Puzzle Pack #1 Puzzle 1 Answer

Column A contains Alex at A1, Frank at A2, Keith at A3, Paula at A4, and Vicky at A5. Megan is at B3, so her neighbors in column A are exactly Frank at A2, Keith at A3, and Paula at A4. Frank’s clue says exactly one innocent in column A is neighboring Megan, and Frank is already known to be innocent, so that one innocent must be Frank. That leaves Keith and Paula as not innocent. Therefore, we can determine that A4 Paula is CRIMINAL and A3 Keith is CRIMINAL.

Sofia is at C4, so the people to her left are Paula at A4 and Ryan at B4. Keith’s clue says an odd number of those people are criminals. Paula is already known to be a criminal, so there is already one criminal to Sofia’s left. That means Ryan cannot also be a criminal, or the total would become two instead of odd. Therefore, we can determine that B4 Ryan is INNOCENT.

Keith is at A3, and "above Keith" means the people in the same column above him: Frank at A2 and Alex at A1. Ryan says there is only one innocent among those two. Since Frank is already known to be innocent, Alex cannot also be innocent. Therefore, we can determine that A1 Alex is CRIMINAL.

Alex is at A1, and the people below him in the same column are Frank at A2, Keith at A3, Paula at A4, and Vicky at A5. His clue says that the criminals below him are exactly two people, and those two criminals must be connected by orthogonal adjacency with no break. We already know Keith at A3 and Paula at A4 are both criminals, and they are directly next to each other, so they already form the connected pair of criminals below Alex. That leaves no room for Vicky at A5 to also be a criminal below Alex. Therefore, we can determine that A5 is INNOCENT.

Vicky’s clue says that Xavi is one of the 4 innocents in column C. That directly states Xavi’s identity as one of the innocents there. Therefore, we can determine that C5 is INNOCENT.

Vicky says that Xavi is one of 4 innocents in column C, so in column C exactly four people are innocent. In that column, Xavi at C5 is already innocent, and the only people there who could still be criminal are Chris at C1, Isaac at C2, Nancy at C3, and Sofia at C4. Xavi says there are as many criminal farmers as criminal guards. We already know two criminal farmers, Keith at A3 and Paula at A4, and Paula is a cop, not a farmer. So the criminal farmers are Alex at A1 and Keith at A3, which makes 2 criminal farmers in total. That means there must also be exactly 2 criminal guards. The guards are Isaac at C2, Nancy at C3, and Sofia at C4, all in column C. So among those three, exactly two are criminal, which means the remaining one is innocent. Since column C must contain exactly four innocents total and Xavi is already one of them, there can be only one criminal in the rest of column C besides those guards, so Chris cannot be criminal. Therefore, we can determine that C1 is INNOCENT.

Column C contains Chris at C1, Isaac at C2, Nancy at C3, Sofia at C4, and Xavi at C5. Vicky says Xavi is one of 4 innocents in that column, and we already know Chris and Xavi are innocent, so exactly two of Isaac, Nancy, and Sofia must also be innocent. Chris says exactly one guard has a criminal directly below them. The guards are Isaac at C2, Nancy at C3, and Sofia at C4, and the people directly below them are Nancy, Sofia, and Xavi. Since Xavi at C5 is innocent, Sofia cannot be that guard, so the one qualifying guard must be either Isaac or Nancy. That means at least one of Nancy or Sofia is criminal, leaving only one criminal total among Isaac, Nancy, and Sofia, so Isaac must be one of the two innocents there. Therefore, we can determine that C2 is INNOCENT.

Megan is at B3, so her neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. The clue says exactly 2 of Megan’s 3 criminal neighbors are above Vicky, and since Vicky is at A5, every one of those neighboring spaces is above her. So Megan must have exactly 3 criminal neighbors in total. Among those neighbors, A3 and A4 are already known criminals, while A2, C2, and B4 are known innocents. That means exactly one of B2, C3, and C4 is criminal, so the other two are innocent. Vicky’s clue says there are exactly 4 innocents in column C; with C1, C2, and C5 already innocent, exactly one of C3 and C4 can be innocent. Therefore both C3 and C4 cannot be innocent, so Helen at B2 cannot be the one criminal among B2, C3, and C4. Therefore, we can determine that B2 is INNOCENT.

In row 5, the people already known to be innocent are A5 and C5. For all innocents in that row to be connected by orthogonal adjacency, those two innocents must be linked through the spaces between them in the same row. The only person between A5 and C5 is B5, so B5 has to be innocent to make that connected group possible. Therefore, we can determine that B5 is INNOCENT.

Helen is at B2, so her neighbors in row 1 are A1 Alex, B1 Bonnie, and C1 Chris. Alex is already known to be criminal and Chris is already known to be innocent, so among those three row 1 neighbors, exactly one innocent is already fixed. Wally says the total number of innocents there is odd, so Bonnie must be the other innocent-count change and cannot be innocent. Therefore, we can determine that B1 Bonnie is CRIMINAL.

Vicky says Xavi is one of 4 innocents in column C, so column C must contain exactly 4 innocents. We already know C1, C2, and C5 are innocent, which means the only way for column C to reach 4 innocents is for C4, Sofia, to be innocent too. Bonnie says column C has more innocents than any other column, so no other column can also have 4 innocents. In column B, B1, B2, B4, and B5 are already known, and three of them are innocent, so if B3 Megan were innocent then column B would also have 4 innocents. That is impossible, so Megan cannot be innocent. Therefore, we can determine that B3 is CRIMINAL.

Row 1 contains Alex, Bonnie, Chris, and Ellie. We already know Alex and Bonnie are criminals, Chris is innocent, and Ellie is the only unknown in that row. Since Megan says there are more criminals than innocents in row 1, row 1 must have 3 criminals and 1 innocent, so Ellie has to be the third criminal. Therefore, we can determine that D1 is CRIMINAL.

Chris at C1 has neighbors A1, B1, B2, C2, D2, and D1, so among his known neighbors there are three criminals, two innocents, and Julie is the only unknown. Paula at A4 has neighbors A3, B3, B4, A5, and B5, which gives her exactly two criminal neighbors. Since Ellie says Chris has more criminal neighbors than Paula, Chris must have more than two criminal neighbors, and that is only possible if Julie is criminal. Therefore, we can determine that D2 is CRIMINAL.

Julie’s clue says that Terry is an innocent, and also that he is among at least two innocents on edge spaces. Since everyone tells the truth, that statement directly fixes Terry’s identity. Therefore, we can determine that D4 is INNOCENT.

Zara is at D5, so the people above Zara are D1, D2, D3, and D4. Terry’s clue says the two criminals among those people are both neighbors of Isaac at C2. Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3, so among the people above Zara, D1, D2, and D3 are Isaac’s neighbors, but D4 is not. We already know D1 and D2 are criminals, which gives the full pair of criminals above Zara that Terry’s clue refers to. That means D3 cannot also be a criminal. Therefore, we can determine that D3 is INNOCENT.

Julie is at D2, so the people below her in the same column are Olof at D3, Terry at D4, and Zara at D5. Among those three, Olof and Terry are already known to be innocent, so the only criminal below Julie must be Zara. Olof’s clue says that this only criminal below Julie is Terry’s neighbor, which fits Zara because D5 is directly below Terry at D4. Therefore, we can determine that D5, Zara, is CRIMINAL.

Xavi at C5 has neighbors B4, C4, D4, B5, and D5, while Isaac at C2 has neighbors B1, C1, D1, B2, D2, B3, C3, and D3. From what is already known, Xavi currently has three known innocent neighbors at B4, D4, and B5, and one known criminal neighbor at D5, so Sofia at C4 is the only undecided neighbor in Xavi’s count. Isaac has four known innocent neighbors at C1, B2, C3? no, D3, and B2 again would not count twice; more simply, his known innocents are C1, B2, and D3, while his known criminals are B1, D1, D2, and B3, leaving Nancy at C3 as the only undecided neighbor there. Since Zara says Xavi has more innocent neighbors than Isaac, Xavi must have 4 innocent neighbors and Isaac must have only 3, so Sofia must be innocent and Nancy must be criminal. Therefore, we can determine that C4 is INNOCENT and C3 is CRIMINAL.