A Sharp 9×9 Puzzle Solved with Forced Chain in 20 Steps

This 9×9 board rates as sharp (difficulty score 1031). Solving it from scratch takes 20 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 row 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, every open cell left in this region sits in row 2, 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 rows 3 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.
4. After that, every remaining candidate in this region touches G1 and F3 and G3, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, together, two regions have open cells only in columns H and I; 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.
6. From there, every remaining candidate in this region touches E3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
7. 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 2 touching neighbours.
8. First, place a marker at G2. Region has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region. Placing here also clears its 1 touching neighbour.
9. Next, place a marker at I1. Region has exactly one open cell left. Placing here clears the rest of row 1, column I, and its region.
10. Then, place a marker at D3. Region has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
11. After that, 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.
12. Now, every remaining candidate in this region touches B8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
13. From there, together, two regions have open cells only in columns C and E; 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, if C5 were the marker, it would force A6 (the last open cell in region 1), then B9 (the last open cell in region 8), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, place a marker at A5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region.
16. Next, place a marker at E6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region.
17. Then, 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.
18. After that, place a marker at C9. Region has exactly one open cell left. Placing here clears the rest of row 9, column C, and its region.
19. Now, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region.
20. Finally, place a marker at H8. Region has exactly one open cell left.

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