QED Logic quod erat demonstrandum
Logic Ascent

How to solve this 8×8 logic puzzle

8×8 · sharp · 17 steps

A
B
C
D
E
F
G
H
1
2
3
4
5
6
7
8
1
2
3
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4
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5
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6
7🐱
8🐱
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This 8×8 board rates as sharp (difficulty score 993). Solving it from scratch takes 26 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, 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. Placing here also clears its 4 touching neighbours.
  2. Next, place a marker at A7. Region has exactly one open cell left. Placing here clears the rest of row 7, column A, and its region. Placing here also clears its 1 touching neighbour.
  3. Then, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
  4. After that, together, two regions have open cells only in columns E and C; 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.
  5. Now, 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.
  6. From there, every remaining candidate in this region touches G4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  7. Following that, every open cell left in this region sits in column H, so its marker has to land there; that clears every other open cell in the column.
  8. First, every remaining candidate in this region touches G5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  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, if B1 were the marker, it would force E2 (the last open cell in region 2), then G3 (the last open cell in region 3), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
  11. After that, if E1 were the marker, it would force C8 (the last open cell in region 6), then G3 (the last open cell in row 3), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
  12. Now, place a marker at G1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
  13. From there, place a marker at F3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region. Placing here also clears its 1 touching neighbour.
  14. Following that, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
  15. First, place a marker at B5. Region has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region.
  16. Next, place a marker at H4. Region has exactly one open cell left.
  17. Finally, place a marker at E8. Region has exactly one open cell left.

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