An Expert 10×10 Puzzle Solved with Forced Chain in 38 Steps

This 10×10 board rates as expert (difficulty score 1694). Solving it from scratch takes 38 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, 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.
2. Next, marking D1 would force I2 (the last open cell in region 3), then F3, B4, F4, and 3 more cleared (together, two regions have open cells only in rows 3 and 4), then J5 (the last open cell in region 5), and the chain continues, and then column H would be left with no open cell for its marker. So D1 can't be the marker there.
3. Then, if H1 were the marker, it would force C2 (the last open cell in region 2), then A5, A6, A7, and 3 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then B6, B7, B8, and 2 more cleared (every open cell left in this region sits in column B, so its marker has to land there), and the chain continues, and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at I1 it would force C2 (the last open cell in region 2), then A5, A6, A7, and 3 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then B6, B7, B8, and 2 more cleared (every open cell left in this region sits in column B, so its marker has to land there), and the chain continues, and row 10 would be left with no open cell for its marker.
5. Now, marking J1 would force C2 (the last open cell in region 2), then A5, A6, A7, and 3 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then B6, B7, B8, and 2 more cleared (every open cell left in this region sits in column B, so its marker has to land there), and the chain continues, and then region 5 would be left with no open cell for its marker. So J1 can't be the marker there.
6. From there, place a marker at I2. Region has exactly one open cell left. Placing here clears the rest of row 2, column I, and its region. Placing here also clears its 2 touching neighbours.
7. Following that, if A3 were the marker, it would force E8, F8, H8, and 8 more cleared (every open cell in row 9 belongs to the same region, so that region's marker has to be somewhere in this row), and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, if D3 were the marker, region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, every remaining candidate in this region touches B4, 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 B6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
11. After that, that one can't be the marker: at C1 it would force B7, B8, B9, and 5 more cleared (together, two regions have open cells only in columns B and A), then D8 (the last open cell in region 8), then F6 cleared (every open cell in row 7 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 10 would be left with no open cell for its marker.
12. Now, that one can't be the marker: at E3 it would force A4 (the last open cell in region 1), then J5 (the last open cell in region 5), then C6 (the last open cell in region 6), and region 7 would be left with no open cell for its marker.
13. From there, marking F3 would force A4 (the last open cell in region 1), then D7, D8, D9, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C7, C8, C9, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and then row 10 would be left with no open cell for its marker. So F3 can't be the marker there.
14. Following that, if G3 were the marker, it would force A4 (the last open cell in region 1), then J5 (the last open cell in region 5), then C6 (the last open cell in region 6), and the chain continues, and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, every open cell in row 3 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, 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.
17. Then, every remaining candidate in this region touches E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
18. After that, if E4 were the marker, it would force J5 (the last open cell in region 5), then F6, G6 and H6 cleared (every open cell left in this region sits in row 6, so its marker has to land there), then A7, B7, F7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and row 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, 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.
20. From there, every remaining candidate in this region touches C5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
21. Following that, marking C3 would force J5 cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), then F5, G5, H5, and 5 more cleared (together, two regions have open cells only in rows 5 and 6), then E7 (the last open cell in region 7), and the chain continues, and then region 9 would be left with no open cell for its marker. So C3 can't be the marker there.
22. First, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region.
23. Next, together, two regions have open cells only in columns A and C; 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.
24. Then, if E1 were the marker, it would force D5, J5, D6, and 1 more cleared (together, two regions have open cells only in rows 5 and 6), then D4 (the last open cell in region 4), and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
25. After that, that one can't be the marker: at F4 it would force G1 (the last open cell in region 2), then H5, E6, H6, and 1 more cleared (together, two regions have open cells only in rows 5 and 6), then E7 (the last open cell in region 7), and the chain continues, and column C would be left with no open cell for its marker.
26. Now, marking G4 would force F1 (the last open cell in region 2), then E6, H6 and J6 cleared (together, two regions have open cells only in rows 5 and 6), then E7 (the last open cell in region 7), and the chain continues, and then column C would be left with no open cell for its marker. So G4 can't be the marker there.
27. From there, if H4 were the marker, it would force F5, E6, F6, and 2 more cleared (together, two regions have open cells only in rows 5 and 6), then E7 (the last open cell in region 7), then D5 (the last open cell in region 4), and the chain continues, and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
28. Following that, 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.
29. First, marking J4 would force F5, G5, H5, and 4 more cleared (together, two regions have open cells only in rows 5 and 6), then E7 (the last open cell in region 7), then D5 (the last open cell in region 4), and the chain continues, and then row 9 would be left with no open cell for its marker. So J4 can't be the marker there.
30. Next, place a marker at J5. Region has exactly one open cell left. Placing here clears the rest of row 5, column J, and its region.
31. Then, 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.
32. After that, 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.
33. Now, 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 1 touching neighbour.
34. From there, place a marker at G8. Region has exactly one open cell left. Placing here clears the rest of row 8, column G, and its region. Placing here also clears its 2 touching neighbours.
35. Following that, place a marker at F1. Region has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
36. First, place a marker at H10. Region has exactly one open cell left. Placing here clears the rest of row 10, column H, and its region.
37. 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.
38. Finally, place a marker at C6. Region has exactly one open cell left.

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