An Expert 9×9 Puzzle Solved with Forced Chain in 32 Steps

This 9×9 board rates as expert (difficulty score 1522). Solving it from scratch takes 32 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, marking A1 would force B7, B9, C5, and 3 more cleared (together, two regions have open cells only in columns B and C), then D4, D5, D6, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E2, E3, E4, and 3 more cleared (every open cell left in this region sits in column E, 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 A1 can't be the marker there.
2. Next, if B1 were the marker, it would force C5, C7, C8, and 1 more cleared (together, two regions have open cells only in columns C and A), then D4, D5, D6, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E2, E3, E4, and 3 more cleared (every open cell left in this region sits in column E, so its marker has to land there), 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.
3. Then, that one can't be the marker: at C1 it would force B7 and B9 cleared (together, two regions have open cells only in columns B and A), then D4, D5, D6, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E2, E3, E4, and 3 more cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and region 7 would be left with no open cell for its marker.
4. After that, that one can't be the marker: at F1 it would force I6, G7, I7, and 3 more cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then G3 cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), then G9 (the last open cell in column G), and the chain continues, and region 9 would be left with no open cell for its marker.
5. Now, that one can't be the marker: at A2 it would force G3, I3 and I4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then E4, F4, G4, and 8 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then B7, B9, C5, and 3 more cleared (together, two regions have open cells only in columns B and C), and the chain continues, and row 4 would be left with no open cell for its marker.
6. From there, marking B2 would force G3, I3 and I4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then E4, F4, G4, and 8 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then C5, C7, C8, and 1 more cleared (together, two regions have open cells only in columns C and A), and the chain continues, and then row 5 would be left with no open cell for its marker. So B2 can't be the marker there.
7. Following that, if C2 were the marker, it would force G3, I3 and I4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then E4, F4, G4, and 8 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then B7 and B9 cleared (together, two regions have open cells only in columns B and A), and the chain continues, and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, that one can't be the marker: at D2 it would force G3, I3 and I4 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then E4, F4, G4, and 8 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then A5 and B5 cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 4 would be left with no open cell for its marker.
9. Next, marking G1 would force E2 (the last open cell in row 2), then A4 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then I6, I7, I8, and 1 more cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), and then column I would be left with no open cell for its marker. So G1 can't be the marker there.
10. Then, if H1 were the marker, it would force E3, F3, E4, and 7 more cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then A4 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then G7 and G8 cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and row 5 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 E3, F3, H3, and 10 more cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then A4 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then G7 and G8 cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 9 would be left with no open cell for its marker.
12. Now, 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.
13. From there, 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.
14. Following that, every remaining candidate in this region touches H3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
15. First, if F2 were the marker, it would force D1 (the last open cell in region 1), then I3 (the last open cell in row 3), then A5 and B5 cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
16. Next, that one can't be the marker: at G2 it would force E3 (the last open cell in row 3), then D1 (the last open cell in region 1), then C5 (the last open cell in region 5), and row 4 would be left with no open cell for its marker.
17. Then, marking H2 would immediately leave row 3 with no open cell for its marker, so it can't be the marker there.
18. After that, place a marker at I2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column I, and its region.
19. Now, every open cell in row 3 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.
20. From there, every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
21. Following that, 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. Placing here also clears its 1 touching neighbour.
22. First, 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.
23. Next, place a marker at F3. Column 6 has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region.
24. Then, 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.
25. After that, place a marker at E1. Column 5 has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
26. Now, every open cell left in this region sits in row 6, so its marker has to land there; that clears every other open cell in the row.
27. From there, together, two regions have open cells only in columns D and C; 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.
28. Following that, every remaining candidate in this region touches A5 and B5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
29. First, place a marker at B4. Region has exactly one open cell left. Placing here clears the rest of row 4, column B, and its region. Placing here also clears its 1 touching neighbour.
30. Next, 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.
31. Then, place a marker at A6. Region has exactly one open cell left.
32. 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. ∎