An Expert 8×8 Puzzle Solved with Forced Chain in 27 Steps

This 8×8 board rates as expert (difficulty score 1463). Solving it from scratch takes 27 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 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.
2. Next, marking A1 would force D6, E6, F6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then B3, C3, C4, and 1 more cleared (together, two regions have open cells only in columns B and C), then D5 (the last open cell in region 4), and the chain continues, and then row 2 would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, if B1 were the marker, 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 C6 (the last open cell in region 7), and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at C1 it would force B8 (the last open cell in region 8), and region 7 would be left with no open cell for its marker.
5. Now, marking D1 would force B2, B4, B5, and 2 more cleared (together, two regions have open cells only in columns B and C), then A7 and A8 cleared (every open cell left in this region sits in column A, so its marker has to land there), and then region 7 would be left with no open cell for its marker. So D1 can't be the marker there.
6. From there, that one can't be the marker: at F1 it would force D4, D5, D6, and 2 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E6, E7, G7, and 2 more cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then G5 and G6 cleared (every open cell in column E belongs to the same region, so that region's marker has to be somewhere in this column), and column G would be left with no open cell for its marker.
7. Following that, marking G1 would force E6, D7, E7, and 4 more cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then E3, D4, E4, and 1 more cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), then E2 (the last open cell in column E), and the chain continues, and then row 8 would be left with no open cell for its marker. So G1 can't be the marker there.
8. First, that one can't be the marker: at A2 it would force D6, E6, F6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then C3, C4 and C5 cleared (together, two regions have open cells only in columns B and C), then D5 (the last open cell in region 4), and the chain continues, and region 3 would be left with no open cell for its marker.
9. Next, marking B2 would force C4, C5 and C6 cleared (every open cell left in this region sits in column C, so its marker has to land there), then D3, D4, D7, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E1 (the last open cell in region 2), 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.
10. Then, if C2 were the marker, it would force E1 (the last open cell in region 2), then B8 (the last open cell in region 8), and column A 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 H1 (the last open cell in row 1), then B4, B5, B6, and 1 more cleared (together, two regions have open cells only in columns B and C), then A7 and A8 cleared (every open cell left in this region sits in column A, so its marker has to land there), and region 7 would be left with no open cell for its marker.
12. Now, marking E2 would force H1 (the last open cell in row 1), then A7 and A8 cleared (together, two regions have open cells only in rows 7 and 8), then A6, D6, F6, and 1 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, 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, that one can't be the marker: at G2 it would force E1 (the last open cell in row 1), then D7, F7, D8, and 1 more cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then D4 cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and row 8 would be left with no open cell for its marker.
14. Following that, if A3 were the marker, it would force D6, E6, F6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then F4, G4, H4, and 1 more cleared (together, two regions have open cells only in rows 4 and 5), then H7 and H8 cleared (every open cell left in this region sits in column H, 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.
15. First, if H1 were the marker, it would force F2 (the last open cell in row 2), then C4, C5, D5, 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), then B4, B5, B6, and 1 more cleared (together, two regions have open cells only in columns B and C), 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.
16. Next, place a marker at E1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
17. Then, place a marker at H2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region.
18. After that, together, two regions have open cells only in rows 7 and 8; with one marker each, those two markers have to fill exactly those two rows, so every other region's open cells there can be cleared.
19. Now, 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.
20. From there, every open cell in column A 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.
21. Following that, together, two regions have open cells only in columns B and C; 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.
22. First, place a marker at D5. Region has exactly one open cell left. Placing here clears the rest of row 5, column D, and its region. Placing here also clears its 1 touching neighbour.
23. Next, place a marker at F3. Region has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region.
24. Then, place a marker at A4. Region has exactly one open cell left.
25. After that, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region. Placing here also clears its 1 touching neighbour.
26. Now, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region.
27. Finally, place a marker at G7. Region has exactly one open cell left.

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