An Expert 9×9 Puzzle Solved with Forced Chain in 25 Steps

This 9×9 board rates as expert (difficulty score 1373). Solving it from scratch takes 25 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 remaining candidate in this region touches E1, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, marking A1 would force F2, G2, H2, and 1 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F3 (the last open cell in region 3), and then region 4 would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, if B1 were the marker, it would force F2, G2, H2, and 1 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F3 (the last open cell in region 3), and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at C1 it would force E2 (the last open cell in region 2), and region 3 would be left with no open cell for its marker.
5. Now, that one can't be the marker: at F1 it would force D2 (the last open cell in region 2), then C4, E4, H4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), then C5 (the last open cell in region 5), and the chain continues, and region 1 would be left with no open cell for its marker.
6. From there, every remaining candidate in this region touches G3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
7. Following that, marking G1 would force F3 (the last open cell in region 3), then D2 (the last open cell in region 2), then C4, H4 and I4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), and the chain continues, and then region 7 would be left with no open cell for its marker. So G1 can't be the marker there.
8. First, if H1 were the marker, it would force A2, B2, C2, and 1 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F3 (the last open cell in region 3), then D2 (the last open cell in region 2), and the chain continues, and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, that one can't be the marker: at I1 it would force A2, B2, C2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F3 (the last open cell in region 3), then D2 (the last open cell in region 2), and the chain continues, and region 7 would be left with no open cell for its marker.
10. Then, place a marker at D1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region. Placing here also clears its 1 touching neighbour.
11. After that, that one can't be the marker: at A2 it would force F3 (the last open cell in region 3), and region 4 would be left with no open cell for its marker.
12. Now, marking B2 would force F3 (the last open cell in region 3), and then region 4 would be left with no open cell for its marker. So B2 can't be the marker there.
13. From there, if F2 were the marker, it would force C4, E4, H4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), then C5 (the last open cell in region 5), then B7 (the last open cell in region 8), and the chain continues, and region 1 would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, marking H2 would force G4 (the last open cell in region 4), then E3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then C5 (the last open cell in region 5), and the chain continues, and then region 7 would be left with no open cell for its marker. So H2 can't be the marker there.
15. First, if I2 were the marker, it would force F3 (the last open cell in region 3), then C4 and H4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then C5 (the last open cell in region 5), and the chain continues, and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
16. Next, place a marker at G2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region. Placing here also clears its 1 touching neighbour.
17. Then, place a marker at I3. Region has exactly one open cell left. Placing here clears the rest of row 3, column I, and its region. Placing here also clears its 1 touching neighbour.
18. After that, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
19. Now, every open cell left in this region sits in row 5, so its marker has to land there; that clears every other open cell in the row.
20. From there, place a marker at H6. Region has exactly one open cell left. Placing here clears the rest of row 6, column H, and its region.
21. 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. Placing here also clears its 2 touching neighbours.
22. First, 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.
23. Next, 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.
24. Then, place a marker at F5. Region has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region.
25. Finally, place a marker at E8. Region has exactly one open cell left.

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