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

This 10×10 board rates as hard (difficulty score 1075). Solving it from scratch takes 22 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 J, so its marker has to land there; that clears every other open cell in the column.
2. Next, place a marker at I1. Region has exactly one open cell left. Placing here clears the rest of row 1, column I, and its region. Placing here also clears its 1 touching neighbour.
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 F2. Region has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
5. Now, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
6. From there, every open cell left in this region sits in column E, so its marker has to land there; that clears every other open cell in the column.
7. Following that, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
8. First, every remaining candidate in this region touches C5 and E5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
9. Next, every remaining candidate in this region touches D6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
10. Then, every remaining candidate in this region touches C4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
11. After that, 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.
12. Now, marking A4 would force D5 (the last open cell in region 2), and then row 6 would be left with no open cell for its marker. So A4 can't be the marker there.
13. From there, if B4 were the marker, it would force D5 (the last open cell in region 2), and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, place a marker at D4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region.
15. 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.
16. Next, place a marker at E6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region.
17. Then, that one can't be the marker: at A5 it would force J7 (the last open cell in row 7), then B9 (the last open cell in row 9), and column C would be left with no open cell for its marker.
18. After that, place a marker at B5. Region has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region.
19. Now, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region.
20. From there, place a marker at J7. Region has exactly one open cell left. Placing here clears the rest of row 7, column J, and its region.
21. Following that, place a marker at G10. Region has exactly one open cell left. Placing here clears the rest of row 10, column G, and its region.
22. Finally, place a marker at A9. Region has exactly one open cell left.

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