Puzzle Pack #2 Puzzle 11 Answer

Xavi’s clue says that Freya is one of the 3 innocents in row 2. That directly includes Freya among the innocents in that row, regardless of who the other two are. Therefore, we can determine that B2 is INNOCENT.

Olof is at A4, so the two people to his right are Ryan at B4 and Stella at C4. Stella’s neighbors among those two are both Ryan and Stella is not counted as her own neighbor, so exactly one of those two is Stella’s neighbor only when Ryan is innocent and Stella is not. Since Freya’s clue says only 1 of those 2 innocents is Stella’s neighbor, Stella cannot be one of the innocents to Olof’s right. Therefore, we can determine that C4 is INNOCENT.

Betty at B1 and Vera at B5 are in the same column, so the people in between them are exactly Freya at B2, Luigi at B3, and Ryan at B4. Stella’s clue says exactly 2 innocents among those in-between people are neighbors of Olof at A4. Olof’s neighbors include Luigi and Ryan, but not Freya, because Freya at B2 is too far away. Since Freya is already known innocent and is not one of the counted neighboring people, the two counted innocents must be Luigi and Ryan. Therefore, we can determine that B3 is INNOCENT and B4 is INNOCENT.

Olof is at A4, so the two people to his right are Ryan at B4 and Stella at C4, and both are already known to be innocent. Freya says that only one of those two innocents is Stella's neighbor. Stella is never counted as her own neighbor, but Ryan at B4 is adjacent to Stella at C4, so Ryan is the one innocent who is Stella's neighbor. That means the other innocent to Olof's right, at D4, cannot be innocent, so Thor must be criminal. Therefore, we can determine that D4 is CRIMINAL.

In row 4, the two innocents are already known to be Ryan at B4 and Stella at C4. Thor’s clue says both innocents in that row are connected, and B4 and C4 are directly next to each other, so those two already make the full connected group of row 4 innocents. That means there cannot be any other innocent in row 4, so A4 cannot be innocent. Therefore, we can determine that A4 is CRIMINAL.

Ethan is at A2, so the people below him in the same column are Karen at A3, Olof at A4, and Uma at A5. Olof’s clue says exactly 2 of those 3 people are innocent. Since Olof himself at A4 is already known to be criminal, the two innocents below Ethan must be Karen and Uma. Therefore, we can determine that A3 is INNOCENT and A5 is INNOCENT.

Amy and Uma are in the same column, with Ethan and Karen strictly between them. Karen’s clue says there are exactly 2 innocents in between Amy and Uma, so those two in-between people must both be innocent. Karen already is innocent, so Ethan must be the other innocent in that column segment. Therefore, we can determine that A2 is INNOCENT.

Ethan says column C is the only column with exactly 4 innocents. In column C, Stella at C4 is innocent, so to reach exactly 4 innocents there, the other three unknowns in that column must also be innocent, including Cheryl at C1. That means column A cannot also have exactly 4 innocents, and since A2, A3, and A5 are already innocent, Amy at A1 would make column A also have 4 innocents. Therefore, we can determine that A1 is CRIMINAL.

Column C contains Cheryl at C1, Gabe at C2, Megan at C3, Stella at C4, and Wally at C5. Ethan says column C is the only column with exactly 4 innocents, and Stella is already known innocent, so column C must end up with exactly one criminal and the other four people innocent. Amy says there is an odd number of innocents below Cheryl, and the people below Cheryl are Gabe, Megan, Stella, and Wally. Since Stella is already innocent and those four people must contain either three innocents or four innocents depending on Cheryl, Amy’s clue forces the number below Cheryl to be 3, not 4, so Cheryl must be the fourth innocent in column C. Therefore, we can determine that C1 is INNOCENT.

Column C already has Cheryl and Stella as innocents, and Ethan says it is the only column with exactly 4 innocents. Cheryl also says Column C has more innocents than any other column, so every other column must have fewer innocents than Column C. That means Column C must indeed end up with 4 innocents, because if it had only 3 then some other column could not have more than 2, but column A already has Ethan, Karen, and Uma as 3 innocents. So Column C has 4 innocents, and every other column has at most 3. Column B already has Freya, Luigi, and Ryan as 3 innocents, so it cannot have any more innocents. Therefore both unknowns in column B, Betty at B1 and Vera at B5, must be criminals. Therefore, we can determine that B1 is CRIMINAL and B5 is CRIMINAL.

In row 5, Uma at A5 has two innocent neighbors, Vera at B5 has four innocent neighbors, and Xavi at D5 has two innocent neighbors, so none of those three has exactly one innocent neighbor. Betty’s clue says exactly one person in row 5 does have exactly one innocent neighbor, so that can only be Wally at C5. Wally’s neighbors are Ryan, Stella, Thor, Vera, and Xavi; among those, Ryan, Stella, and Xavi are already innocent, so if Wally were innocent then Wally himself would also have more than one innocent neighbor and the clue could not be true. Therefore, we can determine that C5 is INNOCENT.

Cheryl at C1 and Wally at C5 each already have one known criminal neighbor: Cheryl touches Amy at A1 and Betty at B1, while Wally touches Vera at B5 and Thor at D4. Wally’s other neighbors are all known innocents, so Wally has exactly 2 criminal neighbors in total. Since Wally says he and Cheryl have the same number of criminal neighbors, Cheryl must also have exactly 2 criminal neighbors. Cheryl’s only unknown neighbor is Daniel at D1, so Daniel has to be Cheryl’s second criminal neighbor. Therefore, we can determine that D1 is INNOCENT.

Daniel’s clue compares Isaac at D2 with Xavi at D5. Xavi’s neighbors are C4, D4, C5, and among them only Thor at D4 is criminal, so Xavi has 1 criminal neighbor. Isaac’s neighbors are C1, D1, C2, C3, and D3, and since Cheryl and Daniel are innocent, Isaac can have more than 1 criminal neighbor only if both C3 and D3 are criminal. Therefore, we can determine that D3 is CRIMINAL.

Nicole’s neighbors are C2, D2, C3, C4, C5, B2, B3, and B4, but on the edge at D3 the actual neighboring spaces are C2, D2, C3, C4, D4, C2’s diagonal pair B2 and B4 through C3/C4 do not count; the neighbors of D3 are C2, D2, C3, C4, D4, C2? More simply, the eight-direction neighbors of D3 are C2, D2, C3, C4, D4. Among these, C4 is innocent and D4 is criminal, so Nicole’s clue says the total number of criminals among C2, D2, and C3 must make her neighboring criminal count odd overall. Xavi’s clue says row 2 has exactly 3 innocents, and row 2 is Ethan, Freya, Gabe, and Isaac. Since Ethan and Freya are already innocent, Gabe and Isaac must be one innocent and one criminal. That means among C2 and D2 there is exactly 1 criminal. With D4 already a criminal, Nicole currently has 2 known neighboring criminals if C3 were innocent, but 3 if C3 is criminal. Since her total must be odd, C3 has to be criminal. Therefore, we can determine that C3 is CRIMINAL.

Column C already has Cheryl at C1, Stella at C4, and Wally at C5 as innocents, while Megan at C3 is criminal. For column C to have exactly 4 innocents, the only unknown there, Gabe at C2, must be innocent. This also fits Ethan’s statement that column C is the only column with exactly 4 innocents. Therefore, we can determine that C2 is INNOCENT.

Row 2 contains Ethan, Freya, Gabe, and Isaac. Xavi says Freya is one of 3 innocents in that row, so row 2 must have exactly three innocents in total. Ethan, Freya, and Gabe are already known to be innocent, which already makes those three innocents. Therefore, we can determine that D2 is CRIMINAL.