An Expert 10×10 Puzzle Solved with Forced Chain in 36 Steps

This 10×10 board rates as expert (difficulty score 1790). Solving it from scratch takes 36 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 row 5, so its marker has to land there; that clears every other open cell in the row.
2. Next, every remaining candidate in this region touches G4 and G6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
3. Then, if H1 were the marker, it would force F4, F5, F6, and 4 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then G5 (the last open cell in region 7), then A2, B2, J2, and 5 more cleared (together, two regions have open cells only in rows 2 and 3), and the chain continues, and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, that one can't be the marker: at I1 it would force J6, J7, J8, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A2, B2, A3, and 3 more cleared (together, two regions have open cells only in rows 2 and 3), then J4 (the last open cell in region 4), and the chain continues, and region 10 would be left with no open cell for its marker.
5. Now, marking J1 would force A2, B2, H2, and 5 more cleared (together, two regions have open cells only in rows 2 and 3), then A4, B4, C4, and 3 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then A6, B6, F6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and then region 10 would be left with no open cell for its marker. So J1 can't be the marker there.
6. From there, that one can't be the marker: at A2 it would force G3, H3, I3, and 1 more cleared (together, two regions have open cells only in rows 1 and 3), then J4 (the last open cell in region 4), and region 3 would be left with no open cell for its marker.
7. Following that, marking B2 would force G3, H3, I3, and 1 more cleared (together, two regions have open cells only in rows 1 and 3), then J4 (the last open cell in region 4), and then region 3 would be left with no open cell for its marker. So B2 can't be the marker there.
8. First, that one can't be the marker: at D2 it would force F3 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then A3, B3, A4, and 4 more cleared (together, two regions have open cells only in rows 3 and 4), then A6, B6, F6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and column C would be left with no open cell for its marker.
9. Next, marking E2 would force G1 (the last open cell in region 2), then A3, B3, C3, and 5 more cleared (together, two regions have open cells only in rows 3 and 4), then A6, B6, F6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and then column C would be left with no open cell for its marker. So E2 can't be the marker there.
10. Then, if F2 were the marker, it would force D3 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then A3, B3, C3, and 5 more cleared (together, two regions have open cells only in rows 3 and 4), then A6, B6, H6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and region 10 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 G2 it would force A4, B4, C4, and 4 more cleared (every open cell left in this region sits in row 4, so its marker has to land there), then A3, B3, C3, and 2 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A6, B6, F6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and region 10 would be left with no open cell for its marker.
12. Now, that one can't be the marker: at C1 it would force F3 (the last open cell in region 2), then A4, B4 and D4 cleared (together, two regions have open cells only in rows 2 and 4), then A6, B6, H6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and region 10 would be left with no open cell for its marker.
13. From there, marking D1 would force F3 (the last open cell in region 2), then A4, B4 and C4 cleared (together, two regions have open cells only in rows 2 and 4), then A6, B6, H6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and then column C would be left with no open cell for its marker. So D1 can't be the marker there.
14. Following that, if I2 were the marker, it would force A3, B3, C3, and 1 more cleared (together, two regions have open cells only in rows 1 and 3), then H4 (the last open cell in region 3), then F5 (the last open cell in region 7), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, if A3 were the marker, it would force B1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then C2 (the last open cell in region 1), then J4 (the last open cell in region 4), and region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
16. Next, that one can't be the marker: at B3 it would force A1 (the last open cell in region 1), and region 2 would be left with no open cell for its marker.
17. Then, marking C3 would force A1 and B1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), and then region 1 would be left with no open cell for its marker. So C3 can't be the marker there.
18. After that, if D3 were the marker, it would force J6, J7, J8, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A4, B4 and F4 cleared (together, two regions have open cells only in rows 2 and 4), then A6, B6, F6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, that one can't be the marker: at E3 it would force J6, J7, J8, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A4, B4 and C4 cleared (together, two regions have open cells only in rows 2 and 4), then A6, B6, F6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and column C would be left with no open cell for its marker.
20. From there, that one can't be the marker: at F1 it would force C2 (the last open cell in region 1), then A4, B4, D4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), then A6, B6, H6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and row 3 would be left with no open cell for its marker.
21. Following that, marking G1 would force C2 (the last open cell in region 1), then A4, B4, D4, and 2 more cleared (together, two regions have open cells only in rows 3 and 4), then A6, B6, F6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and then row 3 would be left with no open cell for its marker. So G1 can't be the marker there.
22. First, marking H2 would force E1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then F3 (the last open cell in region 2), then J4 (the last open cell in region 4), and the chain continues, and then region 10 would be left with no open cell for its marker. So H2 can't be the marker there.
23. Next, marking A1 would force F3 (the last open cell in region 2), then J2 (the last open cell in row 2), then B4, C4 and D4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), and the chain continues, and then region 10 would be left with no open cell for its marker. So A1 can't be the marker there.
24. Then, if B1 were the marker, it would force F3 (the last open cell in region 2), then J2 (the last open cell in row 2), then A4, C4 and D4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), and the chain continues, and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
25. After that, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
26. Now, 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.
27. From there, together, two regions have open cells only in rows 3 and 4; 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, place a marker at D6. Region has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region.
29. First, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region. Placing here also clears its 1 touching neighbour.
30. Next, place a marker at H8. Region has exactly one open cell left. Placing here clears the rest of row 8, column H, and its region. Placing here also clears its 2 touching neighbours.
31. Then, every remaining candidate in this region touches I4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
32. After that, 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.
33. Now, place a marker at J4. Region has exactly one open cell left. Placing here clears the rest of row 4, column J, and its region.
34. From there, place a marker at F5. Region has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region.
35. Following 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.
36. 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. ∎