An Expert 11×11 Puzzle Solved with Forced Chain in 41 Steps

This 11×11 board rates as expert (difficulty score 1877). Solving it from scratch takes 41 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, together, two regions have open cells only in rows 1 and 2; 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.
2. Next, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region. Placing here also clears its 1 touching neighbour.
3. Then, every remaining candidate in this region touches B7 and C7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
4. After that, 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.
5. Now, if E1 were the marker, it would force J2 (the last open cell in region 3), then C4, D4, G4, and 1 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then C6 cleared (every remaining candidate in this region touches C6), 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.
6. From there, that one can't be the marker: place it at I1, and row 2 would be left with no open cell for its marker.
7. Following that, that one can't be the marker: at C4 it would force J5 (the last open cell in region 4), then K1 (the last open cell in region 3), and region 9 would be left with no open cell for its marker.
8. First, marking D4 would force J5 (the last open cell in region 4), then K1 (the last open cell in region 3), and then region 9 would be left with no open cell for its marker. So D4 can't be the marker there.
9. Next, marking E2 would force C6 cleared (every remaining candidate in this region touches C6), then B5, B9, B10, and 1 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then C11, D9, D10, and 1 more cleared (together, two regions have open cells only in columns C and D), and then region 11 would be left with no open cell for its marker. So E2 can't be the marker there.
10. Then, if E4 were the marker, it would force J5 (the last open cell in region 4), then K1 (the last open cell in region 3), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, marking G4 would force J5 (the last open cell in region 4), then K1 (the last open cell in region 3), and then region 9 would be left with no open cell for its marker. So G4 can't be the marker there.
12. Now, that one can't be the marker: place it at I4, and region 4 would be left with no open cell for its marker.
13. From there, 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.
14. Following that, marking J4 would force K1 (the last open cell in region 3), then I6 (the last open cell in region 9), then B8 (the last open cell in region 5), and then region 10 would be left with no open cell for its marker. So J4 can't be the marker there.
15. First, if K4 were the marker, it would force J6, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then I6 (the last open cell in region 9), then B8 (the last open cell in region 5), and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
16. Next, every remaining candidate in this region touches G5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
17. Then, that one can't be the marker: at F1 it would force J2 (the last open cell in region 3), then H4 (the last open cell in region 4), then I7, I8, K8, and 6 more cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and column K would be left with no open cell for its marker.
18. After that, if H1 were the marker, it would force J2 (the last open cell in region 3), then F4 (the last open cell in region 4), then I5, I7, I8, and 7 more cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and column K would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, marking J1 would force H5, I5, H6, and 14 more cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), then I6 (the last open cell in column I), and then column K would be left with no open cell for its marker. So J1 can't be the marker there.
20. From there, if F2 were the marker, it would force K1 (the last open cell in region 3), then H4 (the last open cell in region 4), then I7, I8, I9, and 5 more cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and column J would end up with no open cell for its marker, which can't happen. So it can't go there.
21. Following that, marking H2 would force K1 (the last open cell in region 3), then F4 (the last open cell in region 4), then I5, I7, I8, and 6 more cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then column J would be left with no open cell for its marker. So H2 can't be the marker there.
22. First, that one can't be the marker: at F4 it would force I5, G6, G7, and 16 more cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then G1 (the last open cell in column G), then J2 (the last open cell in region 3), and the chain continues, and column K would be left with no open cell for its marker.
23. Next, place a marker at H4. Region has exactly one open cell left. Placing here clears the rest of row 4, column H, and its region. Placing here also clears its 1 touching neighbour.
24. Then, every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
25. After that, marking B5 would immediately leave region 5 with no open cell for its marker, so it can't be the marker there.
26. Now, if C5 were the marker, it would force B8 (the last open cell in region 5), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
27. From there, if F5 were the marker, it would force E8 (the last open cell in column E), then D6 (the last open cell in region 6), and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
28. Following that, if K1 were the marker, it would force I2 (the last open cell in region 2), then D5 (the last open cell in row 5), then E8 (the last open cell in column E), 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.
29. First, place a marker at J2. Region has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region.
30. Next, place a marker at G1. Region has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
31. Then, marking C6 would immediately leave region 6 with no open cell for its marker, so it can't be the marker there.
32. After that, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
33. Now, if D6 were the marker, it would force B8 (the last open cell in region 5), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
34. From there, that one can't be the marker: at E6 it would force B8 (the last open cell in region 5), and region 10 would be left with no open cell for its marker.
35. Following that, place a marker at E8. Column 5 has exactly one open cell left. Placing here clears the rest of row 8, column E, and its region. Placing here also clears its 3 touching neighbours.
36. First, place a marker at D5. Region has exactly one open cell left. Placing here clears the rest of row 5, column D, and its region.
37. Next, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region.
38. Then, place a marker at K7. Region has exactly one open cell left. Placing here clears the rest of row 7, column K, and its region.
39. After that, place a marker at C11. Region has exactly one open cell left. Placing here clears the rest of row 11, column C, and its region.
40. Now, place a marker at F10. Region has exactly one open cell left. Placing here clears the rest of row 10, column F, and its region.
41. Finally, place a marker at I9. Region has exactly one open cell left.

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