An Expert 10×10 Puzzle Solved with Forced Chain in 39 Steps

This 10×10 board rates as expert (difficulty score 1787). Solving it from scratch takes 39 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 1, 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 I2, 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 B9 and B10, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, marking H2 would force J1 (the last open cell in region 2), then I5, I6 and I7 cleared (every open cell left in this region sits in column I, so its marker has to land there), then G6 (the last open cell in region 5), and the chain continues, and then region 6 would be left with no open cell for its marker. So H2 can't be the marker there.
6. From there, that one can't be the marker: at J2 it would force H1 (the last open cell in region 2), then I4 cleared (every open cell left in this region sits in column I, so its marker has to land there), then G3, G7, G8, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), and the chain continues, and region 6 would be left with no open cell for its marker.
7. Following that, marking C3 would force G2 (the last open cell in row 2), then J4 (the last open cell in row 4), then I1 (the last open cell in region 2), and the chain continues, and then column A would be left with no open cell for its marker. So C3 can't be the marker there.
8. First, marking F3 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
9. Next, if G3 were the marker, row 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, that one can't be the marker: at H3 it would force E4, F4, E5, and 2 more cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then I5, I6, I7, and 3 more cleared (together, two regions have open cells only in columns I and J), then G6 (the last open cell in region 5), and the chain continues, and row 5 would be left with no open cell for its marker.
11. After that, marking I3 would force E4, F4, E5, and 2 more cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then B5, C5 and 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 J5 (the last open cell in row 5), and the chain continues, and then region 9 would be left with no open cell for its marker. So I3 can't be the marker there.
12. Now, if C2 were the marker, it would force J3 (the last open cell in row 3), then B7, D7, E7, and 3 more cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), then G8, F9, G9, and 1 more 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 region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, if J3 were the marker, it would force H4, H5 and H7 cleared (together, two regions have open cells only in columns H and I), then G7, G8, G9, and 1 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then F9 (the last open cell in region 8), and the chain continues, and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, if B4 were the marker, it would force E3 (the last open cell in row 3), then G2 (the last open cell in row 2), then H6, I6, J6, and 1 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 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, that one can't be the marker: at C4 it would force E3 (the last open cell in row 3), then G2 (the last open cell in row 2), then H6, I6, J6, and 1 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 A would be left with no open cell for its marker.
16. Next, that one can't be the marker: at E3 it would force G2 (the last open cell in row 2), then J4 (the last open cell in row 4), then I1 (the last open cell in region 2), and the chain continues, and row 6 would be left with no open cell for its marker.
17. Then, 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.
18. After that, if E4 were the marker, it would force B3 (the last open cell in region 3), then G2 (the last open cell in row 2), then H6, I6, J6, and 1 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 6 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 F4 it would force G2 (the last open cell in row 2), then H7 (the last open cell in region 8), then J6 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 D would be left with no open cell for its marker.
20. From there, marking J1 would force G2, G5 and H5 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 E5, F5 and E6 cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then I5 (the last open cell in row 5), and the chain continues, and then row 6 would be left with no open cell for its marker. So J1 can't be the marker there.
21. Following that, that one can't be the marker: at G2 it would force I1 (the last open cell in region 2), then J4 (the last open cell in row 4), then H7 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and region 6 would be left with no open cell for its marker.
22. First, 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.
23. Next, if H4 were the marker, it would force I1 (the last open cell in region 2), then J5 (the last open cell in row 5), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
24. Then, that one can't be the marker: place it at I4, and row 5 would be left with no open cell for its marker.
25. After that, if F2 were the marker, it would force G6, H1, H6, and 3 more cleared (together, two regions have open cells only in columns G and H), then I1 (the last open cell in region 2), then J7, J8, J9, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
26. Now, marking J4 would force B7, D7, E7, and 3 more cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), then H5 and H7 cleared (together, two regions have open cells only in columns H and I), then G5 (the last open cell in region 4), and the chain continues, and then row 6 would be left with no open cell for its marker. So J4 can't be the marker there.
27. From there, place a marker at G4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column G, and its region.
28. 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.
29. 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.
30. Next, that one can't be the marker: at I1 it would force J5 (the last open cell in region 5), then H7 cleared (every open cell left in this region sits in column H, so its marker has to land there), then F9 (the last open cell in region 8), and the chain continues, and row 6 would be left with no open cell for its marker.
31. Then, 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.
32. After that, place a marker at F9. Region has exactly one open cell left. Placing here clears the rest of row 9, column F, and its region. Placing here also clears its 2 touching neighbours.
33. Now, 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.
34. From there, place a marker at E2. Column 5 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.
35. Following that, 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.
36. First, place a marker at D6. Column 4 has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region. Placing here also clears its 1 touching neighbour.
37. Next, 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.
38. Then, 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.
39. Finally, place a marker at I5. Region has exactly one open cell left.

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