Puzzle Pack #1 Puzzle 37 Answer

Ronald is at B4, and Wally is at B5, so Wally is indeed one of Ronald’s neighbors. Sofia’s clue says that Wally is one of Ronald’s 6 criminal neighbors, which directly states that Wally is criminal. Therefore, we can determine that B5 is CRIMINAL.

Ronald is at B4, and his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Sofia says Wally is one of Ronald’s 6 criminal neighbors, so B5 is criminal and Ronald has exactly 6 criminal neighbors among those 8 positions. Since Sofia herself is at C4 and is already known innocent, the only other non-criminal neighbor Ronald can have must be one of those remaining seven spots. Now look at column A: its people are A1, A2, A3, A4, and A5. Wally says all criminals in column A are connected, and A5 is one of Ronald’s neighbors while A4 is directly above A5 and is also one of Ronald’s neighbors. If A4 were innocent, then any criminal at A5 or higher in column A would be separated across that innocent space, so A4 has to be part of the criminal chain in column A. Therefore, we can determine that A4 is CRIMINAL.

Ronald is at B4, and his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Sofia says Wally at B5 is one of Ronald's 6 criminal neighbors, so Ronald has exactly 6 criminal neighbors in total; since C4 is Sofia and she is innocent, the other seven neighboring spaces can contain only one innocent besides Sofia. A4 is Paul and he is criminal, and B5 is Wally and he is criminal, so among A3, B3, C3, A5, and C5 there can be only one innocent. That means in row 3, at least two of A3, B3, and C3 are criminals, and Paul says all criminals in row 3 are connected, so the row 3 criminals there must form one unbroken block; with at least two criminals among A3, B3, and C3, that forces the middle space B3 to be criminal. Therefore, we can determine that B3 is CRIMINAL.

Wally is at B5, so his edge neighbors are A4, A5, B4, C4, and C5. We already know A4 is criminal and C4 is innocent, and Sofia’s clue says Wally is one of Ronald’s 6 criminal neighbors, so Ronald at B4 must be criminal. Larry’s clue says an odd number of those edge neighbors are innocent, and with B4 criminal and C4 innocent, that means A5 and C5 must be one innocent and one criminal, so among Wally’s five edge neighbors there are exactly three criminals. Ronald’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among these, B3, A4, B5, and one of A5/C5 are already criminal, while C4 and the other of A5/C5 are innocent, so Ronald currently has exactly five criminal neighbors before counting A3 and C3. Since Sofia says Ronald has exactly 6 criminal neighbors, A3 and C3 together must contribute exactly one more criminal neighbor. But Larry at B3 is criminal, and his clue says an odd number of innocents on the edges neighbor Wally; that already forced only one of A5 and C5 to be innocent, so the only way Ronald can stop at 6 while still matching all known counts around the B3-C4-B5 area is for both A3 and C3, the two unknowns flanking Larry, to be criminal. Therefore, we can determine that A3 is CRIMINAL and C3 is CRIMINAL.