An Expert 11×11 Puzzle Solved with Forced Chain in 40 Steps

This 11×11 board rates as expert (difficulty score 1737). Solving it from scratch takes 40 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.

1. First, every remaining candidate in this region touches C10, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, if B1 were the marker, it would force A9, E9, F9, and 5 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then A10, E10, F10, and 4 more cleared (together, two regions have open cells only in rows 10 and 11), then A8, G8, H8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), 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.
3. Then, marking D1 would force E11, F11, G11, and 4 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then A9, E9, F9, and 7 more cleared (together, two regions have open cells only in rows 9 and 10), then A8, B8, G8, and 4 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and then column F would be left with no open cell for its marker. So D1 can't be the marker there.
4. After that, marking J1 would force G2, G3, H3, and 6 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 A3, B3, C3, and 3 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), and then row 3 would be left with no open cell for its marker. So J1 can't be the marker there.
5. Now, marking B2 would force A9, E9, F9, and 5 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then K3 and K4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then A4, C4, D4, and 9 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), and the chain continues, and then column F would be left with no open cell for its marker. So B2 can't be the marker there.
6. From there, if C2 were the marker, it would force K3 and K4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then A4, B4, D4, and 9 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 J4 (the last open cell in row 4), 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.
7. Following that, that one can't be the marker: at D2 it would force E11, F11, G11, and 4 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then K3 and K4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then A4, B4, C4, and 10 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), and the chain continues, and column F would be left with no open cell for its marker.
8. First, if F2 were the marker, it would force A3, B3, C3, and 1 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then A4, B4, C4, and 9 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 J4 (the last open cell in row 4), 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.
9. Next, marking H2 would force A3, B3, C3, and 3 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then J3 (the last open cell in row 3), and then row 4 would be left with no open cell for its marker. So H2 can't be the marker there.
10. Then, if I2 were the marker, it would force A3, B3, C3, and 3 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then G3 (the last open cell in row 3), then J4 (the last open cell in row 4), and the chain continues, and column H would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, that one can't be the marker: at J2 it would force K4 (the last open cell in row 4), then A3, B3, C3, and 3 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), and row 3 would be left with no open cell for its marker.
12. Now, that one can't be the marker: at F1 it would force A2 (the last open cell in row 2), then B4, C4, D4, 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 J4 (the last open cell in row 4), and the chain continues, and row 5 would be left with no open cell for its marker.
13. From there, marking G1 would force A3, B3, C3, and 3 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 A4, B4, C4, and 10 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 J4 (the last open cell in row 4), and the chain continues, and then column I would be left with no open cell for its marker. So G1 can't be the marker there.
14. Following that, if H1 were the marker, it would force A3, B3, C3, and 3 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 A4, B4, C4, and 10 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 J4 (the last open cell in row 4), and the chain continues, and column I would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, that one can't be the marker: at A2 it would force K3 and K4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then B4, C4, D4, and 8 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 J4 (the last open cell in row 4), and the chain continues, and region 11 would be left with no open cell for its marker.
16. Next, if E1 were the marker, it would force G2 (the last open cell in row 2), then J4 (the last open cell in row 4), then D5 (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.
17. Then, marking K2 would force A3, B3, C3, and 3 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then A4, B4, C4, and 11 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 J4 (the last open cell in row 4), and the chain continues, and then column G would be left with no open cell for its marker. So K2 can't be the marker there.
18. After that, marking A1 would force G2 (the last open cell in row 2), then K3 (the last open cell in row 3), and then row 4 would be left with no open cell for its marker. So A1 can't be the marker there.
19. Now, that one can't be the marker: at C1 it would force G2 (the last open cell in row 2), then K3 (the last open cell in row 3), and row 4 would be left with no open cell for its marker.
20. From there, every open cell in row 1 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.
21. Following that, that one can't be the marker: at G2 it would force J4 (the last open cell in row 4), then E6, F6, H6, and 2 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 A7, B7, C7, and 5 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 column K would be left with no open cell for its marker.
22. First, place a marker at E2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region.
23. Next, every open cell in row 3 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.
24. Then, place a marker at J4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column J, and its region. Placing here also clears its 1 touching neighbour.
25. After that, every open cell in row 5 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.
26. Now, every open cell in row 6 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.
27. From there, place a marker at A11. Column 1 has exactly one open cell left. Placing here clears the rest of row 11, column A, and its region. Placing here also clears its 1 touching neighbour.
28. Following that, place a marker at K1. Column 11 has exactly one open cell left. Placing here clears the rest of row 1, column K, and its region.
29. First, every open cell left in this region sits in row 9, so its marker has to land there; that clears every other open cell in the row.
30. 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.
31. Then, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
32. After that, place a marker at I10. Column 9 has exactly one open cell left. Placing here clears the rest of row 10, column I, and its region.
33. Now, place a marker at H3. Column 8 has exactly one open cell left. Placing here clears the rest of row 3, column H, and its region.
34. From there, every remaining candidate in this region touches F8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
35. Following that, place a marker at F7. Column 6 has exactly one open cell left. Placing here clears the rest of row 7, column F, and its region.
36. First, 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.
37. Next, every remaining candidate in this region touches C9 and D9, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
38. Then, place a marker at B9. Region has exactly one open cell left. Placing here clears the rest of row 9, column B, and its region. Placing here also clears its 1 touching neighbour.
39. After that, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region.
40. Finally, place a marker at C6. Region has exactly one open cell left.

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