Puzzle Pack #2 Puzzle 4 Answer

Daniel is at C1, and the people below him in column C are Hilda at C2, Mark at C3, Tom at C4, and Xavi at C5. Tom says Hilda is one of 2 innocents below Daniel, so among those four people below Daniel, Hilda must be innocent, and there are exactly two innocents in total there. Therefore, we can determine that C2 Hilda is INNOCENT.

Daniel is at C1, so the people below him in the same column are Hilda at C2, Mark at C3, Tom at C4, and Xavi at C5. Tom says Hilda is one of 2 innocents below Daniel, so exactly two of those four people are innocent. We already know Hilda and Tom are innocent, which uses up those two innocent spots. That means the other two people below Daniel in that column, Mark and Xavi, cannot be innocent. Therefore, we can determine that C5 is CRIMINAL and C3 is CRIMINAL.

Jane is at D2, so her neighbors are C1 Daniel, D1 Eve, C2 Hilda, C3 Mark, D3 Noah, and no others because she is on the right edge. Hilda’s clue says Jane has exactly 4 innocent neighbors. Among those five possible neighbors besides Jane herself, Hilda at C2 is already innocent and Mark at C3 is already criminal, so the remaining three neighbors must all be innocent to make Jane’s total exactly 4 innocent neighbors. Therefore, we can determine that D1 Eve is INNOCENT, D3 Noah is INNOCENT, and C1 Daniel is INNOCENT.

Daniel’s clue says every person must have at least two innocent neighbors. Uma at D4 has exactly five neighbors: C3 Mark, D3 Noah, C4 Tom, C5 Xavi, and D5 Zane. Among those, Mark and Xavi are already known criminals, while Noah and Tom are innocent, so Uma can reach the required minimum of two innocent neighbors only if Zane is innocent. Therefore, we can determine that D5 is INNOCENT.

Row 3 contains Kumar at A3, Linda at B3, Mark at C3, and Noah at D3. Uma’s clue says there are exactly 3 innocents in that row, and we already know Mark is criminal while Noah is innocent. That means the other two people in row 3, Kumar and Linda, must both be innocents to make the total exactly 3 innocents in the row. Therefore, we can determine that A3 is INNOCENT and B3 is INNOCENT.

In row 5, we already know Xavi at C5 is a criminal, and Kumar’s clue says there are exactly two criminals in that row and those two criminals must be connected by orthogonal adjacency. Since row 5 is A5, B5, C5, D5, the only positions connected to C5 within that row are B5 and D5. That means the second criminal in row 5 must be B5 or D5, so A5 cannot be one of the two criminals in that row. Therefore, we can determine that A5 is INNOCENT.

Alice at A1 and Hilda at C2 can only share two neighbors at all: Barb at B1 and Frank at A2. Vera says they have only one innocent neighbor in common, so exactly one of Barb and Frank is innocent. Daniel says everyone has at least 2 innocent neighbors, and Alice’s only neighbors are Barb, Frank, and Gus; since Hilda is already innocent, Alice must have at least one more innocent neighbor among Barb and Frank. That makes exactly one of Barb and Frank innocent, and in this step that forces Frank to be the innocent one. Therefore, we can determine that A2 is INNOCENT.

Tom is at C4, so his neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Among them, B3, D3, and D4 are already innocent, while C3 and C5 are already criminals, so Tom currently has exactly two criminal neighbors plus B4 and D5 still unknown. Frank says the number of criminals neighboring Tom is odd, so among B4 and D5 exactly one must be a criminal. Kumar says both criminals in row 5 are connected. Since Xavi at C5 is already a criminal, the other criminal in row 5 must be next to C5 in that row, so it has to be either B5 or D5. But Vera at A5 is innocent, so B5 cannot be the second criminal because then the two row 5 criminals would not be connected through row 5. That forces D5 to be the other criminal in row 5. Since exactly one of B4 and D5 is criminal, B4 cannot be criminal. Therefore, we can determine that B4 is INNOCENT.

Scott says row 4 has more innocents than any other row, so row 4 must have uniquely the highest innocent count. Row 3 already has four innocents, because only Mark at C3 is criminal there. In row 4, Scott, Tom, and Uma are already innocent, so for row 4 to have more innocents than row 3, Pam at A4 must also be innocent and make row 4 total four innocents; otherwise row 4 would have only three and could not be the most. Therefore, we can determine that A4 is INNOCENT.

Alice is at A1 and Eve is at D1, so Eve’s innocent neighbors are only C1, C2, and D2. Since C1 and C2 are already innocent, Pam’s clue says Alice must also have exactly two innocent neighbors among her own neighbors, which are B1, A2, and B2; because A2 is already innocent, that means exactly one of B1 and B2 is innocent. Vera’s clue looks at the neighbors Alice and Hilda share in common. Alice’s neighbors are B1, A2, and B2, while Hilda’s neighbors are B1, C1, B2, C2, B3, and C3, so their common neighbors are exactly B1 and B2. Vera says they have only one innocent neighbor in common, so exactly one of B1 and B2 is innocent. Those two clues agree that among B1 and B2, one is innocent and one is criminal. Since Eve must have the same number of innocent neighbors as Alice, and Eve’s neighbors are C1, C2, and D2 with C1 and C2 already innocent, D2 cannot be innocent. Therefore, we can determine that D2 is CRIMINAL.

Jane says row 5 is the only row with exactly 2 criminals. Row 5 already has Xavi as a criminal, Vera as innocent, and Wanda and Zane unknown, so for row 5 to have exactly 2 criminals, exactly one of Wanda or Zane must also be criminal. That means row 2 cannot also have exactly 2 criminals; but row 2 already has Frank and Hilda innocent and Jane criminal, so if Gus were criminal then row 2 would also have exactly 2 criminals. Therefore, we can determine that B2 is INNOCENT.

Alice is at A1 and Hilda is at C2, so their common neighbors are exactly B1 and B2. Vera’s clue says that among those shared neighbors, only one is innocent. Since B2, Gus, is already known to be innocent, the other shared neighbor, B1 Barb, cannot also be innocent. Therefore, we can determine that B1 is CRIMINAL.

Row 5 already has exactly two known criminals, Xavi at C5 and one other spot still unknown among B5 and D5, so Jane’s clue tells us no other row can end up with exactly two criminals. In row 1, Barb at B1 is a criminal while Daniel at C1 and Eve at D1 are innocent, so the only undecided person in that row is Alice at A1. If Alice were criminal, row 1 would also have exactly two criminals, which Jane says cannot happen because row 5 is the only such row. Therefore, we can determine that A1 is INNOCENT.

Eve at D1 has three neighbors, and all three are already known innocents: C1 Daniel, C2 Hilda, and D2 Jane is criminal, so Eve has exactly two innocent neighbors. Uma at D4 has five neighbors: C3 Mark, D3 Noah, C4 Tom, C5 Xavi, and D5 Zane, and among these only Noah and Tom are already known innocents while Mark and Xavi are criminals. Since Barb says Uma has more innocent neighbors than Eve, Uma must have more than two innocent neighbors, so the only way that can happen is if Zane is also innocent. Therefore, we can determine that D5 is INNOCENT.

Row 5 contains Vera at A5, Wanda at B5, Xavi at C5, and Zane at D5, and we already know Xavi is a criminal. Kumar’s clue says both criminals in row 5 are connected, so row 5 must contain exactly two criminals and they must touch through a continuous left-right chain with no innocent between them. The only way for Xavi at C5 to be connected to the other criminal is if that other criminal is right next to him at B5 or D5, and D5 is already innocent. Therefore, we can determine that B5 is CRIMINAL.