A Hard 9×9 Puzzle Solved with Forced Chain in 17 Steps

This 9×9 board rates as hard (difficulty score 1020). Solving it from scratch takes 17 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 9, so its marker has to land there; that clears every other open cell in the row.
2. Next, place a marker at I8. Region has exactly one open cell left. Placing here clears the rest of row 8, column I, and its region. Placing here also clears its 1 touching neighbour.
3. Then, 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.
4. After that, 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 2 touching neighbours.
5. Now, together, two regions have open cells only in rows 3 and 4; 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.
6. From there, together, two regions have open cells only in columns H and G; 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.
7. Following that, every remaining candidate in this region touches E2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
8. First, if B1 were the marker, it would force F2 (the last open cell in region 2), then A4 (the last open cell in region 4), then H3 (the last open cell in region 3), and row 6 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 C1 it would force F2 (the last open cell in region 2), then E7 (the last open cell in region 7), then H6 (the last open cell in region 5), and the chain continues, and column A would be left with no open cell for its marker.
10. Then, if E1 were the marker, it would force C7 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 B3 cleared (every remaining candidate in this region touches B3), then H3 (the last open cell in row 3), and the chain continues, and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, place a marker at E7. Column 5 has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region.
12. 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.
13. From there, place a marker at G4. Region has exactly one open cell left. Placing here clears the rest of row 4, column G, and its region.
14. Following that, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region. Placing here also clears its 2 touching neighbours.
15. First, place a marker at A1. Region has exactly one open cell left. Placing here clears the rest of row 1, column A, and its region.
16. Next, place a marker at F2. Region has exactly one open cell left.
17. Finally, place a marker at C9. Region has exactly one open cell left.

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