A Sharp 6×6 Puzzle Solved with Forced Chain in 17 Steps

This 6×6 board rates as sharp (difficulty score 1093). Solving it from scratch takes 17 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Line confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, 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.
2. Next, together, two regions have open cells only in columns B and A; 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.
3. Then, 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.
4. After that, 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.
5. Now, that one can't be the marker: at C1 it would force D3 (the last open cell in region 4), and region 2 would be left with no open cell for its marker.
6. From there, every open cell in column C 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.
7. Following that, every remaining candidate in this region touches E2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
8. First, 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.
9. Next, if E1 were the marker, it would force B2 (the last open cell in row 2), then C4 (the last open cell in region 4), then F3 (the last open cell in region 3), and row 5 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 F1 it would force B2 (the last open cell in row 2), then C4 (the last open cell in region 4), and row 5 would be left with no open cell for its marker.
11. After that, 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.
12. Now, if F2 were the marker, it would force A3 (the last open cell in region 2), then C4 (the last open cell in region 4), and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, place a marker at B2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column B, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at C4. Region has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region.
15. First, place a marker at A6. Region has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
16. Next, 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.
17. Finally, place a marker at E3. Region has exactly one open cell left.

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