Puzzle Pack #2 Puzzle 28 Answer

Gabe at B2 is innocent, so his clue is true and applies to every person on the board. Ben at B1 and Flora at A2 are corner-adjacent edge people, and each of them has only five neighbors in total; one of those neighbors is Gabe himself. Since everyone must have at least 2 criminal neighbors, Ben and Flora each need at least 2 criminals among their own neighbors, and with Gabe being an innocent common neighbor, the only way this local requirement is forced in this step is that Ben and Flora themselves must be criminal. Therefore, we can determine that B1 is CRIMINAL and A2 is CRIMINAL.

Olive at D3 and Xia at C5 have only two common neighbors: C4 and D4. Flora’s clue says they have only one innocent neighbor in common, so among C4 and D4 exactly one is innocent and the other is criminal. Gabe’s clue says everyone has at least 2 criminal neighbors. Xia’s only neighbors are B4, C4, D4, B5, and D5, and because exactly one of C4 and D4 is criminal, Xia can reach the required total of at least 2 criminal neighbors only if Xia is criminal herself or if enough of B4, B5, and D5 are criminal. But the clue about Olive and Xia already fixes the common pair C4 and D4 to contain exactly one innocent, so Xia cannot be the one innocent needed in that neighborhood arrangement. Therefore, we can determine that C5 is CRIMINAL.

Ben is at B1, so the people below Ben are B2, B3, B4, and B5. The clue says both criminals below Ben are connected, so among those four people there are exactly two criminals, and they must touch through a continuous up-and-down chain in that same column. We already know B2 is innocent, so the two criminals below Ben cannot include B2. That means the only way to place exactly two connected criminals in B2 through B5 is B3 and B4 together, or B4 and B5 together. In either case, B4 has to be one of the criminals. Therefore, we can determine that B4 is CRIMINAL.

Olive is at D3 and Xia is at C5, so their common neighbors are only C4 and D4. Flora’s clue says Olive and Xia have only one innocent neighbor in common, which means exactly one of C4 Salil and D4 Uma is innocent. Rob’s clue says row 4 has an odd number of criminals, and since B4 Rob is already a criminal, the other three people in row 4 must contain an even number of criminals. That forces A4 Paula to be a criminal, with C4 and D4 contributing one criminal and one innocent between them. Therefore, we can determine that A4 is CRIMINAL.