An Expert 9×9 Puzzle Solved with Forced Chain in 26 Steps

This 9×9 board rates as expert (difficulty score 1221). Solving it from scratch takes 26 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 open cell left in this region sits in row 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, every remaining candidate in this region touches G2 and H2, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, that one can't be the marker: at A2 it would force D3, E3, F3, and 1 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then F4, G4, H4, and 1 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then G1, G3, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), and the chain continues, and row 3 would be left with no open cell for its marker.
4. After that, marking B2 would force D3, E3, F3, and 1 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then F4, G4, H4, and 1 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then G1, G3, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), and the chain continues, and then row 3 would be left with no open cell for its marker. So B2 can't be the marker there.
5. Now, that one can't be the marker: at D2 it would force A3, B3, F3, and 1 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then G4, G5, G6, and 9 more cleared (together, two regions have open cells only in columns G and H), then F4 (the last open cell in region 5), and the chain continues, and row 8 would be left with no open cell for its marker.
6. From there, if F2 were the marker, it would force H1 (the last open cell in region 3), then G7, G8 and G9 cleared (every open cell left in this region sits in column G, so its marker has to land there), then I7, I8 and I9 cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and row 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
7. Following that, every open cell left in this region sits in row 3, so its marker has to land there; that clears every other open cell in the row.
8. First, together, two regions have open cells only in columns G 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.
9. Next, every open cell left in this region sits in column I, so its marker has to land there; that clears every other open cell in the column.
10. Then, together, two regions have open cells only in rows 2 and 4; with one marker each, those two markers have to fill exactly those two rows, so every other region's open cells there can be cleared.
11. After that, 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.
12. Now, place a marker at I6. Region has exactly one open cell left. Placing here clears the rest of row 6, column I, and its region.
13. From there, together, two regions have open cells only in columns E and F; 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, every remaining candidate in this region touches C8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
15. First, if C5 were the marker, region 2 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 A5. Region has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region.
17. Then, if B7 were the marker, region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
18. After that, every open cell in column B 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.
19. Now, every remaining candidate in this region touches C9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
20. From there, place a marker at C2. Column 3 has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
21. Following that, place a marker at F4. Region has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region. Placing here also clears its 1 touching neighbour.
22. First, place a marker at H3. Region has exactly one open cell left. Placing here clears the rest of row 3, column H, and its region.
23. Next, place a marker at G1. Region has exactly one open cell left.
24. Then, place a marker at E9. Region has exactly one open cell left. Placing here clears the rest of row 9, column E, and its region. Placing here also clears its 1 touching neighbour.
25. After that, place a marker at D7. Region has exactly one open cell left.
26. Finally, place a marker at B8. Region has exactly one open cell left.

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