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

This 9×9 board rates as hard (difficulty score 1048). Solving it from scratch takes 18 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 I, so its marker has to land there; that clears every other open cell in the column.
2. Next, place a marker at H1. Region has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region. Placing here also clears its 1 touching neighbour.
3. Then, together, two regions have open cells only in columns A and B; 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 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.
5. Now, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
6. From there, every remaining candidate in this region touches D7 and D9, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
7. Following that, place a marker at C9. Region has exactly one open cell left. Placing here clears the rest of row 9, column C, and its region. Placing here also clears its 1 touching neighbour.
8. 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. Placing here also clears its 1 touching neighbour.
9. Next, that one can't be the marker: at D2 it would force A3 (the last open cell in region 1), and region 4 would be left with no open cell for its marker.
10. Then, if D3 were the marker, region 4 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: at A2 it would force B7 (the last open cell in region 7), then G3 and G6 cleared (every open cell left in this region sits in column G, so its marker has to land there), then F3 (the last open cell in row 3), and the chain continues, and column D would be left with no open cell for its marker.
12. Now, marking B2 would force G3, G6 and G7 cleared (every open cell left in this region sits in column G, so its marker has to land there), then F3 (the last open cell in row 3), then G5 (the last open cell in region 2), and then column D would be left with no open cell for its marker. So B2 can't be the marker there.
13. From there, 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.
14. Following that, 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.
15. First, place a marker at F2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
16. Next, place a marker at G6. Column 7 has exactly one open cell left. Placing here clears the rest of row 6, column G, and its region.
17. Then, place a marker at D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region.
18. Finally, place a marker at I5. Region has exactly one open cell left.

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