Puzzle Pack #2 Puzzle 19 Answer

Adam is at A1, so the people below him are Freya, Joyce, Pam, and Vera. Xavi says that exactly 2 of those 4 are innocent, and exactly 1 of those 2 innocents is a neighbor of Kumar at B3. Among those four people, Kumar’s neighbors are only Freya at A2, Joyce at A3, and Pam at A4, while Vera at A5 is not Kumar’s neighbor. So if exactly one innocent below Adam is Kumar’s neighbor, the other innocent below Adam must be the only one there who is not Kumar’s neighbor: Vera. Therefore, we can determine that A5 is INNOCENT.

In row 5, we already know Vera at A5 and Xavi at C5 are innocent. Vera’s clue says every row has at least 2 criminals, so row 5 must still contain at least 2 criminals among the remaining people in that row. The only unknowns left in row 5 are Wally at B5 and Ziad at D5, so both of them have to be the criminals that satisfy the clue. Therefore, we can determine that B5 is CRIMINAL and D5 is CRIMINAL.

Vera is in A5, so the people above her are Pam at A4, Joyce at A3, Freya at A2, and Adam at A1. Ziad says exactly 2 of those 4 are innocents. Adam’s neighbors are Freya at A2, Chris at B1, and Gus at B2, so the innocents below Adam who are also Kumar’s neighbors can only be Freya and Gus, because Kumar at B3 is not adjacent to Adam. Xavi says that among the innocents below Adam, exactly 1 is Kumar’s neighbor, so there cannot be any other innocents below Adam besides those two possible people. That means all innocents below Adam must be contained within A2 and B2. Since there are exactly 2 innocents above Vera, Adam himself has to be one of them. Therefore, we can determine that A1 is INNOCENT.

Debra is at C1, so the people below her are Hope at C2, Laura at C3, Sue at C4, and Xavi at C5. Adam’s clue says the criminals among those four are exactly two people, and those two must be connected by orthogonal adjacency. Xavi at C5 is already known to be innocent, so the two criminals below Debra must be chosen from C2, C3, and C4. In one column, the only way for two criminals to be connected is for them to be next to each other, so the criminal pair must be either C2 and C3 or C3 and C4. Either way, C3 is one of the two criminals. Therefore, we can determine that C3 is CRIMINAL.

Ziad is at D5, so the people above Ziad in the same column are Eli at D1, Isaac at D2, Olivia at D3, and Tina at D4. Laura’s clue says there are exactly 2 innocents among those four people, and that Isaac is one of them. Since everyone tells the truth, Isaac must be innocent. Therefore, we can determine that D2 Isaac is INNOCENT.

In column D, the people who neighbor Tina at D4 are Olivia at D3 and Ziad at D5. Isaac’s clue says there are no innocents in column D who neighbor Tina, so every column D neighbor of Tina must be criminal. Ziad already fits that, which means Olivia must as well. Therefore, we can determine that D3 is CRIMINAL.

Adam says both criminals below Debra are connected, so in column C the two criminals below C1 must be the pair C3 and C4, since C3 and C5 are separated by innocent Xavi at C5 and so are not connected. That means Sue at C4 is criminal, leaving only one innocent below Debra in column C, namely Hope at C2. Olivia says columns A and C contain the same number of innocents. Column A already has Adam and Vera innocent, so it has 2 innocents. In column C, Laura is criminal, Sue is now criminal, and Xavi is innocent, so to also total 2 innocents there, Debra and Hope must both be innocent. Therefore, we can determine that C1 is INNOCENT.

In row 1, Adam at A1 and Debra at C1 are already known to be innocent. Vera’s clue says every row must have at least 2 criminals, so row 1 still needs 2 criminals somewhere among its four people. That only leaves B1 and D1, so both of those spots must be criminals. Therefore, we can determine that B1 is CRIMINAL and D1 is CRIMINAL.

Ziad is at D5, so the people above him in the same column are Eli at D1, Isaac at D2, Olivia at D3, and Tina at D4. Laura’s clue says there are exactly 2 innocents above Ziad, and Isaac is one of them. We already know Eli and Olivia are criminals, and Isaac is innocent, so the only way to reach exactly 2 innocents above Ziad is for Tina to be the other innocent. Therefore, we can determine that D4 is INNOCENT.

Rob is at B4, so his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those, the people already known to be innocent are A5 and C5, and both of them are not in row 4. Tina says that of Rob's four innocent neighbors, only one is in row 4. Since Rob already has two innocent neighbors that are outside row 4, his total of four innocent neighbors leaves exactly two criminal neighbors among those eight surrounding spaces. Vera says every row has at least 2 criminals. In row 4, Tina at D4 is innocent, so row 4 must get its two criminals from A4, B4, and C4. But Tina's clue allows only one innocent among Rob's row 4 neighbors A4 and C4, so at least one of A4 and C4 is criminal; together with the need for two criminals in row 4, that forces B4 to be criminal as well. Therefore, we can determine that B4 is CRIMINAL.

Tina is at D4, so her neighbors are C3, C4, C5, D3, and D5. Among those, Laura at C3, Olivia at D3, and Ziad at D5 are criminal, while Xavi at C5 is innocent. Since Rob’s clue says Tina has exactly 2 innocent neighbors, the only remaining neighbor, Sue at C4, must be the second innocent one. Therefore, we can determine that C4 is INNOCENT.

Row 4 contains Pam at A4, Rob at B4, Sue at C4, and Tina at D4. Vera’s clue says every row has at least 2 criminals, and in row 4 we already know Rob is a criminal while Sue and Tina are innocent. That means Pam must be the second criminal in that row. Therefore, we can determine that A4 is CRIMINAL.

Debra is at C1, so the people below her are C2, C3, C4, and C5. Among those, C3 is criminal, C4 is innocent, and C5 is innocent, so for Adam’s clue to be true, the two criminals below Debra must be C2 and C3. Those two are directly adjacent in the same column, so they are connected. Therefore, we can determine that C2 is CRIMINAL.

The two innocents below Adam are Joyce at A3 and Vera at A5. Kumar is at B3, and both of those people are his neighbors, so the number of those innocents who are Kumar's neighbors is 2, not 1. That means Xavi's statement cannot be true, and since everyone tells the truth, Xavi must be a criminal for making a false statement. Therefore, we can determine that B3, Kumar, is CRIMINAL.

Rob is at B4, so his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among these, the people already known innocent are C4, A5, and C5, and of those three only C4 is in row 4. Tina says that only 1 of Rob's 4 innocent neighbors is in row 4, so Rob must have exactly 4 innocent neighbors total. Since only three of Rob's neighboring innocents are identified so far, the remaining needed innocent neighbor must be A3, Joyce. Therefore, we can determine that A3 is INNOCENT.

Adam is in A1, and the people below him in the same column are Freya at A2, Joyce at A3, Pam at A4, and Vera at A5. Of those, the innocents are Joyce and Vera, so the clue says exactly one of Joyce and Vera is a neighbor of Kumar at B3. Kumar’s neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. That means Joyce at A3 is Kumar’s neighbor, while Vera at A5 is not. So among the two innocents below Adam, exactly one is Kumar’s neighbor as the clue says, and this only works if Freya is not innocent. Therefore, we can determine that A2 is CRIMINAL.

Freya says row 3 is the only row with exactly 3 criminals. Row 3 does have exactly 3 criminals: Kumar, Laura, and Olivia. Row 2 already has Freya and Hope as criminals, so if Gus were also a criminal then row 2 would also have exactly 3 criminals, which the clue rules out. Therefore, we can determine that B2 is INNOCENT.