QED Logic quod erat demonstrandum

A Sharp 8×8 Puzzle Solved with Forced Chain in 27 Steps

8×8 · sharp · 19 steps

Play this board yourself →
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This 8×8 board rates as sharp (difficulty score 1017). Solving it from scratch takes 27 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, together, two regions have open cells only in rows 1 and 2; 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.
  2. Next, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region. Placing here also clears its 2 touching neighbours.
  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, every remaining candidate in this region touches G2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  5. Now, that one can't be the marker: at F1 it would force H2 (the last open cell in region 3), then E4 (the last open cell in region 5), then G7 (the last open cell in region 7), and the chain continues, and column D would be left with no open cell for its marker.
  6. From there, marking G1 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
  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 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.
  9. Next, every open cell left in this region sits in row 7, so its marker has to land there; that clears every other open cell in the row.
  10. Then, 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.
  11. After that, place a marker at G7. Region has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region.
  12. Now, every open cell left in this region sits in row 5, so its marker has to land there; that clears every other open cell in the row.
  13. From there, every remaining candidate in this region touches C6 and D6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  14. Following that, place a marker at A6. Region has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
  15. First, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region.
  16. Next, place a marker at D5. Region has exactly one open cell left. Placing here also clears its 1 touching neighbour.
  17. Then, place a marker at F4. Region has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region.
  18. After that, place a marker at E1. Region has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
  19. Finally, place a marker at H2. Region has exactly one open cell left.

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