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

This 10×10 board rates as expert (difficulty score 1632). 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 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 open cell in row 10 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.
4. After that, 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.
5. Now, every remaining candidate in this region touches B2 and B3, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
6. From there, marking E2 would force A3 (the last open cell in region 1), and then region 2 would be left with no open cell for its marker. So E2 can't be the marker there.
7. Following that, if F2 were the marker, it would force A3 (the last open cell in region 1), and region 2 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 G2 it would force A3 (the last open cell in region 1), then H6, H7, H8, and 2 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then I7, I8, I10, and 3 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and row 9 would be left with no open cell for its marker.
9. Next, marking H2 would force J1 (the last open cell in region 3), then A3 (the last open cell in region 1), and then region 2 would be left with no open cell for its marker. So H2 can't be the marker there.
10. Then, that one can't be the marker: at J2 it would force H1 (the last open cell in region 3), then A3 (the last open cell in region 1), and region 2 would be left with no open cell for its marker.
11. After that, marking C3 would force A2 (the last open cell in region 1), then B7, B8, B9, and 1 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then F7, G7, H7, and 4 more cleared (every open cell in row 9 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then row 4 would be left with no open cell for its marker. So C3 can't be the marker there.
12. Now, if D3 were the marker, it would force B4 (the last open cell in region 2), and region 7 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: at E3 it would force A2 (the last open cell in region 1), then B4 (the last open cell in region 2), and region 7 would be left with no open cell for its marker.
14. Following that, marking F3 would force A2 (the last open cell in region 1), then B4 (the last open cell in region 2), and then region 7 would be left with no open cell for its marker. So F3 can't be the marker there.
15. First, if G3 were the marker, it would force A2 (the last open cell in region 1), then H5 (the last open cell in region 5), then J1, 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 the chain continues, and row 9 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 A2 it would force B4 (the last open cell in region 2), and region 7 would be left with no open cell for its marker.
17. Then, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region. Placing here also clears its 1 touching neighbour.
18. After that, every remaining candidate in this region touches G4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
19. Now, if H1 were the marker, it would force G5 (the last open cell in region 5), then I4, J4, E6, and 2 more cleared (together, two regions have open cells only in rows 4 and 6), then J7 (the last open cell in region 6), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
20. From there, that one can't be the marker: at C4 it would force D2 (the last open cell in region 2), then F7 and F10 cleared (every open cell left in this region sits in column F, so its marker has to land there), then E5, F5, I5, and 1 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and region 9 would be left with no open cell for its marker.
21. Following that, every remaining candidate in this region touches C6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
22. First, every remaining candidate in this region touches E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
23. Next, if E4 were the marker, it would force B5, C5, I5, and 1 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then B6 (the last open cell in region 7), then J7 (the last open cell in region 6), and region 8 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: at F4 it would force H5 (the last open cell in region 5), then B6 (the last open cell in region 7), then J7 (the last open cell in region 6), and row 9 would be left with no open cell for its marker.
25. After that, that one can't be the marker: at D2 it would force F7 and F10 cleared (every open cell left in this region sits in column F, so its marker has to land there), then G5, H5, I5, and 6 more cleared (together, two regions have open cells only in rows 5 and 6), then H4 (the last open cell in region 5), and the chain continues, and region 9 would be left with no open cell for its marker.
26. Now, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
27. From there, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
28. Following that, every open cell in row 9 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, place a marker at E8. Row 8 has exactly one open cell left. Placing here clears the rest of row 8, column E, and its region.
30. Next, place a marker at J7. Row 7 has exactly one open cell left. Placing here clears the rest of row 7, column J, and its region.
31. Then, place a marker at I1. Region has exactly one open cell left. Placing here clears the rest of row 1, column I, and its region.
32. After that, place a marker at H9. Region has exactly one open cell left. Placing here clears the rest of row 9, column H, and its region. Placing here also clears its 1 touching neighbour.
33. Now, place a marker at G5. Region has exactly one open cell left. Placing here clears the rest of row 5, column G, and its region. Placing here also clears its 1 touching neighbour.
34. From there, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region.
35. Following that, place a marker at D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region.
36. Finally, place a marker at F10. Region has exactly one open cell left.

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