Puzzle Pack #1 Puzzle 15 Answer

Ronald’s clue says that Helen is one of the 11 criminals on the edges. Since everyone tells the truth, that directly tells us Helen is a criminal. Therefore, we can determine that D2 is CRIMINAL.

Alex and Tom are in column A, so the people in between them are Eve at A2, Isaac at A3, and Maria at A4. Helen says the only innocent among those three is Isaac’s neighbor. Isaac’s neighbors are A2, B2, B3, A4, and B4, so among the three in-between people, only Eve and Maria are Isaac’s neighbors, while Isaac is not his own neighbor. That means the one innocent in between must be either Eve or Maria, so Isaac cannot be the innocent one in between. Therefore, we can determine that A3 is CRIMINAL.

Gabe is at C2, so his neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. The clue says exactly 3 of the 6 criminals neighboring Gabe are in row 1, so Gabe has 6 criminal neighbors in total, and exactly three of those must be B1, C1, and D1, since those are his only neighboring people in row 1. That means all three row 1 neighbors of Gabe have to be criminals. Therefore, we can determine that D1 is CRIMINAL, B1 is CRIMINAL, and C1 is CRIMINAL.

Column C contains Carol at C1, Gabe at C2, Kyle at C3, Paula at C4, and Xia at C5. Bobby says the two innocents in that column are both neighbors of Linda at D3, so they must be in column C squares that touch D3 diagonally or orthogonally: C2, C3, or C4. Since C1 and C5 are not neighbors of D3, they cannot be those two innocents. Carol at C1 is already known to be criminal, so the only remaining non-neighbor in column C is Xia at C5. Therefore, we can determine that C5 is CRIMINAL.

Flora is at B2, so the people below her are Joyce at B3, Nick at B4, and Will at B5. Xia’s clue says that the criminals among those three must form one continuous vertical group, with no innocent or unknown gap between any two criminals. Since B3 is not known to be a criminal, but B5 cannot be a criminal unless B4 is also criminal to keep the criminals below Flora connected, the clue forces the middle position to be criminal. Therefore, we can determine that B4 is CRIMINAL.

The mechs are Joyce at B3, Kyle at C3, and Paula at C4. Donald’s clue says exactly one mech has an innocent directly to the left, so among those three, only one can have an innocent immediately left of them. Joyce is the only mech whose left neighbor is still undecided, because the person directly left of Kyle is Joyce and the person directly left of Paula is Nick, who is already criminal. That means Kyle and Paula cannot be the one described by the clue, so the only possible mech is Joyce. For Joyce to fit the clue, the person directly left of her, Isaac at A3, would have to be innocent, but Isaac is already criminal, so Joyce cannot be that mech either. Since Joyce cannot have an innocent directly to her left, Joyce herself cannot be criminal here, leaving her innocent. Therefore, we can determine that B3 is INNOCENT.

Flora is at B2, so the people below her are Joyce at B3, Nick at B4, and Will at B5. The clue says both criminals below Flora are connected, so among those three people there must be exactly two criminals, and they must form one unbroken vertical group in column B. Joyce is already known to be innocent and Nick is already known to be criminal, so the only way to have both criminals below Flora is for Will to be the second criminal, connected to Nick. Therefore, we can determine that B5 is CRIMINAL.

To the right of Maria at A4 are Nick at B4, Paula at C4, and Ronald at D4. We already know Nick is a criminal and Ronald is innocent, so among those three, the only undecided person is Paula. For there to be more innocents than criminals in that group, Paula must be innocent, because if Paula were criminal the totals would be two criminals and one innocent instead. Therefore, we can determine that C4 is INNOCENT.

Isaac is at A3, so the only people in column B who neighbor him are Bobby at B1, Flora at B2, Joyce at B3, and Nick at B4. Paula says an odd number of those four are innocent. We already know Bobby and Nick are criminals, and Joyce is innocent, so among those four there is currently exactly one innocent unless Flora is also innocent. Since the total must stay odd, Flora cannot be innocent. Therefore, we can determine that B2, Flora, is CRIMINAL.

Alex and Tom are in column A, so the people between them are Eve at A2, Isaac at A3, and Maria at A4. Helen says the only innocent among those in-between people is a neighbor of Isaac, and since Isaac is at A3, both A2 and A4 are neighbors of him while A3 is not considered in between as Isaac’s neighbor in that way. Isaac is already criminal, so the single innocent in that in-between group must be either Eve or Maria, which means Tom cannot be the only innocent mentioned by that clue and the clue forces the remaining person in column A’s end pair to be criminal. Flora’s clue matches this, because Xia is already a criminal farmer, so to keep the number of criminal coders equal to the number of criminal farmers, Tom must also be criminal rather than adding another innocent at A5 while the coder side is already accounted for by Isaac. Therefore, we can determine that A5 is CRIMINAL.

Linda is at D3, so the people below her are Ronald at D4 and Zoe at D5. Ronald is already known to be innocent, and Tom’s clue says the people below Linda contain the same number of criminals and innocents. That means Zoe must be criminal to balance Ronald’s innocence. Therefore, we can determine that D5 is CRIMINAL.

The two innocents in column C are already Paula at C4 and one of Gabe at C2, Kyle at C3, or Xia at C5. Bobby says both of those innocents are Linda’s neighbors, so besides Paula, the other innocent in column C must also touch Linda at D3; among C2, C3, and C5, only C2 and C3 are Linda’s neighbors, so Xia cannot be innocent. That makes the two innocents in column C Gabe and Paula. Now count innocent neighbors for Linda and Tom. Tom at A5 has neighbors A4, B4, and B5, and only B4 is already innocent, while Linda at D3 has neighbors C2, C3, C4, D2, and D4, where C2 and C4 and D4 are innocent and D2 is criminal. Zoe says Linda and Tom have 4 innocent neighbors in total, so since Linda already has 3 innocent neighbors, Tom must have exactly 1, which means A4 cannot be criminal. Therefore, we can determine that A4 is INNOCENT.

Alex and Tom are in column A, so the people in between them are Eve at A2, Isaac at A3, and Maria at A4. The clue says the only innocent among those three is a neighbor of Isaac. Both Eve and Maria are neighbors of Isaac, but Isaac himself is already criminal, and Maria is already known to be innocent. That makes Maria the one innocent person in between Alex and Tom, so Eve cannot be innocent. Therefore, we can determine that A2 is CRIMINAL.

In row 2, the known criminals are A2, B2, and D2, while C2 is the only undecided space between B2 and D2. The clue says that all criminals in that row form one connected group using left-right adjacency. Since D2 cannot be connected to A2 and B2 unless the space at C2 is also criminal, C2 must be part of that chain. Therefore, we can determine that C2 is CRIMINAL.

In column C, the two people already known there are Carol at C1 and Gabe at C2 as criminals, Paula at C4 as innocent, and Xia at C5 as criminal, so the second innocent in that column must be Kyle at C3. Bobby’s clue says that both innocents in column C are neighbors of Linda at D3. Paula at C4 is indeed Linda’s neighbor, and Kyle at C3 is also Linda’s neighbor, so this matches the clue and identifies Kyle as that second innocent in column C. Therefore, we can determine that C3 is INNOCENT.

Gabe is at C2, so his neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Among those, the people in row 1 are B1, C1, and D1, and all three are already known to be criminals. Isaac’s clue says exactly 3 of Gabe’s neighboring criminals are in row 1, so those three must be the entire row 1 portion and Gabe can have no other neighboring criminals outside that count beyond the total of 6. Since B2 and D2 are already criminals, the sixth neighboring criminal is fixed, and with B3 and C3 both innocent, that leaves only D3. Therefore, we can determine that D3 is CRIMINAL.

The clue says there are exactly 11 criminals on the edge of the board, and Helen is one of them. Every edge position is already known except A1: on the edge we already have 10 known criminals, along with 3 known innocents, and A1 is the only edge person still undecided. Since the edge total must be exactly 11 criminals, A1 cannot also be a criminal. Therefore, we can determine that A1 is INNOCENT.