Puzzle Pack #2 Puzzle 20 Answer

Paula is in column B and is already known to be innocent. Her clue says column B is the only column with exactly 4 criminals. Since column B has five people total and Paula is the one innocent in that column, the other four people in column B must be the four criminals there. Therefore, we can determine that B1 is CRIMINAL, B2 is CRIMINAL, B3 is CRIMINAL, and B5 is CRIMINAL.

Alice is at A1, and the people below her in the same column are Freya at A2, Kyle at A3, Olof at A4, and Tom at A5. Gary says that Kyle is one of 3 criminals below Alice, so among those four people below Alice, exactly three are criminals, and Kyle is one of them. That directly fixes Kyle's identity. Therefore, we can determine that A3, Kyle, is CRIMINAL.

Gary says that below Alice there are exactly 3 criminals, and the people below Alice are Freya at A2, Kyle at A3, Olof at A4, and Tom at A5. Since Kyle is already one of those criminals, exactly two of the other three people in column A below Alice must also be criminals. That means column A has exactly 3 criminals below A1, so if Alice were also a criminal then column A would have 4 criminals in total. Paula says column B is the only column with exactly 4 criminals, so no other column can also have 4 criminals. Therefore, we can determine that A1 is INNOCENT.

Alice is at A1, so the people below her are Freya at A2, Kyle at A3, Olof at A4, and Tom at A5. Gary says Kyle is one of 3 criminals below Alice, and Kyle is already known to be one of them, so among A2, A3, A4, and A5 there must be exactly 3 criminals in total. Bonnie says there is only one innocent below Kyle, and the people below Kyle are Olof at A4 and Tom at A5, so exactly one of those two is innocent and the other is criminal. That means below Alice, besides Kyle, A4 and A5 contribute exactly 1 criminal, so Freya at A2 must be the third criminal. Therefore, we can determine that A2 is CRIMINAL.

Paula’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. We already know A3, B3, and B5 are criminals, and Alice says exactly 2 of Paula’s neighbors are innocent, with only 1 of those 2 on the edge. Column B already has Bonnie, Gary, Logan, and Vera as criminals, and Paula is innocent, so column B has exactly 4 criminals. Since Paula says column B is the only column with exactly 4 criminals, no other column can end with exactly 4 criminals. In column C, the only unknowns are C1, C2, C3, C4, and C5. If C5 were innocent, then Paula’s two innocent neighbors could be C5 together with only one of A4 or C4, which would force the other one to be criminal; that would leave column C with exactly 4 criminals, matching column B, which Paula says cannot happen. So C5 cannot be innocent. Therefore, we can determine that C5 is CRIMINAL.

Xena’s clue says every person on the board must have at least 2 criminal neighbors. Olof at A4 has only five neighbors: A3, B3, B4, A5, and B5. Among those, A3, B3, and B5 are already known criminals, and B4 is already known innocent. Since Olof must have at least 2 criminal neighbors no matter who we look at, A5 cannot be the only remaining uncertainty in Olof’s neighborhood, so Olof himself has to be criminal to keep this requirement satisfied for the nearby people who depend on that cluster of criminals. Therefore, we can determine that A4 is CRIMINAL.

Kyle is at A3, and the people below Kyle in the same column are Olof at A4 and Tom at A5. Bonnie says there is only one innocent below Kyle, so among A4 and A5 exactly one is innocent. Olof is already known to be criminal, so the one innocent below Kyle must be Tom. Therefore, we can determine that A5 is INNOCENT.

Paula’s clue says column B is the only column with exactly 4 criminals. Column B already has Bonnie, Gary, Logan, and Vera as criminals, with Paula innocent, so column B does have exactly 4 criminals. That means no other column can also have exactly 4 criminals. In column A, Alice and Tom are innocent while Freya, Kyle, and Olof are criminals, so column A has only 3 criminals. In column C, Xena’s clue forces Rob to be criminal, because as a corner Xena has only three neighbors and already needs at least two criminal neighbors; with Vera already criminal and Paula innocent, Rob must be criminal. That gives column C exactly 3 criminals as well, since Chase, Hank, and Maria are still not all criminal. So the only remaining way another column could tie column B at 4 criminals would be column D. But in column D, if Joyce were innocent, then D would have at most 3 criminals among Eli, Nancy, Salil, and Zoe, because Joyce herself would not be one of them. Therefore, we can determine that D2 is CRIMINAL.

Paula’s neighbors are Logan at B3, Maria at C3, Olof at A4, Rob at C4, Tom at A5, Vera at B5, and Xena at C5. We already know Logan, Olof, Vera, and Xena are criminals, and Tom is innocent, so Paula’s only possible innocent neighbors are Maria and Rob. Alice says exactly one of Paula’s two innocent neighbors is on an edge, and among Maria and Rob only Rob is on an edge while Maria is not. So Maria and Rob must be Paula’s two innocent neighbors, with Rob being the edge one and Maria the non-edge one. Therefore, we can determine that C3 is INNOCENT.

Paula is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those, the people already known to be innocent are Maria at C3 and Tom at A5, and both of those positions are on the edge of the board. Alice’s clue says that of Paula’s two innocent neighbors, only one is on an edge, so Paula cannot have both Maria and Tom as her two innocent neighbors. Since Maria and Tom are already innocent, the only way to satisfy the clue is for C4, Rob, not to be innocent. Therefore, we can determine that C4 is CRIMINAL.

Above Rob in column C are Chase at C1, Hank at C2, and Maria at C3, and Maria is already known to be innocent. Logan says an odd number of those three are criminals, so among Chase and Hank there must be exactly one criminal. Now look at the people in column D and count criminal neighbors for each using the known surrounding statuses plus that fact. Eli at D1 has Bonnie, Gary, Hank, and Joyce as neighbors, so he has either 3 or 4 criminal neighbors depending on Hank. Joyce at D2 has Chase, Hank, Maria, and Nancy around her, so she cannot have exactly 2 criminal neighbors because Chase and Hank contribute only 1 in total and Maria is innocent. Nancy at D3 has Hank, Joyce, Maria, Rob, Salil, and Zoe around her, so she also cannot have exactly 2 because Joyce and Rob are already criminal and exactly one of Chase and Hank is not relevant there. Zoe at D5 has Rob, Salil, and Xena as neighbors, so she has exactly 2 criminal neighbors only if Salil is innocent. Rob says only one person in column D has exactly 2 criminal neighbors. If Salil were innocent, then both Nancy and Zoe would have exactly 2, which is impossible, so Salil must be criminal. Therefore, we can determine that D4 is CRIMINAL.

In column D, D2 and D4 are already criminals, while D1, D3, and D5 are still the ones that could affect how many criminals are in that column. Paula says column B is the only column with exactly 4 criminals, so column D cannot also end up with exactly 4 criminals. That means among D1, D3, and D5, either exactly one is criminal or all three are criminal. Rob’s clue says only one person in column D has exactly 2 criminal neighbors. If only one of D1, D3, and D5 were criminal, then D1, D3, and D5 would each have exactly 2 criminal neighbors, giving more than one such person in column D. So that option is impossible, and all three of D1, D3, and D5 must be criminals. Therefore, we can determine that D1 is CRIMINAL, D3 is CRIMINAL, and so on.

Chase is at C1, so the people below him are Hank at C2, Maria at C3, Rob at C4, and Xena at C5. Eli’s clue says that all of the criminals among those four must form one continuous vertical group, with no innocent separating any two of them. But Maria at C3 is already known to be innocent, while Rob at C4 and Xena at C5 are criminals, so if Hank at C2 were also criminal, the criminals below Chase would be at C2, C4, and C5, split apart by Maria and therefore not connected. Therefore, we can determine that C2 is INNOCENT.

Rob is at C4, so the people above him in the same column are Chase at C1, Hank at C2, and Maria at C3. Hank and Maria are already known to be innocent, so among those three, the only possible criminal above Rob is Chase. Since Logan says the number of criminals above Rob is odd, that count must be 1, not 0. Therefore, we can determine that C1 Chase is CRIMINAL.

Paula says column B is the only column with exactly 4 criminals. Column B already has Bonnie, Gary, Logan, and Vera as criminals, with only Paula innocent, so column B does have exactly 4 criminals. Column D currently has Eli, Joyce, Nancy, and Salil as criminals already, so for Paula's statement to stay true as the only such column, Zoe at D5 cannot make column D also reach exactly 4 criminals. That means column D must have 5 criminals instead, so Zoe must be criminal. Therefore, we can determine that D5 is CRIMINAL.