QED Logic quod erat demonstrandum
Logic Ascent

How to solve this 8×8 logic puzzle

8×8 · sharp · 19 steps

A
B
C
D
E
F
G
H
1
2
3
4
5
6
7
8
1
🐱
2🐱
3
4
5
🐱
6
🐱
🐱
7🐱
🐱
8
🐱

This 8×8 board rates as sharp (difficulty score 1056). Solving it from scratch takes 28 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 G3, 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 G7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  4. After that, marking A1 would force D6 (the last open cell in column D), then C8 (the last open cell in region 6), then H7 (the last open cell in region 8), and the chain continues, and then column F would be left with no open cell for its marker. So A1 can't be the marker there.
  5. Now, every open cell in column A 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.
  6. From there, if B1 were the marker, it would force D6 (the last open cell in column D), then C8 (the last open cell in region 6), then H7 (the last open cell in region 8), and the chain continues, and column F would end up with no open cell for its marker, which can't happen. So it can't go 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, place a marker at C1. Column 3 has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
  9. Next, place a marker at D6. Column 4 has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region. Placing here also clears its 1 touching neighbour.
  10. Then, every open cell in column E 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.
  11. After that, place a marker at A5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region.
  12. Now, every remaining candidate in this region touches F3 and F4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  13. From there, every open cell left in this region sits in row 2, so its marker has to land there; that clears every other open cell in the row.
  14. Following that, that one can't be the marker: place it at G2, and region 3 would be left with no open cell for its marker.
  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. Placing here also clears its 1 touching neighbour.
  16. Next, place a marker at E4. Region has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region.
  17. Then, place a marker at H3. Region has exactly one open cell left. Placing here clears the rest of row 3, column H, and its region.
  18. After that, place a marker at G8. Region has exactly one open cell left. Placing here clears the rest of row 8, column G, and its region.
  19. Finally, place a marker at B7. Region has exactly one open cell left.

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