A Severe 6×6 Puzzle Solved with Forced Chain in 27 Steps

This 6×6 board rates as severe (difficulty score 1444). Solving it from scratch takes 27 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, marking D1 would force F2 (the last open cell in row 2), then E4 (the last open cell in region 5), then C3 (the last open cell in region 3), and the chain continues, and then row 6 would be left with no open cell for its marker. So D1 can't be the marker there.
2. Next, if E1 were the marker, it would force F5 (the last open cell in region 5), then D6 (the last open cell in region 6), then C3 (the last open cell in region 3), and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
3. Then, 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.
4. After that, every remaining candidate in this region touches D5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, that one can't be the marker: at A2 it would force F1 (the last open cell in row 1), and row 4 would be left with no open cell for its marker.
6. From there, that one can't be the marker: at F1 it would force B2 (the last open cell in region 1), then C5 (the last open cell in column C), then E4 (the last open cell in region 5), and row 3 would be left with no open cell for its marker.
7. Following that, 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.
8. First, every remaining candidate in this region touches E3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
9. Next, if C2 were the marker, it would force A1 (the last open cell in region 1), then B6 (the last open cell in region 4), and region 6 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 D2 it would force E4 (the last open cell in region 5), and column F would be left with no open cell for its marker.
11. After that, marking E2 would force F4 (the last open cell in region 2), and then region 5 would be left with no open cell for its marker. So E2 can't be the marker there.
12. Now, place a marker at F2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
13. From there, if A3 were the marker, it would force C4 (the last open cell in region 3), then B1 (the last open cell in region 1), then E5 (the last open cell in region 5), and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, that one can't be the marker: at C1 it would force D3 (the last open cell in row 3), then E5 (the last open cell in region 5), and column B would be left with no open cell for its marker.
15. First, that one can't be the marker: place it at B3, and region 3 would be left with no open cell for its marker.
16. Next, 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.
17. Then, every remaining candidate in this region touches D4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
18. After that, 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.
19. Now, every remaining candidate in this region touches B5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
20. From there, 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.
21. Following that, marking A1 would force B6 (the last open cell in region 4), and then region 6 would be left with no open cell for its marker. So A1 can't be the marker there.
22. First, place a marker at B1. Region has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
23. Next, if D3 were the marker, it would force E5 (the last open cell in region 5), and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
24. Then, place a marker at C3. Region has exactly one open cell left. Placing here clears the rest of row 3, column C, and its region.
25. After 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. Placing here also clears its 1 touching neighbour.
26. Now, 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.
27. Finally, place a marker at A5. Region has exactly one open cell left.

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