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

This 10×10 board rates as expert (difficulty score 1645). Solving it from scratch takes 34 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region 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 column H, 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 G7 and I7 and G8 and I8, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, marking A1 would force B7 (the last open cell in region 6), then H8 (the last open cell in region 9), then J9, J10, I2, and 2 more cleared (together, two regions have open cells only in columns J and I), and the chain continues, and then row 3 would be left with no open cell for its marker. So A1 can't be the marker there.
4. After that, if B1 were the marker, it would force J8, J9, J10, and 4 more cleared (together, two regions have open cells only in columns J and I), then G10 (the last open cell in region 10), then F2 (the last open cell in region 2), 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.
5. Now, if E1 were the marker, it would force J8, J9, J10, and 2 more cleared (together, two regions have open cells only in columns J and I), then G10 (the last open cell in region 10), then F3, F4, F5, and 1 more cleared (every open cell left in this region sits in column F, 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.
6. From there, marking G1 would force I6, I9 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 then region 10 would be left with no open cell for its marker. So G1 can't be the marker there.
7. Following that, 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.
8. First, place a marker at G10. Region has exactly one open cell left. Placing here clears the rest of row 10, column G, and its region. Placing here also clears its 1 touching neighbour.
9. Next, marking J1 would force F2 (the last open cell in region 2), then I6 (the last open cell in region 5), then H8 (the last open cell in region 9), and the chain continues, and then row 5 would be left with no open cell for its marker. So J1 can't be the marker there.
10. Then, that one can't be the marker: at A2 it would force B7 (the last open cell in region 6), then H8 (the last open cell in region 9), then E9 (the last open cell in region 8), and the chain continues, and column I would be left with no open cell for its marker.
11. After that, if C2 were the marker, it would force F1 (the last open cell in region 2), then E3, E4, E5, and 1 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then D5 (the last open cell in region 4), 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.
12. Now, marking E2 would force C1 (the last open cell in region 2), then F6 cleared (every open cell left in this region sits in column F, so its marker has to land there), then A7, B7, D7, and 4 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and then column I would be left with no open cell for its marker. So E2 can't be the marker there.
13. From there, that one can't be the marker: at C1 it would force J2 (the last open cell in row 2), then I6 (the last open cell in region 5), then H8 (the last open cell in region 9), and the chain continues, and region 8 would be left with no open cell for its marker.
14. Following that, that one can't be the marker: at J2 it would force I6 (the last open cell in region 5), then H8 (the last open cell in region 9), then E7, F7 and E9 cleared (together, two regions have open cells only in rows 7 and 9), and region 8 would be left with no open cell for its marker.
15. First, marking C3 would force F2 (the last open cell in row 2), then I1 (the last open cell in row 1), then E4 (the last open cell in row 4), and then region 8 would be left with no open cell for its marker. So C3 can't be the marker there.
16. Next, if D3 were the marker, it would force F2 (the last open cell in row 2), then I1 (the last open cell in row 1), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
17. Then, that one can't be the marker: at E3 it would force B2 (the last open cell in row 2), then I1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then F1 and F6 cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and row 4 would be left with no open cell for its marker.
18. After that, marking F3 would force D1 (the last open cell in region 2), then B2 (the last open cell in row 2), then E6 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 4 would be left with no open cell for its marker. So F3 can't be the marker there.
19. Now, marking D1 would force B2 (the last open cell in row 2), then J3 (the last open cell in row 3), then I6 (the last open cell in region 5), and the chain continues, and then row 5 would be left with no open cell for its marker. So D1 can't be the marker there.
20. From there, every open cell left in this region sits in column F, so its marker has to land there; that clears every other open cell in the column.
21. Following that, 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.
22. First, every remaining candidate in this region touches D8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
23. Next, marking B2 would force F1 (the last open cell in region 2), then J3 (the last open cell in row 3), then I6 (the last open cell in region 5), and the chain continues, and then row 5 would be left with no open cell for its marker. So B2 can't be the marker there.
24. Then, that one can't be the marker: at B3 it would force I1 and I5 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 F1 (the last open cell in row 1), then D2 (the last open cell in row 2), and the chain continues, and row 4 would be left with no open cell for its marker.
25. After that, if J3 were the marker, it would force I6 (the last open cell in region 5), then H8 (the last open cell in region 9), then F1 (the last open cell in row 1), 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.
26. Now, place a marker at A3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
27. From there, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region. Placing here also clears its 2 touching neighbours.
28. Following that, place a marker at H8. Region has exactly one open cell left. Placing here clears the rest of row 8, column H, and its region.
29. First, 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.
30. Next, place a marker at D6. Region has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region. Placing here also clears its 1 touching neighbour.
31. Then, 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.
32. After that, place a marker at J5. Region has exactly one open cell left. Placing here clears the rest of row 5, column J, and its region.
33. Now, 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.
34. Finally, place a marker at F2. Region has exactly one open cell left.

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