Puzzle Pack #1 Puzzle 3 Answer

Scott’s clue says that Karen is a criminal, and also that she is below Gabe. Since Karen is at C3, this clue directly identifies her status. Therefore, we can determine that C3 is CRIMINAL.

Karen’s clue is specifically about Bonnie and row 1. It says Bonnie is one of 2 criminals in that row, which directly includes Bonnie among the criminals there. Therefore, we can determine that B1 is CRIMINAL.

Karen’s clue says row 1 contains exactly 2 criminals, and we already know Bonnie at B1 is one of them. In row 1, the people to the right of Adam are Bonnie at B1, Carol at C1, and Denis at D1. Bonnie says there is only one innocent to Adam’s right, so among those three people exactly one is innocent and the other two are criminals. Since Bonnie is already a criminal, that means row 1 has exactly 2 criminals among Bonnie, Carol, and Denis, leaving no room for Adam to be a criminal as well. Therefore, we can determine that A1 is INNOCENT.

Scott’s clue is about the people below Gabe in column C. Since Karen at C3 is one of two or more criminals below Gabe, Gabe must be above Karen, so Gabe is at C1 or C2; but C1 is Carol, so Gabe must be C2. The people below Gabe are then Karen at C3, Olof at C4, and Wally at C5, and because Karen is one of at least two criminals among them, at least one of Olof or Wally is criminal. Adam’s clue says Xavi has an odd number of innocent neighbors. Xavi is at D5, so his only neighbors are C4 Olof, C5 Wally, and D4 Scott. Scott is already innocent, so Olof and Wally together must contribute an even number of innocents. Since at least one of them is criminal, they cannot be one innocent and one criminal, so both must be criminal. Therefore, we can determine that C4 is CRIMINAL and C5 is CRIMINAL.

Row 4 contains Mary at A4, Nick at B4, Olof at C4, and Scott at D4. Wally’s clue says there are more criminals than innocents in that row, and we already know Olof is a criminal while Scott is innocent. That leaves Mary and Nick as the only undecided people in row 4, and with one criminal and one innocent already there, the row can have more criminals than innocents only if both of those remaining spaces are criminals. Therefore, we can determine that A4 is CRIMINAL and B4 is CRIMINAL.

Mary’s clue is about guards who have a criminal immediately above them. The guards on the board are Emily at A2, Karen at C3, Nick at B4, and Mary herself at A4, and among them only Karen and Nick have someone directly above them at all. The person directly above Karen is Gabe at C2, and Gabe must be criminal because Karen is one of the two guards Mary is counting; the person directly above Nick is Jane at B3, and Jane must also be criminal because Nick is the other one. Therefore, we can determine that B3, Jane is CRIMINAL and C2, Gabe is CRIMINAL.

In column B, the known criminals are B1, B3, and B4. For all criminals in that column to be connected, they must form one continuous vertical group with no innocent person between them. Since B2 is the only person between B1 and B3, B2 has to be a criminal to keep those column-B criminals connected. Therefore, we can determine that B2 is CRIMINAL.

Emily at A2 and Flora at B2 have exactly two common neighbors: Adam at A1 and Isaac at A3. Flora says she and Emily share an odd number of innocent neighbors, so among those two shared neighbors, an odd number must be innocent. Adam is already known to be innocent, so Isaac cannot also be innocent, because that would make two innocent shared neighbors instead of an odd number. Therefore, we can determine that A3 is CRIMINAL.

Isaac’s clue compares Emily at A2 and Vicky at B5. Emily’s neighbors are Adam, Bonnie, Flora, Isaac, and Jane, and among those only Adam is innocent, so Emily has exactly 1 innocent neighbor. Vicky’s neighbors are Mary, Nick, Olof, Tyler, and Wally, and since Mary, Nick, Olof, and Wally are all criminal, Vicky can have 1 innocent neighbor only if Tyler is innocent. Therefore, we can determine that A5 Tyler is INNOCENT.

Tyler is at A5 and Wally is at C5, so their common neighbors are only B4 and B5. Tyler says those common neighbors include no innocents. B4 is already known to be criminal, so the only way for the clue to hold is for B5 not to be innocent. Therefore, we can determine that B5 is CRIMINAL.

Jane is at B3. Her neighbors are A2, B2, C2, A3, C3, A4, B4, and C4, and among those everyone is already known to be criminal except A2. Vicky’s clue says Jane has the most criminal neighbors, and “the most” must be unique. So Jane must have more criminal neighbors than anyone else. If A2 were innocent, Jane would have only 7 criminal neighbors, but Karen at C3 already has 7 known criminal neighbors around her, so Jane would not have uniquely the most. That means A2 must also be criminal, giving Jane 8 criminal neighbors and making her uniquely highest. Therefore, we can determine that A2 is CRIMINAL.

In row 1, Bonnie is said to be one of exactly 2 criminals, and Bonnie is already known to be a criminal. Since A1 is innocent, that means exactly one of C1 and D1 is criminal, so row 1 has 2 innocents. Emily’s clue says row 1 has more innocents than row 5, so row 5 must have fewer than 2 innocents, meaning at most 1 innocent. But A5 is already innocent, so D5 cannot also be innocent. Therefore, we can determine that D5 is CRIMINAL.

Linda is at D3, so her four criminal neighbors are Gabe at C2, Karen at C3, Olof at C4, and Xavi at D5. Karen is at C3, and among those four, Gabe, Olof, and Xavi all also neighbor Karen, while Helen at D2 is the only remaining neighbor of Linda who does not. Since the clue says exactly three of Linda's four criminal neighbors also neighbor Karen, those three must be Gabe, Olof, and Xavi, so Linda's fourth criminal neighbor has to be Helen. Therefore, we can determine that D2 is CRIMINAL.

Denis and Xavi are in the same column, so the people in between them are Linda at D3 and Scott at D4. Helen says the only innocent among those in-between people is Linda’s neighbor. Scott is already known to be innocent, so Scott must be that “only innocent in between Denis and Xavi.” Scott is a neighbor of Linda, which leaves Linda as the other person between Denis and Xavi, and Linda cannot be innocent. Therefore, we can determine that D3 is CRIMINAL.

Flora is at B2, so her neighbors in row 1 are exactly A1, B1, and C1. The clue says both innocents in row 1 are Flora's neighbors, so the two innocents in that row must be chosen from those three positions. A1 is already innocent and B1 is already criminal, so the only other row 1 position that can be an innocent neighbor of Flora is C1. That gives the two innocents in row 1 as A1 and C1, leaving D1 not innocent. Therefore, we can determine that D1 is CRIMINAL and C1 is INNOCENT.