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

This 11×11 board rates as expert (difficulty score 1477). Solving it from scratch takes 31 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 column A, so its marker has to land there; that clears every other open cell in the column.
2. Next, every remaining candidate in this region touches C4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, every remaining candidate in this region touches B6 and B7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches D7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, if B1 were the marker, it would force C11 (the last open cell in region 8), then D6 (the last open cell in region 9), and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
6. From there, that one can't be the marker: at C1 it would force D6 (the last open cell in region 9), then A7 (the last open cell in region 7), then B9, B10 and B11 cleared (every open cell left in this region sits in column B, so its marker has to land there), and region 8 would be left with no open cell for its marker.
7. Following that, marking D1 would force C5, C9, C10, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then B8, B9, B10, and 1 more cleared (every open cell left in this region sits in column B, so its marker has to land there), and then region 8 would be left with no open cell for its marker. So D1 can't be the marker there.
8. First, marking J1 would force K5, K6, K7, and 4 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then I11 (the last open cell in region 11), then H7 (the last open cell in region 6), and the chain continues, and then region 8 would be left with no open cell for its marker. So J1 can't be the marker there.
9. Next, if K1 were the marker, it would force I2, I6, I7, and 8 more cleared (together, two regions have open cells only in columns I and J), then H7 (the last open cell in region 6), then A6 (the last open cell in region 7), and the chain continues, and region 11 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, 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.
11. After that, marking B2 would force C11 (the last open cell in region 8), then D6 (the last open cell in region 9), and then region 4 would be left with no open cell for its marker. So B2 can't be the marker there.
12. Now, if C2 were the marker, it would force D6 (the last open cell in region 9), then A7 (the last open cell in region 7), then I1 (the last open cell in row 1), and the chain continues, and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, that one can't be the marker: at D2 it would force I1 (the last open cell in row 1), then C5, C9, C10, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then B8, B9, B10, and 1 more cleared (every open cell left in this region sits in column B, so its marker has to land there), and region 8 would be left with no open cell for its marker.
14. Following that, marking E2 would force I1 (the last open cell in row 1), then K5, K6, J5, and 4 more cleared (together, two regions have open cells only in columns K and J), then H7 (the last open cell in region 6), and the chain continues, and then region 4 would be left with no open cell for its marker. So E2 can't be the marker there.
15. First, if F2 were the marker, it would force G11 (the last open cell in column G), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
16. Next, marking H2 would immediately leave row 1 with no open cell for its marker, so it can't be the marker there.
17. Then, if I2 were the marker, row 1 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 J2 it would force K4 (the last open cell in region 3), then I11 (the last open cell in region 11), then H7 (the last open cell in region 6), and the chain continues, and region 8 would be left with no open cell for its marker.
19. Now, every open cell left in this region sits in column I, so its marker has to land there; that clears every other open cell in the column.
20. From there, together, two regions have open cells only in columns K and J; 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, 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. Placing here also clears its 2 touching neighbours.
22. First, place a marker at A6. Region has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region. Placing here also clears its 1 touching neighbour.
23. Next, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region. Placing here also clears its 2 touching neighbours.
24. Then, 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.
25. After that, place a marker at J4. Column 10 has exactly one open cell left. Placing here clears the rest of row 4, column J, and its region. Placing here also clears its 2 touching neighbours.
26. Now, place a marker at I1. Region has exactly one open cell left. Placing here clears the rest of row 1, column I, and its region.
27. From there, 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. Placing here also clears its 1 touching neighbour.
28. Following that, place a marker at D3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
29. First, place a marker at E10. Region has exactly one open cell left. Placing here clears the rest of row 10, column E, and its region. Placing here also clears its 1 touching neighbour.
30. Next, place a marker at F5. Region has exactly one open cell left.
31. Finally, place a marker at K9. Region has exactly one open cell left.

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