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

This 9×9 board rates as expert (difficulty score 1277). 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, marking A1 would force H2 (the last open cell in region 3), then G4 (the last open cell in region 2), then E3 (the last open cell in region 4), and then region 5 would be left with no open cell for its marker. So A1 can't be the marker there.
2. Next, if B1 were the marker, it would force H2 (the last open cell in region 3), then G4 (the last open cell in region 2), then E3 (the last open cell in region 4), and the chain continues, and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
3. Then, marking D1 would force H2 (the last open cell in region 3), then G4 (the last open cell in region 2), then E3 (the last open cell in region 4), and the chain continues, and then column C would be left with no open cell for its marker. So D1 can't be the marker there.
4. After that, if E1 were the marker, it would force H2 (the last open cell in region 3), then F3 (the last open cell in region 4), and region 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. Now, that one can't be the marker: at F1 it would force H2 (the last open cell in region 3), then C3, D3 and E3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then E4 (the last open cell in region 4), and the chain continues, and region 8 would be left with no open cell for its marker.
6. From there, that one can't be the marker: at A2 it would force B4 and B9 cleared (every open cell left in this region sits in column B, so its marker has to land there), then C5, C6, C7, and 2 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then F6, I6, F7, and 6 more cleared (every open cell in row 9 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and column C would be left with no open cell for its marker.
7. Following that, marking B2 would force A4, A5, A6, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then C4 (the last open cell in region 5), then G3, H3 and I3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), and then region 2 would be left with no open cell for its marker. So B2 can't be the marker there.
8. First, marking G1 would force C3, D3, E3, and 3 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A4, A5, A6, and 3 more cleared (together, two regions have open cells only in columns A and B), then C4 (the last open cell in region 5), and the chain continues, and then region 4 would be left with no open cell for its marker. So G1 can't be the marker there.
9. Next, every remaining candidate in this region touches I2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
10. Then, if H1 were the marker, it would force C3, D3, E3, and 3 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A4, A5, A6, and 3 more cleared (together, two regions have open cells only in columns A and B), then C4 (the last open cell in region 5), and the chain continues, and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, that one can't be the marker: at I1 it would force C3, D3, E3, and 3 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A4, A5, A6, and 3 more cleared (together, two regions have open cells only in columns A and B), then C4 (the last open cell in region 5), and the chain continues, and region 2 would be left with no open cell for its marker.
12. Now, place a marker at H2. Region has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region. Placing here also clears its 2 touching neighbours.
13. From there, place a marker at G4. Region has exactly one open cell left. Placing here clears the rest of row 4, column G, and its region. Placing here also clears its 2 touching neighbours.
14. Following 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.
15. First, place a marker at C1. Region has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
16. Next, 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.
17. Then, 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.
18. 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.
19. Now, place a marker at I5. Region has exactly one open cell left. Placing here clears the rest of row 5, column I, and its region.
20. From there, place a marker at D7. Region has exactly one open cell left. Placing here clears the rest of row 7, column D, and its region.
21. Following that, 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.
22. First, place a marker at B8. Region has exactly one open cell left. Placing here clears the rest of row 8, column B, and its region.
23. Finally, place a marker at F9. Region has exactly one open cell left.

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