A Sharp 6×6 Puzzle Solved with Forced Chain in 16 Steps

This 6×6 board rates as sharp (difficulty score 1109). Solving it from scratch takes 16 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, 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.
2. Next, every remaining candidate in this region touches E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, 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.
4. After that, marking A1 would force B6 (the last open cell in column B), then F2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F4 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then column E 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 A3 (the last open cell in region 1), then C4 (the last open cell in region 4), and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
6. From there, that one can't be the marker: at C1 it would force D3 (the last open cell in region 4), then F2 (the last open cell in region 3), then B4 (the last open cell in region 1), and row 5 would be left with no open cell for its marker.
7. Following that, if E1 were the marker, it would force F3 (the last open cell in region 3), then C4 (the last open cell in region 4), then A2 (the last open cell in region 1), and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, that one can't be the marker: at F1 it would force E6 (the last open cell in column E), then A2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A4 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 5 would be left with no open cell for its marker.
9. Next, place a marker at D1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region.
10. Then, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
11. After that, every remaining candidate in this region touches B3 and B4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
12. Now, place a marker at B6. Column 2 has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region.
13. From there, place a marker at F5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at E3. Region has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region.
15. First, place a marker at A2. Region has exactly one open cell left.
16. Finally, place a marker at C4. Region has exactly one open cell left.

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