Puzzle Pack #1 Puzzle 23 Answer

Ellie is at D1, so the people below her are Isaac at D2, Maria at D3, Tom at D4, and Zara at D5. Zara’s clue says that the criminals among those four form one connected group, and since Zara herself is already known to be innocent, any criminals in that column below Ellie must be linked through the remaining spaces. That forces the middle space D3 to be criminal, because without Maria there would be no way to keep criminals on both sides of her in one continuous connected chain. Therefore, we can determine that D3 is CRIMINAL.

In column D, Maria at D3 is a criminal and Zara at D5 is innocent. Maria says there are more criminals than innocents in column D, so among D1, D2, and D4 there must be at least two criminals to make criminals the majority in that column. Zara says both criminals below Ellie are connected, which means there are exactly two criminals below D1 and they must form one orthogonally connected group in column D. Since one of them is already Maria at D3, the other must be either D2 or D4, and D2 and D4 cannot both be criminals. That leaves only one criminal below Ellie, so Ellie herself must be the second criminal needed for column D to have more criminals than innocents. Therefore, we can determine that D1 is CRIMINAL.

Below Ellie in column D, the criminals are already Ellie at D1 and Maria at D3, so Zara’s clue means every criminal strictly between them must form one unbroken vertical group with Maria. The only people between D1 and D5 who can connect that lower criminal group upward are D2 and D4, and since D5 is innocent, that clue forces the criminal group below Ellie to extend through the matching middle positions in that band rather than break apart. Ellie also says Maria has only one innocent neighbor. Maria’s neighbors are C2, D2, C3, C4, and D4, because D5 is already known innocent. So among C2, D2, C3, C4, and D4, exactly one can be innocent. That means the cells next to Maria on the C side cannot leave gaps in the connected criminal group while also keeping Maria to only one innocent neighbor. The only way to satisfy both clues together is for C2, C3, and C4 all to be criminals. Therefore, we can determine that C2 is CRIMINAL, C3 is CRIMINAL, and C4 is CRIMINAL.

The key people here are the criminals below Ellie in column D: Maria at D3 is already known to be a criminal, and the only spaces below Ellie besides her are D2 and D4. Zara’s clue says both criminals below Ellie are connected, so there must be exactly one more criminal among D2 and D4, and it must connect to Maria by orthogonal adjacency. Since D2 is not directly connected to D3 but D4 is, that second criminal has to be Tom at D4. Now look at Larry’s clue comparing Maria’s and Vicky’s innocent neighbors. Maria’s neighbors are Hilda, Isaac, Larry, Tom, Peter, and Zara; with Hilda, Larry, Peter, and now Tom all criminal, and Zara innocent, Maria can have at most two innocent neighbors. Vicky’s neighbors are Nancy, Wanda, and Olive, so for Maria to have more innocent neighbors than Vicky, Vicky must have fewer than two innocent neighbors. Since Vicky’s three neighbors are only Nancy, Wanda, and Olive, that forces all three of them to be criminals. Therefore, we can determine that A4 is CRIMINAL, B4 is CRIMINAL, and B5 is CRIMINAL.

Peter is at C4, so his neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Hilda says Maria is one of Peter's 6 criminal neighbors, so Peter has exactly 6 criminal neighbors in total. Among those eight neighbors, C3 Larry, D3 Maria, B4 Olive, B5 Wanda, and D5 Zara are already known, with four criminals and one innocent, so the remaining three spots B3, D4, and C5 must contain exactly two criminals. Olive says she and Peter share an odd number of innocent neighbors. The common neighbors of Olive at B4 and Peter at C4 are B3, C3, B5, C5, and D5. Among these, C3 and B5 are criminals and D5 is innocent, so the number of innocent common neighbors depends only on B3 and C5, and it must be odd. Since B3 and C5 together also must account for exactly one of the two criminals among B3, D4, and C5, B3 and C5 cannot both be criminals or both be innocent; exactly one of them is innocent. That gives Peter exactly two innocent neighbors total, namely D5 and one of B3 or C5, so the only way to reach 6 criminal neighbors is for D4 to be criminal. Therefore, we can determine that D4 is CRIMINAL.

Ellie is at D1, so the people below her in the same column are Isaac at D2, Maria at D3, Tom at D4, and Zara at D5. Zara’s clue says that the criminals among those people must form one continuous vertical group with no innocent breaking them apart. Since Maria at D3 and Tom at D4 are criminals, that group already exists there, and Zara at D5 is innocent, so Isaac at D2 cannot be a criminal above them or the criminals below Ellie would be split by an innocent at D5 from any lower extension and, more importantly here, the criminal set below Ellie is exactly the connected block at D3 and D4. Therefore, we can determine that D2 is INNOCENT.

Olive is at B4, so her criminal neighbors are the seven surrounding spots A3, B3, C3, A4, C4, B5, and C5. Karen is at B3, and among those seven, the ones who also neighbor Karen are A3, C3, A4, C4, and B5, while C5 does not neighbor Karen. Isaac says exactly 4 of Olive’s 7 criminal neighbors also neighbor Karen, so among those five shared-neighbor positions exactly four are criminal. Since C3, A4, C4, and B5 are already known criminals, A3 must be the fourth one. Therefore, we can determine that A3 is CRIMINAL.

Olive is at B4 and Peter is at C4. Their common neighbors are B3, C3, B5, C5, and D5, and among those we already know B3, C3, and B5 are criminal while D5 is innocent, so the only uncertainty there is C5. That means the number of innocent neighbors shared by Olive and Peter is either 1 if C5 is criminal or 2 if C5 is innocent, and Olive says that number is odd, so C5 must be criminal. Isaac says exactly 4 of Olive's 7 criminal neighbors also neighbor Karen. Olive's criminal neighbors are A3, B3, C3, A4, C4, B5, and now C5; among these, the ones that neighbor Karen are A3, C3, A4, C4, B5, and C5, while A3, A4, C4, B5, and C5 are the only ones among them that can make the total exactly 4 only if A5 is criminal rather than innocent. Therefore, we can determine that A5 is CRIMINAL.

Column C already has three known criminals: Hilda at C2, Larry at C3, and Peter at C4. Wanda’s clue says there are exactly four criminals in column C, and only one of those four is on the edge; since C2, C3, and C4 are all not edge spaces, the fourth criminal in that column must be C5, so Xavi is criminal. Jose’s clue says the number of criminal pilots equals the number of criminal sleuths. With Hilda and now Xavi, there are two criminal sleuths, and among pilots Vicky is already criminal while Gabe is the only other pilot whose status is still unknown. So Gabe must also be criminal to make the criminal pilots total two. Therefore, we can determine that B2 is CRIMINAL.

Peter is at C4, so his neighbors are B3 Karen, C3 Larry, D3 Maria, B4 Olive, D4 Tom, B5 Wanda, C5 Xavi, and D5 Zara. Hilda says Maria is one of Peter's 6 criminal neighbors, so Peter has exactly 6 criminal neighbors in total. Among those eight neighbors, Larry, Maria, Olive, Tom, and Wanda are already criminal, and Zara is innocent, so the only way to reach exactly 6 criminal neighbors is for exactly one of Karen and Xavi to be criminal. Gabe says Karen has the most criminal neighbors, and Karen already touches six known criminals around her, so if Xavi were the extra criminal then Karen would have 7 criminal neighbors, which would tie Olive's 7 and fail to be uniquely the most. Therefore the extra criminal neighbor must be Karen, which makes Xavi innocent, and since Karen is criminal, Frank also reaches 7 criminal neighbors around him and ties her unless Frank is criminal too. Therefore, we can determine that A2 is CRIMINAL.

Vicky at A5 has three neighbors: Nancy at A4, Olive at B4, and Wanda at B5, and all three are criminals. So Nancy’s clue says Alice at A1 must have fewer than three criminal neighbors. Alice’s neighbors are Bobby at B1, Frank at A2, and Gabe at B2. Frank and Gabe are already criminals, so if Bobby were also criminal then Alice would also have three criminal neighbors. That would not be fewer than Vicky’s. Therefore, we can determine that B1 is INNOCENT.

Wanda says there are exactly 4 criminals in column C, and we already know three of them there: Hilda at C2, Larry at C3, and Peter at C4. So the remaining space in that column, either Carl at C1 or Xavi at C5, must also be criminal. Wanda also says only 1 of those 4 column-C criminals is on an edge, and since C2, C3, and C4 are not edge spaces, that means the fourth criminal is the only edge one, so either C1 or C5 is criminal. Now use Bobby’s clue that column A has more criminals than any other column. Column B already has 4 criminals, and column D also has 4 criminals, so column A must have more than 4 criminals. Since a column only has 5 spaces and column A already has four known criminals at A2, A3, A4, and A5, Alice at A1 must be the fifth criminal there. Therefore, we can determine that A1 is CRIMINAL.

Alice says there are 4 innocents in total. We already know three innocents for sure: Bobby at B1, Isaac at D2, and Zara at D5, so there is room for only one more innocent on the whole board. Olive and Peter are B4 and C4. Olive’s innocent neighbors are only Karen at B3 if Karen is innocent, because everyone else around B4 is criminal. Peter’s innocent neighbors are Karen at B3, Xavi at C5, and Zara at D5, so Peter has either 1, 2, or 3 innocent neighbors depending on Karen and Xavi. Since Olive says she and Peter each have an odd number of innocent neighbors, Olive must have 1 innocent neighbor, so Karen is innocent, and then Peter must also have an odd count, which forces Xavi to be innocent as well. That uses up the only remaining innocent slot and more, so the only unresolved person left among the unknowns cannot be innocent. Therefore, we can determine that C1 is CRIMINAL.

Column C has four criminals total, and the edge squares in that column are C1 and C5. We already know C1, Carl, is a criminal, so the clue’s one edge criminal in column C is already accounted for. That means the other edge square in column C, C5, cannot also be criminal. Therefore, we can determine that C5 is INNOCENT.

Olive is at B4 and Peter is at C4, so we only need to compare the neighbors around those two cells. Their common neighbors are B3 Karen, C3 Larry, D3 Maria, B4 Olive, D4 Tom, B5 Wanda, C5 Xavi, and D5 Zara, and among those the only known innocents are Xavi and Zara. Since Olive says that she and Peter share an odd number of innocent neighbors, Karen must also be innocent to make that shared innocent count odd. Therefore, we can determine that B3 is CRIMINAL.