cell for its marker. So B5 can't be the marker there.
Following that, if I5 were the marker, it would force J7, J8, J9, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then H11 (the last open cell in region 11), and the chain continues, and column C 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 J5 it would force I3 (the last open cell in region 3), then K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then H11 (the last open cell in region 11), and the chain continues, and row 6 would be left with no open cell for its marker.
Next, marking K5 would force J8 (the last open cell in region 7), then I3 (the last open cell in region 3), then H11 (the last open cell in region 11), and the chain continues, and then column F would be left with no open cell for its marker. So K5 can't be the marker there.
Then, marking G1 would force A4, C4, A5, and 1 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 D5, E5, A6, and 12 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 H5 (the last open cell in row 5), and the chain continues, and then region 11 would be left with no open cell for its marker. So G1 can't be the marker there.
After that, marking G4 would force A6, B6, C6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then J7, J8, J9, and 7 more cleared (together, two regions have open cells only in columns J and I), then K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and then region 8 would be left with no open cell for its marker. So G4 can't be the marker there.
Now, marking A1 would force E4, E5, E6, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then E3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then E2 (the last open cell in region 2), and the chain continues, and then column D would be left with no open cell for its marker. So A1 can't be the marker there.
From there, if B1 were the marker, it would force E4, E5, E6, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then D4 (the last open cell in row 4), then E2 (the last open cell in region 2), and the chain continues, and column A would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, that one can't be the marker: at C1 it would force E4, E5, E6, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then E3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then E2 (the last open cell in region 2), and the chain continues, and column D would be left with no open cell for its marker.
First, if H1 were the marker, it would force A4, C4, A5, and 1 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 D5, E5, A6, and 12 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 G5 (the last open cell in row 5), 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, every remaining candidate in this region touches D2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, marking B2 would force D5, E5, A6, and 9 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 C11 (the last open cell in column C), and then column A would be left with no open cell for its marker. So B2 can't be the marker there.
After that, if C2 were the marker, it would force E4, E5, E6, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then D5, A6, B6, and 5 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 I6 and J6 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking E2 would force J1 (the last open cell in row 1), and then row 3 would be left with no open cell for its marker. So E2 can't be the marker there.
From there, marking A4 would force G2 (the last open cell in row 2), then B6, C6, D6, and 8 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then C11 (the last open cell in column C), and then column B would be left with no open cell for its marker. So A4 can't be the marker there.
Following that, that one can't be the marker: at C4 it would force G2 (the last open cell in row 2), then E5 (the last open cell in row 5), then D1 (the last open cell in region 2), and the chain continues, and column B would be left with no open cell for its marker.
First, 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.
Next, if E1 were the marker, it would force C5 cleared (every remaining candidate in this region touches C5), then C11 (the last open cell in column C), then B4 (the last open cell in column B), and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, marking J1 would force E3 (the last open cell in region 2), then B4 (the last open cell in region 5), then A2 (the last open cell in region 1), and the chain continues, and then column D would be left with no open cell for its marker. So J1 can't be the marker there.
After that, 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.
Now, that one can't be the marker: at G2 it would force B4 cleared (every remaining candidate in this region touches B4), then E4 (the last open cell in region 5), then H8, H9, H10, 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 column C would be left with no open cell for its marker.
From there, 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.
Following that, place a marker at C11. Column 3 has exactly one open cell left. Placing here clears the rest of row 11, column C, and its region.
First, place a marker at B4. Column 2 has exactly one open cell left. Placing here clears the rest of row 4, column B, and its region.
Next, every open cell left in this region sits in row 10, so its marker has to land there; that clears every other open cell in the row.
Then, place a marker at F9. Region has exactly one open cell left. Placing here clears the rest of row 9, column F, and its region. Placing here also clears its 2 touching neighbours.
After that, place a marker at E7. Column 5 has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region.
Now, place a marker at G5. Column 7 has exactly one open cell left. Placing here clears the rest of row 5, column G, and its region.
From there, place a marker at K6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column K, and its region.
Following that, place a marker at J10. Region has exactly one open cell left. Placing here clears the rest of row 10, column J, and its region.
First, place a marker at I3. Region has exactly one open cell left. Placing here clears the rest of row 3, column I, and its region.
Finally, place a marker at H8. Region has exactly one open cell left.

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