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

This 10×10 board rates as expert (difficulty score 1756). 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 column A, so its marker has to land there; that clears every other open cell in the column.
2. Next, 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.
3. Then, every remaining candidate in this region touches I9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, if B1 were the marker, it would force F2, G2 and H2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H4, I4, J4, 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), then C5, D5, E5, and 3 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 region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. Now, that one can't be the marker: at C1 it would force F2, G2 and H2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H4, I4, J4, 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), then B5, D5, E5, and 3 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 region 10 would be left with no open cell for its marker.
6. From there, marking D1 would force B2, F2, G2, and 1 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H4, I4, J4, 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), then B5, C5, E5, and 3 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 region 10 would be left with no open cell for its marker. So D1 can't be the marker there.
7. Following that, if E1 were the marker, it would force B2, C2, G2, and 1 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H4, I4, J4, 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), then B5, C5, D5, and 3 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 region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, that one can't be the marker: at F1 it would force B2, C2 and D2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then E4 and G4 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 B5, C5, D5, and 5 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 region 10 would be left with no open cell for its marker.
9. Next, 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.
10. Then, if H1 were the marker, it would force I3, I4, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then J3, J4, J8, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), and region 10 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 I1 it would force J3, J4, J8, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H3, H4, H5, and 3 more cleared (every open cell left in this region sits in column H, so its marker has to land there), and region 10 would be left with no open cell for its marker.
12. Now, marking J1 would force I3, I4, I5, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then H2, H3, H4, and 4 more cleared (every open cell left in this region sits in column H, so its marker has to land there), and then region 10 would be left with no open cell for its marker. So J1 can't be the marker there.
13. 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. Placing here also clears its 2 touching neighbours.
14. Following that, if C2 were the marker, it would force F5, F6, D7, and 4 more 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 E9, F9, E10, and 1 more cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), then F4 cleared (every open cell in column E belongs to the same region, so that region's marker has to be somewhere in this column), 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.
15. First, that one can't be the marker: place it at B3, and row 2 would be left with no open cell for its marker.
16. Next, marking C3 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
17. Then, that one can't be the marker: at H3 it would force I7, I8 and I10 cleared (every open cell left in this region sits in column I, so its marker has to land there), then J8, J9 and J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), and region 10 would be left with no open cell for its marker.
18. After that, marking E2 would force H4, I4, J4, 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), then B5, C5, D5, and 3 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 I5, I6, I10, and 3 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and then region 10 would be left with no open cell for its marker. So E2 can't be the marker there.
19. Now, marking I3 would force J8, J9 and J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), then H8, H9 and H10 cleared (every open cell left in this region sits in column H, so its marker has to land there), and then region 10 would be left with no open cell for its marker. So I3 can't be the marker there.
20. From there, if J3 were the marker, it would force I5, I6 and I10 cleared (every open cell left in this region sits in column I, so its marker has to land there), then H8, H9 and H10 cleared (every open cell left in this region sits in column H, so its marker has to land there), and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
21. Following that, marking D4 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
22. First, if E4 were the marker, row 3 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: place it at F4, and row 3 would be left with no open cell for its marker.
24. Then, if H4 were the marker, it would force I6 (the last open cell in region 6), and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
25. After that, marking B2 would force F3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then 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 I5, I6, I10, and 3 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and then region 10 would be left with no open cell for its marker. So B2 can't be the marker there.
26. Now, place a marker at D2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
27. From there, 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.
28. Following that, 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.
29. First, if I5 were the marker, row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
30. Next, place a marker at J5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column J, and its region. Placing here also clears its 1 touching neighbour.
31. Then, place a marker at H6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column H, and its region.
32. After that, place a marker at I10. Region has exactly one open cell left. Placing here clears the rest of row 10, column I, and its region.
33. Now, place a marker at E9. Region has exactly one open cell left. Placing here clears the rest of row 9, column E, and its region.
34. From there, place a marker at A8. Region has exactly one open cell left. Placing here clears the rest of row 8, column A, and its region. Placing here also clears its 1 touching neighbour.
35. Following that, place a marker at C7. Region has exactly one open cell left. Placing here clears the rest of row 7, column C, and its region.
36. Finally, place a marker at B4. Region has exactly one open cell left.

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