A Severe 11×11 Puzzle Solved with Forced Chain in 26 Steps

This 11×11 board rates as severe (difficulty score 1200). Solving it from scratch takes 26 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 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 row 2, so its marker has to land there; that clears every other open cell in the row.
3. Then, every open cell left in this region sits in row 3, so its marker has to land there; that clears every other open cell in the row.
4. After that, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
5. Now, 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.
6. From there, every remaining candidate in this region touches H2 and I4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
7. Following that, every remaining candidate in this region touches J5 and K5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
8. First, if K1 were the marker, it would force J4 (the last open cell in region 5), then H3 (the last open cell in region 4), then F2 (the last open cell in region 2), and the chain continues, and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, that one can't be the marker: at G2 it would force I3 (the last open cell in region 4), then J1 (the last open cell in region 3), then K4 (the last open cell in region 5), and the chain continues, and row 7 would be left with no open cell for its marker.
10. Then, place a marker at F2. Region has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
11. After that, marking J1 would force K4 (the last open cell in region 5), then H5, H6, H7, and 8 more cleared (together, two regions have open cells only in columns H and I), then G6, E5, E6, and 3 more cleared (together, two regions have open cells only in columns G and E), 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.
12. 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.
13. From there, place a marker at H3. Region has exactly one open cell left. Placing here clears the rest of row 3, column H, and its region.
14. Following that, together, two regions have open cells only in columns J 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.
15. First, together, two regions have open cells only in columns G and E; 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.
16. Next, place a marker at C5. Region has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region. Placing here also clears its 2 touching neighbours.
17. Then, if J6 were the marker, it would force K4 (the last open cell in region 5), then B7, A8 and B8 cleared (together, two regions have open cells only in rows 7 and 8), then A7 cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
18. After that, if K4 were the marker, it would force A6 (the last open cell in row 6), then B8 (the last open cell in region 9), then G7 (the last open cell in region 7), and the chain continues, and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, 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.
20. From there, that one can't be the marker: at K6 it would force B7, A8 and B8 cleared (together, two regions have open cells only in rows 7 and 8), then A7 cleared (every open cell left in this region sits in column A, so its marker has to land there), then D9, D10 and D11 cleared (every open cell left in this region sits in column D, so its marker has to land there), and the chain continues, and column B would be left with no open cell for its marker.
21. Following that, place a marker at A6. Row 6 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.
22. First, 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.
23. Next, place a marker at G7. Region has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region.
24. Then, place a marker at E9. Region has exactly one open cell left. Placing here clears the rest of row 9, column E, and its region. Placing here also clears its 1 touching neighbour.
25. After that, place a marker at K10. Region has exactly one open cell left.
26. Finally, place a marker at D11. Region has exactly one open cell left.

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