A Hard 8×8 Puzzle Solved with Forced Chain in 20 Steps

This 8×8 board rates as hard (difficulty score 1097). Solving it from scratch takes 20 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 row 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, 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.
3. Then, 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.
4. After that, every remaining candidate in this region touches E3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, every remaining candidate in this region touches A7 and C7 and A8 and C8, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
6. From there, every remaining candidate in this region touches E7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
7. Following that, that one can't be the marker: at A2 it would force F3 (the last open cell in region 2), and row 4 would be left with no open cell for its marker.
8. First, if C2 were the marker, it would force F3 (the last open cell in region 2), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, 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.
10. Then, every open cell in row 2 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.
11. After that, if F2 were the marker, it would force H1 (the last open cell in region 3), then B8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), then B7 (the last open cell in region 8), and the chain continues, and row 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, place a marker at E2. Region has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region. Placing here also clears its 1 touching neighbour.
13. From there, every open cell left in this region sits in column F, so its marker has to land there; that clears every other open cell in the column.
14. Following that, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region.
15. First, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region. Placing here also clears its 2 touching neighbours.
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 F4. Region has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region.
18. After that, place a marker at A3. Region has exactly one open cell left.
19. Now, place a marker at H6. Region has exactly one open cell left. Placing here clears the rest of row 6, column H, and its region.
20. Finally, place a marker at G1. Region has exactly one open cell left.

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