QED Logic quod erat demonstrandum
Logic Ascent

How to solve this 6×6 logic puzzle

6×6 · severe · 19 steps

A
B
C
D
E
F
1
2
3
4
5
6
1🐱
2
3
🐱
🐱
4
5🐱
6🐱
🐱

This 6×6 board rates as severe (difficulty score 1164). Solving it from scratch takes 26 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 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.
  2. Next, every remaining candidate in this region touches A5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  3. Then, if B1 were the marker, it would force A6 (the last open cell in region 6), then E4 (the last open cell in row 4), and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
  4. After that, that one can't be the marker: at C1 it would force D3 (the last open cell in row 3), and row 4 would be left with no open cell for its marker.
  5. Now, marking D1 would force B2 (the last open cell in row 2), then A6 (the last open cell in region 6), then C5 (the last open cell in region 4), and the chain continues, and then row 3 would be left with no open cell for its marker. So D1 can't be the marker there.
  6. From there, if E1 were the marker, region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
  7. Following that, that one can't be the marker: at F1 it would force D2 (the last open cell in region 3), and row 3 would be left with no open cell for its marker.
  8. First, place a marker at A1. Region has exactly one open cell left. Placing here clears the rest of row 1, column A, and its region. Placing here also clears its 1 touching neighbour.
  9. Next, 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.
  10. Then, every remaining candidate in this region touches C5 and C6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  11. After that, every open cell in column C 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.
  12. Now, 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.
  13. From there, if C2 were the marker, it would force D4 (the last open cell in row 4), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
  14. Following 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.
  15. First, place a marker at C3. Row 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.
  16. Next, place a marker at E4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region. Placing here also clears its 1 touching neighbour.
  17. Then, place a marker at F2. Region has exactly one open cell left.
  18. After that, place a marker at D6. Region has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region.
  19. 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.