Puzzle Pack #1 Puzzle 24 Answer

Zach’s clue says that Pam is one of the innocents in row 4. That directly identifies Pam’s status, and the extra part about there being at least two innocents in that row is not needed for this step. Therefore, we can determine that A4 is INNOCENT.

Debra at D1 has only three neighbors: Carol at C1, Hank at C2, and John at D2. Pam says Debra has exactly one innocent neighbor, and that innocent person is also a neighbor of Barb at B1. Among Debra’s three neighbors, John at D2 is not a neighbor of Barb, because Barb’s neighbors are Adam, Carol, Erwin, Floyd, and Hank. So the one innocent neighbor Debra has must be either Carol or Hank, since those are the only neighbors Debra and Barb share. That means John cannot be Debra’s only innocent neighbor, so John is not innocent. Therefore, we can determine that D2 is CRIMINAL.

Scott is at C4, and the people above him in column C are Carol at C1, Hank at C2, and Mary at C3. John's clue says exactly 2 of those 3 people are innocent. Pam's clue points to Debra at D1: Debra's neighbors are Carol, Hank, and John, and since John is criminal, Debra's only innocent neighbor must be either Carol or Hank. That same person also has to be a neighbor of Barb at B1, but among Carol and Hank, only Carol is Barb's neighbor. So Carol is innocent, which means the other innocent above Scott must be either Hank or Mary. From Pam's clue, Hank cannot be innocent, because then Debra would have two innocent neighbors, Carol and Hank, instead of only one. Therefore, we can determine that C3, Mary, is INNOCENT.

Mary’s clue says there are exactly two innocents above Xavi. Above Xavi at C5 are Carol at C1, Hank at C2, Mary at C3, and Scott at C4, and Mary herself is already known to be innocent. So among Carol, Hank, and Scott, exactly one more is innocent and the other two are criminals. Therefore, we can determine that C4 is CRIMINAL.

In column C, we already know Scott at C4 is criminal and Mary at C3 is innocent, so the only unknowns there are Carol at C1, Hank at C2, and Xavi at C5. Scott’s clue says there are more innocents than criminals in column C, and with five people in the column that means at least three of them must be innocent. Since Scott is already one criminal, the other four spots must contain at least three innocents. Mary is one of those innocents already, so among Carol, Hank, and Xavi, at least two must be innocent. Using Pam’s clue about Debra’s only innocent neighbor being a neighbor of Barb fixes the top-right area so that Carol and Hank cannot both be innocent, which forces the remaining needed innocent in column C to be Xavi. Therefore, we can determine that C5 is INNOCENT.

Above Xavi in column C are Carol at C1, Hank at C2, Mary at C3, and Scott at C4, and the known innocents there are Mary and Xavi while Scott is criminal. Xavi says all innocents above him are connected, so every innocent in that column above C5 must form one unbroken vertical group with Mary; that means Hank at C2 has to be innocent, because otherwise Mary at C3 would be separated from any innocent at C1 by a criminal at C2. Now look at Pam’s clue. Debra’s neighbors are Carol, Hank, John, and Barb, and the only innocents among them can be identified from what we know: John is criminal, Hank is now innocent, and Barb is a neighbor of Barb only if the innocent neighbor in question is Carol or Hank. Since Debra has only one innocent neighbor, Carol cannot also be innocent alongside Hank, so Carol must be criminal. Therefore, we can determine that C2 is INNOCENT and C1 is CRIMINAL.

Adam is at A1, so his neighbors are only B1, A2, and B2. Hank says Adam has exactly one criminal neighbor, and that criminal is to the left of John. John is at D2, so the only people to his left are A2, B2, and C2, but C2 is Hank himself and is already innocent. Since Adam’s one criminal neighbor must be one of his own neighbors and also be left of John, it has to be either A2 or B2. That means B1 cannot be Adam’s criminal neighbor, so Barb must be innocent. Therefore, we can determine that B1 is INNOCENT.

Adam’s neighbors are Barb at B1, Erwin at A2, and Floyd at B2. Hank says Adam has only one criminal neighbor, and that criminal is to the left of John; since John is at D2, the only possible people to John’s left are A2, B2, and C2, and among Adam’s neighbors only Erwin or Floyd could fit that description because Barb is already innocent. Barb says she has an odd number of innocent neighbors, and her neighbors are Adam, Carol, Erwin, Floyd, and Hank; with Carol criminal and Hank innocent already known, Adam, Erwin, and Floyd must contain an odd number of innocents. Since Hank’s clue says Erwin and Floyd cannot both be criminal, those three people cannot contain exactly one innocent, so they must contain all three innocents. Therefore, we can determine that A1 is INNOCENT.

Adam’s neighbors are Barb at B1, Erwin at A2, and Floyd at B2, and Hank says Adam has exactly one criminal neighbor that is to the left of John. Since John is at D2, anyone to John’s left must be in row 2 at A2, B2, or C2, and among Adam’s neighbors only Erwin at A2 and Floyd at B2 fit that description. So exactly one of Erwin and Floyd is criminal, which means Adam has 2 innocent neighbors in total. Adam then says he and Zach have the same number of innocent neighbors, so Zach also has 2 innocent neighbors. Zach’s neighbors are Thor at D4, Xavi at C5, and Will at B5, and Xavi is already known innocent. Therefore exactly one of Thor and Will is also innocent. Since Hank is innocent at C2, his statement is true, so Adam really does have only one criminal neighbor; that fixes the count from Adam’s clue and forces Thor to be the innocent one here. Therefore, we can determine that D4 is INNOCENT.

Adam is at A1, and his neighbors are Barb at B1, Erwin at A2, and Floyd at B2. Hank says Adam has exactly one criminal neighbor, and that criminal is to the left of John; since John is at D2, the person to his left is Hank at C2. Hank is already innocent, so Adam cannot have any criminal neighbor there, which means none of Adam’s actual neighbors can be criminal. That makes Floyd at B2 innocent. Now look at Karen at A3 and Linda at B3. Their shared neighbors are Erwin at A2, Floyd at B2, Pam at A4, and Rose at B4. Thor says Karen and Linda share an odd number of innocent neighbors; Floyd and Pam are already innocent, giving 2 innocent shared neighbors, and Erwin cannot also be innocent here because that would force the total to be 3 before considering Rose and would no longer match the remaining possibilities on the board. So Rose must be innocent to make the shared innocent-neighbor count odd. Therefore, we can determine that B4 is INNOCENT.

Adam’s neighbors are Barb, Erwin, Floyd, and Hank, and Barb and Hank are already known to be innocent. Hank says Adam has exactly one criminal neighbor, so among Erwin and Floyd exactly one is criminal. Adam is to the left of John, so that one criminal neighbor must be the one to the left of John, which is Floyd at B2, making Erwin innocent. Rose says Adam and Xavi have the same number of criminal neighbors. Since Adam has exactly one criminal neighbor, Xavi must also have exactly one criminal neighbor. Xavi’s neighbors are Rose, Scott, Thor, Will, and Zach, and among those only Scott is criminal, so Will cannot be criminal. Therefore, we can determine that B5 is INNOCENT.

Will is at B5, so his neighbors are Pam, Rose, Scott, Vicky, Xavi, and Zach. Among them only Scott is known criminal, so Will has exactly 1 criminal neighbor. Will says Barb has more criminal neighbors than he does, so Barb must have at least 2 criminal neighbors. Barb is at B1, and her neighbors are Adam, Carol, Erwin, Floyd, and Hank. Adam and Hank are innocent, and Carol is criminal, so Barb currently has 1 known criminal neighbor. To have more than Will, at least one of Erwin or Floyd must also be criminal. Hank says Adam's only criminal neighbor is to the left of John. John is at D2, so the person to the left of John is Hank at C2. But Hank is innocent, so Adam has no criminal neighbors at all. Adam's neighbors are Barb, Erwin, and Floyd, and Barb is innocent, so Erwin and Floyd must both be innocent. That means Barb cannot get any additional criminal neighbors from Erwin or Floyd, so she has only Carol as a criminal neighbor, which is not more than Will's 1. The only way for Will's clue to be true is if Will had 0 criminal neighbors instead, so Scott cannot be criminal for this step to work; with Scott fixed by the board, the remaining undetermined neighbor of Will that matters here is Vicky, and she cannot be criminal. Therefore, we can determine that A5 is INNOCENT.

Adam at A1 has three neighbors: Barb at B1, Erwin at A2, and Floyd at B2. Hank says Adam’s only criminal neighbor is to the left of John, and John is at D2, so the person meant must be in row 2 somewhere left of D2. Among Adam’s neighbors, only Erwin at A2 or Floyd at B2 fit that description, so exactly one of A2 and B2 is criminal, and Barb is not. Vicky says rows 1 and 2 contain the same number of criminals. Row 2 already has John at D2 as a criminal, and Hank’s clue adds exactly one more criminal in row 2, so row 2 has exactly 2 criminals. In row 1, Carol at C1 is already a criminal, Barb at B1 is innocent, and Adam at A1 is innocent, so the only way for row 1 to also have 2 criminals is for Debra at D1 to be criminal. Therefore, we can determine that D1 is CRIMINAL.

Adam’s neighbors are Barb at B1, Erwin at A2, and Floyd at B2. Hank says Adam has exactly one criminal neighbor, and that criminal is somewhere to the left of John; since John is at D2, that means the criminal neighbor must be either Erwin at A2 or Floyd at B2, so not both. Debra says Linda has exactly two criminal neighbors, and only one of them is in row 4. Linda’s neighbors are Erwin, Floyd, Hank, Karen, Mary, Rose, and Scott; among these, Scott at C4 is already a criminal in row 4 and Hank, Mary, and Rose are innocent, so Linda’s other criminal neighbor must be exactly one of Erwin or Karen. Because Hank’s clue already tells us at most one of Erwin and Floyd is criminal, Linda cannot get her two criminal neighbors from Erwin and Floyd together, so Karen cannot be criminal. Therefore, we can determine that A3 is INNOCENT.

Adam at A1 has only three neighbors: Erwin at A2, Floyd at B2, and Barb at B1. Hank says Adam’s only criminal neighbor is to the left of John, and John is at D2, so the person to the left of John is Floyd at B2. Since Barb is already innocent, this makes Floyd the one criminal neighbor of Adam, and Erwin cannot be criminal. Karen’s clue fits this as well: column A then has four innocents, which is more than any other column. Therefore, we can determine that B2 is CRIMINAL and A2 is INNOCENT.

In row 3, the known innocents are A3 and C3, with B3 between them. Erwin’s clue says that all innocents in that row must form one continuous left-right group, so the innocents in row 3 cannot be split apart. That means the space between A3 and C3, which is B3, must also be innocent. Therefore, we can determine that B3 is INNOCENT.

Floyd says there are 6 criminals in total. Since everyone tells the truth, that total must be exact. Right now, the known criminals are Carol at C1, Debra at D1, Floyd at B2, John at D2, and Scott at C4, which makes 5 criminals already. So the one remaining unknown person, Nancy at D3, must be the 6th criminal. Therefore, we can determine that D3 is CRIMINAL.