An Expert 10×10 Puzzle Solved with Forced Chain in 31 Steps

This 10×10 board rates as expert (difficulty score 1528). Solving it from scratch takes 31 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Forced chain. Every step below is forced: nothing here is a guess.

1. First, if B1 were the marker, it would force D6, E6, F6, and 4 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then E3, F3, G3, and 9 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, C4, D4, 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), 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.
2. Next, that one can't be the marker: at C1 it would force B3, B4, B7, and 8 more cleared (together, two regions have open cells only in columns B and A), then 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 E8 (the last open cell in region 7), and the chain continues, and region 10 would be left with no open cell for its marker.
3. Then, that one can't be the marker: at F1 it would force A3, B3, C3, and 9 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 E4, H4, I4, and 4 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 G4 (the last open cell in row 4), and the chain continues, and column J would be left with no open cell for its marker.
4. After that, that one can't be the marker: at A2 it would force J3 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 E4, H4, I4, and 4 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 E5, F5, G5, 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 the chain continues, and region 7 would be left with no open cell for its marker.
5. Now, marking B2 would force E6, F6, G6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then J3 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 E4, H4, I4, and 4 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), 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.
6. From there, marking D1 would force E3, F3, G3, and 9 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, C4, 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 E5, F5, G5, 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 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.
7. Following that, that one can't be the marker: at D2 it would force A1 (the last open cell in row 1), then E4, H4, I4, and 4 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 E5, F5, G5, 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 the chain continues, and region 7 would be left with no open cell for its marker.
8. First, marking H2 would force A1 (the last open cell in row 1), then E4, I4, J4, 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 E5, F5, G5, 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 then region 7 would be left with no open cell for its marker. So H2 can't be the marker there.
9. Next, if I2 were the marker, it would force A1 (the last open cell in row 1), then E4, H4, J4, 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 E5, F5, G5, 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 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, that one can't be the marker: at J2 it would force A1 (the last open cell in row 1), then E4, H4, I4, 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 E5, F5, G5, and 4 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 7 would be left with no open cell for its marker.
11. After that, marking A1 would force J3 (the last open cell in region 2), then E4, H4, H5, 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 E5, F5, G5, and 4 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 7 would be left with no open cell for its marker. So A1 can't be the marker there.
12. Now, 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.
13. From there, marking E2 would force A4, B4, C4, 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 F5, G5, G6, 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), then B5 (the last open cell in row 5), 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.
14. Following that, marking G1 would force C2 (the last open cell in row 2), then E4, H4, I4, and 4 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 F4 (the last open cell in row 4), and the chain continues, and then row 7 would be left with no open cell for its marker. So G1 can't be the marker there.
15. First, if F2 were the marker, it would force A4, B4, C4, 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 G4 (the last open cell in row 4), then B5 (the last open cell in row 5), 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.
16. Next, if H1 were the marker, it would force C2 (the last open cell in row 2), then E4, I4, J4, 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 E5, F5, G5, 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 region 10 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 G2 it would force A4, B4, C4, 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 F4 (the last open cell in row 4), then B5 (the last open cell in row 5), and the chain continues, and column J would be left with no open cell for its marker.
18. After that, place a marker at C2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
19. Now, 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.
20. From there, 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.
21. Following that, place a marker at B5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region.
22. First, place a marker at A9. Region has exactly one open cell left. Placing here clears the rest of row 9, column A, and its region.
23. Next, 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.
24. Then, place a marker at J1. Column 10 has exactly one open cell left. Placing here clears the rest of row 1, column J, and its region.
25. After that, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
26. Now, place a marker at D7. Region has exactly one open cell left. Placing here clears the rest of row 7, column D, and its region. Placing here also clears its 1 touching neighbour.
27. From there, place a marker at F6. Region has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region.
28. Following that, place a marker at G4. Region has exactly one open cell left. Placing here clears the rest of row 4, column G, and its region. Placing here also clears its 1 touching neighbour.
29. First, place a marker at E10. Region has exactly one open cell left. Placing here clears the rest of row 10, column E, and its region.
30. Next, place a marker at I3. Region has exactly one open cell left. Placing here clears the rest of row 3, column I, and its region.
31. Finally, place a marker at H8. Region has exactly one open cell left.

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