Puzzle Pack #1 Puzzle 39 Answer

Bobby’s clue says that the two criminals above Vicky are connected. Above Vicky are Bobby, Gabe, Kay, and Peter, and Bobby is already known to be innocent, so the clue can only be affecting Gabe, Kay, and Peter. With exactly two criminals needed there, the connected pair has to include the middle person so the two criminals can touch within that group. So Kay must be criminal.

Above Vicky are Bobby, Gabe, Kay, and Peter, and among them Bobby is innocent while Kay is criminal. Bobby’s clue says both criminals above Vicky are connected, so there must be exactly two criminals in that group above Vicky. Kay is already one of them, which means the only possible second criminal above Vicky has to come from Gabe or Peter. Kay’s clue says column B has an odd number of criminals, and column B currently has Bobby innocent, Kay criminal, and only Gabe, Peter, and Vicky still unknown. If Vicky were innocent, Gabe and Peter would be the only unknowns left to satisfy both clues at once, and that is impossible. So Vicky must be criminal.

Vicky’s clue says Lucy has exactly 2 innocent neighbors, and only 1 of those 2 is also Kay’s neighbor. Among Lucy’s neighbors, the ones who are also Kay’s neighbors are B2 Gabe, C2 Hal, B4 Peter, and C4 Ronald, so the other innocent neighbor must come from Lucy’s neighbors who are not Kay’s neighbors. That means C2 Hal and C4 Ronald cannot be that “not Kay’s neighbor” innocent. Since the clue allows only 1 innocent among Lucy’s neighbors who is also Kay’s neighbor, and Hal and Ronald are both in that group, this step identifies them as not innocent. So Hal and Ronald must be criminals.

Hal’s clue directly says that Nancy is one of Jane’s three criminal neighbors. That identifies Nancy as a criminal.

Jane has exactly 3 criminal neighbors, and two of them are already known: Kay and Nancy. So among Jane's still-unknown neighbors, A2 Emma, B2 Gabe, and B4 Peter, exactly one is the remaining criminal neighbor. But in that group, the only people who can still be that criminal are Gabe and Peter, not Emma. So Emma must be innocent.

Nancy’s clue says the people to the left of Martin contain exactly one innocent, and among those people the only unknowns are Jane and Lucy. Emma’s clue says row 3 has an odd number of criminals, and row 3 already has Kay as one known criminal, with only Jane, Lucy, and Martin still unknown there. So the one required innocent to Martin’s left has to come from Jane or Lucy, while the row still has to keep an odd criminal count with Kay already contributing one. That makes Martin criminal.

Martin’s clue says Gabe has more criminal than innocent neighbors. Among Gabe’s neighbors, Bobby and Emma are already known innocents, and Hal and Kay are already known criminals, so the known count there is 2 innocents and 2 criminals. To have more criminals than innocents, the unknown neighbors must add more criminals than innocents overall, and that requires Anna and Cheryl to be criminals. So Anna and Cheryl must be criminals.

Among the teachers, Ronald is already a known criminal, and the only other teachers are Wally and Xia. Anna’s clue says the total number of criminal teachers is odd. Since there is already 1 criminal teacher, adding exactly one of Wally or Xia would make the total even, so neither of them can be criminal under that clue. So Wally and Xia must be innocent.

Xia’s clue is the one that determines this directly: column D has more innocents than any other column. The clue is being applied to Debra in column D, so Debra must be innocent.

Debra’s clue says row 5 has more innocents than any other row. In row 5, Vicky is criminal and Wally and Xia are innocent, so row 5 already has 2 innocents and Thor is the only unknown person there. For row 5 to have more innocents than every other row, Thor has to be innocent too. So Thor must be innocent.

Vicky’s clue says Lucy has exactly two innocent neighbors, and exactly one of those two is also a neighbor of Kay. Among Kay’s neighbors, the people who are also neighbors of Lucy are B2 Gabe, C2 Hal, B4 Peter, and C4 Ronald, with Hal and Ronald already criminal, so the one innocent there must be either Gabe or Peter. That means the other innocent neighbor of Lucy must come from the part of Kay’s neighbors that is not also neighboring Lucy: A2 Emma, A3 Jane, C3 Lucy, and A4 Nancy. In that group, Emma is already innocent and Nancy is already criminal, so Emma is the one innocent there. That makes Jane criminal.

Nancy’s clue says the people to the left of Martin contain exactly one innocent. In that group, there are currently no known innocents, and the only unknown person left there is Lucy. So Lucy must be innocent.

Jane’s clue says Martin and Nancy have the same number of innocent neighbors. The computed counts already show 1 known innocent among Martin’s neighbors and 1 known innocent among Nancy’s neighbors. The only unknown neighbor on Nancy’s side is Peter, while Martin’s unknown neighbors are Isaac and Steve, so matching Nancy’s neighbor count keeps Peter as the innocent needed there. Xia’s clue is the other piece used here: column D has more innocents than any other column. With Debra and Xia already innocent in column D, that clue fixes the remaining balance in column D through Isaac and Steve, and then the neighbor-count clue above leaves Peter’s status determined. Once Peter is set, the remaining undetermined target in this step is Gabe. So Peter must be innocent, and Gabe must be criminal.

Ronald’s clue says the number of innocent judges and innocent sleuths must be the same. The judges already give 2 innocents, while the sleuths currently have only 1 innocent, and the only unknown person this can still change is Isaac. So Isaac must be innocent.

Lucy’s neighbors already contain exactly 2 innocents in total, as this clue requires. Those two known innocents are Isaac and Peter, and Steve is the only unknown among Lucy’s neighbors. Since the full total of 2 innocents in Lucy’s neighboring group is already used up, Steve cannot be innocent. So Steve must be criminal.