A Hard 6×6 Puzzle Solved with Forced Chain in 21 Steps

This 6×6 board rates as hard (difficulty score 1114). Solving it from scratch takes 21 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Paired regions, 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 column A, so its marker has to land there; that clears every other open cell in the column.
2. Next, every remaining candidate in this region touches C2, 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, 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.
5. Now, every remaining candidate in this region touches B4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
6. From there, 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.
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 C5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
9. Next, if B1 were the marker, it would force E4 and E5 cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), then E3 (the last open cell in column E), and row 2 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 C1 it would force D5 (the last open cell in region 3), then E3 (the last open cell in region 2), and row 2 would be left with no open cell for its marker.
11. After that, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
12. Now, 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.
13. From there, together, two regions have open cells only in columns C and D; 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.
14. Following that, every remaining candidate in this region touches F2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
15. First, 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.
16. Next, place a marker at F3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region.
17. Then, place a marker at E1. Region has exactly one open cell left.
18. After that, 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.
19. Now, 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.
20. From there, place a marker at A5. Region has exactly one open cell left.
21. Finally, place a marker at D6. Region has exactly one open cell left.

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