A Severe 9×9 Puzzle Solved with Forced Chain in 32 Steps

This 9×9 board rates as severe (difficulty score 1131). Solving it from scratch takes 32 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region 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 column I, so its marker has to land there; that clears every other open cell in the column.
2. Next, 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.
3. Then, 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.
4. After that, place a marker at F5. Region has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region. Placing here also clears its 3 touching neighbours.
5. Now, place a marker at G1. Column 7 has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
6. From there, every remaining candidate in this region touches C3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
7. Following that, that one can't be the marker: at A2 it would force D3 (the last open cell in region 2), and region 3 would be left with no open cell for its marker.
8. First, marking B2 would force D3 (the last open cell in region 2), and then region 3 would be left with no open cell for its marker. So B2 can't be the marker there.
9. Next, that one can't be the marker: place it at D2, and region 3 would be left with no open cell for its marker.
10. Then, marking E2 would immediately leave region 2 with no open cell for its marker, so it can't be the marker there.
11. After that, place a marker at C2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region. Placing here also clears its 1 touching neighbour.
12. Now, if A3 were the marker, it would force D6 (the last open cell in region 3), and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
14. Following that, marking I3 would force D6 (the last open cell in region 3), then A4 (the last open cell in region 1), and then region 7 would be left with no open cell for its marker. So I3 can't be the marker there.
15. First, place a marker at E3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region. Placing here also clears its 1 touching neighbour.
16. Next, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
17. Then, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region.
18. After that, place a marker at I7. Region has exactly one open cell left. Placing here clears the rest of row 7, column I, and its region. Placing here also clears its 1 touching neighbour.
19. Now, 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.
20. Finally, place a marker at D8. Region has exactly one open cell left.

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