Puzzle Pack #1 Puzzle 14 Answer

Barb’s clue says that Wally is an innocent, and also that Wally is among at least two innocent neighbors of Nicole. Since everyone tells the truth, that directly fixes Wally’s identity. Therefore, we can determine that B5 is INNOCENT.

In row 5, Wally at B5 is already known to be innocent, and his clue says the two innocents in that row are connected. That means row 5 has exactly two innocents, and they must form one continuous left-right group with no criminal between them. Since B5 is one of those innocents, the only places the other innocent could be are A5 or C5, because those are the only row-5 spaces directly extending a connected group with B5. D5 is not adjacent to B5 in the row, so if D5 were innocent, there would have to be another innocent at C5 as well, which would give row 5 more than two innocents. Therefore, we can determine that D5 is CRIMINAL.

Vicky is at A5, so the people to her right are B5, C5, and D5. Zach’s clue says an odd number of those three are innocent, and we already know B5 is innocent while D5 is criminal. That means C5 cannot be innocent, because then there would be two innocents to Vicky’s right, which is even, not odd. Therefore, we can determine that C5 is CRIMINAL.

In row 5, we already know Wally at B5 is innocent, while Xia at C5 and Zach at D5 are criminals. Wally’s clue says the two innocents in row 5 are connected, so the only other innocent in that row must be orthogonally linked with B5. Since B5’s only orthogonal neighbor in row 5 is A5, the second innocent has to be Vicky at A5. Therefore, we can determine that A5 is INNOCENT.

Freya is at B2, so her neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. The clue says that both criminals in row 3 are Freya's neighbors, so every criminal in row 3 must be among A3, B3, and C3. Lucy at D3 is in row 3 but is not Freya's neighbor, so she cannot be one of those two criminals. Therefore, we can determine that D3 is INNOCENT.

Row 3 is A3 Isaac, B3 Jose, C3 Keith, and D3 Lucy, and we already know Lucy at D3 is innocent. Xia says both criminals in row 3 are Freya's neighbors, and Freya is at B2, so the row 3 people who neighbor her are only A3, B3, and C3. That means the two criminals in row 3 must be chosen from A3, B3, and C3, so C3 must be a criminal because Lucy is innocent and there are exactly two criminals in that row. Lucy also says all criminals in row 3 are connected, so the two criminals in that row must be adjacent; among A3, B3, and C3, the only adjacent pair that can make up both criminals is B3 and C3. Therefore, we can determine that B3 is CRIMINAL.

Below Barb in column B, the people are Freya at B2, Jose at B3, Nicole at B4, and Wally at B5. Among those, the innocents are Freya, Nicole, and Wally, since Jose is criminal. Gus is at C2, so the people below Barb who neighbor Gus are exactly Freya at B2 and Jose at B3, and only Freya is innocent. That matches the clue that exactly one innocent below Barb is neighboring Gus, so Freya must be innocent. Therefore, we can determine that B2 is INNOCENT.

Carol is at C1, and her neighbors are B1, B2, C2, D1, and D2. In column D, the two criminals must be exactly two of D1, D2, D4, and D5, since D3 is already innocent. Among those four, the only ones who are Carol's neighbors are D1 and D2, while D4 and D5 are not. Freya says exactly one of the two criminals in column D is Carol's neighbor. Since D5 is already known to be a criminal and is not Carol's neighbor, the other criminal in column D must be one of D1 or D2, not D4. Therefore, we can determine that D4 is INNOCENT.

The cops are Gus at C2 and Zach at D5. Sofia’s clue says exactly one cop has a criminal directly above them, so among those two cops, exactly one of the squares above them must contain a criminal. The square directly above Zach is D4, and Sofia at D4 is already innocent, so Zach does not have a criminal directly above him. That means the one cop who does have a criminal directly above them must be Gus at C2, so the person directly above Gus at C1 must be criminal. Therefore, we can determine that C1, Carol, is CRIMINAL.

Freya is at B2, so the people below her are Jose at B3, Nicole at B4, and Wally at B5. Carol’s clue says exactly 2 of those 3 people are innocents. We already know Jose is criminal and Wally is innocent, so Nicole must be the second innocent below Freya. Therefore, we can determine that B4 is INNOCENT.

Nicole is a sleuth, and Xia at C5 is already known to be a criminal sleuth, so there is at least one criminal sleuth on the board. Nicole’s clue says there are more criminal cops than criminal sleuths, so there must be at least two criminal cops. The only cops are Gus at C2 and Zach at D5, and Zach is already known to be a criminal, so Gus must be the other criminal cop. Therefore, we can determine that C2 is CRIMINAL.

Freya is at B2, so her neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. The clue from Xia says that both criminals in row 3 are Freya's neighbors, and in row 3 the only people who are Freya's neighbors are A3, B3, and C3, so the two criminals in that row must be among those three; since Jose at B3 is already a criminal, exactly one of A3 and C3 is also criminal. Now compare the innocent neighbors of Freya and Jose. Freya and Jose have the same definite innocent neighbors B1 and B4, and the same definite criminal neighbors C1 and C2, so the shared-neighbor count depends on A2, A3, and C3. Gus says Freya and Jose share an odd number of innocent neighbors, so among A2, A3, and C3 an odd number must be innocent. But from the first clue, exactly one of A3 and C3 is criminal, so exactly one of them is innocent; that contributes 1 innocent already, meaning A2 cannot also be innocent or the total would become even. Therefore, we can determine that A2 is CRIMINAL.

Carol is at C1, so her neighbors are B1, B2, C2, D1, D2, and only the people in column D among those are D1 and D2. Freya says that only 1 of the 2 criminals in column D is Carol's neighbor, so exactly one of D1 and D2 is a criminal. Vicky at A5 has exactly two neighbors, A4 and B4, and B4 is already innocent, so Vicky’s number of innocent neighbors is either 1 or 2 depending on A4. Carol at C1 has neighbors B1, B2, C2, D1, and D2, and among those B1 and B2 are innocent while C2 is criminal, so for Carol to have the same number of innocent neighbors as Vicky, Carol must have exactly two innocent neighbors total. That means both D1 and D2 cannot be innocent, so exactly one of them is innocent and the other is criminal, making Carol’s innocent-neighbor count 3. Therefore Vicky must also have 3 innocent neighbors counted the same way only if A4 is innocent, so we can determine that A4 is INNOCENT.

Zach is at D5, so his only neighbors are C4 Rob, C5 Xia, and D4 Sofia. We already know Xia is criminal and Sofia is innocent. For Zach to have more criminal than innocent neighbors, Rob must be criminal, because if Rob were innocent then Zach would have one criminal neighbor and two innocent neighbors instead. Therefore, we can determine that C4 is CRIMINAL.

Barb is at B1, and her neighbors are A1, A2, B2, C1, and C2. We already know three of those are criminals: Ethan at A2, Carol at C1, and Gus at C2. Since Rob says Carol is one of Barb's 3 criminal neighbors, Barb must have exactly those three criminal neighbors, so the other two neighbors, Alex at A1 and Freya at B2, are innocent. Therefore, we can determine that A1 is INNOCENT.

Alex says row 2 has uniquely the most criminals. Row 2 already has Ethan and Gus as criminals, so it has at least 2 criminals. But row 5 already also has 2 criminals, Xia and Zach, so for row 2 to have more criminals than every other row, row 2 must have 3 criminals. The only unknown person in row 2 is Helen at D2, so she must be the third criminal. Therefore, we can determine that D2 is CRIMINAL.

Column D already has two known criminals, Helen at D2 and Zach at D5, so Freya’s clue is talking about those two people. Carol is at C1, and her neighbors are B1, B2, C2, D1, and D2. Of the two criminals in column D, Helen at D2 is Carol’s neighbor, but Zach at D5 is not, so the clue is satisfied only if David at D1 is not a criminal in column D. Therefore, we can determine that D1 is INNOCENT.

Isaac is at A3, so the people to his right are Jose at B3, Keith at C3, and Lucy at D3. Among those three, Jose is criminal and Lucy is innocent, so the only innocent to Isaac’s right is Lucy. David’s clue says that this only innocent is Keith’s neighbor, and Lucy at D3 is indeed directly next to Keith at C3, so the clue identifies Lucy as that neighbor of Keith. Therefore, we can determine that C3 Keith is CRIMINAL.

Freya is at B2. Her neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. In row 3, the only people who are Freya’s neighbors are A3, B3, and C3. Xia’s clue says that both criminals in row 3 are Freya’s neighbors. We already know B3 and C3 are the two criminals in row 3, so any other person in that row cannot be criminal. That leaves A3 as innocent. Therefore, we can determine that A3 is INNOCENT.