A Hard 11×11 Puzzle Solved with Forced Chain in 19 Steps

This 11×11 board rates as hard (difficulty score 1021). Solving it from scratch takes 19 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, 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.
3. Then, place a marker at I11. Region has exactly one open cell left. Placing here clears the rest of row 11, column I, and its region. Placing here also clears its 1 touching neighbour.
4. After that, 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.
5. Now, place a marker at F3. Region has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region. Placing here also clears its 3 touching neighbours.
6. From there, place a marker at C4. Region has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region. Placing here also clears its 2 touching neighbours.
7. Following that, place a marker at K2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column K, and its region.
8. First, every open cell in row 5 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.
9. Next, place a marker at J6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column J, and its region.
10. Then, every remaining candidate in this region touches B8 and B9, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
11. After that, that one can't be the marker: at G5 it would force H1 (the last open cell in region 3), then B10 cleared (every open cell left in this region sits in row 10, so its marker has to land there), then D9 (the last open cell in region 10), and row 10 would be left with no open cell for its marker.
12. Now, marking H5 would force G1 (the last open cell in region 3), then B10 cleared (every open cell left in this region sits in row 10, so its marker has to land there), then D9 (the last open cell in region 10), and then row 10 would be left with no open cell for its marker. So H5 can't be the marker there.
13. From there, place a marker at E5. Region has exactly one open cell left. Placing here clears the rest of row 5, column E, and its region.
14. Following that, if B7 were the marker, it would force A9 (the last open cell in region 9), and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, place a marker at B10. Column 2 has exactly one open cell left. Placing here clears the rest of row 10, column B, and its region. Placing here also clears its 1 touching neighbour.
16. Next, place a marker at A8. Region has exactly one open cell left. Placing here clears the rest of row 8, column A, and its region.
17. Then, place a marker at D7. Column 4 has exactly one open cell left. Placing here clears the rest of row 7, column D, and its region.
18. After that, 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.
19. Finally, place a marker at G1. Region has exactly one open cell left.

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