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

This 11×11 board rates as hard (difficulty score 1005). Solving it from scratch takes 19 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Paired regions, 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, together, two regions have open cells only in columns A and B; 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 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 3 touching neighbours.
4. After that, 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. Placing here also clears its 2 touching neighbours.
5. Now, place a marker at A2. Region has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region.
6. From there, place a marker at E7. Region has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region. Placing here also clears its 2 touching neighbours.
7. Following that, place a marker at F4. Region has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region.
8. First, place a marker at B11. Region has exactly one open cell left. Placing here clears the rest of row 11, column B, and its region.
9. Next, every open cell left in this region sits in row 6, so its marker has to land there; that clears every other open cell in the row.
10. Then, every open cell left in this region sits in row 9, so its marker has to land there; that clears every other open cell in the row.
11. After that, together, two regions have open cells only in columns G and H; 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.
12. Now, if G6 were the marker, it would force H9 (the last open cell in region 10), and column I 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 H6. Region has exactly one open cell left. Placing here clears the rest of row 6, column H, and its region.
14. Following that, place a marker at G9. Region has exactly one open cell left.
15. First, that one can't be the marker: place it at J8, and row 10 would be left with no open cell for its marker.
16. Next, if K1 were the marker, it would force I10 (the last open cell in region 11), and row 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
17. Then, 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.
18. After that, place a marker at I8. Region has exactly one open cell left. Placing here clears the rest of row 8, column I, and its region.
19. Finally, place a marker at K10. Region has exactly one open cell left.

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