Puzzle Pack #2 Puzzle 24 Answer

Luigi’s clue says that Bonnie is one of Hazel’s 2 criminal neighbors. That directly tells us Bonnie is a neighbor of Hazel and is criminal. Bonnie at B1 is indeed next to Hazel at B2, so the clue applies to her. Therefore, we can determine that B1 is CRIMINAL.

Julie is at D2, so the people in row 3 who neighbor Julie are only Maria at C3 and Noah at D3. Bonnie’s clue says exactly 2 innocents in row 3 are neighboring Julie. Since those are the only two row 3 neighbors Julie has, both of them must be innocents. Therefore, we can determine that D3 is INNOCENT, C3 is INNOCENT, and so on.

Isaac is at C2, so his neighbors are Bonnie, Carol, Julie, Hazel, Luigi, Maria, Noah, and Sam. We already know Bonnie is criminal, while Luigi, Maria, and Noah are innocent, so among Isaac’s neighbors the six innocents must include Hazel, Carol, Julie, and Sam together with those three known innocents, leaving exactly one of Bonnie or the unknowns out as the only criminal neighbor. The ones among Isaac’s neighbors who are above Will at B5 are exactly Bonnie, Carol, Julie, Hazel, Luigi, Maria, Noah, and Sam, and the clue says exactly 2 of the 6 innocent neighbors are above Will; since Luigi, Maria, and Noah are already three innocent neighbors above Will, Hazel cannot be criminal without making the count impossible. Therefore, we can determine that B2 is INNOCENT.

Vince is at A5, so the people above Vince are A1, A2, A3, and A4. Hazel says all innocents among those four are connected, so any innocent people in column A must form one unbroken vertical group with no gaps. Luigi says Bonnie is one of Hazel's two criminal neighbors. Hazel's neighbors are A1, B1, C1, A2, C2, A3, B3, and C3, and we already know B1 is Bonnie the criminal while B3 and C3 are innocent. That means Hazel has exactly one more criminal neighbor among A1, C1, A2, C2, and A3. Since A3 is next to Hazel and also lies in the group above Vince, making A3 criminal would split any innocents in column A above Vince into separate pieces unless A2 stayed innocent to keep the innocent group connected. Therefore, we can determine that A2 is INNOCENT.

The people above Vince are Gary at A2, Kay at A3, Olive at A4, and Alex at A1. Hazel’s clue says that every innocent among those four must form one orthogonally connected group. Gary at A2 is already innocent, so if Kay at A3 were not innocent, then the innocents above Vince would have to skip over A3. That would leave Gary separated from any innocent higher up in column A, because A3 is the only square connecting A2 to A4 and A1. So the innocents above Vince could not all be connected unless Kay is also innocent. Therefore, we can determine that A3 is INNOCENT.

The teachers are Alex at A1, Maria at C3, Thor at D4, and Sam at C4. For a teacher to have an innocent directly above them, the person immediately above must exist and be innocent, so Maria qualifies because Isaac is directly above her at C2, while Thor does not because Noah above him at D3 is innocent but Thor is a teacher with an innocent above only if D3 is above D4, and Sam does not because Maria above him at C3 is innocent, while Alex cannot qualify because nobody is above A1. Kay says exactly 2 teachers have an innocent directly above them, and one of them is already Sam because Maria at C3 is innocent. The only way to make the second one is for Maria to qualify as well, which requires Isaac at C2 to be innocent. Therefore, we can determine that C2 is INNOCENT.

Hazel’s neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. We already know A2, C2, A3, B3, and C3 are innocent, and Luigi’s clue says Bonnie at B1 is one of Hazel’s exactly 2 criminal neighbors. That leaves only A1 and C1 as the possible second criminal neighbor, so among Hazel’s row 1 neighbors, exactly one of A1 and C1 is criminal and the other is innocent. Isaac’s clue says row 1 has exactly 2 innocents total. Since A1 and C1 contain exactly one innocent between them, the second innocent in row 1 must be D1. Therefore, we can determine that D1 is INNOCENT.

Alex at A1 has only three neighbors: Bonnie, Gary, and Hazel. Since Bonnie is criminal and Gary and Hazel are innocent, Alex has exactly one criminal neighbor. Emma’s clue says Alex has more criminal neighbors than Thor, so Thor must have fewer than one, which means Thor has zero criminal neighbors. Thor at D4 is next to Maria, Noah, Sam, Xia, and Zed. Maria and Noah are already innocent, so for Thor to have zero criminal neighbors, Sam, Xia, and Zed must also all be innocent. Therefore, we can determine that C4 is INNOCENT, C5 is INNOCENT, and D5 is INNOCENT.

Will is at B5, so the people above Will are Bonnie at B1, Hazel at B2, Luigi at B3, and Paula at B4. Zed says exactly 3 of those 4 people are innocents. Hazel and Luigi are already known to be innocent, and Bonnie is already known to be criminal, so the third innocent above Will must be Paula at B4. Therefore, we can determine that B4 is INNOCENT.

Column B already contains Bonnie at B1 as a criminal, while Hazel at B2, Luigi at B3, and Paula at B4 are all innocent. Since Paula says column B is the only column with exactly 2 criminals, column B must have one more criminal besides Bonnie. The only remaining person in that column is Will at B5, so he has to be that second criminal. Therefore, we can determine that B5 is CRIMINAL.

Column B already has exactly 2 criminals, because Bonnie at B1 and Will at B5 are criminals while Hazel, Luigi, and Paula are innocent. Paula’s clue says column B is the only column with exactly 2 criminals, so no other column can also have exactly 2 criminals. In column A, Gary at A2 and Kay at A3 are innocent, so the only unknowns there are Alex at A1, Olive at A4, and Vince at A5. That means column A would have exactly 2 criminals if Olive were criminal together with exactly one of Alex or Vince, so Olive cannot be criminal here without making column A fit Paula’s forbidden count. Therefore, we can determine that A4 is INNOCENT.

Row 3 and row 4 are already all innocents, so they do not have exactly 2 innocents. Row 2 already has three innocents, and because Hazel at B2 has Bonnie at B1 as one of her two criminal neighbors, Hazel’s other criminal neighbor must be either Alex at A1 or Carol at C1, which makes row 1 have at least two criminals and therefore at most two innocents. Since Olive says only one row has exactly 2 innocents, row 1 must be that row, so row 5 cannot also have exactly 2 innocents. But row 5 already has Will criminal and Xia and Zed innocent, so Vince at A5 cannot be criminal, because that would make row 5 have exactly 2 innocents. Therefore, we can determine that A5 is INNOCENT.

Gary is at A2 and Julie is at D2. Gary’s neighbors are Alex, Bonnie, Hazel, Kay, and Luigi, and among them only Bonnie is criminal, so Gary has 1 criminal neighbor. Julie’s neighbors are Carol, Emma, Isaac, Maria, Noah, Sam, Thor, and Zed. Since Vince says Gary and Julie have the same number of criminal neighbors, Julie must also have exactly 1 criminal neighbor. In Julie’s neighborhood, Emma, Isaac, Maria, Noah, Sam, and Zed are already innocent, so the only possible criminal neighbors there are Carol and Thor, and exactly one of them must be criminal. Thor is also a neighbor of Zed at D5, and Zed is innocent. Zed’s other neighbors are Sam, Will, and Xia, with only Will criminal, so Zed has 1 criminal neighbor already. That means Thor cannot also be criminal, or Zed would have 2 criminal neighbors. So Thor is innocent, leaving Carol as Julie’s one criminal neighbor. Then Gary’s one criminal neighbor must remain only Bonnie, so Alex cannot be criminal. Therefore, we can determine that C1 is CRIMINAL and A1 is INNOCENT.

Isaac is at C2, so his six neighboring innocents are the six known innocent neighbors around him: B1 Bonnie is not innocent, but B2 Hazel, D1 Emma, B3 Luigi, C3 Maria, D3 Noah, and C1 Carol is not innocent, leaving A2 irrelevant because it is not adjacent. Among Isaac’s actual neighboring innocents, the ones above Will at B5 are simply those in rows 1 to 4, so every innocent neighbor of Isaac except possibly Julie at D2 is already above Will. Since Noah says exactly 2 of Isaac’s neighboring innocents are above Will, Julie cannot be another neighbor who changes that count upward, so she must be innocent in the only way that keeps the statement true for the already fixed set. Therefore, we can determine that D2 is INNOCENT.

Luigi is at B3, so his neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. In the current board, all eight of those people are already known to be innocent, so Luigi has 8 innocent neighbors. The clue says Luigi has the most innocent neighbors, and “the most” must be unique. Thor at D4 has neighbors C3, D3, C4, C5, D5, C4, C5, and D5 around him only once each, namely C3, D3, C4, C5, D5, C4? More simply, Thor’s actual neighboring spaces are C3, D3, C4, C5, and D5, plus C4 and C5 are already counted, and also C3 and D3 above, and C5 and D5 below, with C4 to the left and C3/C5 diagonally; among these, every neighbor except D4 itself is known innocent if Thor were innocent. That would give Thor 8 innocent neighbors as well: C3, D3, C4, C5, D5, plus the three on the left side B? No—looking correctly, D4’s neighbors are C3, D3, C4, C5, D5, and also D? Since D4 is on the edge, it has exactly five neighbors, so Thor cannot tie Luigi. The only way Gary’s clue matters here is through the people who could challenge Luigi’s unique maximum, and the full interior positions are B2, C2, B3, and C3. B2, C2, and C3 each already have at least one criminal neighbor because Bonnie at B1 and Carol at C1 are criminals, so none of them can reach 8 innocent neighbors. The remaining interior position is C4’s neighbor set around D4’s area: if Thor were innocent, then Sam at C4 would have all eight neighbors innocent, tying Luigi. Since Luigi must be uniquely highest, Thor cannot be innocent. Therefore, we can determine that D4 is CRIMINAL.