An Expert 11×11 Puzzle Solved with Forced Chain in 29 Steps

This 11×11 board rates as expert (difficulty score 1383). Solving it from scratch takes 29 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 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.
2. Next, 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.
3. Then, every remaining candidate in this region touches H9 and H10, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
4. After that, marking A1 would force J5 and J6 cleared (every open cell left in this region sits in column J, so its marker has to land there), then E2, G2, H2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H4 (the last open cell in region 3), and the chain continues, and then column K would be left with no open cell for its marker. So A1 can't be the marker there.
5. Now, marking G1 would force J5 and J6 cleared (every open cell left in this region sits in column J, so its marker has to land there), then A2, B2, E2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 and K4 cleared (every remaining candidate in this region touches K3 and K4), and the chain continues, and then region 5 would be left with no open cell for its marker. So G1 can't be the marker there.
6. From there, if H1 were the marker, it would force J5 and J6 cleared (every open cell left in this region sits in column J, so its marker has to land there), then A2, B2, E2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 and K4 cleared (every remaining candidate in this region touches K3 and K4), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
7. Following that, every remaining candidate in this region touches G3 and H3, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
8. First, marking J1 would force A2, B2, E2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H4 (the last open cell in region 3), then K3, A5, B5, and 2 more cleared (together, two regions have open cells only in rows 3 and 5), and then column K would be left with no open cell for its marker. So J1 can't be the marker there.
9. Next, that one can't be the marker: at A2 it would force H4 (the last open cell in region 3), then K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then J3 (the last open cell in region 4), and the chain continues, and column K would be left with no open cell for its marker.
10. Then, marking B2 would force H4 (the last open cell in region 3), then K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then J3 (the last open cell in region 4), and the chain continues, and then column K would be left with no open cell for its marker. So B2 can't be the marker there.
11. After that, that one can't be the marker: at D2 it would force H4 (the last open cell in region 3), then K1 (the last open cell in row 1), then F6, F7, F8, and 3 more cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and region 8 would be left with no open cell for its marker.
12. Now, marking E2 would force H4 (the last open cell in region 3), then K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then J3 (the last open cell in region 4), and the chain continues, and then region 1 would be left with no open cell for its marker. So E2 can't be the marker there.
13. From there, if F2 were the marker, it would force H4 (the last open cell in region 3), then D3 (the last open cell in region 5), then K1 (the last open cell in region 4), 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.
14. Following that, that one can't be the marker: at G2 it would force K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then J5 and J6 cleared (every open cell left in this region sits in column J, so its marker has to land there), then K3 and K4 cleared (every remaining candidate in this region touches K3 and K4), and the chain continues, and region 5 would be left with no open cell for its marker.
15. First, every open cell left in this region sits in column H, so its marker has to land there; that clears every other open cell in the column.
16. Next, marking H2 would force K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then J5 and J6 cleared (every open cell left in this region sits in column J, so its marker has to land there), then K3 and K4 cleared (every remaining candidate in this region touches K3 and K4), and the chain continues, and then region 5 would be left with no open cell for its marker. So H2 can't be the marker there.
17. Then, place a marker at H4. Region has exactly one open cell left. Placing here clears the rest of row 4, column H, and its region. Placing here also clears its 1 touching neighbour.
18. After that, together, two regions have open cells only in rows 3 and 5; 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.
19. Now, 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.
20. From there, place a marker at J6. Region has exactly one open cell left. Placing here clears the rest of row 6, column J, and its region.
21. Following that, place a marker at K1. Region has exactly one open cell left. Placing here clears the rest of row 1, column K, and its region.
22. First, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region. Placing here also clears its 2 touching neighbours.
23. Next, place a marker at D7. Region has exactly one open cell left. Placing here clears the rest of row 7, column D, and its region. Placing here also clears its 1 touching neighbour.
24. Then, 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.
25. After that, place a marker at F5. Region has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region.
26. Now, 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.
27. From there, 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.
28. Following that, place a marker at I10. Region has exactly one open cell left. Placing here clears the rest of row 10, column I, and its region.
29. Finally, place a marker at G11. Region has exactly one open cell left.

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