An Expert 10×10 Puzzle Solved with Forced Chain in 34 Steps

This 10×10 board rates as expert (difficulty score 1523). Solving it from scratch takes 34 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, every remaining candidate in this region touches B7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, marking A1 would force E10, F10, G10, and 3 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then C9, C10, B9, and 1 more cleared (together, two regions have open cells only in columns C and B), then D10 (the last open cell in region 10), and then region 8 would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, if B1 were the marker, it would force C9, C10, A9, and 1 more cleared (together, two regions have open cells only in columns C and A), then D10 (the last open cell in region 10), and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at C1 it would force D5, F5, G5, and 3 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then A9, A10, B9, and 1 more cleared (together, two regions have open cells only in columns A and B), then D10 (the last open cell in region 10), and region 8 would be left with no open cell for its marker.
5. Now, marking D1 would force E3, E4, E5, and 3 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then A9, F9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then F10, G10, H10, and 2 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), and the chain continues, and then column I would be left with no open cell for its marker. So D1 can't be the marker there.
6. From there, if E1 were the marker, it would force D3, D7, D8, and 2 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then A9, F9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then F10, G10, H10, and 2 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), and the chain continues, and column I would end up with no open cell for its marker, which can't happen. So it can't go there.
7. Following that, marking J1 would force B9, C9, D9, and 2 more cleared (together, two regions have open cells only in rows 9 and 10), then D2, D3, D4, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E2, E3, E4, and 2 more cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and then row 2 would be left with no open cell for its marker. So J1 can't be the marker there.
8. First, 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.
9. Next, that one can't be the marker: at A2 it would force E10, F10, G10, and 3 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then H4, I4, J4, and 6 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 C9, C10, B9, and 1 more cleared (together, two regions have open cells only in columns C and B), and the chain continues, and region 8 would be left with no open cell for its marker.
10. Then, marking B2 would force H4, I4, J4, and 6 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 C9, C10, A9, and 1 more cleared (together, two regions have open cells only in columns C and A), then D10 (the last open cell in region 10), and then region 8 would be left with no open cell for its marker. So B2 can't be the marker there.
11. After that, if C2 were the marker, it would force D5, F5, G5, and 3 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then H4, I4, J4, 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 A9, A10, B9, and 1 more cleared (together, two regions have open cells only in columns A and B), and the chain continues, and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, that one can't be the marker: at I1 it would force A3, B3, C3, and 8 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 H4, J4, H5, and 2 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 D7, D8, D9, and 4 more cleared (together, two regions have open cells only in columns D and E), and the chain continues, and column G would be left with no open cell for its marker.
13. From there, that one can't be the marker: at D2 it would force E8, E9 and E10 cleared (every open cell left in this region sits in column E, so its marker has to land there), then A9, F9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then F10, G10, H10, and 2 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), and the chain continues, and row 10 would be left with no open cell for its marker.
14. Following that, marking E2 would force D7, D8, D9, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then A9, F9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then F10, G10, H10, and 2 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), and the chain continues, and then column F would be left with no open cell for its marker. So E2 can't be the marker there.
15. 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.
16. Next, 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.
17. Then, marking G1 would force J2 (the last open cell in region 3), then B9, C9, D9, and 2 more cleared (together, two regions have open cells only in rows 9 and 10), then D3, D4, D5, and 2 more cleared (every open cell left in this region sits in column D, so its marker has to land there), and the chain continues, and then column D would be left with no open cell for its marker. So G1 can't be the marker there.
18. After that, marking H2 would force F1 (the last open cell in region 2), then G9, I9, G10, and 1 more cleared (every open cell in column J belongs to the same region, so that region's marker has to be somewhere in this column), then I8 (the last open cell in column I), and then column G would be left with no open cell for its marker. So H2 can't be the marker there.
19. Now, place a marker at J2. Region has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region.
20. From there, 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.
21. Following that, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
22. First, every open cell left in this region sits in column E, so its marker has to land there; that clears every other open cell in the column.
23. Next, every remaining candidate in this region touches F6 and D7 and F7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
24. Then, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region. Placing here also clears its 2 touching neighbours.
25. After that, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region. Placing here also clears its 1 touching neighbour.
26. Now, place a marker at A7. Region has exactly one open cell left. Placing here clears the rest of row 7, column A, and its region.
27. From there, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
28. Following that, every open cell left in this region sits in row 10, so its marker has to land there; that clears every other open cell in the row.
29. First, place a marker at I9. Column 9 has exactly one open cell left. Placing here clears the rest of row 9, column I, and its region.
30. Next, place a marker at H1. Column 8 has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region.
31. Then, place a marker at F3. Column 6 has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region. Placing here also clears its 1 touching neighbour.
32. After that, 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.
33. Now, place a marker at C4. Region has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region.
34. Finally, place a marker at B10. Region has exactly one open cell left.

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