An Expert 11×11 Puzzle Solved with Forced Chain in 44 Steps

This 11×11 board rates as expert (difficulty score 1922). Solving it from scratch takes 44 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 columns A and B; 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.
2. Next, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
3. Then, every remaining candidate in this region touches D9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, that one can't be the marker: place it at F1, and region 2 would be left with no open cell for its marker.
5. Now, marking H2 would force G4 (the last open cell in region 3), then A1, J1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then A3, B3, J3, and 1 more 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 column B would be left with no open cell for its marker. So H2 can't be the marker there.
6. From there, if G3 were the marker, it would force D5, D6, D7, and 10 more cleared (together, two regions have open cells only in columns D and E), then F5, F6, F9, and 2 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then J7, K7, I8, and 8 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), 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.
7. Following that, marking A1 would force G5, G6, G7, and 4 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then F3 and H3 cleared (every remaining candidate in this region touches F3 and H3), then D2, I2, J2, and 7 more cleared (together, two regions have open cells only in rows 2 and 4), and the chain continues, and then row 3 would be left with no open cell for its marker. So A1 can't be the marker there.
8. First, marking J1 would force G5, G6, G7, and 4 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then F3 and H3 cleared (every remaining candidate in this region touches F3 and H3), then A2, D2, A4, and 5 more cleared (together, two regions have open cells only in rows 2 and 4), and the chain continues, and then row 3 would be left with no open cell for its marker. So J1 can't be the marker there.
9. Next, if K1 were the marker, it would force G5, G6, G7, and 4 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then F3 and H3 cleared (every remaining candidate in this region touches F3 and H3), then A2, D2, I2, and 7 more cleared (together, two regions have open cells only in rows 2 and 4), 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.
10. Then, that one can't be the marker: at G2 it would force F4 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 D5, D6, D7, and 10 more cleared (together, two regions have open cells only in columns D and E), then F5, F6, F9, and 2 more cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and row 6 would be left with no open cell for its marker.
11. After that, marking D1 would force G4 (the last open cell in region 3), then E3 (the last open cell in region 5), then A7, A8, A9, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and then column B would be left with no open cell for its marker. So D1 can't be the marker there.
12. Now, if E1 were the marker, it would force G4 (the last open cell in region 3), then D3 (the last open cell in region 5), then A7, A8, A9, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
13. 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.
14. Following 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.
15. First, that one can't be the marker: at A2 it would force F5, F6, F7, and 4 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then H3, I3, J3, and 5 more cleared (together, two regions have open cells only in rows 3 and 4), then K5 (the last open cell in region 4), and region 6 would be left with no open cell for its marker.
16. Next, if I2 were the marker, it would force G1 (the last open cell in region 3), then F7, F8, F9, and 2 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and column B 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 J2 it would force F7, F8, F9, and 2 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A3, B3, A4, and 3 more cleared (together, two regions have open cells only in rows 3 and 4), and the chain continues, and column B would be left with no open cell for its marker.
18. After that, marking K2 would force F5, F6, F7, and 4 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then A3, B3, H3, and 5 more cleared (together, two regions have open cells only in rows 3 and 4), then A7, A8, A9, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and then column B would be left with no open cell for its marker. So K2 can't be the marker there.
19. Now, every remaining candidate in this region touches J4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
20. From there, if A3 were the marker, it would force K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then H4, I4 and K4 cleared (together, two regions have open cells only in rows 2 and 4), then K5 (the last open cell in region 4), 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.
21. Following that, if D3 were the marker, it would force F2 (the last open cell in row 2), then K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A4, B4, I4, and 3 more cleared (together, two regions have open cells only in rows 4 and 5), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
22. First, if F2 were the marker, it would force A4, B4, H4, and 2 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then B3, A5, D5, and 2 more cleared (together, two regions have open cells only in rows 3 and 5), then A6 (the last open cell in region 1), and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
23. Next, marking F3 would force D2 (the last open cell in row 2), then K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A4, B4, I4, and 3 more cleared (together, two regions have open cells only in rows 4 and 5), and the chain continues, and then column B would be left with no open cell for its marker. So F3 can't be the marker there.
24. Then, together, two regions have open cells only in rows 2 and 4; 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.
25. After that, that one can't be the marker: at D2 it would force F4 (the last open cell in region 2), then H1, H6, H7, and 4 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then B3, A5, I5, and 1 more cleared (together, two regions have open cells only in rows 3 and 5), and the chain continues, and column B would be left with no open cell for its marker.
26. Now, 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.
27. From there, place a marker at D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region.
28. Following that, every open cell in row 11 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.
29. First, marking G1 would force F5, F6 and F11 cleared (every open cell left in this region sits in column F, so its marker has to land there), then H6, H7, H8, and 3 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then A9 (the last open cell in row 9), and the chain continues, and then column H would be left with no open cell for its marker. So G1 can't be the marker there.
30. Next, that one can't be the marker: at H3 it would force I1 (the last open cell in region 3), then K5 (the last open cell in region 4), and region 7 would be left with no open cell for its marker.
31. Then, every remaining candidate in this region touches G6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
32. After that, marking I3 would force H1 (the last open cell in region 3), then A9 (the last open cell in row 9), and then column B would be left with no open cell for its marker. So I3 can't be the marker there.
33. Now, if J3 were the marker, it would force A7, A8, A9, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then H9 (the last open cell in row 9), then I1 (the last open cell in region 3), 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.
34. From there, every open cell left in this region sits in column K, so its marker has to land there; that clears every other open cell in the column.
35. Following that, that one can't be the marker: at K3 it would force A7, A8, A9, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then H9 (the last open cell in row 9), then I1 (the last open cell in region 3), and the chain continues, and row 7 would be left with no open cell for its marker.
36. First, place a marker at K5. Region has exactly one open cell left. Placing here clears the rest of row 5, column K, and its region. Placing here also clears its 1 touching neighbour.
37. 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. Placing here also clears its 1 touching neighbour.
38. Then, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region.
39. After that, place a marker at G8. Region has exactly one open cell left. Placing here clears the rest of row 8, column G, and its region. Placing here also clears its 2 touching neighbours.
40. Now, 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.
41. From there, place a marker at I7. Region has exactly one open cell left. Placing here clears the rest of row 7, column I, and its region.
42. Following that, place a marker at H1. Region has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region.
43. First, place a marker at A9. Region has exactly one open cell left.
44. Finally, place a marker at J11. Region has exactly one open cell left.

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