# Solution to The Greatest Jigsaw

#### Back to Puzzle

#### Answer: CAPTAIN HOOK

#### by Anderson Wang, Lillian McKinley, Mitchell Lee

Art by Lillian McKinley

Production by Up In Pieces

We first solve the jigsaw puzzle. The assembled image looks like this:

The background image is quite literally a red herring, suggesting that it’s not important, so we turn to the digits and symbols. Each one is located in the center of a jigsaw piece, suggesting that they should map to the individual pieces. From here, we see that there are 11 unique symbols: the 10 digits from 0–9, plus a capital sigma. The jigsaw puzzle itself has dimensions 11×11, which means that this could be a logic puzzle where you have to assign a symbol to each jigsaw piece so that every row and column contains all symbols (i.e., a Latin square). This is also encouraged by the fact that no existing symbol repeats in a row or column.

If we try to solve the Latin square as is, we’ll quickly find that there is not nearly enough information to have a unique solution (indeed, we can’t even place a single symbol), so there must be additional constraints somewhere. The additional constraints aren’t coming from the red herring, and the only other data from the jigsaw is the shapes of the pieces and how they connect to each other. This leads to the first main aha: the directions the pieces “poke into” each other is important! In particular, the pieces with 9's have all 4 sides “poking out”, while the pieces with 0's have all 4 sides “poking in”, implying that this is on the right track.

We also notice that the smaller numbers like 1 and 2 generally have more sides poking in, while the larger numbers generally have more sides poking out. This, combined with the title (and perhaps the layout of the given numbers, which looks like a less-than sign), leads to the second aha, which is that if piece X pokes into piece Y, then the digit on X must be greater than the digit on Y. This property holds for the existing adjacent digits, and additionally there are no “cycles” of pieces that point into each other, strongly suggesting that this is the correct interpretation.

The only uncertainty left in the rules is how sigma works. It can’t represent a single number, because among the given symbols, a 3 points into a sigma (in the 4th row), and a sigma also points into a 3 (in the 10th row). So, the safest assumption is that there are no constraints on the sigmas: they can point in any direction. We can make additional assumptions like “sigma is always less than 9 or greater than 0” or “if X points into sigma which points into Y, then X is greater than Y”—it turns out all of these are true due to how extraction works, but they are not needed to uniquely solve the logic puzzle and we don’t expect the typical solve path to use them (though if you do accidentally make such an assumption, it won’t hurt you).

We now solve the logic puzzle—a walkthrough is included in the video below, and if you want to try just the logic puzzle itself, here’s a pen-pa link with all the data transcribed (sigmas are represented by “E”).

This is the solution:

We see that every sigma can be replaced with at least one number such that the adjacent inequalities are satisfied—for example, the sigma in the second row is greater than 2, 3, and 5 and less than 8 so it can be 6 or 7. The fact that the symbol is a sigma suggests summing, so we add up the possible replacements for each sigma (e.g., for the second row, 6+7=13). In row order, these are:

Possibilities | Sum | Letter |
---|---|---|

1,2,3,4 | 10 | J |

6,7 | 13 | M |

2 | 2 | B |

0,1 | 1 | A |

5,6,7 | 18 | R |

3,4,5,6 | 18 | R |

2,3,4 | 9 | I |

5 | 5 | E |

1,2,3 | 6 | F |

4,5,6 | 15 | O |

2,3 | 5 | E |

Converting these sums to letters alphanumerically gives the phrase JMBARRIEFOE. J. M. Barrie is most known for being the creator of Peter Pan, whose foe is CAPTAIN HOOK.

(Note: another possibility is to sum all the neighbors instead of the possible digit replacements, but this is ruled out by the sigma in the 8th row having sum 9+8+6+4=27 which is larger than 26.)