A Hard 9×9 Puzzle Solved with Forced Chain in 18 Steps

This 9×9 board rates as hard (difficulty score 1036). Solving it from scratch takes 18 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 remaining candidate in this region touches H3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, together, two regions have open cells only in rows 1 and 2; 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, 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.
4. After that, place a marker at B5. Region has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region. Placing here also clears its 3 touching neighbours.
5. Now, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
6. From there, 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.
7. Following that, every remaining candidate in this region touches C8 and D8, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
8. First, every open cell in column C 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.
9. Next, place a marker at D1. Column 4 has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region.
10. Then, 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.
11. After that, that one can't be the marker: at I4 it would force C9 cleared (every open cell left in this region sits in row 9, so its marker has to land there), then C7 (the last open cell in region 7), then F9 cleared (every remaining candidate in this region touches F9), and the chain continues, and row 9 would be left with no open cell for its marker.
12. Now, 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.
13. From there, marking I6 would force C9 cleared (every open cell left in this region sits in row 9, so its marker has to land there), then C7 (the last open cell in region 7), then E9 and F9 cleared (every remaining candidate in this region touches E9 and F9), and then row 9 would be left with no open cell for its marker. So I6 can't be the marker there.
14. Following that, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region. Placing here also clears its 1 touching neighbour.
15. First, that one can't be the marker: place it at I7, and row 8 would be left with no open cell for its marker.
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. Then, place a marker at I9. Region has exactly one open cell left. Placing here clears the rest of row 9, column I, and its region.
18. Finally, place a marker at C7. Region has exactly one open cell left.

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