A Sharp 9×9 Puzzle Solved with Forced Chain in 23 Steps

This 9×9 board rates as sharp (difficulty score 1109). Solving it from scratch takes 23 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, 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, every open cell left in this region sits in column H, so its marker has to land there; that clears every other open cell in the column.
3. Then, 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.
4. After that, 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.
5. Now, every open cell left in this region sits in column E, so its marker has to land there; that clears every other open cell in the column.
6. From there, 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.
7. Following that, every remaining candidate in this region touches I8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
8. First, 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. Placing here also clears its 1 touching neighbour.
9. Next, 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.
10. Then, 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.
11. After that, 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.
12. Now, every remaining candidate in this region touches C3 and C4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
13. From there, marking A1 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
14. Following that, if B1 were the marker, row 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, that one can't be the marker: at C1 it would force F3 (the last open cell in region 3), then D4 (the last open cell in region 2), and column B would be left with no open cell for its marker.
16. Next, place a marker at F1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
17. Then, every open cell left in this region sits in row 2, so its marker has to land there; that clears every other open cell in the row.
18. After that, every remaining candidate in this region touches B3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
19. Now, if C2 were the marker, it would force D4 (the last open cell in region 2), and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
20. From there, place a marker at B2. Region has exactly one open cell left. Placing here clears the rest of row 2, column B, and its region. Placing here also clears its 1 touching neighbour.
21. Following that, 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.
22. First, place a marker at D3. Region has exactly one open cell left.
23. Finally, place a marker at C8. Region has exactly one open cell left.

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