An Expert 10×10 Puzzle Solved with Forced Chain in 36 Steps

This 10×10 board rates as expert (difficulty score 1557). Solving it from scratch takes 36 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, 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 I2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, marking J1 would force I3, I4, I5, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then C9, D9, E9, and 8 more cleared (together, two regions have open cells only in rows 9 and 10), then H8 (the last open cell in region 8), and the chain continues, and then region 7 would be left with no open cell for its marker. So J1 can't be the marker there.
4. After that, every remaining candidate in this region touches H2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, that one can't be the marker: at G2 it would force I1 (the last open cell in region 3), then C3 and D3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then J4, J5, J6, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and column C would be left with no open cell for its marker.
6. From there, that one can't be the marker: at J2 it would force H1 (the last open cell in region 3), then C3 and D3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then I4, I5, I6, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and row 5 would be left with no open cell for its marker.
7. Following that, that one can't be the marker: at H3 it would force I1 (the last open cell in region 3), then C2 and D2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then J4, J5, J6, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and row 5 would be left with no open cell for its marker.
8. First, marking I3 would force H1 (the last open cell in region 3), then C2 and D2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then J5, J6 and J7 cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and then row 5 would be left with no open cell for its marker. So I3 can't be the marker there.
9. Next, if J3 were the marker, it would force C2 and D2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then I1, I5, I6, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then H1 (the last open cell in region 3), and the chain continues, and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, if H4 were the marker, it would force I1 (the last open cell in region 3), then C9, D9, D10, and 2 more cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), then E9, F9, J9, and 1 more cleared (together, two regions have open cells only in rows 9 and 10), and the chain continues, and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, that one can't be the marker: at I4 it would force H1 (the last open cell in region 3), then E9, F9, J9, and 1 more cleared (together, two regions have open cells only in rows 9 and 10), then J8 (the last open cell in region 9), and the chain continues, and column D would be left with no open cell for its marker.
12. Now, marking J4 would force I1 and I8 cleared (every open cell left in this region sits in column I, so its marker has to land there), then H1 (the last open cell in region 3), then A7, B7, E9, and 1 more cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then region 8 would be left with no open cell for its marker. So J4 can't be the marker there.
13. From there, if C2 were the marker, it would force F3 and G3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then F5, G5, H5, and 3 more 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 G6, H6, I6, and 6 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and column F would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, that one can't be the marker: at D2 it would force F3 and G3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then F5, G5, H5, and 3 more 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 G6, H6, I6, and 6 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 9 would be left with no open cell for its marker.
15. First, if F2 were the marker, it would force C3 and D3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A5, B5, C5, and 3 more 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 G6, H6, I6, and 6 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
16. Next, every open cell in row 2 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.
17. Then, marking C3 would force F5, G5, H5, and 3 more 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 G6, H6, I6, and 6 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then D7 and E7 cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then column G would be left with no open cell for its marker. So C3 can't be the marker there.
18. After that, if D3 were the marker, it would force F5, G5, H5, and 3 more 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 G6, H6, I6, and 6 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then C7 and E7 cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 9 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 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.
20. From there, every open cell in row 4 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.
21. Following that, 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.
22. First, every open cell in row 6 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.
23. Next, every open cell in row 7 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.
24. Then, 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.
25. After 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.
26. Now, place a marker at J8. Row 8 has exactly one open cell left. Placing here clears the rest of row 8, column J, and its region.
27. From there, place a marker at I5. Region has exactly one open cell left.
28. Following that, 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.
29. First, place a marker at A7. Region has exactly one open cell left. Placing here clears the rest of row 7, column A, and its region. Placing here also clears its 1 touching neighbour.
30. Next, place a marker at C6. Region has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region.
31. Then, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
32. After that, place a marker at E2. Region has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region. Placing here also clears its 1 touching neighbour.
33. Now, 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.
34. From there, place a marker at D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region.
35. Following that, place a marker at F10. Region has exactly one open cell left. Placing here clears the rest of row 10, column F, and its region.
36. Finally, place a marker at B9. Region has exactly one open cell left.

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