How to solve this 8×8 logic puzzle

This 8×8 board rates as severe (difficulty score 1142). Solving it from scratch takes 31 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 column A, so its marker has to land there; that clears every other open cell in the column.
2. Next, every remaining candidate in this region touches B6 and B7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
4. After that, together, two regions have open cells only in rows 6 and 7; 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.
5. Now, every open cell in column E 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.
6. From there, every remaining candidate in this region touches B4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
7. Following that, if H1 were the marker, it would force G6 (the last open cell in region 8), and column F would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, 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. Placing here also clears its 2 touching neighbours.
9. Next, every remaining candidate in this region touches G3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
10. Then, if C2 were the marker, it would force B5 (the last open cell in region 4), then G4 (the last open cell in region 5), and row 3 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 G2 it would force F5 (the last open cell in column F), and column C would be left with no open cell for its marker.
12. Now, marking H2 would force G6 (the last open cell in region 8), and then column F would be left with no open cell for its marker. So H2 can't be the marker there.
13. From there, place a marker at B2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column B, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at D8. Column 4 has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region.
15. First, every open cell in row 3 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.
16. Next, every remaining candidate in this region touches G4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
17. Then, place a marker at C4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region.
18. 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.
19. Now, place a marker at G5. Column 7 has exactly one open cell left. Placing here clears the rest of row 5, column G, and its region. Placing here also clears its 1 touching neighbour.
20. From there, place a marker at H7. Region has exactly one open cell left. Placing here clears the rest of row 7, column H, and its region.
21. Following that, place a marker at F3. Region has exactly one open cell left.
22. Finally, place a marker at A6. Region has exactly one open cell left.

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