QED Logic quod erat demonstrandum

A Hard 7×7 Puzzle Solved with Forced Chain in 16 Steps

7×7 · hard · 16 steps

Play this board yourself →
A
B
C
D
E
F
G
1
2
3
4
5
6
7
1
×
×
2×
×
×
×
×
×
×
×
×
3×
×
4×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
5×
×
×
6
×
×
×
×
7×
×
×
×
×
×
×

This 7×7 board rates as hard (difficulty score 1010). Solving it from scratch takes 16 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 6 and 7; 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, 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.
  3. Then, every remaining candidate in this region touches F3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  4. After that, every remaining candidate in this region touches C3 and C4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  5. Now, every remaining candidate in this region touches F4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  6. From there, if B1 were the marker, it would force D4 (the last open cell in region 4), then A7 (the last open cell in region 6), and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
  7. Following that, that one can't be the marker: at C1 it would force A7 (the last open cell in column A), then G6 (the last open cell in region 7), then E4 (the last open cell in region 3), and row 5 would be left with no open cell for its marker.
  8. First, 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.
  9. Next, 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 2 touching neighbours.
  10. Then, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
  11. After that, place a marker at F5. Region has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region.
  12. Now, place a marker at B6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region. Placing here also clears its 1 touching neighbour.
  13. From there, place a marker at E7. Region has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region.
  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 G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region.
  16. Finally, place a marker at A1. Region has exactly one open cell left.

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