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

This 10×10 board rates as expert (difficulty score 1340). 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 in row 10 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.
2. Next, 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.
3. Then, marking A1 would force H2 (the last open cell in region 3), and then region 4 would be left with no open cell for its marker. So A1 can't be the marker there.
4. After that, if B1 were the marker, it would force H2 (the last open cell in region 3), and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. Now, that one can't be the marker: at C1 it would force H2 (the last open cell in region 3), and region 4 would be left with no open cell for its marker.
6. From there, marking D1 would force H2 (the last open cell in region 3), and then region 4 would be left with no open cell for its marker. So D1 can't be the marker there.
7. Following that, if E1 were the marker, it would force H2 (the last open cell in region 3), and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, that one can't be the marker: at F1 it would force H2 (the last open cell in region 3), and region 4 would be left with no open cell for its marker.
9. Next, marking G1 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
10. Then, 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.
11. After that, together, two regions have open cells only in rows 2 and 3; 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.
12. Now, that one can't be the marker: at I1 it would force E3 and F3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then F5, F6, F7, and 7 more cleared (together, two regions have open cells only in columns F and G), then H3 (the last open cell in region 4), and region 9 would be left with no open cell for its marker.
13. From there, marking J1 would force E3 and F3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then F5, F6, F7, and 7 more cleared (together, two regions have open cells only in columns F and G), then H8, H9, H10, and 5 more cleared (together, two regions have open cells only in columns H and I), and then region 9 would be left with no open cell for its marker. So J1 can't be the marker there.
14. Following that, place a marker at H1. Region has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region. Placing here also clears its 1 touching neighbour.
15. First, if F2 were the marker, it would force I3 (the last open cell in region 4), then J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), then G8, G9 and G10 cleared (every open cell left in this region sits in column G, so its marker has to land there), and region 9 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 J2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region.
17. Then, place a marker at I4. Region has exactly one open cell left. Placing here clears the rest of row 4, column I, and its region.
18. After that, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
19. Now, every open cell left in this region sits in column F, so its marker has to land there; that clears every other open cell in the column.
20. From there, place a marker at E3. Region has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region.
21. Following that, place a marker at D7. Region has exactly one open cell left. Placing here clears the rest of row 7, column D, and its region. Placing here also clears its 2 touching neighbours.
22. First, place a marker at F8. Region has exactly one open cell left. Placing here clears the rest of row 8, column F, and its region.
23. Next, every open cell in row 9 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.
24. Then, every remaining candidate in this region touches A10 and B10, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
25. After that, place a marker at C10. Region has exactly one open cell left. Placing here clears the rest of row 10, column C, and its region. Placing here also clears its 1 touching neighbour.
26. Now, 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.
27. From there, place a marker at G5. Region has exactly one open cell left.
28. Finally, place a marker at A9. Region has exactly one open cell left.

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