An Expert 10×10 Puzzle Solved with Forced Chain in 35 Steps

This 10×10 board rates as expert (difficulty score 1608). Solving it from scratch takes 35 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 columns B and A; 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.
2. Next, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
3. Then, 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.
4. After that, every remaining candidate in this region touches A9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, marking D1 would force J2 (the last open cell in row 2), then F9, G9, H9, and 2 more cleared (together, two regions have open cells only in rows 9 and 10), then G3, G4, H4, and 6 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then row 4 would be left with no open cell for its marker. So D1 can't be the marker there.
6. From there, marking J1 would force D2 and E2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then G3 cleared (every remaining candidate in this region touches G3), then D4, E4, F4, and 1 more cleared (every open cell in row 3 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 J1 can't be the marker there.
7. Following that, that one can't be the marker: at D2 it would force I1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then F9, G9, H9, and 4 more cleared (together, two regions have open cells only in rows 9 and 10), then C7, E7, F7, and 6 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and column A would be left with no open cell for its marker.
8. First, if F2 were the marker, it would force I1 (the last open cell in row 1), then D3 (the last open cell in row 3), then G5, H5, J5, and 4 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), and the chain continues, and column A would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, marking H2 would immediately leave row 1 with no open cell for its marker, so it can't be the marker there.
10. Then, if I2 were the marker, row 1 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, if D3 were the marker, it would force F9, G9, H9, and 4 more cleared (together, two regions have open cells only in rows 9 and 10), then C7, E7, F7, and 6 more cleared (together, two regions have open cells only in rows 7 and 8), then E1, E9 and E10 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and column A would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, that one can't be the marker: at I1 it would force G2 (the last open cell in region 2), then E3 (the last open cell in row 3), then F5, F6 and F10 cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and row 6 would be left with no open cell for its marker.
13. From there, every open cell in row 1 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.
14. Following that, that one can't be the marker: at E3 it would force J2 (the last open cell in row 2), then F5, G5, H5, and 7 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 C5 (the last open cell in row 5), and row 6 would be left with no open cell for its marker.
15. First, that one can't be the marker: at F1 it would force J2 (the last open cell in row 2), then G3 (the last open cell in row 3), then E4, E9 and E10 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and row 5 would be left with no open cell for its marker.
16. Next, that one can't be the marker: at H3 it would force E2 (the last open cell in row 2), then G1 (the last open cell in region 2), then J4 (the last open cell in row 4), and the chain continues, and row 6 would be left with no open cell for its marker.
17. Then, marking I3 would force E2 (the last open cell in row 2), then G4 (the last open cell in row 4), then H1 (the last open cell in region 2), and the chain continues, and then row 6 would be left with no open cell for its marker. So I3 can't be the marker there.
18. After that, if J3 were the marker, it would force E2 (the last open cell in row 2), then F5, G5, H5, and 7 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 C5 (the last open cell in row 5), and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, marking E2 would force I4 (the last open cell in region 3), then G3 (the last open cell in row 3), then H1 (the last open cell in region 2), and the chain continues, and then row 6 would be left with no open cell for its marker. So E2 can't be the marker there.
20. From there, place a marker at J2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region.
21. Following that, if G3 were the marker, it would force D5 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 H7, H8 and H9 cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), then D9, E9, F9, and 4 more cleared (together, two regions have open cells only in rows 9 and 10), and the chain continues, and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
22. First, place a marker at F3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region. Placing here also clears its 1 touching neighbour.
23. Next, place a marker at H4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column H, and its region.
24. Then, every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
25. After that, every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
26. Now, 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.
27. From there, together, two regions have open cells only in rows 7 and 8; 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.
28. Following that, place a marker at G1. Column 7 has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
29. First, marking E5 would force C6 (the last open cell in region 4), and then row 7 would be left with no open cell for its marker. So E5 can't be the marker there.
30. Next, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region. Placing here also clears its 1 touching neighbour.
31. Then, place a marker at C5. Region has exactly one open cell left.
32. After that, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region.
33. Now, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region.
34. From there, place a marker at A10. Region has exactly one open cell left. Placing here clears the rest of row 10, column A, and its region.
35. Finally, place a marker at I9. Region has exactly one open cell left.

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