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

This 8×8 board rates as expert (difficulty score 1322). 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 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 F2 and G2, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, every remaining candidate in this region touches A3, 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 G6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, that one can't be the marker: at F1 it would force A5, B5, C5, and 2 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then D4, E4, G4, and 1 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then H2, G3 and H3 cleared (together, two regions have open cells only in rows 2 and 3), and region 4 would be left with no open cell for its marker.
6. From there, 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. Placing here also clears its 1 touching neighbour.
7. Following that, 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.
8. First, 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.
9. Next, marking B2 would force D4, E4, F4, and 1 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then H3 (the last open cell in region 4), then E5 (the last open cell in region 2), and the chain continues, and then column D would be left with no open cell for its marker. So B2 can't be the marker there.
10. Then, if C2 were the marker, region 1 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, that one can't be the marker: at D2 it would force B3 (the last open cell in region 1), and region 5 would be left with no open cell for its marker.
12. Now, marking E2 would force B3 (the last open cell in region 1), and then region 5 would be left with no open cell for its marker. So E2 can't be the marker there.
13. From there, place a marker at A2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region.
14. Following that, every remaining candidate in this region touches D4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
15. First, every remaining candidate in this region touches C5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
16. Next, marking C3 would force B5 (the last open cell in region 5), then E8 (the last open cell in row 8), and then row 7 would be left with no open cell for its marker. So C3 can't be the marker there.
17. Then, 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.
18. After that, that one can't be the marker: at E3 it would force H4 (the last open cell in region 4), then B5 (the last open cell in region 5), then D7 (the last open cell in region 8), and row 8 would be left with no open cell for its marker.
19. Now, that one can't be the marker: at H3 it would force D7 (the last open cell in column D), then B8 (the last open cell in row 8), then C4 (the last open cell in region 5), and the chain continues, and row 6 would be left with no open cell for its marker.
20. From there, place a marker at D3. Row 3 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.
21. Following that, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
22. First, 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.
23. Next, place a marker at F6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region. Placing here also clears its 1 touching neighbour.
24. Then, place a marker at H4. Region has exactly one open cell left.
25. After 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.
26. Finally, place a marker at C7. Region has exactly one open cell left.

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