QED Logic quod erat demonstrandum

A Hard 7×7 Puzzle Solved with Forced Chain in 17 Steps

7×7 · hard · 17 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 1051). Solving it from scratch takes 17 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, every remaining candidate in this region touches B1, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  2. Next, every remaining candidate in this region touches G6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
  3. Then, marking A1 would force F4, F5, F6, and 1 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then G7 (the last open cell in region 7), then C5, D5 and E5 cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then row 4 would be left with no open cell for its marker. So A1 can't be the marker there.
  4. After that, 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.
  5. Now, together, two regions have open cells only in rows 1 and 3; 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.
  6. From there, marking D1 would force F3 (the last open cell in region 3), then G5 (the last open cell in region 4), and then region 7 would be left with no open cell for its marker. So D1 can't be the marker there.
  7. Following that, if E1 were the marker, row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
  8. First, together, two regions have open cells only in columns F and G; 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, 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.
  10. Then, every remaining candidate in this region touches E3 and D5 and E5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
  11. After that, 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. Placing here also clears its 1 touching neighbour.
  12. Now, place a marker at A2. Region has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region.
  13. From there, place a marker at F3. Region 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 D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region.
  15. First, 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.
  16. Next, 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.
  17. Finally, place a marker at E6. Region has exactly one open cell left.

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