An Expert 10×10 Puzzle Solved with Forced Chain in 29 Steps

This 10×10 board rates as expert (difficulty score 1330). Solving it from scratch takes 29 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 A, so its marker has to land there; that clears every other open cell in the column.
2. Next, together, two regions have open cells only in rows 9 and 10; 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.
3. 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.
4. After that, place a marker at G7. Region has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region. Placing here also clears its 2 touching neighbours.
5. Now, place a marker at A6. Region has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region. Placing here also clears its 1 touching neighbour.
6. From there, every open cell left in this region sits in row 9, so its marker has to land there; that clears every other open cell in the row.
7. Following that, every remaining candidate in this region touches I2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
8. First, if B1 were the marker, it would force J2 (the last open cell in region 3), then I5 (the last open cell in region 4), and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, that one can't be the marker: at C1 it would force J2 (the last open cell in region 3), then I5 (the last open cell in region 4), then B8 (the last open cell in region 8), and the chain continues, and row 10 would be left with no open cell for its marker.
10. Then, marking D1 would force J2 (the last open cell in region 3), then I5 (the last open cell in region 4), then H10 (the last open cell in region 10), and the chain continues, and then row 3 would be left with no open cell for its marker. So D1 can't be the marker there.
11. After that, that one can't be the marker: at F1 it would force J2 (the last open cell in region 3), then I5 (the last open cell in region 4), then H9 (the last open cell in region 9), and row 10 would be left with no open cell for its marker.
12. Now, marking B2 would force C8 (the last open cell in region 8), then D3 and D4 cleared (every open cell in column E belongs to the same region, so that region's marker has to be somewhere in this column), then D5 (the last open cell in column D), and then column E would be left with no open cell for its marker. So B2 can't be the marker there.
13. From there, if C2 were the marker, it would force B8 (the last open cell in region 8), then D4 cleared (every open cell in column E belongs to the same region, so that region's marker has to be somewhere in this column), then D5 (the last open cell in column D), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, if H1 were the marker, it would force F9 (the last open cell in region 9), then E3 (the last open cell in region 2), and row 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, that one can't be the marker: at I1 it would force J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), then H10 (the last open cell in region 10), then F9 (the last open cell in region 9), and the chain continues, and row 2 would be left with no open cell for its marker.
16. Next, every open cell left in this region sits in column J, so its marker has to land there; that clears every other open cell in the column.
17. Then, 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.
18. After 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.
19. Now, every open cell left in this region sits in column E, so its marker has to land there; that clears every other open cell in the column.
20. From there, every remaining candidate in this region touches D2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
21. Following that, every remaining candidate in this region touches C4 and C5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
22. First, marking J1 would force E3 (the last open cell in region 2), and then region 1 would be left with no open cell for its marker. So J1 can't be the marker there.
23. Next, place a marker at J2. Region has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region. Placing here also clears its 1 touching neighbour.
24. Then, place a marker at I5. Region has exactly one open cell left. Placing here clears the rest of row 5, column I, and its region.
25. After that, 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.
26. Now, 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.
27. From there, place a marker at E1. Region has exactly one open cell left.
28. Following that, place a marker at C8. Region has exactly one open cell left.
29. Finally, place a marker at H10. Region has exactly one open cell left.

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