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

This 6×6 board rates as expert (difficulty score 1133). Solving it from scratch takes 19 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, every remaining candidate in this region touches B2, 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 B5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, marking A1 would force C2 (the last open cell in region 2), then B4 (the last open cell in region 5), and then region 6 would be left with no open cell for its marker. So A1 can't be the marker there.
4. After that, if B1 were the marker, region 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. 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.
6. From there, that one can't be the marker: at C1 it would force D6, E6 and F6 cleared (every open cell left in this region sits in row 6, so its marker has to land there), then E4 and E5 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 F2, F3, F4, and 1 more cleared (every open cell in column E belongs to the same region, so that region's marker has to be somewhere in this column), and column F would be left with no open cell for its marker.
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, if E1 were the marker, it would force A2 (the last open cell in row 2), then B4 (the last open cell in region 5), and region 6 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 A2 it would force B4 (the last open cell in region 5), and region 6 would be left with no open cell for its marker.
10. Then, every open cell left in this region sits in row 3, so its marker has to land there; that clears every other open cell in the row.
11. After that, every open cell in row 2 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.
12. Now, every open cell in column E 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.
13. From there, every open cell in row 6 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.
14. Following that, place a marker at C3. Column 3 has exactly one open cell left. Placing here clears the rest of row 3, column C, and its region. Placing here also clears its 2 touching neighbours.
15. First, place a marker at F2. Region has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
16. Next, place a marker at D1. Region has exactly one open cell left.
17. Then, 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. Placing here also clears its 1 touching neighbour.
18. After that, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
19. Finally, place a marker at E5. Region has exactly one open cell left.

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