An Expert 11×11 Puzzle Solved with Forced Chain in 25 Steps

This 11×11 board rates as expert (difficulty score 1232). Solving it from scratch takes 25 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, 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 K, so its marker has to land there; that clears every other open cell in the column.
2. Next, every open cell in row 11 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.
3. Then, every remaining candidate in this region touches G1, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches J2 and J3, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, marking A1 would force I10 (the last open cell in row 10), then J4 (the last open cell in region 3), then K2 (the last open cell in region 4), and then region 2 would be left with no open cell for its marker. So A1 can't be the marker there.
6. From there, if B1 were the marker, it would force I2 and K2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and region 3 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 C1 it would force I2 and K2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and region 3 would be left with no open cell for its marker.
8. First, marking D1 would force I2 and K2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and then region 3 would be left with no open cell for its marker. So D1 can't be the marker there.
9. Next, if E1 were the marker, it would force D6 (the last open cell in region 9), then I2 and K2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and region 3 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 F1 it would force H2 (the last open cell in region 2), then K3 (the last open cell in region 4), and region 3 would be left with no open cell for its marker.
11. After that, that one can't be the marker: at I1 it would force A10 (the last open cell in row 10), then B2, C2, D2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and the chain continues, and region 9 would be left with no open cell for its marker.
12. Now, marking J1 would force K3 (the last open cell in region 4), then A2, B2, C2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then E4 (the last open cell in region 1), and the chain continues, and then region 9 would be left with no open cell for its marker. So J1 can't be the marker there.
13. From there, place a marker at H1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, 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 3 touching neighbours.
15. First, 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.
16. Next, place a marker at I6. Region has exactly one open cell left. Placing here clears the rest of row 6, column I, and its region.
17. Then, place a marker at A10. Row 10 has exactly one open cell left. Placing here clears the rest of row 10, column A, and its region. Placing here also clears its 1 touching neighbour.
18. After that, 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.
19. Now, every open cell left in this region sits in row 5, so its marker has to land there; that clears every other open cell in the row.
20. From there, place a marker at B7. Column 2 has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region.
21. Following that, place a marker at C11. Column 3 has exactly one open cell left. Placing here clears the rest of row 11, column C, and its region.
22. First, place a marker at G8. Column 7 has exactly one open cell left. Placing here clears the rest of row 8, column G, and its region.
23. Next, 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.
24. Then, place a marker at D3. Region has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
25. Finally, place a marker at F5. Region has exactly one open cell left.

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