An Expert 9×9 Puzzle Solved with Forced Chain in 23 Steps

This 9×9 board rates as expert (difficulty score 1294). Solving it from scratch takes 23 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Line confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, together, two regions have open cells only in columns I and H; 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.
2. Next, every remaining candidate in this region touches F6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, marking A1 would force B8 (the last open cell in region 6), then G7 and I7 cleared (every open cell left in this region sits in row 7, so its marker has to land there), then H9 and I9 cleared (every open cell left in this region sits in row 9, so its marker has to land there), and the chain continues, and then column H would be left with no open cell for its marker. So A1 can't be the marker there.
4. After that, if B1 were the marker, it would force A3, A4 and A5 cleared (every open cell left in this region sits in column A, so its marker has to land there), then D2 (the last open cell in region 1), then I3 (the last open cell in row 3), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. Now, marking D1 would force A5, B5, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), then C7, C8 and C9 cleared (every open cell left in this region sits in column C, so its marker has to land there), then H5, I5, A6, and 1 more cleared (together, two regions have open cells only in rows 5 and 6), and the chain continues, and then region 6 would be left with no open cell for its marker. So D1 can't be the marker there.
6. From there, marking G1 would force A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then H5 (the last open cell in column H), then E6 (the last open cell in region 5), and the chain continues, and then row 2 would be left with no open cell for its marker. So G1 can't be the marker there.
7. Following that, that one can't be the marker: at I1 it would force H9 (the last open cell in region 8), then G7 (the last open cell in region 9), then A8 and B8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and row 2 would be left with no open cell for its marker.
8. First, every open cell in row 1 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.
9. Next, that one can't be the marker: at A2 it would force B8 (the last open cell in region 6), then I3 (the last open cell in row 3), then H9 (the last open cell in region 8), and the chain continues, and region 7 would be left with no open cell for its marker.
10. Then, that one can't be the marker: at C1 it would force I2 (the last open cell in row 2), then H9 (the last open cell in region 8), then G7 (the last open cell in region 9), and the chain continues, and region 7 would be left with no open cell for its marker.
11. After that, if C2 were the marker, it would force I3 (the last open cell in row 3), then H9 (the last open cell in region 8), then G7 (the last open cell in region 9), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, that one can't be the marker: at D2 it would force F1 (the last open cell in region 2), then I3 (the last open cell in row 3), and row 4 would be left with no open cell for its marker.
13. From there, together, two regions have open cells only in columns B and A; 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.
14. Following that, if I2 were the marker, it would force H9 (the last open cell in region 8), then G7 (the last open cell in region 9), then A3 (the last open cell in row 3), 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.
15. First, place a marker at B2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column B, and its region.
16. Next, place a marker at I3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column I, and its region.
17. Then, place a marker at H9. Region has exactly one open cell left. Placing here clears the rest of row 9, column H, and its region. Placing here also clears its 1 touching neighbour.
18. After that, place a marker at G7. Region has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region. Placing here also clears its 1 touching neighbour.
19. Now, place a marker at D4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region. Placing here also clears its 1 touching neighbour.
20. From there, place a marker at F5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region.
21. Following that, place a marker at E1. Region has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
22. First, 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.
23. Finally, place a marker at A6. Region has exactly one open cell left.

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