A Hard 9×9 Puzzle Solved with Forced Chain in 19 Steps

This 9×9 board rates as hard (difficulty score 1013). Solving it from scratch takes 19 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 A, so its marker has to land there; that clears every other open cell in the column.
2. Next, every open cell left in this region sits in column G, 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 rows 8 and 9; 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 H7. Region has exactly one open cell left. Placing here clears the rest of row 7, column H, and its region. Placing here also clears its 2 touching neighbours.
5. Now, place a marker at G9. Region has exactly one open cell left. Placing here clears the rest of row 9, column G, and its region. Placing here also clears its 1 touching neighbour.
6. From there, 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.
7. Following that, 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.
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.
9. 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.
10. Then, every remaining candidate in this region touches D5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
11. After that, every remaining candidate in this region touches C4 and C5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
12. Now, that one can't be the marker: at F1 it would force D6 cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), then C6 (the last open cell in row 6), then D2 (the last open cell in region 1), and the chain continues, and column B would be left with no open cell for its marker.
13. From there, place a marker at I1. Region has exactly one open cell left. Placing here clears the rest of row 1, column I, and its region.
14. Following that, that one can't be the marker: at A2 it would force C3 (the last open cell in region 1), and region 6 would be left with no open cell for its marker.
15. First, 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. Placing here also clears its 1 touching neighbour.
16. Next, 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 1 touching neighbour.
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. After that, place a marker at F6. Region has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region.
19. Finally, place a marker at C2. Region has exactly one open cell left.

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