A Hard 10×10 Puzzle Solved with Forced Chain in 20 Steps

This 10×10 board rates as hard (difficulty score 1007). Solving it from scratch takes 20 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 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, every remaining candidate in this region touches H2 and I2, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, every remaining candidate in this region touches B5, 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 B3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, every remaining candidate in this region touches J7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
6. From there, that one can't be the marker: at I1 it would force J6 (the last open cell in region 8), then A5 (the last open cell in region 7), then C4 (the last open cell in region 5), and the chain continues, and region 2 would be left with no open cell for its marker.
7. Following that, place a marker at H1. Region 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.
8. First, together, two regions have open cells only in rows 3 and 4; 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, together, two regions have open cells only in rows 5 and 6; 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.
10. Then, 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.
11. After that, every open cell left in this region sits in row 7, so its marker has to land there; that clears every other open cell in the row.
12. Now, 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. Placing here also clears its 1 touching neighbour.
13. From there, 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 1 touching neighbour.
14. Following that, place a marker at F7. Region has exactly one open cell left. Placing here clears the rest of row 7, column F, and its region. Placing here also clears its 1 touching neighbour.
15. First, place a marker at J10. Column 10 has exactly one open cell left. Placing here clears the rest of row 10, column J, and its region.
16. Next, place a marker at A9. Region has exactly one open cell left. Placing here clears the rest of row 9, column A, and its region.
17. Then, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region.
18. After that, 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.
19. Now, 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.
20. Finally, place a marker at D2. Region has exactly one open cell left.

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