An Expert 11×11 Puzzle Solved with Forced Chain in 34 Steps

This 11×11 board rates as expert (difficulty score 1523). Solving it from scratch takes 34 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Paired regions, 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, every remaining candidate in this region touches G10, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, marking A1 would force C3, C5, C6, and 4 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then C4 (the last open cell in column C), then K2 (the last open cell in row 2), and then row 3 would be left with no open cell for its marker. So A1 can't be the marker there.
4. After that, that one can't be the marker: at C1 it would force A2, A3, A4, and 4 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then E5, D6, E6, and 6 more cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), then E2, E3, F3, and 2 more cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 10 would be left with no open cell for its marker.
5. Now, marking D1 would force A2, A3, C3, and 11 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then K2 (the last open cell in row 2), then C4 (the last open cell in column C), and then row 3 would be left with no open cell for its marker. So D1 can't be the marker there.
6. From there, if E1 were the marker, it would force A2, A3, C3, and 11 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then K2 (the last open cell in row 2), then C4 (the last open cell in column C), and row 3 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 F1 it would force A2, A3, C3, and 11 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then K2 (the last open cell in row 2), then C4 (the last open cell in column C), and row 3 would be left with no open cell for its marker.
8. First, marking G1 would force A2, A3, C3, and 11 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then C4 (the last open cell in column C), then 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.
9. Next, if H1 were the marker, it would force A2 and E2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F2, F3, F6, and 8 more cleared (together, two regions have open cells only in columns F and G), then E3, E4, E5, and 3 more cleared (every open cell left in this region sits in column E, 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.
10. Then, marking J1 would force A2 and E2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K8, K9 and K10 cleared (every open cell left in this region sits in column K, so its marker has to land there), then H2, H4, H6, and 3 more cleared (together, two regions have open cells only in columns H and I), 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.
11. After that, if K1 were the marker, it would force A2 and E2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then J6, J8, J9, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H2, H4, H6, and 4 more cleared (together, two regions have open cells only in columns H and I), 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.
12. Now, that one can't be the marker: place it at A2, and column B would be left with no open cell for its marker.
13. From there, marking B2 would force I1 (the last open cell in row 1), then C4, D4, E4, and 1 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 A4, A5, A6, and 2 more 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 then region 10 would be left with no open cell for its marker. So B2 can't be the marker there.
14. Following that, that one can't be the marker: at I1 it would force E2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A3, C3, A4, and 10 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then E5, D6, E6, and 6 more cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and column D would be left with no open cell for its marker.
15. First, place a marker at B1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
16. Next, marking E2 would force A3, A4, A5, and 3 more cleared (every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column), then D6, D7, D8, and 2 more cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), and then column D would be left with no open cell for its marker. So E2 can't be the marker there.
17. Then, place a marker at K2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column K, and its region.
18. After that, every open cell left in this region sits in column J, so its marker has to land there; that clears every other open cell in the column.
19. Now, together, two regions have open cells only in columns H and I; 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.
20. From there, together, two regions have open cells only in columns F and G; 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.
21. Following that, every open cell left in this region sits in column E, so its marker has to land there; that clears every other open cell in the column.
22. First, together, two regions have open cells only in rows 9 and 10; 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.
23. Next, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
24. Then, place a marker at C4. Region 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 A3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
26. Now, every remaining candidate in this region touches I6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
27. From there, if F5 were the marker, it would force J7 (the last open cell in region 6), and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
28. Following that, place a marker at F11. Column 6 has exactly one open cell left. Placing here clears the rest of row 11, column F, and its region. Placing here also clears its 1 touching neighbour.
29. First, place a marker at E9. Region has exactly one open cell left. Placing here clears the rest of row 9, column E, and its region. Placing here also clears its 1 touching neighbour.
30. Next, place a marker at I8. Row 8 has exactly one open cell left. Placing here clears the rest of row 8, column I, and its region. Placing here also clears its 1 touching neighbour.
31. Then, place a marker at J5. Region has exactly one open cell left. Placing here clears the rest of row 5, column J, and its region.
32. After that, place a marker at G6. Region has exactly one open cell left. Placing here clears the rest of row 6, column G, and its region.
33. Now, place a marker at D7. Region has exactly one open cell left.
34. Finally, place a marker at H10. Region has exactly one open cell left.

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