Puzzle Pack #1 Puzzle 2 Answer

In column A, the innocents we need to consider are Aaron at A1, Gabe at A2, Keith at A3, Olivia at A4, and Tyler at A5, and Paula is at B4. The people in column A who neighbor Paula are only Keith at A3, Olivia at A4, and Tyler at A5, because neighbors include diagonals. Olivia’s clue says exactly one innocent in column A is neighboring Paula, and Olivia herself is already known to be innocent, so she is that one innocent neighbor. That leaves Keith and Tyler as not innocent. Therefore, we can determine that A5 is CRIMINAL and A3 is CRIMINAL.

Tyler is at A5, so his neighbors are A4, B4, and B5. The clue says there are exactly 2 criminals in row 4, and only 1 of those 2 is Tyler's neighbor. In row 4, A4 is Olivia, B4 is Paula, C4 is Ruby, and D4 is Sarah, and we already know A4 is innocent. That means the only row 4 person who is Tyler's neighbor is B4, so for Tyler to have exactly 1 neighboring criminal in row 4, B4 must be that criminal. Therefore, we can determine that B4 is CRIMINAL.

Tyler is at A5, so the people above Tyler are Olivia at A4, Keith at A3, Gabe at A2, and Aaron at A1. Paula’s clue says that the criminals among those four are exactly two people, and those two criminals must be connected in that column with no innocent between them. We already know Olivia at A4 is innocent and Keith at A3 is criminal, so one of the remaining two, Gabe or Aaron, must be the second criminal. Aaron cannot be that second criminal, because then the two criminals above Tyler would be A3 and A1, and Gabe at A2 would be between them, so they would not be connected. Therefore Gabe must be the other criminal, and Aaron must be innocent. Therefore, we can determine that A2 is CRIMINAL and A1 is INNOCENT.

Tyler is at A5, so his neighbors are A4, B4, and B5. We already know A4 is innocent and B4 is criminal, so among Tyler's known neighbors there is exactly one innocent. Gabe says the total number of innocents neighboring Tyler is odd, so B5 cannot be innocent, because that would make the total 2. Therefore, we can determine that B5 is CRIMINAL.

Paula is at B4, so her edge neighbors are A5, B5, and C5. We already know A5 Tyler and B5 Vince are criminals, so among those edge neighbors the only one who could be an innocent is C5 Wally. Vince says an odd number of edge innocents neighbor Paula, so that count cannot be 0 and must be 1 here. Therefore, we can determine that C5 is CRIMINAL.

The guards are Tyler at A5, Vince at B5, and Wally at C5. Tyler’s person directly to the right is Vince, and Vince’s person directly to the right is Wally, and both Vince and Wally are already criminals, so neither Tyler nor Vince can be a guard with an innocent directly to the right. Since Wally says exactly one guard has an innocent directly to the right, that only leaves Wally himself, whose right-hand neighbor is Xia at D5. Therefore, we can determine that D5 is INNOCENT.

Tyler’s neighbors are A4, B4, and B5. The two criminals in row 4 are Paula at B4 and one of Ruby at C4 or Sarah at D4, and Tyler says only one of those two is his neighbor. Since B4 is definitely Tyler’s neighbor, the other row 4 criminal cannot also be his neighbor, so C4 cannot be criminal and the second criminal in row 4 must be D4 Sarah. That means Olivia’s neighbors are Keith at A3, Paula at B4, Logan at B3, and Tyler at A5, with only Logan still unknown among them. Xia says she has more innocent neighbors than Olivia, and Xia’s neighbors are Sarah at D4, Wally at C5, and Ruby at C4, where Sarah and Wally are criminals, so Xia can have at most 1 innocent neighbor. Therefore Olivia must have 0 innocent neighbors, which forces Logan at B3 to be criminal. Therefore, we can determine that B3 is CRIMINAL.

The only guards on the board are Tyler, Vince, Wally, and Xia, and among them only Xia is innocent, so there is exactly 1 innocent guard. The only pilots are Keith, Logan, and Martin, with Keith and Logan already criminal, so the clue means there must also be exactly 1 innocent pilot. Since Martin is the only pilot whose status was still open, he has to be that one innocent pilot. Therefore, we can determine that C3 is INNOCENT.

Gabe is at A2, so the people to his right are B2 Helen, C2 Isaac, and D2 Joyce. Martin’s clue says Helen is one of two or more criminals to the right of Gabe, which means Helen herself must be among those criminals. Therefore, we can determine that B2 is CRIMINAL.

Gabe is at A2, so the people to his right in row 2 are B2, C2, and D2. Martin says Helen is one of two or more criminals to the right of Gabe, so among B2, C2, and D2 there must be at least two criminals, and Helen at B2 is one of them. Helen also says that all criminals in row 2 are connected. Since B2 is already a criminal, any other criminal to the right of Gabe must stay connected to B2 in that same row, so the next space C2 has to be criminal. Therefore, we can determine that C2 is CRIMINAL.

Column B contains Brian at B1, Helen at B2, Logan at B3, Paula at B4, and Vince at B5. We already know that Helen, Logan, Paula, and Vince are all criminals, and Isaac’s clue says there is only one innocent in that whole column. Since those four cannot be the innocent one, the only person in column B who can be innocent is Brian at B1. Therefore, we can determine that B1 is INNOCENT.

Helen is at B2, and her neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Brian says Keith is one of Helen's 6 criminal neighbors, so exactly 6 of those 8 neighbors must be criminals. We already know that A2, C2, A3, B3, and C3 are criminals, while A1 and B1 are innocents, so the only way for Helen to have 6 criminal neighbors is for C1 to be criminal as well. Therefore, we can determine that C1 is CRIMINAL.

To the right of Aaron in row 1 are Brian, Carol, and Flora. We already know Brian is innocent and Carol is criminal, so among those three people the counts are currently one innocent, one criminal, and Flora unknown. For there to be more criminals than innocents there, Flora must be criminal, because if Flora were innocent the totals would be two innocents and one criminal instead. Therefore, we can determine that D1 is CRIMINAL.

Tyler’s clue says there are exactly 2 criminals in row 4, and only 1 of them is his neighbor. Tyler at A5 neighbors A4, B4, and B5, so among the row 4 people he touches only A4 and B4. Since A4 is already innocent, the one row 4 criminal who is Tyler’s neighbor must be B4, Paula, which means the other row 4 criminal is either C4 or D4. That leaves at most one innocent in the pair C4 and D4. Nicole at D3 neighbors C2, D2, C3, C4, D4, C5, and D5. Among those, C2, C3, and C5 are criminal, D5 is innocent, and from the row 4 result at most one of C4 and D4 can be innocent. So if Joyce at D2 were innocent, Nicole would have at least two innocent neighbors, D2 and D5, and possibly more, which cannot fit Flora’s clue that Nicole has only one innocent neighbor. Therefore, we can determine that D2, Joyce, is CRIMINAL.

Vince is at B5, so his neighbors are A4, B4, C4, A5, C5, and no one else. In row 4, Olivia at A4 is already known to be innocent, and the clue says both innocents in row 4 are Vince's neighbors. That means the other innocent in row 4 must also be one of Vince's neighbors, so it has to be C4 Ruby, because D4 Sarah is not next to Vince at all. Therefore, we can determine that D4 is CRIMINAL and C4 is INNOCENT.

Sarah’s clue says the whole board contains exactly 15 criminals. Counting the people already known, there are currently 14 confirmed criminals. Since D3, Nicole, is the only person still unknown, she must be the 15th criminal. Therefore, we can determine that D3 is CRIMINAL.