An Expert 8×8 Puzzle Solved with Forced Chain in 26 Steps

This 8×8 board rates as expert (difficulty score 1386). Solving it from scratch takes 26 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 remaining candidate in this region touches G2, 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 B7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, every remaining candidate in this region touches G7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, marking A1 would force B8 (the last open cell in region 7), then C6 (the last open cell in region 5), then H7 (the last open cell in region 8), and then region 3 would be left with no open cell for its marker. So A1 can't be the marker there.
5. Now, if B1 were the marker, it would force H4, H5, H6, and 2 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then G8 (the last open cell in region 8), then A5 cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
6. From there, marking D1 would force H4, H5, H6, and 2 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then G8 (the last open cell in region 8), then A2, A3, A4, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and then column C would be left with no open cell for its marker. So D1 can't be the marker there.
7. Following that, if E1 were the marker, it would force H4, H5, H6, and 2 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then G8 (the last open cell in region 8), then F4 cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and row 3 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 F1 it would force A3, B3 and G3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A4, B4, C4, 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 C5, D5, E5, 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 column D would be left with no open cell for its marker.
9. Next, together, two regions have open cells only in columns G and H; 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.
10. Then, that one can't be the marker: at A2 it would force B8 (the last open cell in region 7), then C6 (the last open cell in region 5), then H7 (the last open cell in region 8), and the chain continues, and row 4 would be left with no open cell for its marker.
11. After that, marking B2 would force H3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A5 cleared (every open cell left in this region sits in column A, so its marker has to land there), then C5 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and then row 4 would be left with no open cell for its marker. So B2 can't be the marker there.
12. Now, if C2 were the marker, it would force H3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then D5, E5, D6, and 2 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 column D 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 D2 it would force F3 (the last open cell in region 2), and row 4 would be left with no open cell for its marker.
14. Following that, marking G1 would force E2 (the last open cell in row 2), then F4 cleared (every open cell left in this region sits in column F, so its marker has to land there), then A4, B4, C4, and 1 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 then row 4 would be left with no open cell for its marker. So G1 can't be the marker there.
15. First, place a marker at G8. Column 7 has exactly one open cell left. Placing here clears the rest of row 8, column G, and its region. Placing here also clears its 1 touching neighbour.
16. Next, every open cell left in this region sits in column A, so its marker has to land there; that clears every other open cell in the column.
17. Then, 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.
18. After that, if H1 were the marker, it would force C3, D3, E3, 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 B3 (the last open cell in row 3), then C5 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, place a marker at C1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
20. From there, place a marker at B5. Region has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region. Placing here also clears its 1 touching neighbour.
21. Following that, 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.
22. First, every open cell left in this region sits in row 6, so its marker has to land there; that clears every other open cell in the row.
23. Next, place a marker at D3. Column 4 has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region. Placing here also clears its 1 touching neighbour.
24. Then, place a marker at H2. Region has exactly one open cell left.
25. 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.
26. Finally, place a marker at E6. Region has exactly one open cell left.

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