A Hard 8×8 Puzzle Solved with Forced Chain in 19 Steps

This 8×8 board rates as hard (difficulty score 1080). Solving it from scratch takes 19 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 column A, so its marker has to land there; that clears every other open cell in the column.
2. Next, 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.
3. 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.
4. After that, place a marker at F2. Region has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region. Placing here also clears its 2 touching neighbours.
5. Now, every remaining candidate in this region touches B7 and B8, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
6. From there, that one can't be the marker: at B3 it would force H1 (the last open cell in row 1), then D4 (the last open cell in row 4), and row 5 would be left with no open cell for its marker.
7. Following that, marking C3 would force H1 (the last open cell in row 1), and then row 4 would be left with no open cell for its marker. So C3 can't be the marker there.
8. First, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
9. Next, if H1 were the marker, it would force B4 (the last open cell in region 1), then C8 (the last open cell in region 6), then A7 (the last open cell in region 8), 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.
10. Then, place a marker at B1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
11. After that, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
12. Now, every remaining candidate in this region touches H4 and H5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
13. From there, that one can't be the marker: at H3 it would force G5 (the last open cell in region 3), then D4 (the last open cell in region 4), and row 6 would be left with no open cell for its marker.
14. Following that, 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.
15. First, 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.
16. Next, place a marker at C5. Region has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region.
17. Then, place a marker at G4. Region has exactly one open cell left.
18. After 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.
19. Finally, place a marker at H6. Region has exactly one open cell left.

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