Puzzle Pack #2 Puzzle 28 Answer

Gabe’s clue says everyone has at least 2 criminal neighbors, so Alice’s neighbor group can have at most 1 innocent. Alice’s neighbors already include 1 known innocent, Gabe. The only other people there whose status is not fixed are Ben and Flora, so neither of them can be innocent without putting more than 1 innocent in Alice’s neighbors. So Ben and Flora must be criminal.

Flora’s clue says Olive and Xia have exactly one innocent neighbor in common, and that shared group is only Salil and Uma. So among Salil and Uma, exactly one must be innocent. Now test Xia as innocent. Gabe’s clue says everyone has at least 2 criminal neighbors, so Xia would need at least two criminal neighbors around her while Salil and Uma still have to supply exactly one innocent in that shared pair. Those requirements cannot all be met together. So Xia must be criminal.

Xia's clue says that the two criminals below Ben are connected. Below Ben are Gabe, Lisa, Rob, and Wally, and Gabe is already known to be innocent, so the clue can only act on Lisa, Rob, and Wally. With exactly two criminals in that group needing to form one orthogonally connected pair, Rob is the person that has to be one of them. So Rob must be criminal.

Olive and Xia’s common neighbors are exactly Salil and Uma, and Flora’s clue says exactly one of those two is innocent. Rob’s clue says row 4 has an odd number of criminals, and row 4 currently already has 1 known criminal, with Paula, Salil, and Uma still open. If Paula were innocent, then Salil and Uma would have to satisfy both clues by themselves, but that cannot be done. So Paula cannot be innocent. That makes Paula criminal.