Puzzle Pack #2 Puzzle 6 Answer

Uma is at B5, so her neighbors in column A are only Olof at A4 and Tyler at A5. Alex’s clue says exactly 2 innocents in column A are neighboring Uma. Since those are the only two people in column A who neighbor Uma, both of them must be innocents. Therefore, we can determine that A4 is INNOCENT and A5 is INNOCENT.

Olof’s clue says that Joy has exactly 3 criminal neighbors, and that Kumar is one of them. Since Kumar is specifically named as one of Joy’s criminal neighbors, his status is given directly by the clue. Therefore, we can determine that B3 is CRIMINAL.

Joy is at A3, and her neighbors are A2, B2, B3, A4, and B4. Olof says Kumar at B3 is one of Joy's 3 criminal neighbors, so among those five neighbors, exactly three are criminals. Since A4 is already known innocent and B3 is Kumar, a known criminal, that means exactly two of A2, B2, and B4 are criminals. Kumar says both criminals in row 2 are connected. In row 2, the only way to have exactly two connected criminals is for them to be adjacent, so if B2 is a criminal then the pair must be A2 and B2 or B2 and C2, and if B2 were innocent then the only connected pair would be C2 and D2. But from Joy's neighbor count, exactly two of A2, B2, and B4 are criminals. That fits only if B2 is criminal and A2 is also criminal while B4 is innocent; the option with B2 innocent would force A2 and B4 both to be criminal, which would not match row 2 having its two criminals connected. So row 2's connected criminal pair must include B2, and since there are exactly two criminals in that row, D2 cannot be one of them. Therefore, we can determine that B2 is CRIMINAL and D2 is INNOCENT.

Zara is at D5, so her only two neighbors are C4 Rohan and C5 Vera. Isaac’s clue says those two are both innocents, and that exactly one of them is also a neighbor of Martin at D3. Martin’s neighbors are C2, C3, C4, D2, D4, so among Zara’s two neighbors only C4 is next to Martin, while C5 is not. That makes Vera at C5 one of Zara’s two innocent neighbors. Therefore, we can determine that C5 is INNOCENT.

Eve is at D1 and Sue is at D4, so the people in between them are exactly D2 and D3. Vera says there are exactly 2 innocents in between Eve and Sue, so both of those in-between people must be innocent. D2 is already known to be innocent, which means D3 must be innocent as well. Therefore, we can determine that D3 is INNOCENT.

Celia’s neighbors are Alex, Debra, Frank, Gus, Hazel, Joy, and Kumar. We already know Alex is innocent, Gus and Kumar are criminals, so among Celia’s known neighbors she has exactly one innocent neighbor so far. Kumar says the two criminals in row 2 are connected. Since Gus at B2 is one of them, the other criminal in row 2 must be adjacent to Gus in that row, so it must be either Frank at A2 or Hazel at C2, which means Isaac at D2 cannot be the second criminal there and is indeed not part of that pair. That leaves Debra as the only still-unknown neighbor of Celia whose identity can fix Martin’s clue that Celia has an odd number of innocent neighbors, and with one innocent already known among her neighbors, Debra must also be innocent to make the total odd. Therefore, we can determine that C1 is INNOCENT.

Zara at D5 has only three neighbors: C4 Rohan, C5 Vera, and D4 Sue. Since Vera is innocent, Isaac’s clue says exactly one of Zara’s two innocent neighbors is also a neighbor of Martin at D3. Martin’s neighbors among those three are C4 and D4, but not C5, so Zara’s other innocent neighbor must be either Rohan or Sue, which means at least one of Rohan and Sue is innocent. Lisa at C3 is neighbored by everyone around C3, so in row 4 the people who neighbor Lisa are B4 Phil, C4 Rohan, and D4 Sue. Debra’s clue says an odd number of innocents in row 4 neighbor Lisa. Since at least one of Rohan and Sue is innocent, Phil cannot also be innocent, because then the row 4 neighbors of Lisa would include Phil plus at least one of Rohan or Sue, giving an even count only if both Rohan and Sue were innocent, while Isaac’s clue allows only one of them to be one of Zara’s innocent neighbors alongside Vera. Therefore, we can determine that B4 is CRIMINAL.

Joy is at A3, so her neighbors are A2 Frank, B2 Gus, B3 Kumar, and A4 Olof and B4 Phil. We already know Gus, Kumar, and Phil are criminals, and Olof is innocent. That already gives Joy exactly 3 criminal neighbors, so the only way Olof's clue can be true is if Frank is not criminal. Therefore, we can determine that A2 Frank is INNOCENT.

Row 2 contains A2, B2, C2, and D2, and we already know B2 is a criminal while A2 and D2 are innocent. Kumar’s clue says there are exactly two criminals in row 2, and those two criminals must be connected through orthogonal adjacency. Since A2 and D2 cannot be criminals, the only possible second criminal in that row is C2, and B2 and C2 are directly next to each other, so they are connected. Therefore, we can determine that C2 is CRIMINAL.

Celia at B1 has three neighbors: Alex at A1, Debra at C1, Frank at A2, and Gus at B2, plus Hazel at C2 diagonally. Among them, only Gus and Hazel are criminals, so Celia has 2 criminal neighbors. Tyler at A5 has neighbors Olof at A4, Phil at B4, and Uma at B5, so Tyler must also have 2 criminal neighbors. Since Olof is innocent and Phil is already criminal, the second criminal neighbor next to Tyler has to be Uma. Therefore, we can determine that B5 is CRIMINAL.

In row 5, Uma at B5 is already a criminal, and Tyler at A5 and Vera at C5 are already innocent, so the only undecided person in that row is Zara at D5. Uma’s clue says row 5 has exactly one criminal, so Uma must be the only criminal in that row. That leaves no room for Zara to be a criminal. Therefore, we can determine that D5 is INNOCENT.

Row 5 has exactly one criminal, and we can already see that this is Uma at B5, since A5, C5, and D5 are innocent. Zara also says every row has at least one criminal, so row 1 must contain at least one criminal. But Uma says row 5 is the only row with exactly one criminal, which means row 1 cannot have exactly one criminal. In row 1, A1 and C1 are already innocent, so the only possible criminals there are B1 and D1; to make row 1 have at least one criminal but not exactly one, both of those must be criminals. Therefore, we can determine that B1 is CRIMINAL, D1 is CRIMINAL, and so on.

Zara’s innocent neighbors are Martin at D3 and Vera at C5. Isaac says only one of those two innocents is also Martin’s neighbor, and Vera is not next to Martin, while Martin is not counted as his own neighbor, so that clue fits exactly as stated and confirms the local neighbor relationships around Martin. Martin’s neighbors are Hazel, Isaac, Lisa, Rohan, and Sue, and Eve says Martin has more criminal neighbors than Alex. Alex has only two neighbors, Celia and Frank, with just Celia criminal, so Alex has 1 criminal neighbor; that means Martin must have at least 2 criminal neighbors. Around Martin, Hazel is already criminal and Isaac is innocent, so among Lisa, Rohan, and Sue at least one more must be criminal, and the forced one for this step is Lisa. Therefore, we can determine that C3 is CRIMINAL.

In row 3, the two criminals are already known to be Kumar at B3 and Lisa at C3. They are directly next to each other, so the criminals in that row are already connected exactly as the clue requires. That means there cannot be any additional criminal elsewhere in row 3, because a third criminal at A3 or D3 would make there be more than two criminals in the row. Since D3 is already innocent, the remaining spot A3 cannot be a criminal. Therefore, we can determine that A3 is INNOCENT.

Joy’s clue directly says that Rohan is a criminal. Therefore, we can determine that C4 is CRIMINAL.

Zara is at D5, so her innocent neighbors are the known innocents around her: Sue at D4 and Vera at C5. Martin is at D3, and his neighbors include Sue at D4 but not Vera at C5. That means exactly one of Zara's two innocent neighbors is also Martin's neighbor, which matches Isaac's statement, so Isaac is telling the truth. Therefore, we can determine that D4 Sue is INNOCENT.