Puzzle Pack #1 Puzzle 10 Answer

David’s clue says that Zoe is one of the criminals on the edge, and D5 is an edge position. Since the clue directly identifies Zoe as a criminal, nothing else is needed here. Therefore, we can determine that D5 is CRIMINAL.

Xavi is at C5, so his neighbors are B4, C4, D4, B5, and D5. In row 4, the only people who are Xavi’s neighbors are B4, C4, and D4, because A4 is too far away. Since the clue says both criminals in row 4 are Xavi’s neighbors, the two criminals in row 4 must be among B4, C4, and D4, which leaves A4 unable to be one of them. Therefore, we can determine that A4 is INNOCENT.

Sam is at D4, so the people to his left are Olive at A4, Pam at B4, and Ryan at C4. Olive says exactly 2 of those 3 are innocents, and Olive herself is already known to be innocent, so exactly one of Pam and Ryan is innocent and the other is criminal. Zoe says both criminals in row 4 are neighbors of Xavi at C5; in row 4, Pam, Ryan, and Sam are all neighbors of Xavi, but Olive at A4 is not, so Olive cannot be one of the two criminals in that row. That means the two criminals in row 4 must be the criminal among Pam and Ryan, together with Sam. Therefore, we can determine that D4 is CRIMINAL.

In row 4, one criminal is already known to be Sam at D4. Zoe says both criminals in row 4 are neighbors of Xavi at C5, and Xavi’s neighbors in row 4 are only B4, C4, and D4. Since there are exactly two criminals in row 4 and D4 is one of them, the other row 4 criminal must be either B4 or C4. Sam also says all criminals in row 4 are connected, and B4 is not orthogonally connected to D4 because C4 sits between them, so the second criminal cannot be B4 and must be C4. That leaves B4 as not criminal. Therefore, we can determine that C4 is CRIMINAL and B4 is INNOCENT.

Pam’s clue is about column B, which contains Bobby at B1, Frank at B2, John at B3, Pam at B4, and Wanda at B5. She says Bobby is one of 3 criminals in that column, so this clue directly includes Bobby among the criminals in column B. Therefore, we can determine that B1 is CRIMINAL.

Pam’s clue says Bobby is one of exactly 3 criminals in column B. In that column, Bobby at B1 is already a criminal and Pam at B4 is already innocent, so among B2, B3, and B5, exactly two must be criminals. Ryan’s clue talks about the people below Frank, which are B3, B4, and B5. Since B4 is innocent, having more innocents than criminals among those three means there must be at least two innocents there, so at most one of B3 and B5 can be criminal. That means B3 and B5 cannot both be criminals, but Pam’s clue requires exactly two criminals among B2, B3, and B5. So B2 has to be one of those criminals. Therefore, we can determine that B2 is CRIMINAL.

Pam says there are exactly 3 criminals in column B. Since Bobby at B1 and Frank at B2 are already criminals and Pam at B4 is innocent, the third criminal in that column must be B3 John, so B5 Wanda is innocent. Bobby says that among Pam’s 3 criminal neighbors, only 1 is in row 3. Pam’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5, and we already know B3 and C4 are criminals while A4, B4, and B5 are innocent. That means exactly one of A3 and C3 can be criminal, and exactly one of A5 and C5 can be criminal. Frank says Olive has an odd number of criminal neighbors. Olive’s neighbors are A3, B3, B4, and A5, and since B3 is criminal and B4 is innocent, A3 and A5 must contribute an even number of additional criminals between them. Combined with the earlier result that exactly one of A3 and A5 would have to be criminal if either were criminal, the only way to keep Olive’s total odd is for neither A3 nor A5 to be criminal. Therefore, we can determine that A3 is INNOCENT and A5 is INNOCENT.

Hilda is at D2, so her four neighbors are C1 Cheryl, C2 Gary, C3 Karen, and D3 Nancy. Isaac says exactly one of Hilda’s innocent neighbors is also a neighbor of Ryan at C4. Ryan’s neighbors are B3, C3, D3, B4, D4, B5, C5, and D5, so among Hilda’s four neighbors only C3 Karen and D3 Nancy are Ryan’s neighbors. That means the one innocent among Hilda’s neighbors who is also Ryan’s neighbor must be either Karen or Nancy, so Cheryl and Gary cannot be that person. Since Isaac says Hilda has four innocent neighbors in total, Cheryl and Gary must be innocent too. Therefore, we can determine that C1 is INNOCENT and C2 is INNOCENT.

Pam’s clue says column B contains exactly 3 criminals. In column B, Bobby at B1 and Frank at B2 are already criminals, while Pam at B4 is innocent, so the only way to reach 3 criminals in that column is for exactly one of B3 and B5 to be criminal. That means column B has 3 criminals total. Gary’s clue says column D has more criminals than any other column, so column D must have more than 3 criminals. In column D, D1 is innocent, while D4 and D5 are already criminals, so the only way for column D to exceed 3 criminals is for both remaining unknowns there, Hilda at D2 and Nancy at D3, to be criminals. Therefore, we can determine that D2 is CRIMINAL and D3 is CRIMINAL.

Hilda is at D2, so her four innocent neighbors are C1 Cheryl, D1 David, C2 Gary, and C3 Karen. Ryan is at C4, and among those four people, only C3 is neighboring Ryan, because Cheryl, David, and Gary are too far away while Karen is directly above Ryan. Since Isaac says only 1 of Hilda’s four innocent neighbors is Ryan’s neighbor, that one person must be Karen, so Karen has to be one of Hilda’s innocent neighbors. Therefore, we can determine that C3 Karen is INNOCENT.

Pam is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those, the people already known to be criminals are C4 only, and the row 3 neighbors are A3, B3, and C3, with A3 and C3 already known innocent. Bobby’s clue says that among Pam’s three criminal neighbors, exactly one is in row 3. Since C4 is a criminal neighbor of Pam but is in row 4, Pam must have one criminal in row 3, and the only possible person there is B3, John. Therefore, we can determine that B3 is CRIMINAL.

Pam’s clue is about column B, which contains Bobby at B1, Frank at B2, John at B3, Pam at B4, and Wanda at B5. We already know Bobby, Frank, and John are criminals, so column B already has 3 criminals. Since Pam says Bobby is one of 3 criminals in that column, there cannot be any fourth criminal there, and Pam herself is already innocent. Therefore, we can determine that B5 is INNOCENT.

Pam is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those, the criminals already known are B3 John and C4 Ryan, while A3, C3, A4, A5, and B5 are innocent, so the only unknown neighbor there is C5 Xavi. Bobby’s clue says Pam has exactly 3 criminal neighbors, and exactly 1 of those 3 is in row 3. John at B3 is already that one row-3 criminal neighbor, so Ryan at C4 is another criminal neighbor and the third criminal neighbor must be Xavi at C5. Therefore, we can determine that C5 is CRIMINAL.

Karen says row 2 has more criminals than any other row, so row 2 must have the uniquely highest criminal count. Rows 3, 4, and 5 already each have exactly 2 criminals, while row 1 has only 1. In row 2, Frank and Hilda are already known criminals, so row 2 already has 2 criminals and must have more than that to be higher than every other row. The only unknown in row 2 is Eve at A2, so she must be the extra criminal. Therefore, we can determine that A2 is CRIMINAL.

Frank is at B2, and his neighbors in row 1 are A1 and C1. Eve says there is an odd number of innocents among those row 1 neighbors. Since C1, Cheryl, is already known to be innocent, the total can be odd only if A1, Austin, is not innocent. Therefore, we can determine that A1 is CRIMINAL.