An Expert 10×10 Puzzle Solved with Forced Chain in 37 Steps

This 10×10 board rates as expert (difficulty score 1744). Solving it from scratch takes 37 logical steps, using 7 techniques: Single cell, Row/column exclusion, 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 I8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, 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.
4. After that, that one can't be the marker: at I1 it would force A9 (the last open cell in row 9), then J3, J4, J5, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then G3, G4, G6, and 6 more cleared (together, two regions have open cells only in columns G and H), and the chain continues, and region 6 would be left with no open cell for its marker.
5. Now, marking J1 would force I7 (the last open cell in region 8), then A9 (the last open cell in row 9), then C8 (the last open cell in region 6), and the chain continues, and then region 4 would be left with no open cell for its marker. So J1 can't be the marker there.
6. From there, every remaining candidate in this region touches G2 and H2, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
7. Following that, that one can't be the marker: at A2 it would force I9 (the last open cell in row 9), then J7 (the last open cell in region 8), then D8, E8, F8, and 1 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and column E would be left with no open cell for its marker.
8. First, that one can't be the marker: at D2 it would force F8 (the last open cell in column F), then E6 (the last open cell in column E), then J3 (the last open cell in row 3), and row 5 would be left with no open cell for its marker.
9. Next, marking E2 would force F8 (the last open cell in column F), then A7, B7, C7, and 1 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), then A6 (the last open cell in region 6), and the chain continues, and then row 3 would be left with no open cell for its marker. So E2 can't be the marker there.
10. Then, if I2 were the marker, it would force G1 (the last open cell in region 3), then A9 (the last open cell in row 9), then J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), 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.
11. After that, that one can't be the marker: at J2 it would force I7 (the last open cell in region 8), then A9 (the last open cell in row 9), then C8 (the last open cell in region 6), and the chain continues, and row 10 would be left with no open cell for its marker.
12. Now, if A3 were the marker, it would force I9 (the last open cell in row 9), then J7 (the last open cell in region 8), then C8 and C10 cleared (every open cell left in this region sits in column C, so its marker has to land there), 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.
13. From there, that one can't be the marker: place it at B3, and region 1 would be left with no open cell for its marker.
14. Following that, that one can't be the marker: place it at E3, and row 2 would be left with no open cell for its marker.
15. First, if G3 were the marker, it would force H1 (the last open cell in region 3), then F8 (the last open cell in column F), then E6 (the last open cell in column E), 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.
16. Next, that one can't be the marker: at H3 it would force G1 (the last open cell in region 3), then F8 (the last open cell in column F), then E6 (the last open cell in column E), and the chain continues, and row 5 would be left with no open cell for its marker.
17. Then, marking I3 would force A9 (the last open cell in row 9), then F8 (the last open cell in column F), then J7 (the last open cell in region 8), and then region 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 I7 (the last open cell in region 8), then A9 (the last open cell in row 9), then C8 (the last open cell in region 6), 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.
19. Now, marking A4 would force I9 (the last open cell in row 9), then J7 (the last open cell in region 8), then D5, E5 and F5 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 region 6 would be left with no open cell for its marker. So A4 can't be the marker there.
20. From there, if C2 were the marker, it would force F3 (the last open cell in row 3), then G5, H5, I5, 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), then B6, D6, E6, 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 D would end up with no open cell for its marker, which can't happen. So it can't go there.
21. Following that, if F2 were the marker, it would force H1 (the last open cell in region 3), then B4 and C4 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then G5, I5, J5, and 1 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 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 B2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column B, and its region.
23. Next, 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.
24. Then, 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.
25. After 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.
26. Now, if H1 were the marker, it would force I6, I9, I10, and 1 more cleared (together, two regions have open cells only in columns I and J), then G10 (the last open cell in region 9), then A9 (the last open cell in region 10), and row 6 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 G1. Region has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
28. Following that, that one can't be the marker: at I4 it would force A9 (the last open cell in row 9), then C5 (the last open cell in region 5), then H6 (the last open cell in row 6), and the chain continues, and row 8 would be left with no open cell for its marker.
29. First, marking J4 would force I7 (the last open cell in region 8), then H10 (the last open cell in region 9), then A9 (the last open cell in region 10), and then row 6 would be left with no open cell for its marker. So J4 can't be the marker there.
30. Next, place a marker at H4. Region has exactly one open cell left. Placing here clears the rest of row 4, column H, and its region.
31. 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.
32. After that, place a marker at A6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
33. Now, place a marker at C5. Region has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region.
34. From there, place a marker at D10. Region has exactly one open cell left. Placing here clears the rest of row 10, column D, and its region.
35. Following that, place a marker at E8. Region has exactly one open cell left. Placing here clears the rest of row 8, column E, and its region.
36. First, place a marker at I9. Region has exactly one open cell left. Placing here clears the rest of row 9, column I, and its region.
37. Finally, place a marker at J7. Region has exactly one open cell left.

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