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 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, together, two regions have open cells only in rows 2 and 3; 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.
3. Then, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region. Placing here also clears its 1 touching neighbour.
4. After that, place a marker at G5. Region has exactly one open cell left. Placing here clears the rest of row 5, column G, and its region. Placing here also clears its 2 touching neighbours.
5. 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.
6. From there, place a marker at I7. Region has exactly one open cell left. Placing here clears the rest of row 7, column I, and its region. Placing here also clears its 1 touching neighbour.
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, marking D1 would force C3 (the last open cell in region 1), then B6 (the last open cell in region 6), and then region 9 would be left with no open cell for its marker. So D1 can't be the marker there.
9. Next, every remaining candidate in this region touches F2, 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 E3 (the last open cell in region 3), then F1 (the last open cell in region 2), then B6 (the last open cell in region 6), 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.
11. After that, that one can't be the marker: place it at D2, and row 3 would be left with no open cell for its marker.
12. Now, place a marker at C3. Region has exactly one open cell left. Placing here clears the rest of row 3, column C, and its region.
13. From there, place a marker at H2. Region has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region.
14. Following 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.
15. First, place a marker at D9. Region has exactly one open cell left. Placing here clears the rest of row 9, column D, and its region. Placing here also clears its 1 touching neighbour.
16. Next, place a marker at F8. Region has exactly one open cell left. Placing here clears the rest of row 8, column F, and its region.
17. Finally, place a marker at E1. Region has exactly one open cell left.

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