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

This 10×10 board rates as expert (difficulty score 1755). 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 7, 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 E6 and F6 and E8 and F8, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, every remaining candidate in this region touches B9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, marking A1 would force B10 (the last open cell in region 10), then C5, C6 and C8 cleared (every open cell left in this region sits in column C, so its marker has to land there), then D6 (the last open cell in region 5), and the chain continues, and then column E would be left with no open cell for its marker. So A1 can't be the marker there.
5. Now, if B1 were the marker, it would force A4, A5 and A6 cleared (every open cell left in this region sits in column A, so its marker has to land there), then C5, C6, C8, and 2 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then D6 (the last open cell in region 5), 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.
6. From there, that one can't be the marker: at C1 it would force B8 cleared (together, two regions have open cells only in columns A and B), then D6 (the last open cell in region 5), then F7 (the last open cell in region 8), and the chain continues, and region 10 would be left with no open cell for its marker.
7. Following that, that one can't be the marker: at A2 it would force B10 (the last open cell in region 10), then C5, C6 and C8 cleared (every open cell left in this region sits in column C, so its marker has to land there), then D6 (the last open cell in region 5), and the chain continues, and column E would be left with no open cell for its marker.
8. First, marking B2 would force A4, A5 and A6 cleared (every open cell left in this region sits in column A, so its marker has to land there), then C4 (the last open cell in region 3), then D6 (the last open cell in region 5), and the chain continues, and then column E would be left with no open cell for its marker. So B2 can't be the marker there.
9. Next, if C2 were the marker, it would force I6, J6, A8, and 3 more cleared (together, two regions have open cells only in rows 6 and 8), then A9, A10 and B10 cleared (together, two regions have open cells only in rows 9 and 10), and region 10 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 D2 it would force H3, G4, H4, and 1 more 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 J4, G5, H5, and 4 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), then A5, B5, A6, 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 column F would be left with no open cell for its marker.
11. After that, marking J1 would force A3, B3, D3, and 9 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 C3, A5, 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), then I3 (the last open cell in row 3), and the chain continues, and then row 9 would be left with no open cell for its marker. So J1 can't be the marker there.
12. Now, if I2 were the marker, it would force A4, B4, C4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then A3, B3, D3, and 6 more 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 C3 (the last open cell in row 3), and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, if A3 were the marker, it would force B10 (the last open cell in region 10), then C4 (the last open cell in region 3), then D6 (the last open cell in region 5), 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.
14. Following that, that one can't be the marker: at B3 it would force A8, A9 and A10 cleared (every open cell left in this region sits in column A, so its marker has to land there), and region 10 would be left with no open cell for its marker.
15. First, marking C3 would force B10 cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), then B8 (the last open cell in column B), then A10 (the last open cell in region 10), and the chain continues, and then region 7 would be left with no open cell for its marker. So C3 can't be the marker there.
16. Next, marking G1 would force J2 (the last open cell in row 2), then H3 (the last open cell in row 3), then I8 (the last open cell in region 6), and the chain continues, and then row 6 would be left with no open cell for its marker. So G1 can't be the marker there.
17. Then, that one can't be the marker: at I1 it would force D3, E3, F3, and 7 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 J3 (the last open cell in row 3), then A5, B5, A6, 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 9 would be left with no open cell for its marker.
18. After that, marking E2 would force F7 (the last open cell in region 8), then H1 (the last open cell in row 1), then I8 (the last open cell in region 6), and the chain continues, and then region 7 would be left with no open cell for its marker. So E2 can't be the marker there.
19. Now, that one can't be the marker: place it at G2, and row 1 would be left with no open cell for its marker.
20. From there, marking H2 would force D3, E3, F3, and 6 more 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 J3 (the last open cell in row 3), then A5, B5, A6, 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 then row 8 would be left with no open cell for its marker. So H2 can't be the marker there.
21. Following that, marking D1 would force J2 (the last open cell in row 2), then H3 (the last open cell in row 3), then A5, B5, A6, 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 then row 8 would be left with no open cell for its marker. So D1 can't be the marker there.
22. First, if E1 were the marker, it would force F7 (the last open cell in region 8), then J2 (the last open cell in row 2), then H3 (the last open cell in row 3), and the chain continues, and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
23. Next, that one can't be the marker: at F1 it would force E7 (the last open cell in region 8), then J2 (the last open cell in row 2), then H3 (the last open cell in row 3), and the chain continues, and column D would be left with no open cell for its marker.
24. Then, place a marker at H1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region.
25. After that, that one can't be the marker: at J2 it would force D4, E4, F4, and 3 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), then A5, B5, A6, 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), then C5 (the last open cell in row 5), and the chain continues, and row 8 would be left with no open cell for its marker.
26. Now, place a marker at F2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
27. From there, place a marker at E7. Region has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region. Placing here also clears its 2 touching neighbours.
28. Following that, 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.
29. First, 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.
30. Next, place a marker at C5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region. Placing here also clears its 1 touching neighbour.
31. Then, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
32. After that, place a marker at B10. Region has exactly one open cell left. Placing here clears the rest of row 10, column B, and its region.
33. Now, place a marker at G6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column G, and its region.
34. From there, place a marker at D9. Region has exactly one open cell left. Placing here clears the rest of row 9, column D, and its region.
35. Following that, place a marker at J8. Region has exactly one open cell left. Placing here clears the rest of row 8, column J, and its region.
36. Finally, place a marker at I3. Region has exactly one open cell left.

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