An Expert 11×11 Puzzle Solved with Forced Chain in 45 Steps

This 11×11 board rates as expert (difficulty score 1943). Solving it from scratch takes 45 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, every open cell left in this region sits in column A, so its marker has to land there; that clears every other open cell in the column.
2. Next, 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.
3. Then, together, two regions have open cells only in columns B and C; with one marker each, those two markers have to fill exactly those two columns, so every other region's open cells there can be cleared.
4. After that, every remaining candidate in this region touches E3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, every remaining candidate in this region touches B10, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
6. From there, marking D1 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
7. Following that, if E1 were the marker, it would force D3 (the last open cell in region 3), then B6, C6, H6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then A9, B9, C9, and 4 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), 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.
8. First, that one can't be the marker: at F1 it would force D4, D5, D6, and 5 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then J3, I4, J4, and 6 more cleared (every open cell in column K belongs to the same region, so that region's marker has to be somewhere in this column), then E10, E11, G3, and 6 more cleared (together, two regions have open cells only in columns E and G), and region 11 would be left with no open cell for its marker.
9. Next, that one can't be the marker: at I1 it would force J3, J4, J5, and 2 more cleared (every open cell in column K belongs to the same region, so that region's marker has to be somewhere in this column), then G4, H4, H5, and 10 more cleared (every open cell in column J belongs to the same region, so that region's marker has to be somewhere in this column), then H3 (the last open cell in column H), and the chain continues, and region 11 would be left with no open cell for its marker.
10. Then, marking B2 would force D3 (the last open cell in region 3), then C10 (the last open cell in region 10), then A11, H11, I11, and 2 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), and the chain continues, and then column H would be left with no open cell for its marker. So B2 can't be the marker there.
11. After that, if C2 were the marker, region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, if F2 were the marker, it would force D3 (the last open cell in region 3), then E10, E11, G4, and 4 more cleared (together, two regions have open cells only in columns E and G), and region 11 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, that one can't be the marker: at G2 it would force D3 (the last open cell in region 3), then I6, J6, F7, and 7 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 J4, J5, K5, and 8 more cleared (every open cell in column I 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.
14. Following that, that one can't be the marker: at J2 it would force D3 (the last open cell in region 3), then I4, I5, I7, and 1 more cleared (every open cell in column K belongs to the same region, so that region's marker has to be somewhere in this column), then H5, H6, I6, 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), and column I would be left with no open cell for its marker.
15. First, marking G1 would force J3, I4, J4, and 6 more cleared (every open cell in column K belongs to the same region, so that region's marker has to be somewhere in this column), then H4, H5, H6, and 11 more cleared (every open cell in column J belongs to the same region, so that region's marker has to be somewhere in this column), then H3 (the last open cell in column H), and then column I would be left with no open cell for its marker. So G1 can't be the marker there.
16. Next, if H1 were the marker, it would force D3 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 J3, I4, J4, and 6 more cleared (every open cell in column K belongs to the same region, so that region's marker has to be somewhere in this column), then G4, I6, F7, and 6 more cleared (every open cell in column J belongs to the same region, so that region's marker has to be somewhere in this column), 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, that one can't be the marker: at B3 it would force C10 (the last open cell in region 10), then I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A11, H11, I11, and 2 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), and the chain continues, and column K would be left with no open cell for its marker.
18. After that, marking F3 would force D2 (the last open cell in region 3), then I6, J6, G7, and 7 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 J4, J5, K5, and 8 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then column K would be left with no open cell for its marker. So F3 can't be the marker there.
19. Now, if G3 were the marker, it would force I6, J6, F7, and 7 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 J4, J5, K5, and 8 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), then J1 (the last open cell in column J), and column K would end up with no open cell for its marker, which can't happen. So it can't go there.
20. From there, marking I3 would force G4, J6, F7, and 6 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 D5, E5, D6, and 4 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 D7, D8, D9, and 6 more cleared (together, two regions have open cells only in columns D and E), and the chain continues, and then region 11 would be left with no open cell for its marker. So I3 can't be the marker there.
21. Following that, marking J1 would force D3 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 H3 (the last open cell in row 3), then D5, E5, D6, and 4 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 row 8 would be left with no open cell for its marker. So J1 can't be the marker there.
22. First, place a marker at K1. Region has exactly one open cell left. Placing here clears the rest of row 1, column K, and its region.
23. Next, if J3 were the marker, it would force G4, I6, F7, and 6 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), and column I would end up with no open cell for its marker, which can't happen. So it can't go there.
24. Then, marking D4 would force E2 (the last open cell in region 3), then H3 (the last open cell in row 3), then A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), and the chain continues, and then region 11 would be left with no open cell for its marker. So D4 can't be the marker there.
25. After that, marking E2 would force H3 (the last open cell in row 3), then J4 (the last open cell in row 4), then F7, G7, F8, and 1 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then row 8 would be left with no open cell for its marker. So E2 can't be the marker there.
26. Now, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
27. 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.
28. Following that, if E4 were the marker, it would force D2 (the last open cell in region 3), then H3 (the last open cell in row 3), then A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), 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.
29. First, that one can't be the marker: at D2 it would force H3 (the last open cell in row 3), then J4 (the last open cell in row 4), then F7, G7 and G8 cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and row 8 would be left with no open cell for its marker.
30. Next, place a marker at D3. Region has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
31. Then, place a marker at I2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column I, and its region.
32. After that, every open cell left in this region sits in column J, so its marker has to land there; that clears every other open cell in the column.
33. Now, every open cell in column H 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.
34. From there, if H4 were the marker, it would force E7 (the last open cell in row 7), then A9, B9, C9, and 1 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then B5, B8 and C8 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 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
35. Following that, 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.
36. First, marking E5 would force H7 (the last open cell in row 7), then A9, B9 and C9 cleared (every open cell left in this region sits in row 9, so its marker has to land there), then B8 and C8 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 then row 8 would be left with no open cell for its marker. So E5 can't be the marker there.
37. Next, marking H5 would force E7 (the last open cell in row 7), then A9, B9 and C9 cleared (every open cell left in this region sits in row 9, so its marker has to land there), then B8 and C8 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 then row 8 would be left with no open cell for its marker. So H5 can't be the marker there.
38. Then, place a marker at B5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region.
39. After that, place a marker at C10. Region has exactly one open cell left. Placing here clears the rest of row 10, column C, and its region.
40. Now, every open cell left in this region sits in row 11, so its marker has to land there; that clears every other open cell in the row.
41. From there, place a marker at A9. Region has exactly one open cell left. Placing here clears the rest of row 9, column A, and its region.
42. Following that, place a marker at E8. Region has exactly one open cell left. Placing here clears the rest of row 8, column E, and its region.
43. First, place a marker at H7. Row 7 has exactly one open cell left. Placing here clears the rest of row 7, column H, and its region. Placing here also clears its 1 touching neighbour.
44. Next, place a marker at F6. Region has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region.
45. Finally, place a marker at G11. Region has exactly one open cell left.

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