How to solve this 6×6 logic puzzle

6×6 · severe · 19 steps

🐱

This 6×6 board rates as severe (difficulty score 1120). Solving it from scratch takes 24 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.

First, 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.
Next, every open cell in row 5 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.
Then, 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.
After that, every remaining candidate in this region touches C3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, 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.
From there, every open cell left in this region sits in column F, so its marker has to land there; that clears every other open cell in the column.
Following that, every remaining candidate in this region touches B4 and A6 and B6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
First, marking A1 would force B5 (the last open cell in region 5), and then column C would be left with no open cell for its marker. So A1 can't be the marker there.
Next, that one can't be the marker: at C1 it would force D3 (the last open cell in region 2), and region 4 would be left with no open cell for its marker.
Then, marking D1 would force A2 (the last open cell in region 1), then C4 (the last open cell in region 4), and then row 5 would be left with no open cell for its marker. So D1 can't be the marker there.
After that, if E1 were the marker, it would force F3 (the last open cell in row 3), then B5 (the last open cell in column B), then D4 (the last open cell in region 4), and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, place a marker at E6. Column 5 has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region.
From there, 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.
Following that, 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.
First, place a marker at C4. Column 3 has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region. Placing here also clears its 2 touching neighbours.
Next, place a marker at D2. Region has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
Then, place a marker at B1. Region has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
After that, place a marker at F3. Region has exactly one open cell left.
Finally, place a marker at A5. Region has exactly one open cell left.

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