An Expert 8×8 Puzzle Solved with Forced Chain in 23 Steps

This 8×8 board rates as expert (difficulty score 1260). Solving it from scratch takes 23 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line 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 row 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, every remaining candidate in this region touches G2 and H2, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, that one can't be the marker: at A2 it would force G3 and H3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then D4, E4 and H4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then C5, D5, C6, and 2 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), and column C would be left with no open cell for its marker.
4. After that, marking B2 would force G3 and H3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then D4, E4 and H4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then C5, D5, C6, and 2 more 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 then column C would be left with no open cell for its marker. So B2 can't be the marker there.
5. Now, marking E2 would immediately leave region 2 with no open cell for its marker, so it can't be the marker there.
6. From there, if F2 were the marker, it would force H1 (the last open cell in region 4), then D3 (the last open cell in region 2), then G7 and G8 cleared (every open cell left in this region sits in column G, so its marker has to land there), 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.
7. Following that, 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.
8. First, every remaining candidate in this region touches C3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
9. Next, that one can't be the marker: at B3 it would force D2 (the last open cell in region 2), then E4 and H4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then A8 and C8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and column C would be left with no open cell for its marker.
10. Then, if G3 were the marker, it would force H1 (the last open cell in region 4), then F5 (the last open cell in region 6), then D2 cleared (together, two regions have open cells only in columns D and E), 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.
11. After that, if H1 were the marker, it would force A3 (the last open cell in row 3), then D4 and E4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then C5, D5, C6, and 2 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), 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.
12. Now, place a marker at G1. Region has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
13. From there, together, two regions have open cells only in columns H and F; 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.
14. Following that, together, two regions have open cells only in columns D and E; 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.
15. First, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
16. Next, every remaining candidate in this region touches E7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
17. Then, that one can't be the marker: at H3 it would force F5 (the last open cell in region 6), then D4 (the last open cell in region 5), then B6 (the last open cell in region 1), and row 7 would be left with no open cell for its marker.
18. After that, 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. Placing here also clears its 1 touching neighbour.
19. Now, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region. Placing here also clears its 1 touching neighbour.
20. 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.
21. Following that, 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.
22. First, 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.
23. Finally, place a marker at H5. Region has exactly one open cell left.

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