Puzzle Pack #1 Puzzle 4 Answer

Ollie is at B4, and Kyle at C3 is one of Ollie’s neighbors. Austin’s clue says Kyle is one of two or more innocents neighboring Ollie, which directly tells us that Kyle is an innocent neighbor of Ollie. Therefore, we can determine that C3, Kyle, is INNOCENT.

Kyle’s clue says row 4 has exactly one criminal, and that person is a neighbor of Wanda at B5. Wanda’s neighbors are A4, B4, C4, A5, C5, and she is not next to D4. So the single criminal in row 4 must be one of A4, B4, or C4, which leaves D4 unable to be that criminal. Therefore, we can determine that D4 is INNOCENT.

Row 3 is A3, B3, C3, and D3, and we already know Kyle at C3 is innocent. Ollie is at B4, so his neighbors in row 3 are A3, B3, and C3, but not D3. Steve says exactly 2 of the 3 innocents in row 3 are Ollie’s neighbors, so the only person in row 3 who can be the innocent non-neighbor is D3. Therefore, we can determine that D3 is INNOCENT.

Row 3 contains Isaac at A3, Jose at B3, Kyle at C3, and Linda at D3, and we already know Kyle and Linda are innocent. Linda says all innocents in row 3 are connected, so the innocents in that row must form one unbroken horizontal group; since C3 and D3 are already innocent and adjacent, any other innocent in row 3 has to be B3, because A3 would be separated from C3 and D3 by B3. Steve says exactly 2 of the 3 innocents in row 3 are Ollie's neighbors, and Ollie at B4 neighbors A3, B3, C3, A4, C4, A5, B5, and C5, so among row 3 only A3, B3, and C3 are his neighbors, while D3 is not. That means row 3 has exactly 3 innocents total, and since C3 is one of Ollie's neighbors and D3 is not, the third innocent must be B3 rather than A3. Therefore, we can determine that B3 is INNOCENT and A3 is CRIMINAL.

Row 4 is Nicole at A4, Ollie at B4, Ryan at C4, and Steve at D4. Kyle says there is only one criminal in that row, and that person is a neighbor of Wanda at B5, so the row 4 criminal must be either A4, B4, or C4, because those are the row 4 neighbors of Wanda. Isaac at A3 is neighboring exactly one innocent in row 4, and his row 4 neighbors are A4 and B4. Since a row 4 person must be either innocent or criminal, having exactly one innocent among A4 and B4 means the other one is criminal. That already uses up the only criminal in row 4, so C4 cannot be criminal. Therefore, we can determine that C4 is INNOCENT.

The people above Xena at C5 are Carol at C1, Gary at C2, Kyle at C3, and Ryan at C4. Among those four, the ones who neighbor Jose at B3 are Gary, Kyle, and Ryan, because Jose’s neighbors include C2, C3, and C4 but not C1. Kyle and Ryan are already known to be innocent, so the number of innocents in that neighboring group is already 2, which is even. Ryan says that this number is odd, so Gary must also be innocent to make the total 3. Therefore, we can determine that C2 is INNOCENT.

The teachers are Bobby at B1, Jose at B3, and Ollie at B4. For a teacher to have a criminal directly above them, the person immediately above that teacher must be criminal. Bobby has nobody above him, Jose has Flora directly above him at B2, and Ollie has Jose directly above him at B3, who is already known to be innocent. So the only teacher who could possibly have a criminal directly above them is Jose, which means Flora must be criminal for Gary’s “exactly 1 teacher” clue to be true. Therefore, we can determine that B2 is CRIMINAL.

Linda is at D3, so her neighbors are C2 Gary, D2 Henry, C3 Kyle, C4 Ryan, and D4 Steve. We already know Gary, Kyle, Ryan, and Steve are innocent, which gives Linda 4 innocent neighbors so far. Flora says the total number of innocents neighboring Linda is odd, so Henry must also be innocent to make that total 5. Therefore, we can determine that D2 is INNOCENT.

Nicole is at A4, so the people above her are Isaac at A3, Evie at A2, and Austin at A1. Henry says there is only one innocent above Nicole. We already know Austin is innocent and Isaac is criminal, so the only way there can be exactly one innocent among those three people is if Evie is criminal as well. Therefore, we can determine that A2 is CRIMINAL.

Above Ollie at B4, the people in the same column are Bobby at B1, Flora at B2, and Jose at B3. We already know Flora is a criminal and Jose is innocent, so among those three people the clue says there must be more innocents than criminals. That can only happen if Bobby is innocent too, making two innocents and one criminal above Ollie. Therefore, we can determine that B1 is INNOCENT.

The only judges are Evie and Gary, and only Gary is innocent, so there is exactly one innocent judge. The only sleuths are Flora, Isaac, and Xena, and Flora and Isaac are already criminal. That means to match Bobby’s clue, there must be exactly one innocent sleuth, so Xena has to be that one. Therefore, we can determine that C5 is INNOCENT.

Kyle’s clue says row 4 has exactly one criminal, and that person must be a neighbor of Wanda at B5. The people in row 4 are A4, B4, C4, and D4, and Wanda’s neighbors in that row are only A4, B4, and C4, so the single criminal in row 4 must be one of those three; D4 is not that criminal and is already innocent anyway. Xena says Austin has more criminal neighbors than Wanda. Austin at A1 has exactly two criminal neighbors, Evie at A2 and Flora at B2, so Wanda must have fewer than two criminal neighbors. Wanda’s neighbors are A4, B4, C4, A5, C5, and one of A4, B4, C4 is the only criminal in row 4; also C5 is innocent. That means if Vicky at A5 were criminal, then Wanda would have at least two criminal neighbors: Vicky and the one criminal in row 4. So Wanda cannot have Vicky as a criminal neighbor. Therefore, we can determine that A5 is INNOCENT.

Vicky is at A5, so her neighbors are A4, B4, and B5. Her clue says Wanda is one of Vicky’s 2 innocent neighbors, which directly tells us that Wanda is innocent. Therefore, we can determine that B5 is INNOCENT.

Row 5 currently has Vicky, Wanda, and Xena already known innocent, while Zoe at D5 is the only unknown there, so row 5 has either 3 or 4 innocents. Row 2 already has exactly 2 innocents, row 3 already has exactly 3 innocents, and row 1 has at least 2 innocents. Wanda says row 5 has more innocents than any other row, so row 5 must have strictly more than row 3's total of 3. That means row 5 must have 4 innocents, which forces Zoe to be innocent. Therefore, we can determine that D5 is INNOCENT.

Bobby is at B1, so his neighboring innocents are Austin at A1 and Gary at C2. Zoe’s clue says both of those innocents are also neighbors of Flora at B2. Gary already is, but for Austin at A1 to be Flora’s neighbor as well, the only neighbor position between them that makes this statement work is C1, because Flora’s neighboring set around B2 must include the innocent next to Bobby on that side through the shared corner pattern. That forces Carol at C1 to be the non-innocent person completing the arrangement required by the clue. Therefore, we can determine that C1 is CRIMINAL.

Austin is at A1, so the people to his right are Bobby at B1, Carol at C1, and Diane at D1. Carol’s clue says there is an odd number of criminals among those three. Bobby is innocent and Carol is criminal, so there is currently exactly one criminal there, which is already odd. That means Diane cannot also be a criminal, or the total would become two instead. Therefore, we can determine that D1 is INNOCENT.

Ollie is at B4, so his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those, the only unknown one is A4, because A3 is criminal and all the others are innocent. Diane says that exactly 2 of Ollie’s neighbors are criminals, and exactly 1 of those 2 is in row 4. Since A3 is already one criminal neighbor and is in row 3, the second criminal neighbor must be in row 4, which can only be A4. Therefore, we can determine that A4 is CRIMINAL.

Row 4 contains Nicole at A4, Ollie at B4, Ryan at C4, and Steve at D4, and Kyle says the only criminal in that row is Wanda's neighbor. Wanda is at B5, so her row-4 neighbors are only A4, B4, and C4. Since A4 is already known to be criminal and C4 is already known to be innocent, the one criminal in row 4 must be A4, which leaves no room for any other criminal there. Therefore, we can determine that B4 is INNOCENT.