An Expert 10×10 Puzzle Solved with Forced Chain in 30 Steps

This 10×10 board rates as expert (difficulty score 1357). Solving it from scratch takes 30 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, every open cell left in this region sits in row 10, so its marker has to land there; that clears every other open cell in the row.
3. Then, every remaining candidate in this region touches B4 and B5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches D6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, every remaining candidate in this region touches E8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
6. From there, every remaining candidate in this region touches B9 and C9, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
7. Following that, if B1 were the marker, it would force C10 (the last open cell in region 10), then D8 (the last open cell in region 7), then E2 and E4 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and column E 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 C1 it would force B10 (the last open cell in region 10), then D5 and D7 cleared (every open cell left in this region sits in column D, so its marker has to land there), then E2, E4 and E9 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and column E would be left with no open cell for its marker.
9. Next, marking D1 would force G9 and J9 cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then E4, F4, F5, and 1 more cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), then E9 cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then column J would be left with no open cell for its marker. So D1 can't be the marker there.
10. Then, if E1 were the marker, it would force F4, F5 and F6 cleared (every open cell left in this region sits in column F, so its marker has to land there), then G9 cleared (every open cell left in this region sits in column G, so its marker has to land there), then J9 cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and column J would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, that one can't be the marker: at F1 it would force E9 (the last open cell in region 9), then J2, J3, J4, and 4 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and column J would be left with no open cell for its marker.
12. Now, marking G1 would force J9 cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then J2, J3, J4, and 4 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and then column J would be left with no open cell for its marker. So G1 can't be the marker there.
13. From there, if H1 were the marker, it would force D9, E9, F9, and 1 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then F4, F5 and F6 cleared (every open cell left in this region sits in column F, so its marker has to land there), then J2, J3, J4, and 4 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, marking J1 would immediately leave column I with no open cell for its marker, so it can't be the marker there.
15. First, place a marker at I1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column I, and its region. Placing here also clears its 1 touching neighbour.
16. Next, place a marker at B2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column B, and its region.
17. Then, 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.
18. After that, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region. Placing here also clears its 2 touching neighbours.
19. Now, place a marker at J3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column J, and its region.
20. From there, 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.
21. Following that, 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.
22. First, place a marker at A4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
23. Next, 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.
24. Then, every remaining candidate in this region touches H6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
25. After that, together, two regions have open cells only in rows 7 and 9; 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.
26. Now, marking E5 would force G6 (the last open cell in region 5), and then row 7 would be left with no open cell for its marker. So E5 can't be the marker there.
27. From there, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region. Placing here also clears its 1 touching neighbour.
28. Following that, place a marker at G5. Region has exactly one open cell left.
29. First, 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.
30. Finally, place a marker at H7. Region has exactly one open cell left.

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