A Severe 11×11 Puzzle Solved with Forced Chain in 25 Steps

This 11×11 board rates as severe (difficulty score 1243). Solving it from scratch takes 25 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 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 J2, 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 B10, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, if B1 were the marker, it would force K2 (the last open cell in region 4), then J4 (the last open cell in region 5), then H3 (the last open cell in region 3), and the chain continues, and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. Now, that one can't be the marker: at C1 it would force K2 (the last open cell in region 4), then J4 (the last open cell in region 5), then H3 (the last open cell in region 3), and the chain continues, and region 9 would be left with no open cell for its marker.
6. From there, marking D1 would force K2 (the last open cell in region 4), then J4 (the last open cell in region 5), then H3 (the last open cell in region 3), and the chain continues, and then region 9 would be left with no open cell for its marker. So D1 can't be the marker there.
7. Following that, if E1 were the marker, it would force K2 (the last open cell in region 4), then J4 (the last open cell in region 5), then H3 (the last open cell in region 3), and the chain continues, and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, that one can't be the marker: at F1 it would force K2 (the last open cell in region 4), then J4 (the last open cell in region 5), then H3 (the last open cell in region 3), and the chain continues, and region 9 would be left with no open cell for its marker.
9. Next, if H1 were the marker, it would force K2 (the last open cell in region 4), then J4 (the last open cell in region 5), then I7 (the last open cell in region 7), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, that one can't be the marker: at I1 it would force K2 (the last open cell in region 4), then J4 (the last open cell in region 5), and region 7 would be left with no open cell for its marker.
11. After that, marking J1 would force K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then I3, I4, I6, and 4 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then H11 (the last open cell in region 9), and then region 3 would be left with no open cell for its marker. So J1 can't be the marker there.
12. Now, every open cell left in this region sits in column K, so its marker has to land there; that clears every other open cell in the column.
13. From there, together, two regions have open cells only in columns I 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.
14. Following that, place a marker at H11. Region has exactly one open cell left. Placing here clears the rest of row 11, column H, and its region. Placing here also clears its 1 touching neighbour.
15. First, place a marker at G1. Region has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region. Placing here also clears its 1 touching neighbour.
16. Next, place a marker at K2. Region has exactly one open cell left. Placing here clears the rest of row 2, column K, and its region. Placing here also clears its 1 touching neighbour.
17. Then, place a marker at J4. Region has exactly one open cell left. Placing here clears the rest of row 4, column J, and its region. Placing here also clears its 1 touching neighbour.
18. After that, place a marker at I7. Region has exactly one open cell left. Placing here clears the rest of row 7, column I, and its region.
19. Now, place a marker at D10. Region has exactly one open cell left. Placing here clears the rest of row 10, column D, and its region. Placing here also clears its 2 touching neighbours.
20. From there, 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. Placing here also clears its 1 touching neighbour.
21. Following that, every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
22. First, place a marker at C8. Column 3 has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region.
23. Next, 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 1 touching neighbour.
24. Then, place a marker at E3. Region has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region.
25. Finally, place a marker at B5. Region has exactly one open cell left.

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