A Hard 10×10 Puzzle Solved with Forced Chain in 22 Steps

This 10×10 board rates as hard (difficulty score 1102). Solving it from scratch takes 22 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, 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.
2. Next, together, two regions have open cells only in columns I and J; 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 H3. Region has exactly one open cell left. Placing here clears the rest of row 3, column H, and its region. Placing here also clears its 2 touching neighbours.
4. After that, place a marker at J2. Region has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region.
5. Now, 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. Placing here also clears its 2 touching neighbours.
6. From there, place a marker at G1. Region has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
7. Following that, together, two regions have open cells only in columns D and E; 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.
8. First, together, two regions have open cells only in rows 9 and 10; 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.
9. 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.
10. Then, marking B5 would immediately leave region 1 with no open cell for its marker, so it can't be the marker there.
11. After that, if C5 were the marker, it would force B7 (the last open cell in region 1), then E8 (the last open cell in region 6), and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, every remaining candidate in this region touches B7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
13. From there, place a marker at A5. Region has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region.
14. Following that, that one can't be the marker: at C7 it would force E8 (the last open cell in region 6), and row 9 would be left with no open cell for its marker.
15. First, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
16. Next, every remaining candidate in this region touches C9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
17. Then, every open cell left in this region sits in row 10, so its marker has to land there; that clears every other open cell in the row.
18. After that, if E7 were the marker, it would force D9 (the last open cell in region 8), and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, 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.
20. From there, 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.
21. Following that, place a marker at E9. Region has exactly one open cell left.
22. Finally, place a marker at C10. Region has exactly one open cell left.

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