Puzzle Pack #1 Puzzle 13 Answer

Paula is at B4, and Kay is at A3, which is one of Paula’s neighbors. Paula’s clue says that Kay is one of her 3 criminal neighbors, so it directly tells us that Kay is criminal. Therefore, we can determine that A3 is CRIMINAL.

Zoe is at D5, so the people above Zoe are Donna at D1, Jose at D2, Noah at D3, and Steve at D4. Kay’s clue says there are exactly two innocents among those four people, and both of those innocents are below Donna in the same column. Since nobody is above Donna in column D, Donna cannot be one of those innocents. Therefore, we can determine that D1, Donna, is CRIMINAL.

Donna is at D1, so her neighbors are C1, C2, and D2. Since Donna says her only criminal neighbor is Bonnie's neighbor, exactly one of those three people is criminal, and that person must also be next to Bonnie at B1. Among Donna's neighbors, only C1 and C2 are neighbors of Bonnie; D2 is not. So the one criminal neighbor Donna is talking about must be either C1 or C2, which means D2 cannot be that criminal neighbor. Therefore, we can determine that D2 is INNOCENT.

In row 1, Donna at D1 is already known to be a criminal, so the only possible innocents there must come from A1, B1, and C1. Jose’s clue says there are exactly two innocents in that row, and those two innocents must be connected by orthogonal adjacency, so they have to sit next to each other in a continuous group. The only connected pair available among A1, B1, and C1 is either A1 with B1 or B1 with C1, and both possibilities include B1. Therefore, we can determine that B1 is INNOCENT.

The clue says there are exactly 7 criminals on the edge, and none of those edge criminals is a painter. Thor at A5 is on the edge, and Thor is a painter. So Thor cannot be one of the edge criminals mentioned in Bonnie’s clue. Therefore, we can determine that A5 is INNOCENT.

Thor’s clue is specifically about Mary and row 3. It says Mary is one of two or more innocents in that row, so Mary herself must be innocent for that statement to be true. Therefore, we can determine that C3 is INNOCENT.

Donna is at D1, so her neighbors are C1, C2, and D2. Since D2 is innocent, her only criminal neighbor must be exactly one of C1 or C2, and that person must also be a neighbor of Bonnie at B1. Bonnie’s neighbors are A1, A2, B2, C1, and C2, so among Donna’s possible criminal neighbors, the only overlap is C1 or C2. Mary says painters have more innocents than any other profession. The painters are Bonnie at B1, Hank at B2, and Thor at A5, and Bonnie and Thor are already known innocent. So all three painters must be innocent, because if Hank were criminal then painters would have only two innocents, which would not be more than the teachers, who already also have two innocents in Paula and Vicky’s group. Therefore, we can determine that B2 is INNOCENT.

In column B, the people are Bonnie at B1, Hank at B2, Logan at B3, Paula at B4, and Vicky at B5. The ones in that column who neighbor Mary at C3 are Hank, Logan, and Paula, since Mary’s column-B neighbors are B2, B3, and B4. Hank’s clue says an odd number of those column-B neighbors of Mary are innocent. Hank at B2 is innocent, and Paula at B4 is innocent, so there are already two innocents among those three people. For the total to be odd, Logan at B3 must also be innocent. Therefore, we can determine that B3 is INNOCENT.

Below Jose are Noah at D3, Steve at D4, and Zoe at D5. Logan says there are more criminals than innocents among those three, so at least two of them must be criminals. Mary at C3 is innocent and is directly left of Noah, and Paula at B4 is innocent and is diagonally left of Steve, so Noah and Steve each already have an innocent neighbor. Kay says both innocents above Zoe are below Donna, and the only people above Zoe who are below Donna are Jose, Noah, and Steve. Since Jose is already innocent, that clue allows only one of Noah or Steve to be innocent. So among Noah, Steve, and Zoe, at most one is innocent, which means at least two are criminals; because only one of Noah or Steve can be the innocent one, Zoe must be one of those criminals. Therefore, we can determine that D5 is CRIMINAL.

Eve is at A2 and Thor is at A5, so the people in between them are Kay at A3 and Olivia at A4. Zoe says there is an odd number of criminals among those in-between people. Kay is already known to be a criminal, so among A3 and A4 there must be exactly one criminal. That means Olivia cannot also be a criminal. Therefore, we can determine that A4 is INNOCENT.

Olivia’s clue is about row 5, which contains Thor at A5, Vicky at B5, Xavi at C5, and Zoe at D5. We already know Thor is innocent and Zoe is criminal, so for all criminals in that row to be connected, the criminals there must form one unbroken horizontal group reaching Zoe. Since A5 cannot be part of that group, the only way Zoe can be connected to every other criminal in row 5 is for C5 to be criminal. Therefore, we can determine that C5 is CRIMINAL.

Paula at B4 says she has exactly 3 criminal neighbors, and one of them is Kay. Paula’s neighbors are A3 Kay, B3 Logan, C3 Mary, A4 Olivia, C4 Rob, A5 Thor, B5 Vicky, and C5 Xavi. Among those, Kay and Xavi are already criminal, while Logan, Mary, Olivia, and Thor are already innocent, so the third criminal neighbor must be either Rob or Vicky. Xavi at C5 says he has exactly 3 innocent neighbors. Xavi’s neighbors are B4 Paula, C4 Rob, D4 Steve, B5 Vicky, and D5 Zoe. Paula is innocent and Zoe is criminal, so among Rob, Steve, and Vicky, exactly two must be innocent. Since Paula’s clue allows at most one of Rob and Vicky to be criminal, at least one of them is innocent. That means Xavi already has Paula plus at least one of Rob or Vicky as innocent neighbors, so Steve must be the third innocent neighbor. Therefore, we can determine that D4 is INNOCENT.

Zoe is at D5, so the people above Zoe are D1, D2, D3, and D4. The clue says both innocents above Zoe are below Donna, and Donna is at D1, so anyone below Donna in that same column is D2, D3, or D4. Since D2 and D4 are already innocent, those are the two innocents above Zoe, which leaves D3 as not innocent. Therefore, we can determine that D3 is CRIMINAL.

Eve at A2 and Hank at B2 share exactly four common neighbors: Aaron at A1, Kay at A3, Logan at B3, and Isaac at C2. Steve’s clue says that exactly 2 of those shared neighbors are innocent. We already know Kay is criminal and Logan is innocent, so among Aaron and Isaac, exactly one must be innocent. Since Isaac is already determined not to be innocent in the current board state, the second innocent common neighbor has to be Aaron’s alternative, which makes Aaron criminal. Therefore, we can determine that A1 is CRIMINAL.

In row 1, Aaron at A1 and Donna at D1 are already known criminals, while Bonnie at B1 is innocent and Cheryl at C1 is the only unknown. Jose’s clue says both innocents in row 1 are connected, which means row 1 has exactly two innocents and they must touch through orthogonal adjacency. Since Bonnie is already one innocent, the second innocent cannot be A1 or D1, so it must be Cheryl at C1, and B1 and C1 are directly next to each other. Therefore, we can determine that C1 is INNOCENT.

Donna is at D1, so her neighbors are C1, C2, and D2. C1 Cheryl and D2 Jose are already known to be innocent, so Donna’s only criminal neighbor must be C2 Isaac. The clue also says that this only criminal neighbor is Bonnie’s neighbor, and Isaac at C2 is indeed a neighbor of Bonnie at B1, so that fits the clue exactly. Therefore, we can determine that C2 Isaac is CRIMINAL.

Isaac’s clue compares the number of innocent neighbors around Noah at D3 and Eve at A2. Noah’s neighbors are C2, C3, C4, D2, and D4, and among those only C3, D2, and D4 are innocent, so Noah has 3 innocent neighbors. Eve’s neighbors are A1, B1, A2’s diagonal and side neighbors A3, B2, and B3, and among those B1, B2, and B3 are innocent while A1 and A3 are criminal, so Eve also already has 3 innocent neighbors without Rob being involved. Since the clue says their numbers are equal, Rob cannot be an innocent neighbor adding a fourth innocent next to Noah. Therefore, we can determine that C4 is CRIMINAL.

Paula is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those, Kay at A3, Rob at C4, and Xavi at C5 are already known criminals, which already makes exactly 3 criminal neighbors. Since Paula says Kay is one of her 3 criminal neighbors, every other neighbor besides those three must be innocent, so B5 cannot be criminal. Therefore, we can determine that B5 is INNOCENT.

The edge criminals already known are Aaron at A1, Donna at D1, Kay at A3, Noah at D3, Rob at C4, Xavi at C5, and Zoe at D5, which makes 7 edge criminals if Eve is not included. Bonnie’s clue says the 7 criminals on the edges are all not painters, so every edge criminal must be one of those seven people. Eve is on the edge and she is a painter, so she cannot be outside that set by being innocent; she must be the remaining edge criminal count that the clue is referring to. Therefore, we can determine that A2 is CRIMINAL.