QED Logic quod erat demonstrandum

A Hard 6×6 Puzzle Solved with Forced Chain in 15 Steps

6×6 · hard · 15 steps

Play this board yourself →
A
B
C
D
E
F
1
2
3
4
5
6
1×
×
×
2×
×
3×
×
×
×
×
×
×
4×
×
×
5×
×
×
×
×
×
×
6×
×
×
×
×
×
×
×

This 6×6 board rates as hard (difficulty score 999). Solving it from scratch takes 15 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, 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 in column F 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.
  2. Next, 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.
  3. Then, together, two regions have open cells only in columns A and B; 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.
  4. After that, every remaining candidate in this region touches F5 and F6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  5. Now, marking D1 would force B4 cleared (every remaining candidate in this region touches B4), then B3 (the last open cell in column B), then C5 (the last open cell in region 4), and the chain continues, and then column E would be left with no open cell for its marker. So D1 can't be the marker there.
  6. From there, that one can't be the marker: at A2 it would force B4 (the last open cell in region 5), then C1 (the last open cell in region 1), then F3 (the last open cell in region 2), and column D would be left with no open cell for its marker.
  7. Following that, every open cell left in this region sits in row 3, so its marker has to land there; that clears every other open cell in the row.
  8. First, every remaining candidate in this region touches A4 and B4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  9. Next, place a marker at B3. Column 2 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.
  10. Then, 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.
  11. After that, place a marker at D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region. Placing here also clears its 1 touching neighbour.
  12. Now, 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.
  13. From there, place a marker at F2. Region has exactly one open cell left.
  14. Following that, 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.
  15. Finally, place a marker at A5. Region has exactly one open cell left.

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