An Expert 8×8 Puzzle Solved with Forced Chain in 33 Steps

This 8×8 board rates as expert (difficulty score 1608). Solving it from scratch takes 33 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, together, two regions have open cells only in columns F and G; 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.
3. Then, every remaining candidate in this region touches B2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches G1 and G2, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, 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.
6. From there, that one can't be the marker: at C1 it would force H2 (the last open cell in region 4), then F5, F6 and F7 cleared (every open cell left in this region sits in column F, so its marker has to land there), then B4, B5 and B8 cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 8 would be left with no open cell for its marker.
7. Following that, marking D1 would force H2 (the last open cell in region 4), then A3 (the last open cell in region 1), then F4 (the last open cell in region 3), and then column G would be left with no open cell for its marker. So D1 can't be the marker there.
8. First, if E1 were the marker, it would force H2 (the last open cell in region 4), then A3 (the last open cell in region 1), then F4 (the last open cell in region 3), and column G would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, if C2 were the marker, it would force H1 (the last open cell in region 4), then E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), then D7 (the last open cell in region 8), 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.
10. Then, together, two regions have open cells only in columns A and B; 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.
11. After that, that one can't be the marker: at D2 it would force H1 (the last open cell in region 4), then A3 (the last open cell in region 1), then F4 (the last open cell in region 3), and column G would be left with no open cell for its marker.
12. Now, marking E2 would force H1 (the last open cell in region 4), then A3 (the last open cell in region 1), then F4 (the last open cell in region 3), and then column G would be left with no open cell for its marker. So E2 can't be the marker there.
13. From there, marking C3 would force F5, F6 and F7 cleared (every open cell left in this region sits in column F, so its marker has to land there), then D7 cleared (every open cell left in this region sits in column D, so its marker has to land there), then F1 cleared (together, two regions have open cells only in rows 1 and 2), 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.
14. Following that, if D3 were the marker, it would force F5, F6 and F7 cleared (every open cell left in this region sits in column F, so its marker has to land there), then F1 cleared (together, two regions have open cells only in rows 1 and 2), then F4 (the last open cell in region 3), and column G 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 E3 it would force F1 (the last open cell in region 3), then A2 (the last open cell in region 1), and column H would be left with no open cell for its marker.
16. Next, every remaining candidate in this region touches D5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
17. Then, if G3 were the marker, it would force H1 (the last open cell in region 4), then A2 (the last open cell in region 1), then E6 cleared (every remaining candidate in this region touches E6), and the chain continues, and row 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
18. After that, 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.
19. Now, every remaining candidate in this region touches F4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
20. From there, marking A4 would force B1 (the last open cell in region 1), then H2 (the last open cell in region 4), then F3 (the last open cell in region 3), and the chain continues, and then column D would be left with no open cell for its marker. So A4 can't be the marker there.
21. Following that, if B4 were the marker, it would force G5 (the last open cell in region 5), then E6 (the last open cell in region 2), and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
22. First, marking A1 would force H2 (the last open cell in region 4), then F3 (the last open cell in region 3), then G5 (the last open cell in region 5), and the chain continues, and then row 6 would be left with no open cell for its marker. So A1 can't be the marker there.
23. Next, if H1 were the marker, it would force F3 (the last open cell in region 3), then A2 (the last open cell in region 1), then G5 (the last open cell in region 5), 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.
24. Then, place a marker at H2. Region has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region.
25. After that, marking D4 would immediately leave column C with no open cell for its marker, so it can't be the marker there.
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 D6. Column 4 has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region. Placing here also clears its 3 touching neighbours.
28. Following that, place a marker at E8. Region has exactly one open cell left. Placing here clears the rest of row 8, column E, and its region.
29. First, 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.
30. Next, 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.
31. Then, 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.
32. After that, place a marker at B1. Region has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
33. Finally, place a marker at F3. Region has exactly one open cell left.

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