A Severe 11×11 Puzzle Solved with Forced Chain in 48 Steps

This 11×11 board rates as severe (difficulty score 2051). Solving it from scratch takes 48 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 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 left in this region sits in row 11, so its marker has to land there; that clears every other open cell in the row.
3. Then, 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.
4. After that, every remaining candidate in this region touches G9 and H9 and G11, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, if B1 were the marker, it would force E11 (the last open cell in column E), then A9, C9, I9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then D6, D7 and D8 cleared (every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
6. From there, marking G1 would force H10 (the last open cell in region 10), then F8 (the last open cell in region 8), then J3, J4, J5, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and then row 6 would be left with no open cell for its marker. So G1 can't be the marker there.
7. Following that, that one can't be the marker: at I1 it would force J3, J4, J5, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K2, K3, K4, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then C3, D3, E3, and 14 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 the chain continues, and region 5 would be left with no open cell for its marker.
8. First, marking J1 would force I8 (the last open cell in region 9), and then region 8 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 C3, D3, E3, and 14 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 I3, I4, I5, and 4 more cleared (together, two regions have open cells only in columns I and J), then H7, H8 and G8 cleared (together, two regions have open cells only in columns H and G), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, marking B2 would force A9, C9, I9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then F3, G3, F4, 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 G4, H4, I4, 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), and the chain continues, and then row 5 would be left with no open cell for its marker. So B2 can't be the marker there.
11. After that, marking E2 would force H1 (the last open cell in row 1), then G10 (the last open cell in region 10), then F8 (the last open cell in column F), and the chain continues, and then row 6 would be left with no open cell for its marker. So E2 can't be the marker there.
12. Now, if F2 were the marker, it would force H1 (the last open cell in region 2), then G10 (the last open cell in region 10), then I9 (the last open cell in region 8), and the chain continues, and region 6 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 H10 (the last open cell in region 10), then F8 (the last open cell in region 8), then J3, J4, J5, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and row 6 would be left with no open cell for its marker.
14. Following that, marking H2 would force G10 (the last open cell in region 10), then C3, D3, E3, and 12 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 E11 (the last open cell in column E), and the chain continues, and then row 6 would be left with no open cell for its marker. So H2 can't be the marker there.
15. First, if I2 were the marker, it would force J4, J5 and J6 cleared (every open cell left in this region sits in column J, so its marker has to land there), then K3, K4 and K5 cleared (every open cell left in this region sits in column K, so its marker has to land there), then C3, D3, E3, and 14 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 the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
16. Next, that one can't be the marker: at J2 it would force I8 (the last open cell in region 9), and region 8 would be left with no open cell for its marker.
17. Then, that one can't be the marker: at B3 it would force A9, E9, I9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then A7, E7, F7, and 6 more cleared (together, two regions have open cells only in rows 7 and 8), then A6, E6, F6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and row 2 would be left with no open cell for its marker.
18. After that, marking C3 would force K2 (the last open cell in row 2), then F4 and F5 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 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So C3 can't be the marker there.
19. Now, if G3 were the marker, it would force H10 (the last open cell in region 10), then F8 (the last open cell in region 8), then J4, J5 and J6 cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
20. From there, that one can't be the marker: at H3 it would force G10 (the last open cell in region 10), then C1, D1, C4, and 11 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 F4 and F5 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 the chain continues, and column K would be left with no open cell for its marker.
21. Following that, if J3 were the marker, it would force I8 (the last open cell in region 9), and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
22. First, if B4 were the marker, it would force A9, E9, I9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then A7, E7, F7, and 6 more cleared (together, two regions have open cells only in rows 7 and 8), then A6 (the last open cell in region 4), and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
23. Next, if D3 were the marker, it would force K2 (the last open cell in row 2), then F4 and F5 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 (the last open cell in row 4), and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
24. Then, that one can't be the marker: at E3 it would force K2 (the last open cell in row 2), then A4 (the last open cell in row 4), then F5 (the last open cell in row 5), and row 1 would be left with no open cell for its marker.
25. After that, marking F3 would force C2, D2, C4, and 10 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 K2 (the last open cell in row 2), then A4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So F3 can't be the marker there.
26. Now, that one can't be the marker: at C1 it would force K2 (the last open cell in row 2), then A3 (the last open cell in row 3), then F4 (the last open cell in row 4), and the chain continues, and column I would be left with no open cell for its marker.
27. From there, marking D1 would force K2 (the last open cell in row 2), then A3 (the last open cell in row 3), then F4 (the last open cell in row 4), and the chain continues, and then region 7 would be left with no open cell for its marker. So D1 can't be the marker there.
28. Following 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.
29. First, marking K2 would force A3 (the last open cell in row 3), then C5, D5, E5, and 8 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 B5 (the last open cell in row 5), and the chain continues, and then column I would be left with no open cell for its marker. So K2 can't be the marker there.
30. Next, every open cell in row 2 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.
31. Then, if H1 were the marker, it would force G10 (the last open cell in region 10), then E11 (the last open cell in column E), then F8 (the last open cell in column F), 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.
32. After that, marking I3 would force A4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So I3 can't be the marker there.
33. Now, that one can't be the marker: at K3 it would force A4 (the last open cell in row 4), and row 5 would be left with no open cell for its marker.
34. From there, place a marker at A3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
35. Following that, 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.
36. First, 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.
37. Next, 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.
38. Then, 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.
39. After that, every remaining candidate in this region touches J7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
40. Now, 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 2 touching neighbours.
41. From there, place a marker at J8. Region has exactly one open cell left. Placing here clears the rest of row 8, column J, and its region.
42. Following that, place a marker at K6. Region has exactly one open cell left. Placing here clears the rest of row 6, column K, and its region.
43. First, place a marker at G10. Region has exactly one open cell left. Placing here clears the rest of row 10, column G, and its region. Placing here also clears its 2 touching neighbours.
44. Next, place a marker at I4. Region has exactly one open cell left.
45. Then, place a marker at D9. Region has exactly one open cell left. Placing here clears the rest of row 9, column D, and its region.
46. After that, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
47. Now, place a marker at E11. Region has exactly one open cell left. Placing here clears the rest of row 11, column E, and its region.
48. Finally, place a marker at F1. Region has exactly one open cell left.

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