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

This 11×11 board rates as expert (difficulty score 1586). 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 remaining candidate in this region touches J3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, marking A1 would force J2 (the last open cell in region 2), then H3 (the last open cell in row 3), then K4 (the last open cell in region 5), and then region 8 would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, if B1 were the marker, it would force J2 (the last open cell in region 2), then A4, C4, D4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then K4, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at C1 it would force J2 (the last open cell in region 2), then A4, B4, D4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then K4, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and region 4 would be left with no open cell for its marker.
5. Now, marking D1 would force J2 (the last open cell in region 2), then A4, B4, C4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then K4, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and then region 4 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 J2 (the last open cell in region 2), then A4, B4, C4, and 4 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then K4, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and region 4 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 J2 (the last open cell in region 2), then A4, B4, C4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then K4, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and region 4 would be left with no open cell for its marker.
8. First, marking G1 would force J2 (the last open cell in region 2), then A4, B4, C4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then K4, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and then region 4 would be left with no open cell for its marker. So G1 can't be the marker there.
9. Next, 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.
10. Then, that one can't be the marker: at I1 it would force K8, K9, K10, and 4 more cleared (together, two regions have open cells only in columns K and J), and region 10 would be left with no open cell for its marker.
11. After that, marking J1 would force K3, K4, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then I5, I6, I7, and 4 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and then region 10 would be left with no open cell for its marker. So J1 can't be the marker there.
12. Now, if K1 were the marker, it would force J4, J9, J10, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then I5, I6, I7, and 4 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, place a marker at H1. Region has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region. Placing here also clears its 2 touching neighbours.
14. Following that, that one can't be the marker: at A2 it would force B4, C4, D4, and 2 more cleared (together, two regions have open cells only in rows 3 and 4), then I3, I4, I8, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K8, K9, K10, and 4 more cleared (together, two regions have open cells only in columns K and J), and region 10 would be left with no open cell for its marker.
15. First, marking B2 would force C4, D4, F4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), then I3, I4, I8, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K8, K9, K10, and 4 more cleared (together, two regions have open cells only in columns K and J), and then region 10 would be left with no open cell for its marker. So B2 can't be the marker there.
16. Next, if C2 were the marker, it would force B4, D4, F4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), then I3, I4, I8, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K8, K9, K10, and 4 more cleared (together, two regions have open cells only in columns K and J), and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
17. Then, if F2 were the marker, it would force G7, G11, I3, and 3 more cleared (together, two regions have open cells only in columns G and I), then K8, K9, K10, and 4 more cleared (together, two regions have open cells only in columns K and J), and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
18. After that, marking K2 would force J9, J10 and J11 cleared (every open cell left in this region sits in column J, so its marker has to land there), then I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then G4 (the last open cell in region 4), 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.
19. Now, 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.
20. From there, if A3 were the marker, it would force C4, D4, F4, and 1 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then I4, I8, I9, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then J9, J10, J11, and 4 more cleared (together, two regions have open cells only in columns J and K), and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
21. Following that, 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.
22. First, that one can't be the marker: at K3 it would force J9, J10 and J11 cleared (every open cell left in this region sits in column J, so its marker has to land there), then I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then G4 (the last open cell in region 4), and the chain continues, and column I would be left with no open cell for its marker.
23. Next, place a marker at I3. Region has exactly one open cell left. Placing here clears the rest of row 3, column I, and its region.
24. Then, place a marker at G4. Region has exactly one open cell left. Placing here clears the rest of row 4, column G, and its region. Placing here also clears its 1 touching neighbour.
25. After that, place a marker at F6. Region has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region. Placing here also clears its 2 touching neighbours.
26. Now, place a marker at D5. Region has exactly one open cell left. Placing here clears the rest of row 5, column D, and its region.
27. From there, place a marker at E2. Region has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region.
28. Following that, every open cell left in this region sits in row 7, so its marker has to land there; that clears every other open cell in the row.
29. First, 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. Placing here also clears its 1 touching neighbour.
30. Next, place a marker at B9. Region has exactly one open cell left. Placing here clears the rest of row 9, column B, and its region. Placing here also clears its 2 touching neighbours.
31. Then, place a marker at C7. Region has exactly one open cell left. Placing here clears the rest of row 7, column C, and its region.
32. After that, place a marker at A11. Region has exactly one open cell left. Placing here clears the rest of row 11, column A, and its region.
33. Finally, place a marker at K10. Region has exactly one open cell left.

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