An Expert 10×10 Puzzle Solved with Forced Chain in 28 Steps

This 10×10 board rates as expert (difficulty score 1230). Solving it from scratch takes 28 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 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 column J, so its marker has to land there; that clears every other open cell in the column.
3. Then, every open cell left in this region sits in column H, so its marker has to land there; that clears every other open cell in the column.
4. After that, every remaining candidate in this region touches G5 and G6 and I6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, 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.
6. From there, place a marker at H5. Region has exactly one open cell left. Placing here clears the rest of row 5, column H, and its region. Placing here also clears its 1 touching neighbour.
7. Following that, 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.
8. First, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
9. Next, every remaining candidate in this region touches C2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
10. Then, every remaining candidate in this region touches F9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
11. After that, if E1 were the marker, it would force D3 (the last open cell in region 1), then J2 (the last open cell in region 3), then C6 (the last open cell in region 4), 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.
12. Now, that one can't be the marker: at F1 it would force J2 (the last open cell in region 3), then D3 (the last open cell in region 1), then C6 (the last open cell in region 4), and the chain continues, and region 8 would be left with no open cell for its marker.
13. From there, marking G1 would force J2 (the last open cell in region 3), then D3 (the last open cell in region 1), then C6 (the last open cell in region 4), and the chain continues, and then region 8 would be left with no open cell for its marker. So G1 can't be the marker there.
14. Following that, marking J1 would force D4, D6, D8, and 2 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E2, C3 and E3 cleared (every remaining candidate in this region touches E2 and C3 and E3), then F4, B6, E6, and 1 more cleared (together, two regions have open cells only in rows 4 and 6), and the chain continues, and then column F would be left with no open cell for its marker. So J1 can't be the marker there.
15. First, 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.
16. Next, every open cell in row 1 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.
17. Then, marking C3 would force D1 (the last open cell in region 1), then E10, F10 and G10 cleared (every open cell in row 9 belongs to the same region, so that region's marker has to be somewhere in this row), and then row 10 would be left with no open cell for its marker. So C3 can't be the marker there.
18. After that, that one can't be the marker: place it at E3, and region 2 would be left with no open cell for its marker.
19. Now, every open cell in row 3 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.
20. From there, together, two regions have open cells only in rows 4 and 6; 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.
21. Following that, place a marker at B8. Region has exactly one open cell left. Placing here clears the rest of row 8, column B, and its region. Placing here also clears its 1 touching neighbour.
22. First, together, two regions have open cells only in columns C and D; 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.
23. Next, place a marker at E4. Region has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region. Placing here also clears its 1 touching neighbour.
24. Then, place a marker at G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region.
25. After that, place a marker at A6. Region has exactly one open cell left.
26. Now, place a marker at F10. Region has exactly one open cell left. Placing here clears the rest of row 10, column F, and its region.
27. From there, place a marker at D9. Region has exactly one open cell left. Placing here clears the rest of row 9, column D, and its region.
28. Finally, place a marker at C1. Region has exactly one open cell left.

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