An Expert 8×8 Puzzle Solved with Forced Chain in 32 Steps

This 8×8 board rates as expert (difficulty score 1647). Solving it from scratch takes 32 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 remaining candidate in this region touches G6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, every remaining candidate in this region touches B6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, that one can't be the marker: at C1 it would force A7 and A8 cleared (every open cell left in this region sits in column A, so its marker has to land there), then B7 (the last open cell in region 6), then F8, G8 and H8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and region 4 would be left with no open cell for its marker.
4. After that, that one can't be the marker: at F1 it would force G4, H2, H3, and 1 more cleared (together, two regions have open cells only in columns G and H), then E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), then A7, B7, C7, and 1 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and region 5 would be left with no open cell for its marker.
5. Now, marking G1 would force H3, H4 and H8 cleared (every open cell left in this region sits in column H, so its marker has to land there), then F4, F5 and F7 cleared (every open cell left in this region sits in column F, so its marker has to land there), then E7 and E8 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 region 6 would be left with no open cell for its marker. So G1 can't be the marker there.
6. From there, if H1 were the marker, it would force G5 (the last open cell in region 5), then F8 (the last open cell in region 7), then D7 (the last open cell in region 8), and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
7. Following that, if C2 were the marker, it would force E1 (the last open cell in row 1), then D7, F7, G7, and 1 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), then F8, G8 and H8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, if F2 were the marker, it would force G4, H3 and H4 cleared (together, two regions have open cells only in columns G and H), then E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), then A7, B7, C7, and 1 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, that one can't be the marker: at G2 it would force H4 and H8 cleared (every open cell left in this region sits in column H, so its marker has to land there), then F4, F5 and F7 cleared (every open cell left in this region sits in column F, so its marker has to land there), then E1, E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and region 6 would be left with no open cell for its marker.
10. Then, marking H2 would force G5 (the last open cell in region 5), then F8 (the last open cell in region 7), then D7 (the last open cell in region 8), and then region 6 would be left with no open cell for its marker. So H2 can't be the marker there.
11. After that, if A3 were the marker, it would force D2 and B5 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 E2 (the last open cell in row 2), and row 1 would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, marking C3 would force D7, E7, F7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), then F8, G8 and H8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), then F6 (the last open cell in region 7), and the chain continues, and then column G would be left with no open cell for its marker. So C3 can't be the marker there.
13. From there, if D3 were the marker, it would force C5 (the last open cell in region 3), then H4 and H8 cleared (every open cell left in this region sits in column H, so its marker has to land there), then E7, F7, G7, and 1 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), 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.
14. Following that, that one can't be the marker: at E3 it would force A1, B1, A4, 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 D1 (the last open cell in row 1), then F5 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 4 would be left with no open cell for its marker.
15. First, marking F3 would force H4 cleared (together, two regions have open cells only in columns G and H), then E1, E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), then A7, B7, C7, and 1 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and then region 5 would be left with no open cell for its marker. So F3 can't be the marker there.
16. Next, if G3 were the marker, it would force H8 cleared (every open cell left in this region sits in column H, so its marker has to land there), then F5 and F7 cleared (every open cell left in this region sits in column F, so its marker has to land there), then E1, E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
17. Then, if B1 were the marker, it would force H3 (the last open cell in row 3), then G5 (the last open cell in region 5), then F8 (the last open cell in region 7), and the chain continues, and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
18. After that, marking D1 would force H3 (the last open cell in row 3), then G5 (the last open cell in region 5), then F8 (the last open cell in region 7), and then region 8 would be left with no open cell for its marker. So D1 can't be the marker there.
19. Now, if E1 were the marker, it would force H3 (the last open cell in row 3), then G5 (the last open cell in region 5), then F8 (the last open cell in region 7), 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.
20. From there, place a marker at A1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column A, and its region.
21. Following that, that one can't be the marker: at D2 it would force H3 (the last open cell in row 3), then G5 (the last open cell in region 5), then F8 (the last open cell in region 7), and region 8 would be left with no open cell for its marker.
22. First, place a marker at E2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region.
23. Next, 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.
24. Then, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
25. After that, every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
26. Now, every remaining candidate in this region touches D7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
27. From there, 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 1 touching neighbour.
28. Following that, place a marker at C6. Region has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region. Placing here also clears its 1 touching neighbour.
29. First, place a marker at G7. Region has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region.
30. Next, place a marker at H5. Region has exactly one open cell left. Placing here clears the rest of row 5, column H, and its region.
31. Then, place a marker at F4. Region has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region.
32. Finally, place a marker at B3. Region has exactly one open cell left.

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