An Expert 11×11 Puzzle Solved with Forced Chain in 26 Steps

This 11×11 board rates as expert (difficulty score 1303). Solving it from scratch takes 26 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 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.
2. Next, marking A1 would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then F4, G4, H4, and 15 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then B7, C7, D7, and 4 more cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then column C would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, if B1 were the marker, it would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then H2, I2, G3, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then I4 (the last open cell in region 2), and the chain continues, and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at C1 it would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then H2, I2, G3, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then I4 (the last open cell in region 2), and the chain continues, and region 10 would be left with no open cell for its marker.
5. Now, marking D1 would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then H2, I2, G3, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then I4 (the last open cell in region 2), and the chain continues, and then region 10 would be left with no open cell for its marker. So D1 can't be the marker there.
6. From there, if E1 were the marker, it would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then H2, I2, G3, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then I4 (the last open cell in region 2), and the chain continues, and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
7. Following that, that one can't be the marker: at F1 it would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A2, B2, C2, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then J5, J6, J9, and 6 more cleared (together, two regions have open cells only in columns J and I), and the chain continues, and region 4 would be left with no open cell for its marker.
8. First, marking G1 would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A2, B2, C2, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then J5, J6, J9, and 6 more cleared (together, two regions have open cells only in columns J and I), and the chain continues, and then column C would be left with no open cell for its marker. So G1 can't be the marker there.
9. Next, if H1 were the marker, it would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A2, B2, C2, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then J5, J6, J9, and 6 more cleared (together, two regions have open cells only in columns J and I), 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, that one can't be the marker: at I1 it would force K4, K5, K6, and 5 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then J5, J6, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and region 10 would be left with no open cell for its marker.
11. After that, every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
12. Now, if K1 were the marker, it would force J3, J5, J6, and 10 more cleared (together, two regions have open cells only in columns J and I), then H9 (the last open cell in region 11), then G3 (the last open cell in region 2), and the chain continues, and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, place a marker at J1. Region has exactly one open cell left. Placing here clears the rest of row 1, column J, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, together, two regions have open cells only in columns K and I; 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, place a marker at H9. Region has exactly one open cell left. Placing here clears the rest of row 9, column H, and its region. Placing here also clears its 3 touching neighbours.
16. Next, place a marker at G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region. Placing here also clears its 2 touching neighbours.
17. Then, place a marker at I7. Column 9 has exactly one open cell left. Placing here clears the rest of row 7, column I, and its region.
18. After that, 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.
19. Now, every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
20. From there, place a marker at A5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region. Placing here also clears its 1 touching neighbour.
21. Following that, place a marker at F6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region.
22. First, place a marker at K8. Row 8 has exactly one open cell left. Placing here clears the rest of row 8, column K, and its region.
23. Next, place a marker at B11. Column 2 has exactly one open cell left. Placing here clears the rest of row 11, column B, and its region.
24. Then, 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.
25. After that, place a marker at D2. Region has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
26. Finally, place a marker at C4. Region has exactly one open cell left.

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