An Expert 11×11 Puzzle Solved with Forced Chain in 33 Steps

This 11×11 board rates as expert (difficulty score 1610). Solving it from scratch takes 33 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 row 11, so its marker has to land there; that clears every other open cell in the row.
2. Next, 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. Placing here also clears its 3 touching neighbours.
3. Then, 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.
4. After that, every remaining candidate in this region touches I2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, marking A1 would force J2 (the last open cell in region 3), then E3 (the last open cell in region 2), then B9 (the last open cell in region 9), and the chain continues, and then row 5 would be left with no open cell for its marker. So A1 can't be the marker there.
6. From there, if B1 were the marker, it would force J2 (the last open cell in region 3), then E3 (the last open cell in region 2), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
7. Following that, that one can't be the marker: at A2 it would force C1, E1, F1, and 1 more cleared (every open cell left in this region sits in row 1, so its marker has to land there), then E3 (the last open cell in region 2), then B9 (the last open cell in region 9), and the chain continues, and row 5 would be left with no open cell for its marker.
8. First, marking B2 would force E8 (the last open cell in region 9), then G9 (the last open cell in region 6), then F1 (the last open cell in region 2), and then region 3 would be left with no open cell for its marker. So B2 can't be the marker there.
9. Next, if C2 were the marker, it would force E1, F1 and G1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then E3 (the last open cell in region 2), then B9 (the last open cell in region 9), and the chain continues, and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, marking H2 would force J1 (the last open cell in region 3), then I4, I5 and I11 cleared (every open cell left in this region sits in column I, so its marker has to land there), then A4, B4, C4, and 7 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 then row 4 would be left with no open cell for its marker. So H2 can't be the marker there.
11. After that, that one can't be the marker: at B3 it would force E8 (the last open cell in region 9), then G9 (the last open cell in region 6), then H5, I5, J5, and 2 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 row 5 would be left with no open cell for its marker.
12. Now, marking C3 would force G5, H5, I5, and 5 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 then column A would be left with no open cell for its marker. So C3 can't be the marker there.
13. From there, that one can't be the marker: at E3 it would force B9 (the last open cell in region 9), then J2 (the last open cell in row 2), and row 1 would be left with no open cell for its marker.
14. Following that, marking F3 would force J2 (the last open cell in row 2), then I4 (the last open cell in row 4), then B5 (the last open cell in row 5), and then column A would be left with no open cell for its marker. So F3 can't be the marker there.
15. First, if G3 were the marker, it would force H5, I5, J5, and 2 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 column A 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 H3 it would force A5, C5, E5, and 1 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 column A would be left with no open cell for its marker.
17. Then, if E1 were the marker, it would force J2 (the last open cell in region 3), then B9 (the last open cell in region 9), then A3 (the last open cell in row 3), and the chain continues, and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
18. After that, that one can't be the marker: at F1 it would force J2 (the last open cell in region 3), then A3 (the last open cell in row 3), then I4 (the last open cell in row 4), and the chain continues, and column C would be left with no open cell for its marker.
19. Now, marking G1 would force J2 (the last open cell in region 3), then A3 (the last open cell in row 3), then I4 (the last open cell in row 4), and the chain continues, and then column C would be left with no open cell for its marker. So G1 can't be the marker there.
20. From there, marking I3 would force C1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then A5, C5, E5, and 1 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 then column A 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 A3 (the last open cell in row 3), then I4 (the last open cell in row 4), then B5 (the last open cell in row 5), and then column C would be left with no open cell for its marker. So J1 can't be the marker there.
22. First, if J3 were the marker, it would force C1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then A5, C5, E5, and 1 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 column A would end up with no open cell for its marker, which can't happen. So it can't go there.
23. Next, 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.
24. Then, 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.
25. After that, 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.
26. Now, 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. Placing here also clears its 2 touching neighbours.
27. From there, place a marker at G9. Region has exactly one open cell left. Placing here clears the rest of row 9, column G, and its region.
28. Following that, place a marker at C1. Column 3 has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
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 I4. Region has exactly one open cell left. Placing here clears the rest of row 4, column I, and its region.
31. Then, place a marker at H7. Region has exactly one open cell left. Placing here clears the rest of row 7, column H, and its region.
32. After that, place a marker at K6. Region has exactly one open cell left.
33. Finally, place a marker at F11. Region has exactly one open cell left.

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