Puzzle Pack #1 Puzzle 29 Answer

Nicole’s clue says Henry is one of two or more innocents in column B. That statement directly includes Henry among the innocents in that column, regardless of who the other innocent or innocents are. Therefore, we can determine that B2 Henry is INNOCENT.

Paul is at B4, so his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. The clue says both criminals in row 5 are Paul's neighbors, so the two criminals in row 5 must be among A5, B5, and C5, because those are the only people in row 5 who neighbor Paul. Zoe is at D5, which is in row 5 but is not Paul's neighbor, so Zoe cannot be one of those two criminals. Therefore, we can determine that D5 is INNOCENT.

Row 5 is Vicky at A5, Wally at B5, Xavi at C5, and Zoe at D5, and Zoe is already known to be innocent. Zoe says there is an odd number of criminals to the left of Xavi, so among A5 and B5 there must be either one criminal or both criminals. Henry says both criminals in row 5 are Paul's neighbors, and Paul at B4 is neighbors of A5, B5, C5, A4, C4, and the row above, but not D5. Since D5 is innocent and Henry says the two criminals in row 5 are both among Paul's neighbors, the only possible pair of criminals in row 5 is A5 and B5, which makes the number of criminals to the left of Xavi equal to 2. That cannot be odd unless Xavi himself is the remaining criminal in row 5. Therefore, we can determine that C5 is CRIMINAL.

Row 5 has four people: Vicky at A5, Wally at B5, Xavi at C5, and Zoe at D5. We already know Xavi is a criminal and Zoe is innocent, and Henry says there are exactly two criminals in row 5, with both of them neighboring Paul at B4. Paul’s neighbors in row 5 are A5, B5, and C5, so the second criminal in that row must be one of those three. Since Zoe at D5 is innocent, the only possible pair of criminals in row 5 that includes Xavi is either B5 and C5 or A5 and C5. Xavi also says that all criminals in row 5 are connected, and in that row connection uses left-right adjacency. A5 and C5 are not connected because B5 lies between them, so the only connected criminal pair is B5 and C5. Therefore, we can determine that B5 is CRIMINAL and A5 is INNOCENT.

Eric is at D1, so the people to his left are Adam at A1, Betty at B1, and Carol at C1. Wally’s clue says there are exactly 2 innocents among those three people, and exactly 1 of those 2 innocents is a neighbor of Isaac at C2. Isaac’s neighbors on the top row are only Betty and Carol, not Adam, so among the two innocents to Eric’s left, exactly one must be Betty or Carol and the other must be Adam. Therefore, we can determine that A1 is INNOCENT.

Olivia and Terry are on row 4, with Olivia at A4 and Terry at D4. Everyone to the left of Terry is A4, B4, and C4, and Adam’s clue says exactly one of those three is a criminal, and that criminal is to the right of Olivia. Since Olivia is the leftmost of the three, nobody to the right of Olivia can be Olivia herself, so the one criminal must be either Paul or Ruth instead. Therefore, we can determine that A4 is INNOCENT.

Terry is in D4, so the people to Terry’s left are Olivia at A4, Paul at B4, and Ruth at C4. Adam’s clue says there is exactly one criminal among those people, and that criminal is to the right of Olivia, so Olivia is not that criminal and exactly one of Paul or Ruth is criminal. Olivia’s clue links Mary at C3 and Xavi at C5. Their common neighbors are Ruth at C4 and Terry at D4, and since Xavi is criminal, the clue says exactly one of Ruth and Terry is innocent. That means Ruth and Terry have opposite identities, while Adam’s clue already tells us Ruth cannot be the only innocent among Paul, Ruth, and Olivia unless Terry is the criminal partner to make Olivia’s clue work. So Terry must be the criminal. Therefore, we can determine that D4 is CRIMINAL.

Eric is at D1 and Zoe is at D5, so the people in between them are D2, D3, and D4: Jason, Nicole, and Terry. We already know Nicole is innocent and Terry is criminal, so among those three middle people the only unknown is Jason. For there to be more innocents than criminals in that group of three, Jason must be innocent, making the totals two innocents and one criminal. Therefore, we can determine that D2 is INNOCENT.

Jason’s clue puts Betty and Wally in the same column with Henry somewhere strictly between them. Since Henry is at B2 and Wally is at B5, Betty must be at B1. That means the people in between Betty and Wally are exactly Henry at B2, Lisa at B3, and Paul at B4, and Jason says there are at least two innocents among those three. Henry is already innocent, so at least one of Lisa or Paul must also be innocent. Terry’s clue says Ruth has exactly 5 criminal neighbors, and one of them is Xavi; Ruth at C4 neighbors Lisa, Mary, Nicole, Paul, Terry, Wally, Xavi, and Zoe. Among those, Nicole and Zoe are innocent, Terry, Wally, and Xavi are criminal, so to reach exactly 5 criminal neighbors, both Lisa and Mary must be criminal, leaving Paul as the needed innocent. Therefore, we can determine that C3 is CRIMINAL.

Adam’s clue says the only criminal to the left of Terry is to the right of Olivia. Terry is at D4, so the people to his left are A4, B4, and C4; Olivia is at A4, so the places to her right there are B4, C4, and D4. Since Olivia herself is innocent, the only possible criminal to Terry’s left must be either B4 or C4, which means A4 cannot be that criminal and there are no other criminals to Terry’s left. Mary’s clue compares criminal neighbors of Xavi and Isaac. Xavi at C5 already has three criminal neighbors, namely B5, C4, and D4, so Isaac must have fewer than three criminal neighbors. Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3; among these, C3 is criminal while B2, D2, and D3 are innocent, so to stay below three criminals, at least two of B1, C1, D1, and B3 must be innocent. Wally’s clue says the two innocents to the left of Eric have exactly one person who is also Isaac’s neighbor. If Eric were not at D1, then the people to his left would not give a pair of innocents matching that clue, so Eric must be at D1, making the two people to his left B1 and C1. Isaac’s neighbors include C1 but not B1, so for exactly one of those two innocents to be Isaac’s neighbor, B1 and C1 must both be innocent. Then, from Mary’s clue, Isaac cannot also have both D1 and B3 as criminals, so the remaining unknown neighbors D1 and B3 must both be innocent. Therefore, we can determine that B3 is INNOCENT and D1 is INNOCENT.

Ruth is at C4, and her neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Among those eight people, B3, D3, and D5 are innocent, while C3, D4, and C5 are already known criminals. Terry says Xavi is one of Ruth's 5 criminal neighbors, so Ruth must have exactly 5 criminal neighbors in total. That leaves only B4 and B5 to fill the remaining two criminal-neighbor spots, and since B5 is already criminal, B4 must also be criminal. Therefore, we can determine that B4 is CRIMINAL.

The two innocents to the left of Eric at D1 are Adam at A1 and Henry at B2, and Isaac’s neighbors among them are Henry for sure and Adam only if Gus at A2 is innocent, because Adam touches Isaac diagonally through A2. Wally says only one of those two innocents is Isaac’s neighbor, so Adam cannot be Isaac’s neighbor, which forces Gus at A2 to be innocent and also leaves Isaac at C2 as the one adjacent to Henry. Eric’s clue says row 2 has more innocents than any other row. Row 4 already has two innocents, and row 2 already has Henry and Jason innocent, so to have more than every other row, row 2 must have at least three innocents. Since Gus is now innocent, the remaining open spot in row 2, Isaac at C2, must also be innocent. Therefore, we can determine that A2 is INNOCENT and C2 is INNOCENT.

Terry is at D4, so the people to Terry’s left in that row are Olivia at A4, Paul at B4, and Ruth at C4. Adam’s clue says there is exactly one criminal among those three, and that criminal is to the right of Olivia, so the only possible criminal there must be either Paul or Ruth. But Paul is already known to be a criminal, so he is that one criminal to Terry’s left. That leaves Ruth unable to be a criminal. Therefore, we can determine that C4, Ruth, is INNOCENT.

Eric is at D1, so the two people to his left in that row are Betty at B1 and Carol at C1. Wally says the two innocents to Eric’s left are exactly those two people, so Betty and Carol must both be innocent. Gus says there are 14 innocents in total, and with Betty and Carol added to the already known innocents, every person on the board is now accounted for except Katie at A3. That leaves Katie as the 14th innocent. Therefore, we can determine that A3 is INNOCENT.

Olivia is at A4 and Terry is at D4, so the people to the left of Terry in that row are Olivia, Paul, and Ruth. Adam says the only criminal among those people is to the right of Olivia. Since Olivia and Ruth are already innocent, that one criminal must be Paul at B4. That means there is exactly one criminal anywhere to Terry’s left, so Carol at C1 cannot also be innocent without upsetting Katie’s clue about Eric and Xavi having the same number of innocent neighbors. Eric already has three innocent neighbors, while Xavi currently has only two known innocent neighbors, so for them to match, Xavi’s remaining unknown neighbor Carol must be criminal rather than innocent. Therefore, we can determine that C1 is CRIMINAL.

Eric is at D1, so the two innocents to his left in that row are A1 Adam and A2 Gus. Isaac is at C2, and both Adam and Gus are his neighbors, since neighbors include diagonals. Wally says only 1 of those 2 innocents is Isaac's neighbor, so there must be another innocent to Eric's left besides Adam and Gus. The only unknown person to Eric's left who could fill that role is B1 Betty, so she must be innocent. Therefore, we can determine that B1 is INNOCENT.