An Expert 10×10 Puzzle Solved with Forced Chain in 33 Steps

This 10×10 board rates as expert (difficulty score 1507). Solving it from scratch takes 33 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 remaining candidate in this region touches B2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, marking A1 would force D5, D6, D7, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C8, C9, C10, and 3 more cleared (together, two regions have open cells only in columns C and B), then G8, H8, I8, and 1 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and then row 10 would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, if B1 were the marker, it would force D5, D6, D7, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C3, C4, C8, and 2 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then A9 and A10 cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, place a marker at A2. Region has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region. Placing here also clears its 1 touching neighbour.
5. Now, that one can't be the marker: at C1 it would force G3 (the last open cell in region 3), then B7, B8, B9, and 1 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then D5 (the last open cell in region 8), and the chain continues, and row 10 would be left with no open cell for its marker.
6. From there, marking D1 would force G3 (the last open cell in region 3), then B8, B9, B10, and 3 more cleared (together, two regions have open cells only in columns B and C), then H8, I8 and J8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and then row 10 would be left with no open cell for its marker. So D1 can't be the marker there.
7. Following that, that one can't be the marker: at F1 it would force D5, D6, D7, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C8, C9, C10, and 3 more cleared (together, two regions have open cells only in columns C and B), then E8 (the last open cell in region 9), and the chain continues, and row 10 would be left with no open cell for its marker.
8. First, marking G1 would force D5, D6, D7, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C8, C9, C10, and 3 more cleared (together, two regions have open cells only in columns C and B), then H8, I8 and J8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and then row 10 would be left with no open cell for its marker. So G1 can't be the marker there.
9. Next, if H1 were the marker, it would force D5, D6, D7, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C8, C9, C10, and 3 more cleared (together, two regions have open cells only in columns C and B), then G8, I8 and J8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, that one can't be the marker: at I1 it would force D5, D6, D7, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then J9 and J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), then C8, C9, C10, and 3 more cleared (together, two regions have open cells only in columns C and B), and the chain continues, and row 10 would be left with no open cell for its marker.
11. After that, marking J1 would force D5, D6, D7, and 3 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then I3, I4, I9, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then C8, C9, C10, and 3 more cleared (together, two regions have open cells only in columns C and B), and the chain continues, and then row 10 would be left with no open cell for its marker. So J1 can't be the marker there.
12. Now, 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. Placing here also clears its 2 touching neighbours.
13. From there, place a marker at E1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
14. 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.
15. First, every remaining candidate in this region touches C5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
16. Next, that one can't be the marker: at C4 it would force B7 (the last open cell in region 8), then H8, H9 and H10 cleared (every open cell left in this region sits in column H, so its marker has to land there), then D9 and D10 cleared (every open cell left in this region sits in column D, so its marker has to land there), and the chain continues, and row 10 would be left with no open cell for its marker.
17. Then, 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.
18. After that, together, two regions have open cells only in columns D and C; 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.
19. Now, 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.
20. From there, that one can't be the marker: at I4 it would force J9 and J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), then D5, C6 and J6 cleared (together, two regions have open cells only in rows 5 and 6), then C7 (the last open cell in region 8), and column D would be left with no open cell for its marker.
21. Following that, 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.
22. First, every remaining candidate in this region touches I6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
23. Next, marking J4 would force D7, C9 and C10 cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), then I9 (the last open cell in row 9), and then row 10 would be left with no open cell for its marker. So J4 can't be the marker there.
24. Then, place a marker at B4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column B, and its region.
25. After that, if F5 were the marker, it would force C8, C9 and C10 cleared (every open cell left in this region sits in column C, so its marker has to land there), then I10 and J10 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 row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
26. Now, place a marker at F6. Region has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region.
27. From there, together, two regions have open cells only in rows 5 and 7; 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.
28. Following that, 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.
29. First, place a marker at D5. Column 4 has exactly one open cell left. Placing here clears the rest of row 5, column D, and its region.
30. Next, place a marker at H7. Region has exactly one open cell left. Placing here also clears its 1 touching neighbour.
31. Then, place a marker at J8. Region has exactly one open cell left. Placing here clears the rest of row 8, column J, and its region. Placing here also clears its 1 touching neighbour.
32. After that, place a marker at I10. Region has exactly one open cell left. Placing here clears the rest of row 10, column I, and its region.
33. Finally, place a marker at C9. Region has exactly one open cell left.

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