A Hard 6×6 Puzzle Solved with Forced Chain in 15 Steps

This 6×6 board rates as hard (difficulty score 1059). Solving it from scratch takes 15 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 columns E and F; 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.
2. Next, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
3. Then, 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.
4. After that, every remaining candidate in this region touches C5 and E5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, if B1 were the marker, it would force F2 (the last open cell in row 2), then E6 (the last open cell in region 6), then D4 (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.
6. From there, that one can't be the marker: at C1 it would force B4 (the last open cell in region 2), then A6 (the last open cell in region 4), then D5 (the last open cell in region 5), and region 6 would be left with no open cell for its marker.
7. Following that, if E1 were the marker, it would force C2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A3 cleared (every remaining candidate in this region touches A3), then C3 (the last open cell in row 3), 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.
8. First, that one can't be the marker: at F1 it would force E6 (the last open cell in region 6), then D4 (the last open cell in region 5), then C2 (the last open cell in region 2), and region 1 would be left with no open cell for its marker.
9. Next, 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.
10. 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.
11. After that, place a marker at D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region. Placing here also clears its 2 touching neighbours.
12. Now, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
13. From there, place a marker at F3. Region has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region.
14. Following that, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region.
15. 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. ∎