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 1083). Solving it from scratch takes 24 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region 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 row 6, so its marker has to land there; that clears every other open cell in the row.
  2. Next, 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.
  3. Then, together, two regions have open cells only in rows 4 and 5; 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.
  4. After 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.
  5. 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.
  6. From there, that one can't be the marker: at C1 it would force F2 (the last open cell in row 2), then D5 (the last open cell in region 5), and row 6 would be left with no open cell for its marker.
  7. Following that, marking D1 would immediately leave region 1 with no open cell for its marker, so it can't be the marker there.
  8. First, if E1 were the marker, it would force C2 (the last open cell in region 1), and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
  9. Next, 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.
  10. Then, place a marker at A4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region. Placing here also clears its 1 touching neighbour.
  11. After that, every remaining candidate in this region touches D6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  12. Now, 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. Placing here also clears its 1 touching neighbour.
  13. From there, place a marker at C5. Region has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region.
  14. Following that, that one can't be the marker: at F1 it would force D2 (the last open cell in region 1), and row 3 would be left with no open cell for its marker.
  15. First, place a marker at B1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
  16. Next, place a marker at D3. Region has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
  17. Finally, place a marker at F2. Region has exactly one open cell left.

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