An Expert 9×9 Puzzle Solved with Forced Chain in 29 Steps

This 9×9 board rates as expert (difficulty score 1514). Solving it from scratch takes 29 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, every remaining candidate in this region touches B5, 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 D6, E6, F6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then B7, B8, B9, and 4 more cleared (together, two regions have open cells only in columns B and C), then D8 (the last open cell in region 9), and then region 7 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 A7, A8, A9, and 4 more cleared (together, two regions have open cells only in columns A and C), then D8 (the last open cell in region 9), then E6 (the last open cell in region 7), 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.
4. After that, marking G1 would force C3, D3, E3, and 1 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 H3, D4, E4, and 5 more cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), then B4 and C4 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then column B would be left with no open cell for its marker. So G1 can't be the marker there.
5. Now, if H1 were the marker, it would force C3, D3, E3, and 1 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 G3, D4, E4, and 6 more cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), then B4 and C4 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), 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.
6. From there, that one can't be the marker: at I1 it would force H3, H9, G2, and 3 more cleared (together, two regions have open cells only in columns H and G), then C3, D3, E3, and 1 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 B4 and C4 cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and column B would be left with no open cell for its marker.
7. Following that, that one can't be the marker: at A2 it would force D4, E4, F4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then D6, E6, F6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then I3 and I5 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and column D would be left with no open cell for its marker.
8. First, marking B2 would force C4 (the last open cell in region 4), then A7, A8 and A9 cleared (every open cell left in this region sits in column A, so its marker has to land there), then D8 (the last open cell in region 9), and the chain continues, and then region 8 would be left with no open cell for its marker. So B2 can't be the marker there.
9. Next, marking D1 would force G2 (the last open cell in row 2), and then column F would be left with no open cell for its marker. So D1 can't be the marker there.
10. Then, that one can't be the marker: at F1 it would force C2 (the last open cell in row 2), then A7, A8, A9, and 3 more cleared (together, two regions have open cells only in columns A and B), then D8 (the last open cell in region 9), and the chain continues, and column A would be left with no open cell for its marker.
11. After that, if C2 were the marker, it would force E1 (the last open cell in row 1), then D4, D8 and D9 cleared (every open cell left in this region sits in column D, so its marker has to land there), then G3, H3, G4, and 2 more cleared (every open cell in column F 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 end up with no open cell for its marker, which can't happen. So it can't go there.
12. Now, if E1 were the marker, it would force G2 (the last open cell in row 2), and column F would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, place a marker at C1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, together, two regions have open cells only in columns A and B; 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.
15. First, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region. Placing here also clears its 2 touching neighbours.
16. Next, 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. Placing here also clears its 2 touching neighbours.
17. Then, 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.
18. After that, every remaining candidate in this region touches A4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
19. Now, place a marker at A5. Region has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region. Placing here also clears its 1 touching neighbour.
20. From there, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region.
21. Following that, that one can't be the marker: place it at G2, and column F would be left with no open cell for its marker.
22. First, every open cell in row 2 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.
23. Next, marking H2 would force I7 (the last open cell in region 6), and then region 8 would be left with no open cell for its marker. So H2 can't be the marker there.
24. Then, that one can't be the marker: at F4 it would force I2 (the last open cell in region 2), and row 9 would be left with no open cell for its marker.
25. After that, marking G4 would force I7 (the last open cell in region 6), and then row 9 would be left with no open cell for its marker. So G4 can't be the marker there.
26. Now, place a marker at H4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column H, and its region.
27. From there, place a marker at G7. Row 7 has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region.
28. Following that, place a marker at F9. Region has exactly one open cell left. Placing here clears the rest of row 9, column F, and its region.
29. Finally, place a marker at I2. Region has exactly one open cell left.

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