A Hard 9×9 Puzzle Solved with Forced Chain in 21 Steps

This 9×9 board rates as hard (difficulty score 1036). Solving it from scratch takes 21 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 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 B, so its marker has to land there; that clears every other open cell in the column.
3. Then, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
4. After that, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
5. Now, every open cell in row 9 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
6. From there, together, two regions have open cells only in columns E and F; 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, every remaining candidate in this region touches D7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
9. Next, if E1 were the marker, it would force F5 (the last open cell in region 4), then I3 (the last open cell in row 3), then H7 (the last open cell in region 6), and the chain continues, and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, every open cell left in this region sits in column F, so its marker has to land there; that clears every other open cell in the column.
11. After that, every remaining candidate in this region touches G1 and G2, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
12. Now, place a marker at G9. Column 7 has exactly one open cell left. Placing here clears the rest of row 9, column G, and its region.
13. From there, if H1 were the marker, it would force F2 (the last open cell in region 2), and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, place a marker at H7. Column 8 has exactly one open cell left. Placing here clears the rest of row 7, column H, and its region.
15. First, place a marker at A6. Region has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
16. Next, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region.
17. Then, 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 1 touching neighbour.
18. After that, place a marker at E3. Region has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region. Placing here also clears its 1 touching neighbour.
19. Now, place a marker at F1. Region has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
20. From there, place a marker at I2. Region has exactly one open cell left. Placing here clears the rest of row 2, column I, and its region.
21. Finally, place a marker at B4. Region has exactly one open cell left.

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