A Severe 6×6 Puzzle Solved with Forced Chain in 19 Steps

This 6×6 board rates as severe (difficulty score 1160). Solving it from scratch takes 19 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, 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 6, so its marker has to land there; that clears every other open cell in the row.
2. Next, every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
3. Then, 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.
4. After that, if B1 were the marker, region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. Now, every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
6. From there, that one can't be the marker: at C1 it would force F2 (the last open cell in row 2), then D3 (the last open cell in region 4), then E6 (the last open cell in region 6), and the chain continues, and column A would be left with no open cell for its marker.
7. Following that, marking D1 would force E6 (the last open cell in region 6), then C3 (the last open cell in region 4), then B5 (the last open cell in region 3), and then column A would be left with no open cell for its marker. So D1 can't be the marker there.
8. First, if E1 were the marker, it would force D6 (the last open cell in region 6), then C3 (the last open cell in region 4), and row 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, 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.
10. Then, 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.
11. After that, every remaining candidate in this region touches C4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
12. Now, that one can't be the marker: at F1 it would force B3 cleared (together, two regions have open cells only in rows 2 and 3), then C5 cleared (every remaining candidate in this region touches C5), then D4 (the last open cell in region 5), and column C would be left with no open cell for its marker.
13. From there, place a marker at A1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column A, and its region.
14. Following that, if F2 were the marker, it would force E6 (the last open cell in column E), then B3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then D4 cleared (every remaining candidate in this region touches D4), and the chain continues, and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
15. First, place a marker at D2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
16. Next, place a marker at C5. Region has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region. Placing here also clears its 1 touching neighbour.
17. Then, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region.
18. After that, place a marker at F4. Region has exactly one open cell left.
19. Finally, place a marker at E6. Region has exactly one open cell left.

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