An Expert 11×11 Puzzle Solved with Forced Chain in 32 Steps

This 11×11 board rates as expert (difficulty score 1530). Solving it from scratch takes 32 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 column B, 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 D6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches C7 and C8, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, every remaining candidate in this region touches C10 and D10, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
6. From there, that one can't be the marker: at C1 it would force D11 (the last open cell in region 9), then I5, J5 and K5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then E5, E8 and E9 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and region 10 would be left with no open cell for its marker.
7. Following that, marking D1 would force E5, E8, E9, and 2 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then C4, C5 and C6 cleared (every open cell left in this region sits in column C, so its marker has to land there), then F5 (the last open cell in region 3), and the chain continues, and then region 10 would be left with no open cell for its marker. So D1 can't be the marker there.
8. First, if E1 were the marker, it would force D7 (the last open cell in region 7), then B8 (the last open cell in region 5), then C11 (the last open cell in region 9), and the chain continues, and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, if K1 were the marker, it would force C3, D3, E3, and 10 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 row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, if C2 were the marker, it would force D11 (the last open cell in region 9), then I5, J5 and K5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then E5, E8 and E9 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, that one can't be the marker: at D2 it would force E5, E8, E9, and 2 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then C4, C5 and C6 cleared (every open cell left in this region sits in column C, so its marker has to land there), then F5 (the last open cell in region 3), and the chain continues, and region 10 would be left with no open cell for its marker.
12. Now, marking E2 would force D7 (the last open cell in region 7), then B8 (the last open cell in region 5), then C11 (the last open cell in region 9), and the chain continues, and then region 10 would be left with no open cell for its marker. So E2 can't be the marker there.
13. From there, if F2 were the marker, it would force J3, H4, I4, 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 K3 (the last open cell in row 3), then C4 (the last open cell in row 4), and the chain continues, and column G would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, marking H2 would force J1 (the last open cell in row 1), then K3 (the last open cell in row 3), then C4 (the last open cell in row 4), and the chain continues, and then region 10 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 C3, D3, E3, and 8 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 K3 (the last open cell in row 3), then C4 (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.
16. Next, that one can't be the marker: at J2 it would force C3, D3, E3, and 9 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 row 3 would be left with no open cell for its marker.
17. Then, that one can't be the marker: at F1 it would force K2 (the last open cell in row 2), and row 3 would be left with no open cell for its marker.
18. After that, marking G1 would force K2 (the last open cell in row 2), and then row 3 would be left with no open cell for its marker. So G1 can't be the marker there.
19. Now, if H1 were the marker, it would force K2 (the last open cell in row 2), and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
20. 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.
21. Following that, marking K2 would force I1 (the last open cell in region 2), then D4, E4, F4, and 3 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), then C4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So K2 can't be the marker there.
22. First, place a marker at G2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region.
23. Next, place a marker at K3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column K, and its region.
24. Then, place a marker at C4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region.
25. After that, place a marker at D11. Region has exactly one open cell left. Placing here clears the rest of row 11, column D, and its region. Placing here also clears its 1 touching neighbour.
26. Now, place a marker at I5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column I, and its region.
27. From there, place a marker at J1. Region has exactly one open cell left. Placing here clears the rest of row 1, column J, and its region.
28. Following that, place a marker at H10. Region has exactly one open cell left. Placing here clears the rest of row 10, column H, and its region.
29. First, 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. Placing here also clears its 1 touching neighbour.
30. Next, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region.
31. Then, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region.
32. Finally, place a marker at F8. Region has exactly one open cell left.

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