An Expert 11×11 Puzzle Solved with Forced Chain in 23 Steps

This 11×11 board rates as expert (difficulty score 1162). Solving it from scratch takes 23 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, together, two regions have open cells only in rows 1 and 2; 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.
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, place a marker at B8. Region has exactly one open cell left. Placing here clears the rest of row 8, column B, and its region. Placing here also clears its 2 touching neighbours.
4. After that, marking G1 would force I3, I6, I7, and 9 more cleared (together, two regions have open cells only in columns I and J), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then D9, E9, F9, and 3 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then row 10 would be left with no open cell for its marker. So G1 can't be the marker there.
5. Now, if H1 were the marker, it would force J2 (the last open cell in region 3), then I6, I7, I9, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and row 11 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 I1 it would force J5, J6, J7, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then D9, E9, F9, and 4 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 10 would be left with no open cell for its marker.
7. Following that, marking J1 would force I3, I6, I7, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then D9, E9, F9, and 4 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then row 10 would be left with no open cell for its marker. So J1 can't be the marker there.
8. First, place a marker at F1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
9. Next, every remaining candidate in this region touches I3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
10. Then, if I2 were the marker, it would force J4 (the last open cell in region 7), then A3 (the last open cell in region 1), then H9, H10 and H11 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and row 11 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 J2 it would force I6, I7, I9, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then D9, E9, G9, and 3 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 10 would be left with no open cell for its marker.
12. Now, place a marker at H2. Region has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region. Placing here also clears its 1 touching neighbour.
13. From there, 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.
14. Following that, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
15. First, place a marker at I5. Region has exactly one open cell left. Placing here clears the rest of row 5, column I, and its region. Placing here also clears its 1 touching neighbour.
16. Next, 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.
17. Then, together, two regions have open cells only in rows 6 and 7; 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.
18. After that, place a marker at J9. Region has exactly one open cell left. Placing here clears the rest of row 9, column J, and its region. Placing here also clears its 1 touching neighbour.
19. Now, place a marker at K11. Region has exactly one open cell left. Placing here clears the rest of row 11, column K, and its region.
20. From there, every remaining candidate in this region touches D6 and D7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
21. Following that, place a marker at C6. Region has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region.
22. First, place a marker at E7. Region has exactly one open cell left.
23. Finally, place a marker at D10. Region has exactly one open cell left.

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