How to solve this 6×6 logic puzzle

6×6 · severe · 22 steps

🐱

4🐱

🐱

This 6×6 board rates as severe (difficulty score 1281). Solving it from scratch takes 28 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.

First, every remaining candidate in this region touches E3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, every remaining candidate in this region touches E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, that one can't be the marker: at C1 it would force B3 (the last open cell in column B), then E2 (the last open cell in region 4), and region 2 would be left with no open cell for its marker.
After that, marking D1 would immediately leave region 4 with no open cell for its marker, so it can't be the marker there.
Now, if E1 were the marker, it would force D3 (the last open cell in region 4), and row 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, that one can't be the marker: at F1 it would force E6 (the last open cell in column E), then A5 (the last open cell in region 5), then B2 (the last open cell in region 1), and the chain continues, and row 4 would be left with no open cell for its marker.
Following that, 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.
First, if C2 were the marker, region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, marking E2 would immediately leave region 2 with no open cell for its marker, so it can't be the marker there.
Then, 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.
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 B4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
From there, if F2 were the marker, it would force D3 (the last open cell in region 4), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, place a marker at D2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
First, every remaining candidate in this region touches F4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, 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.
Then, that one can't be the marker: at B3 it would force A1 (the last open cell in region 1), then E4 (the last open cell in region 2), and row 5 would be left with no open cell for its marker.
After 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. Placing here also clears its 1 touching neighbour.
Now, 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.
From there, 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.
Following that, place a marker at A5. Region has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region.
Finally, place a marker at B1. Region has exactly one open cell left.

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