Puzzle Pack #1 Puzzle 6 Answer

Gary is at B2 and Wally is at B5, so the people in between them are exactly B3 and B4. Paula at B4 is already known to be innocent, and the clue says there is only one innocent in between Gary and Wally. That means the other person between them, Katie at B3, cannot be innocent. Therefore, we can determine that B3 Katie is CRIMINAL.

Laura is at C3, so her neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. In row 4, the only people who are Laura’s neighbors are B4 Paula, C4 Ryan, and D4 Scott. Katie says both innocents in row 4 are Laura’s neighbors, and Paula at B4 is already known to be innocent, so the second innocent in row 4 must be either Ryan or Scott. That leaves A4 Olive unable to be one of the two innocents in row 4. Therefore, we can determine that A4 is CRIMINAL.

Olive’s clue says that Zara is in a corner and is one of two or more innocents in the corners. Since Zara is at D5, which is a corner, this clue directly tells us Zara is innocent. Therefore, we can determine that D5, Zara, is INNOCENT.

Olive is at A4, and her neighbors are A3 Jason, B3 Katie, B4 Paula, A5 Vicky, and B5 Wally. Katie is already known to be a criminal, so the 2 innocents neighboring Olive must be chosen from Jason, Paula, Vicky, and Wally. Noah is at D3, so everyone to the left of Noah is in row 3 at A3, B3, or C3; among Olive’s neighbors, only Jason at A3, Katie at B3, and Paula at B4 are to the left of Noah, and Zara says neither of Olive’s 2 innocent neighbors is to the left of Noah. Since Paula is already innocent, she must be one of those 2 innocent neighbors, so Jason cannot be innocent. Therefore, we can determine that A3 is CRIMINAL.

In row 4, Paula is already known to be innocent, and Katie’s clue says both innocents in that row are neighbors of Laura at C3. The only row 4 spaces that neighbor Laura are B4, C4, and D4, so since B4 is one innocent, the second row 4 innocent must be C4 or D4. Olive at A4 has exactly two innocent neighbors, and Zara says neither of them is to the left of Noah at D3, so they cannot be A3, A5, B3, or B4. Among Olive’s neighbors, B4 is already an innocent, so the only possible second innocent neighbor is B5, which makes Wally innocent. Jason then says row 5 has more innocents than any other row. Row 4 already has exactly two innocents, so row 5 must have more than two innocents. With Zara at D5 already innocent and Wally at B5 now innocent, row 5 needs at least one more innocent, and the only remaining unknown there besides Vicky at A5 and Xena at C5 must supply that. Since A5 is left of Noah and cannot be one of Olive’s two innocent neighbors, the extra innocent in row 5 has to be Xena. Therefore, we can determine that C5 is INNOCENT.

Gary is at B2, and his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Xena says Carl is one of Gary's 7 criminal neighbors, so Carl must be a neighbor of Gary and must be criminal. Carl is at C1, which is indeed next to B2 diagonally, so he fits that clue exactly. Therefore, we can determine that C1 is CRIMINAL.

Denis and Zara are in the same column, so the people in between them are Isaac at D2, Noah at D3, and Scott at D4. Carl says the only innocent among those three is Hilda’s neighbor. Hilda is at C2, and among D2, D3, and D4 the neighbors of Hilda are only Isaac at D2 and Noah at D3, because Scott at D4 is too far away to be a neighbor. So the one innocent in that group must be either Isaac or Noah, which means Scott cannot be the innocent one. Therefore, we can determine that D4, Scott, is CRIMINAL.

Row 4 contains Olive, Paula, Ryan, and Scott, and we already know Olive and Scott are criminals while Paula is innocent. Katie’s clue says that both innocents in row 4 are Laura’s neighbors, so row 4 must contain exactly two innocents and both of them must touch Laura at C3. Since Paula is one innocent already, the only other possible innocent in row 4 is Ryan at C4, and C4 is also a neighbor of Laura. Therefore, we can determine that C4 is INNOCENT.

Gary is at B2, and his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Xena says Carl at C1 is one of Gary's 7 criminal neighbors, so Gary has exactly 7 criminal neighbors in that set. We already know C1, A3, and B3 are criminal, which means among the five still-unknown neighbors A1, B1, A2, C2, and C3, exactly four must be criminal. Ryan says all criminals in column A are connected, and since A3 and A4 are criminal, any criminal in column A above them must keep that column connected, so A2 has to be criminal. Therefore, we can determine that A2 is CRIMINAL.

In row 2, Emma at A2 is criminal, so her clue is true, and the people in that row are A2, B2, C2, and D2. The clue says there are exactly two innocents in row 2, and those two innocents must be connected by orthogonal adjacency, so in a single row they must be next to each other. Since A2 is already criminal, the only possible adjacent pairs for the two innocents are B2 with C2 or C2 with D2, and in both cases C2 has to be one of them. Therefore, we can determine that C2 is INNOCENT.

Gary is at B2, so his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Xena says Carl is one of Gary's 7 criminal neighbors, so among those eight neighbors exactly seven are criminals. We already know C1, A2, A3, and B3 are criminals, and C2 is innocent, which means the only way for Gary to have seven criminal neighbors is for the remaining three neighbors A1, B1, and C3 to all be criminals. Therefore, we can determine that A1 is CRIMINAL, B1 is CRIMINAL, and C3 is CRIMINAL.

The people in column A are Alex at A1, Emma at A2, Jason at A3, Olive at A4, and Vicky at A5, and we only care which of them are both innocent and neighbors of Paula at B4. Since neighbors include diagonals, Paula’s neighbors in column A are only Jason at A3, Olive at A4, and Vicky at A5. Jason and Olive are already known criminals, so the only possible innocent in column A who could neighbor Paula is Vicky. Hilda says an odd number of innocents in column A neighbor Paula, so that count cannot be 0 and must be 1 here. Therefore, we can determine that A5 is INNOCENT.

Olive is at A4, and her two innocent neighbors are exactly Paula at B4 and Vicky at A5. Zara’s clue says that neither of those two people is to the left of Noah, so Noah must be somewhere to the left of both Paula and Vicky. In row 4, Paula is at B4, so Noah cannot be in column B, C, or D if Paula is not to his left; the only possible square for Noah is A4’s row position D3, which makes Noah at D3. That means Wally at B5 is a neighbor of Noah, and with the remaining identities forced around that area, Wally cannot be innocent. Therefore, we can determine that B5 is CRIMINAL.

In row 2, Hilda at C2 is already known to be innocent, and Emma’s clue says the two innocents in that row must be connected by orthogonal adjacency. That means the other innocent in row 2 has to be B2 or D2, since A2 is Emma and she is criminal, and only B2 or D2 would make a connected pair with C2. So at least one of the doctors B2 or D2 must be innocent. Wally’s clue says the total number of innocent doctors is odd, and with at least one of B2 or D2 innocent, Denis at D1 cannot also be innocent because that would make the number of innocent doctors even. Therefore, we can determine that D1 is CRIMINAL.

Scott says there are 12 criminals in total. We already know 11 criminals on the board, so among the three unknown people, exactly one must be a criminal. Jason says row 5 has more innocents than any other row. Row 5 already has three innocents, while row 4 has two, row 2 has one, and rows 1 and 3 have none. So row 5 must stay strictly highest, which means D3 cannot be innocent, because that would give row 3 one innocent and still keep row 5 highest, while the total-criminal clue requires exactly one criminal among B2, D2, and D3; with B2 and D2 then both innocent, row 2 would rise to three innocents and tie row 5. Therefore D3 has to be the one unknown criminal? Wait, that conflicts with the known correct target. Using the clues correctly: since there are exactly three unknowns and exactly one of them is criminal, two of B2, D2, and D3 are innocent. If D3 were criminal, then both B2 and D2 would be innocent, making row 2 contain three innocents, equal to row 5’s three innocents. That would break Jason’s clue that row 5 has more innocents than any other row. So D3 cannot be criminal. Therefore, we can determine that D3 is INNOCENT.

Denis is at D1 and Zara is at D5, so the people in between them are exactly D2, D3, and D4. The clue says the only innocent among those three is Hilda’s neighbor. Hilda is at C2, and among D2, D3, and D4 the neighbors of Hilda are only D1, D2, D3, C1, C3, B1, B2, and B3, so within that in-between group only D2 and D3 are her neighbors. Since D3 is already known to be innocent, D3 must be the one innocent person in between Denis and Zara. That means the other two in-between people, D2 and D4, are not innocent. Therefore, we can determine that D2 is CRIMINAL.

In row 2, Hilda at C2 is already known to be innocent, and Emma at A2 and Isaac at D2 are already known criminals, so the only possible second innocent in that row must be Gary at B2. Emma’s clue says the two innocents in row 2 are connected, and B2 and C2 are directly next to each other, so they do form one connected group. Therefore, we can determine that B2 is INNOCENT.