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

This 10×10 board rates as expert (difficulty score 1754). Solving it from scratch takes 36 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 row 10, so its marker has to land there; that clears every other open cell in the row.
2. Next, every open cell left in this region sits in column A, 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 B8 and B9, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches H9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, if B1 were the marker, it would force H2, I2 and J2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F8, G8, H8, and 8 more cleared (together, two regions have open cells only in rows 8 and 9), then D6, E6, F6, and 6 more cleared (together, two regions have open cells only in rows 6 and 7), 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.
6. From there, marking D1 would force H2, I2 and J2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then C8, E8 and C9 cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then C7 (the last open cell in column C), and then column B would be left with no open cell for its marker. So D1 can't be the marker there.
7. Following that, marking G1 would force H3, H4, H5, and 6 more cleared (together, two regions have open cells only in columns H and I), then J5, J6, J7, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and then region 4 would be left with no open cell for its marker. So G1 can't be the marker there.
8. First, if H1 were the marker, it would force F2 (the last open cell in region 2), then I3, I4, I5, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then G10 (the last open cell in region 10), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, that one can't be the marker: at I1 it would force H6 (the last open cell in region 8), then G10 (the last open cell in region 10), then F2 (the last open cell in region 2), and the chain continues, and region 6 would be left with no open cell for its marker.
10. Then, marking J1 would force B2, C2, D2, and 1 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then I10, H7, H8, and 1 more cleared (together, two regions have open cells only in columns I and H), then G10 (the last open cell in region 10), 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.
11. After that, marking B2 would force F8, G8, H8, and 8 more cleared (together, two regions have open cells only in rows 8 and 9), then D6, E6, F6, and 6 more cleared (together, two regions have open cells only in rows 6 and 7), then H5, I5 and J5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and then row 7 would be left with no open cell for its marker. So B2 can't be the marker there.
12. Now, that one can't be the marker: at D2 it would force F1 (the last open cell in region 2), then C8, E8 and C9 cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then C7 (the last open cell in column C), and column B would be left with no open cell for its marker.
13. From there, marking E2 would immediately leave row 1 with no open cell for its marker, so it can't be the marker there.
14. Following that, that one can't be the marker: at G2 it would force C1 (the last open cell in row 1), then D8 and E8 cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then H4, H5, H7, and 4 more cleared (together, two regions have open cells only in columns H and I), and the chain continues, and row 4 would be left with no open cell for its marker.
15. First, marking H2 would force C1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then I4, I5 and I10 cleared (every open cell left in this region sits in column I, so its marker has to land there), then G10 (the last open cell in region 10), and the chain continues, and then row 4 would be left with no open cell for its marker. So H2 can't be the marker there.
16. Next, if I2 were the marker, it would force H6 (the last open cell in region 8), then G10 (the last open cell in region 10), then C1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), and the chain continues, and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
17. Then, that one can't be the marker: at B3 it would force F2 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then J2 (the last open cell in row 2), then I4 (the last open cell in row 4), and the chain continues, and row 7 would be left with no open cell for its marker.
18. After that, if D3 were the marker, it would force F2 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then J2 (the last open cell in row 2), then I4 (the last open cell in row 4), and the chain continues, and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, that one can't be the marker: at E3 it would force F1 (the last open cell in region 2), then J2 (the last open cell in row 2), then I4 (the last open cell in row 4), and the chain continues, and row 5 would be left with no open cell for its marker.
20. From there, marking F3 would force E1 (the last open cell in region 2), then J2 (the last open cell in row 2), then I4 (the last open cell in row 4), and the chain continues, and then row 5 would be left with no open cell for its marker. So F3 can't be the marker there.
21. Following that, if G3 were the marker, it would force C2 (the last open cell in row 2), then I4 (the last open cell in row 4), and column J would end up with no open cell for its marker, which can't happen. So it can't go there.
22. First, that one can't be the marker: at H3 it would force I5 and I10 cleared (every open cell left in this region sits in column I, so its marker has to land there), then G10 (the last open cell in region 10), then C1, C2, B5, and 2 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 row 1 would be left with no open cell for its marker.
23. Next, marking I3 would immediately leave column J with no open cell for its marker, so it can't be the marker there.
24. Then, that one can't be the marker: at C1 it would force F2 (the last open cell in region 2), then J3 (the last open cell in row 3), and row 4 would be left with no open cell for its marker.
25. After that, 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.
26. Now, if C2 were the marker, it would force J3 (the last open cell in row 3), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
27. 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.
28. Following that, place a marker at C3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column C, and its region.
29. First, place a marker at I4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column I, and its region.
30. Next, place a marker at H6. Region has exactly one open cell left. Placing here clears the rest of row 6, column H, and its region. Placing here also clears its 2 touching neighbours.
31. Then, 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. Placing here also clears its 1 touching neighbour.
32. After that, place a marker at B7. Column 2 has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region. Placing here also clears its 1 touching neighbour.
33. Now, 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.
34. From there, place a marker at F8. Region has exactly one open cell left. Placing here clears the rest of row 8, column F, and its region.
35. Following that, place a marker at E1. Region has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
36. Finally, place a marker at D5. Region has exactly one open cell left.

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