Puzzle Pack #1 Puzzle 17 Answer

Row 4 contains exactly two criminals, and Nancy is at D3. Nancy’s neighbors in row 4 are only C4 and D4, so among the two criminals in row 4, exactly one must be in C4 or D4 and the other must be in A4 or B4. Since Olive at A4 is already known to be innocent, the criminal on the left side of row 4 has to be Peter at B4. Therefore, we can determine that B4 is CRIMINAL.

Zed is at D5, so his neighbors are C4, D4, and C5. Peter says Zed has exactly one innocent neighbor, and that innocent neighbor is in row 4, so among those three neighbors the only innocent one must be either C4 or D4. That leaves C5, Xia, unable to be innocent. Therefore, we can determine that C5 is CRIMINAL.

Vicky is at A5, so the people above her are Bonnie at A1, Flora at A2, Katie at A3, and Olive at A4. The clue says that both criminals above Vicky are connected, so there are exactly two criminals among those four, and they must form one unbroken vertical group in column A. Olive is already known to be innocent, so the two criminals cannot be A3 and A4. That means the only possible connected pair above Vicky is A1 and A2. Therefore, we can determine that A2 is CRIMINAL.

Column A contains Bonnie at A1, Flora at A2, Katie at A3, Olive at A4, and Vicky at A5. Flora says there are exactly 2 innocents in column A, and we already know Flora is a criminal while Olive is innocent, so among A1, A3, and A5 there must be exactly one more innocent and two criminals. Xia says both criminals above Vicky are connected, so the criminals above Vicky in column A must be exactly two people with no innocent between them. Above Vicky, A2 is already a criminal, so the second criminal above her must be A3 rather than A1, because A1 and A2 would not be connected with A3 in between. That leaves A5 as the other criminal needed in column A. Therefore, we can determine that A5 is CRIMINAL.

In row 5, we already know that A5 and C5 are criminals. Vicky’s clue says that all criminals in that row must form one continuous left-right group, with no innocent person between any two of them. The only person between A5 and C5 is B5, so B5 has to be a criminal to keep those row 5 criminals connected. Therefore, we can determine that B5 is CRIMINAL.

Nancy is at D3, so her neighbors are C2, D2, C3, C4, and D4. The two criminals in row 4 are Olive at A4 and Peter at B4, and among those two only Peter is Nancy's neighbor, so Olive's clue fits exactly and does not add anyone else around Nancy as a row 4 criminal. Will at B5 says he is one of Xia's four criminal neighbors. Xia is at C5, whose neighbors are B4, C4, D4, B5, and D5. We already know B4 and B5 are criminals, so for Xia to have exactly four criminal neighbors, two more of C4, D4, and D5 must also be criminals. But Olive's clue tells us C4 and D4 cannot both be criminals, because both of them are neighbors of Nancy and then both row 4 criminals next to Nancy would be Nancy's neighbors, which would break the "only 1" statement. So the needed two extra criminal neighbors for Xia must include D5. Therefore, we can determine that D5 is CRIMINAL.

The coders are Isaac at C2, Sofia at C4, and Zed at D5. Zed’s clue says the profession with the most innocents is coders, and it must be uniquely the most. Right now Zed is already known to be a criminal, so the only coders who could be innocent are Isaac and Sofia. That means coders can have at most 2 innocents, so for coders to be the unique profession with the most innocents, they must actually have exactly 2 innocents. So both remaining coders, Isaac and Sofia, have to be innocent. Therefore, we can determine that C4 is INNOCENT and C2 is INNOCENT.

In row 4, the two criminals are already known to be Peter at B4 and one other person, so Olive’s clue is talking about Peter and Tom. Nancy is at D3, and Tom at D4 is definitely Nancy’s neighbor, while Peter at B4 is not a neighbor of Nancy because he is too far away. That means exactly one of the two row 4 criminals is Nancy’s neighbor only if Tom is the other criminal in row 4. Therefore, we can determine that D4, Tom, is CRIMINAL.

Gary is at B2, so his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. The clue says both criminals in row 1 are Gary's neighbors, so every criminal in row 1 must be among A1, B1, and C1, because D1 is not next to Gary. That means D1 cannot be one of the two criminals in row 1. Therefore, we can determine that D1 is INNOCENT.

The coders are Isaac, Sofia, and Zed, and we already know Isaac and Sofia are innocent while Zed is criminal, so coders have exactly 2 innocents. Zed’s clue says the number of innocent coders is more than the number of innocents in any other profession, so every other profession must have at most 1 innocent. The cooks are Chad and Ethan, and Ethan is already known to be innocent, so the cooks cannot have another innocent. Therefore, we can determine that B1, Chad, is CRIMINAL.

Peter is at B4, so his neighbors in column C are C3, C4, and C5. We already know C4 is innocent and C5 is criminal, so among those three column C neighbors, the only unknown is C3. Chad says an odd number of those column C neighbors are innocent, and with C4 already giving one innocent, C3 cannot be innocent as well or the total would be 2, which is even. Therefore, we can determine that C3 is CRIMINAL.

Chad is at B1, so his neighbors are A1, A2, B2, C1, and C2. We already know A2 is criminal and C2 is innocent, and Mary says Chad has more criminal than innocent neighbors, so among the three unknown neighbors A1, B2, and C1, at least two must be criminals. Sofia says both criminals in row 1 are Gary's neighbors. Row 1 already has Chad at B1 as one criminal, and Ethan at D1 is innocent, so the second criminal in row 1 must be either A1 or C1. Since Gary at B2 neighbors A1, B1, and C1 in that row, this means at least one of A1 or C1 is criminal. So among A1, B2, and C1, we already need at least two criminals from Mary's clue, and Sofia's clue guarantees that at least one of A1 or C1 is one of them. That forces the other criminal among those three to be Gary at B2. Therefore, we can determine that B2 is CRIMINAL.

Nancy is at D3, and Vicky is at A5. Vicky’s neighbors are Olive, Peter, and Will, and since Olive is innocent while Peter and Will are criminals, Vicky has exactly 2 criminal neighbors. Nancy’s neighbors are Isaac, John, Mary, Sofia, Tom, and Zed. Among those, Mary, Tom, and Zed are already criminals, so Nancy already has 3 criminal neighbors even before counting John. Since Gary says Nancy has more criminal neighbors than Vicky, and Vicky has 2, that fits only if John is also criminal. Therefore, we can determine that D2 is CRIMINAL.

Zed is at D5, so the people above him are Ethan at D1, John at D2, Nancy at D3, and Tom at D4. We already know Ethan is innocent, while John and Tom are criminal, so among those four, Nancy is the only one not yet fixed. That means the number of innocents above Zed is either 1 if Nancy is criminal or 2 if Nancy is innocent. John says that number is odd, so it must be 1, which forces Nancy to be criminal. Therefore, we can determine that D3 is CRIMINAL.

The guards are Gary at B2, Katie at A3, and Laura at B3. Looking at the squares directly below them, Gary has Peter below him at B4, who is criminal, and Laura has Will below her at B5, who is also criminal. Nancy’s clue says exactly one guard has an innocent directly below them, so the only guard who can satisfy that is Katie, which means Katie is the one innocent guard in that condition. Therefore, we can determine that B3, Laura, is CRIMINAL.

Laura’s clue compares the number of criminal neighbors around John at D2 and Bonnie at A1. John’s neighbors are C1 David, C2 Isaac, C3 Mary, D1 Ethan, and D3 Nancy, and among those we already know Mary and Nancy are criminals while Isaac and Ethan are innocent, so John currently has 2 criminal neighbors plus David if David is criminal. Bonnie’s neighbors are B1 Chad, A2 Flora, and B2 Gary, and all three of those are already known criminals, so Bonnie has 3 criminal neighbors. For John and Bonnie to have equal numbers, David must also be a criminal so that John’s total becomes 3. Therefore, we can determine that C1 David is CRIMINAL.

Row 1 has exactly two criminals, and we already know they are Chad at B1 and David at C1. Sofia’s clue says both of those row 1 criminals are neighbors of Gary at B2. David at C1 is a neighbor of Gary, and Chad at B1 is too, but Bonnie at A1 is also in row 1 and is not a neighbor of Gary because A1 is too far from B2. So Bonnie cannot be one of the criminals in row 1. Therefore, we can determine that A1 is INNOCENT.

Vicky is at A5, so the people above her are A1, A2, A3, and A4. Among them, A1 and A4 are innocent, and A2 is already known to be a criminal, so the clue says the criminals in that column above Vicky must form one connected group. Since A2 is a criminal, any other criminal above Vicky has to be linked to A2 by continuous up-and-down adjacency, and that is only possible if A3 is also criminal. Therefore, we can determine that A3 is CRIMINAL.