An Expert 10×10 Puzzle Solved with Forced Chain in 33 Steps

This 10×10 board rates as expert (difficulty score 1675). Solving it from scratch takes 33 logical steps, using 5 techniques: Single cell, Row/column exclusion, Adjacency clear, Line confinement, Forced chain. Every step below is forced: nothing here is a guess.

1. First, that one can't be the marker: at I1 it would force A8, B8, C8, and 4 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then J9, F10, G10, and 2 more cleared (together, two regions have open cells only in rows 9 and 10), and region 9 would be left with no open cell for its marker.
2. Next, marking J1 would force A10, B10, C10, and 2 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then C8, E8, F8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and then row 10 would be left with no open cell for its marker. So J1 can't be the marker there.
3. Then, if I2 were the marker, it would force J6, J7, J8, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A8, B8, C8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then A10, B10, C10, 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 region 10 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 J2 it would force A10, B10, C10, and 2 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then C8, E8, F8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and row 10 would be left with no open cell for its marker.
5. Now, marking G1 would force A3, B3, C3, 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 C4 (the last open cell in row 4), and the chain continues, and then region 10 would be left with no open cell for its marker. So G1 can't be the marker there.
6. From there, that one can't be the marker: at E3 it would force H1, I4, J4, and 2 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 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 B5, C5, D5, and 6 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 6 would be left with no open cell for its marker.
7. Following that, if G3 were the marker, it would force A1, B1, C1, and 5 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 H1 (the last open cell in row 1), then C4 (the last open cell in row 4), and row 5 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 H3 it would force A2, B2, C2, 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), and row 2 would be left with no open cell for its marker.
9. Next, marking I3 would force A8, B8, C8, and 4 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then J9, F10, G10, and 2 more cleared (together, two regions have open cells only in rows 9 and 10), and then region 9 would be left with no open cell for its marker. So I3 can't be the marker there.
10. Then, if J3 were the marker, it would force A10, B10, C10, and 2 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then C8, E8, F8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, marking A1 would force F3 (the last open cell in row 3), then H2 (the last open cell in row 2), then C4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So A1 can't be the marker there.
12. Now, if B1 were the marker, it would force F3 (the last open cell in row 3), then H2 (the last open cell in row 2), then C4 (the last open cell in row 4), 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, that one can't be the marker: at C1 it would force F3 (the last open cell in row 3), then H2 (the last open cell in row 2), and row 4 would be left with no open cell for its marker.
14. Following that, if E1 were the marker, it would force A2, B2, C2, and 5 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 I4, J4, I5, and 1 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 B5, C5, D5, and 6 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 6 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: at F1 it would force A2, B2, C2, 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 H2 (the last open cell in row 2), 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 row 5 would be left with no open cell for its marker.
16. Next, that one can't be the marker: at A2 it would force F3 (the last open cell in row 3), then H1 (the last open cell in row 1), then C4 (the last open cell in row 4), and row 5 would be left with no open cell for its marker.
17. Then, marking B2 would force F3 (the last open cell in row 3), then H1 (the last open cell in row 1), then C4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So B2 can't be the marker there.
18. After that, if C2 were the marker, it would force H1 (the last open cell in row 1), then F3 (the last open cell in row 3), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, that one can't be the marker: at D2 it would force H1 (the last open cell in row 1), then F3 (the last open cell in row 3), then C4 (the last open cell in row 4), and row 5 would be left with no open cell for its marker.
20. From there, if H1 were the marker, it would force F3, D4, E4, and 1 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 A4, B4, A5, 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 B5, C5, D5, and 7 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 row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
21. Following that, place a marker at D1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region.
22. 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.
23. 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.
24. Then, 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.
25. After that, place a marker at H5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column H, and its region. Placing here also clears its 1 touching neighbour.
26. Now, place a marker at G2. Region has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region.
27. From there, 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. Placing here also clears its 1 touching neighbour.
28. Following that, 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.
29. First, place a marker at E6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region.
30. Next, place a marker at F9. Region has exactly one open cell left. Placing here clears the rest of row 9, column F, and its region.
31. Then, place a marker at I7. Row 7 has exactly one open cell left. Placing here clears the rest of row 7, column I, and its region. Placing here also clears its 1 touching neighbour.
32. After that, place a marker at J10. Region has exactly one open cell left. Placing here clears the rest of row 10, column J, and its region.
33. Finally, place a marker at B8. Region has exactly one open cell left.

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