An Expert 11×11 Puzzle Solved with Forced Chain in 31 Steps

This 11×11 board rates as expert (difficulty score 1321). Solving it from scratch takes 31 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 open cell left in this region sits in row 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, 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.
3. Then, 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.
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, every remaining candidate in this region touches B9 and B10, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
6. From there, if H1 were the marker, it would force J2 (the last open cell in region 3), then I4 (the last open cell in region 5), then K5 (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.
7. Following that, 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.
8. First, every remaining candidate in this region touches G3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
9. Next, that one can't be the marker: at I1 it would force K4, K5, K8, and 10 more cleared (together, two regions have open cells only in columns K and J), then H3, H4, H8, and 3 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then G2 (the last open cell in region 3), and the chain continues, and region 1 would be left with no open cell for its marker.
10. Then, 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. Placing here also clears its 1 touching neighbour.
11. After that, together, two regions have open cells only in columns I and K; 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.
12. Now, 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.
13. From there, place a marker at G2. Region has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at F7. Region has exactly one open cell left. Placing here clears the rest of row 7, column F, and its region. Placing here also clears its 2 touching neighbours.
15. First, every remaining candidate in this region touches D4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
16. Next, 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.
17. Then, marking C3 would force I4 (the last open cell in region 5), then H6 (the last open cell in region 6), and then column K would be left with no open cell for its marker. So C3 can't be the marker there.
18. After that, if D3 were the marker, it would force I4 (the last open cell in region 5), then H6 (the last open cell in region 6), and column K would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, that one can't be the marker: at E3 it would force I4 (the last open cell in region 5), then H6 (the last open cell in region 6), and column K would be left with no open cell for its marker.
20. From there, every open cell in row 3 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, every remaining candidate in this region touches D5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
22. First, marking I3 would force K6 (the last open cell in region 8), then H5 (the last open cell in region 6), then C4 (the last open cell in region 2), and then region 1 would be left with no open cell for its marker. So I3 can't be the marker there.
23. Next, place a marker at K3. Region 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 I8. Region has exactly one open cell left. Placing here clears the rest of row 8, column I, and its region.
25. After that, if E4 were the marker, region 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
26. Now, 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.
27. From there, place a marker at D6. Region has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region.
28. Following that, place a marker at H5. Region has exactly one open cell left.
29. First, 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. Placing here also clears its 1 touching neighbour.
30. Next, 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.
31. Finally, place a marker at E10. Region has exactly one open cell left.

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