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

This 11×11 board rates as expert (difficulty score 1810). Solving it from scratch takes 40 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 remaining candidate in this region touches J2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, marking A1 would force K2 (the last open cell in region 4), then B11 (the last open cell in region 9), then C5, C6, C7, and 6 more cleared (together, two regions have open cells only in columns C and D), and the chain continues, and then region 3 would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, if B1 were the marker, it would force K2 (the last open cell in region 4), then A5 and A6 cleared (every open cell left in this region sits in column A, so its marker has to land there), then C5, C6, C7, and 6 more cleared (together, two regions have open cells only in columns C and D), and the chain continues, and region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at C1 it would force K2 (the last open cell in region 4), then A5, A6, B5, and 5 more cleared (together, two regions have open cells only in columns A and B), then D4, D5, D6, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), and the chain continues, and region 1 would be left with no open cell for its marker.
5. Now, marking D1 would force K2 (the last open cell in region 4), then A5, A6, B5, and 5 more cleared (together, two regions have open cells only in columns A and B), then C5, C6 and C7 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and then region 1 would be left with no open cell for its marker. So D1 can't be the marker there.
6. From there, marking G1 would force K2 (the last open cell in region 4), then H3, I3, D4, and 4 more cleared (together, two regions have open cells only in rows 3 and 4), then H5 (the last open cell in region 5), and the chain continues, and then region 10 would be left with no open cell for its marker. So G1 can't be the marker there.
7. Following that, if H1 were the marker, it would force K2 (the last open cell in region 4), then I3 (the last open cell in region 5), then A4 (the last open cell in region 1), and region 2 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 I1 it would force K2 (the last open cell in region 4), 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 H3, D4, E4, and 2 more cleared (together, two regions have open cells only in rows 3 and 4), and the chain continues, and region 10 would be left with no open cell for its marker.
9. Next, that one can't be the marker: at A2 it would force B11 (the last open cell in region 9), then E1 and F1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then C5, C6, C7, and 6 more cleared (together, two regions have open cells only in columns C and D), and the chain continues, and region 3 would be left with no open cell for its marker.
10. Then, marking B2 would force E1 and F1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then A5 and A6 cleared (every open cell left in this region sits in column A, so its marker has to land there), then D4, D5, D6, and 6 more cleared (together, two regions have open cells only in columns D and C), and the chain continues, and then region 2 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, it would force E1 and F1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B4 (the last open cell in region 2), then D5, D6, D7, and 2 more cleared (every open cell left in this region sits in column D, so its marker has to land there), and the chain continues, and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, 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.
13. From there, marking J1 would force D4, D5, D6, and 6 more cleared (together, two regions have open cells only in columns D and C), then E4 (the last open cell in region 6), then C3, G3, H3, and 2 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 region 3 would be left with no open cell for its marker. So J1 can't be the marker there.
14. Following that, 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.
15. First, every remaining candidate in this region touches I4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
16. Next, if K1 were the marker, it would force D4, D5, D6, and 6 more cleared (together, two regions have open cells only in columns D and C), then E4 (the last open cell in region 6), then C3, G3, H3, and 2 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 region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
17. Then, place a marker at K2. Region has exactly one open cell left. Placing here clears the rest of row 2, column K, and its region. Placing here also clears its 1 touching neighbour.
18. After that, 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.
19. Now, that one can't be the marker: at E3 it would force F1 (the last open cell in region 2), then A4 (the last open cell in region 1), then H5 (the last open cell in region 5), and the chain continues, and region 10 would be left with no open cell for its marker.
20. From there, marking F3 would force E1 (the last open cell in region 2), then A4 (the last open cell in region 1), then H5 (the last open cell in region 5), and the chain continues, and then region 10 would be left with no open cell for its marker. So F3 can't be the marker there.
21. Following that, that one can't be the marker: at H3 it would force A4 (the last open cell in region 1), then B11 (the last open cell in region 9), then G5, I5, J5, and 4 more cleared (together, two regions have open cells only in rows 5 and 6), and the chain continues, and column I would be left with no open cell for its marker.
22. First, marking I3 would force A4 (the last open cell in region 1), then B11 (the last open cell in region 9), then G5, J5, F6, and 3 more cleared (together, two regions have open cells only in rows 5 and 6), and the chain continues, and then row 6 would be left with no open cell for its marker. So I3 can't be the marker there.
23. Next, every open cell left in this region sits in column H, so its marker has to land there; that clears every other open cell in the column.
24. Then, every remaining candidate in this region touches G4 and G5 and I5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
25. After that, marking D4 would force H5 (the last open cell in region 5), then G3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then E6 (the last open cell in region 3), and the chain continues, and then region 8 would be left with no open cell for its marker. So D4 can't be the marker there.
26. Now, if E4 were the marker, it would force F1 (the last open cell in region 2), then H5 (the last open cell in region 5), then G3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), 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.
27. From there, together, two regions have open cells only in columns C and D; 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.
28. Following that, that one can't be the marker: at F4 it would force E1 (the last open cell in region 2), then H5 (the last open cell in region 5), then C6 (the last open cell in region 6), and the chain continues, and region 8 would be left with no open cell for its marker.
29. First, if A3 were the marker, it would force B11 (the last open cell in region 9), then E7, E8, E9, and 6 more cleared (together, two regions have open cells only in columns E and F), then H5, J5, G6, and 2 more cleared (together, two regions have open cells only in rows 5 and 6), 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.
30. Next, that one can't be the marker: at B3 it would force E7, E8, E9, and 8 more cleared (together, two regions have open cells only in columns E and F), then H5, J5, G6, and 2 more cleared (together, two regions have open cells only in rows 5 and 6), then H4 (the last open cell in region 5), and the chain continues, and region 8 would be left with no open cell for its marker.
31. Then, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
32. After that, place a marker at H5. Region has exactly one open cell left. Placing here clears the rest of row 5, column H, and its region. Placing here also clears its 2 touching neighbours.
33. Now, place a marker at C6. Region has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region.
34. From there, place a marker at G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region.
35. Following that, place a marker at J7. Region has exactly one open cell left. Placing here clears the rest of row 7, column J, and its region. Placing here also clears its 1 touching neighbour.
36. First, place a marker at F8. Region has exactly one open cell left. Placing here clears the rest of row 8, column F, and its region. Placing here also clears its 1 touching neighbour.
37. Next, place a marker at E1. Region has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
38. Then, place a marker at B11. Region has exactly one open cell left. Placing here clears the rest of row 11, column B, and its region.
39. After that, place a marker at D10. Region has exactly one open cell left. Placing here clears the rest of row 10, column D, and its region.
40. 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. ∎