An Expert 6×6 Puzzle Solved with Forced Chain in 19 Steps

This 6×6 board rates as expert (difficulty score 1192). Solving it from scratch takes 19 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, together, two regions have open cells only in rows 5 and 6; 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.
2. Next, 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.
3. Then, 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.
4. After that, marking A1 would force E3 and E4 cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), then C3 cleared (every remaining candidate in this region touches C3), then C5 (the last open cell in column C), and then row 4 would be left with no open cell for its marker. So A1 can't be the marker there.
5. Now, if B1 were the marker, it would force C3 (the last open cell in region 3), and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
6. From there, every open cell left in this region sits in row 2, so its marker has to land there; that clears every other open cell in the row.
7. Following that, that one can't be the marker: at C1 it would force A2 (the last open cell in region 1), then B4 (the last open cell in region 3), and region 5 would be left with no open cell for its marker.
8. First, if E1 were the marker, it would force B3 cleared (every remaining candidate in this region touches B3), then C3 cleared (every remaining candidate in this region touches C3), then B4 (the last open cell in region 3), 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.
9. Next, that one can't be the marker: at F1 it would force E6 (the last open cell in region 6), then A2 (the last open cell in column A), then C3 cleared (every remaining candidate in this region touches C3), and the chain continues, and row 5 would be left with no open cell for its marker.
10. Then, place a marker at D1. Region has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region. Placing here also clears its 1 touching neighbour.
11. 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.
12. Now, every remaining candidate in this region touches B3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
13. From there, marking B2 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
14. Following 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.
15. First, that one can't be the marker: at E3 it would force B4 (the last open cell in region 3), and column C would be left with no open cell for its marker.
16. Next, 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. Placing here also clears its 1 touching neighbour.
17. 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.
18. After that, 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.
19. Finally, place a marker at B5. Region has exactly one open cell left.

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