QED Logic quod erat demonstrandum
Logic Ascent

How to solve this 6×6 logic puzzle

6×6 · sharp · 17 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 sharp (difficulty score 1057). Solving it from scratch takes 21 logical steps, using 7 techniques: Single cell, Row/column exclusion, 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 E2, 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 B3 and B4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  4. After that, if B1 were the marker, it would force C3 (the last open cell in region 3), then F2 (the last open cell in region 2), then A4 (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.
  5. Now, marking D1 would force F2 (the last open cell in row 2), then A3 (the last open cell in row 3), then E4 (the last open cell in row 4), and then column C would be left with no open cell for its marker. So D1 can't be the marker there.
  6. From there, marking B2 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
  7. Following that, every open cell in column B 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.
  8. First, 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.
  9. Next, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
  10. Then, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
  11. After that, together, two regions have open cells only in columns E and F; 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.
  12. Now, 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.
  13. From there, 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.
  14. Following that, place a marker at C1. Region has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
  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 E4. Region has exactly one open cell left.
  17. 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. ∎