Puzzle Pack #2 Puzzle 9 Answer

Kay is at B3, so the people to her right in the same row are exactly Noah at C3 and Olof at D3. Freya’s clue says there are exactly 2 innocents to the right of Kay. Since those are the only two people to Kay’s right, both of them must be innocent. Therefore, we can determine that C3 is INNOCENT and D3 is INNOCENT.

Kay is at B3, and Stella is at C4, which is one of Kay's neighboring squares. Olof's clue says Stella is one of Kay's 3 criminal neighbors, so it directly tells us that Stella is criminal. Therefore, we can determine that C4 is CRIMINAL.

Isaac is at D2, so the people to his left are A2 Freya, B2 Gary, and C2 Helen. Barb is at A1, and among those three, the only ones neighboring Barb are Freya at A2 and Gary at B2; Helen at C2 is not a neighbor of A1. Stella says exactly one innocent to the left of Isaac is neighboring Barb, and Freya is already known to be innocent, so Gary cannot also be innocent. Therefore, we can determine that B2 Gary is CRIMINAL.

Noah is at C3, so his four innocent neighbors must be chosen from B2, C2, D2, B3, B4, and D4, since A2 is too far away and D3 is Noah himself. Gary at B2 is criminal, and D3 Olof is already known innocent, so among Noah’s neighbors the innocents include D3? Wait, D3 is not a neighbor of Noah’s clue count because Noah’s neighbors are the eight surrounding spaces, and the actual neighboring innocents currently known are D3 plus possibly C2, D2, B3, B4, and D4, with B2 definitely not innocent. The clue says exactly 2 of Noah’s 4 innocent neighbors are above Wally at B5, and since every neighbor of Noah in row 2 or row 3 is above B5 while B4 is also above B5, the only neighboring innocent not above Wally would have to be D4 if it were innocent; that means Noah’s four innocent neighbors must be Olof, Kay, Ryan, and one of the row 2 neighbors, forcing both B3 and B4 to be innocent. Therefore, we can determine that B3 is INNOCENT and B4 is INNOCENT.

In row 4, Ryan at B4 and Stella at C4 are already known, and the clue says the two innocents in that row must be connected by orthogonal adjacency. Since Stella at C4 is criminal, Ryan cannot be connected to an innocent on the right, so the other innocent in row 4 has to be Paul at A4, directly next to Ryan. That leaves Thor at D4 as the remaining non-innocent in the row. Therefore, we can determine that A4 is INNOCENT and D4 is CRIMINAL.

Xena at C5 has only five neighbors: B4, C4, D4, B5, and D5. Among those, B4 is Ryan, who is innocent, while C4 and D4 are criminal and B5 and D5 are still unknown. So Xena already has at least one innocent neighbor no matter what Xena herself is, and this clue places no restriction on C5 by itself. Therefore, we can determine that C5 is INNOCENT.

In row 5, Xena at C5 is innocent, and the only other possible innocent in that row would have to be adjacent to her by a left-right chain, because the two innocents in the row must be connected. Zara at D5 cannot be that second innocent, since C5 and D5 would then be the only two innocents and Vince at A5 would be separated from them by B5, so the row’s innocents would not fit the clue’s forced connected pair placement. That leaves A5 unable to be one of the two innocents in row 5. Therefore, we can determine that A5 is CRIMINAL.

Olof is at D3, so the only person below Olof is Zara at D5. Vince’s clue says that no one below Olof has more than 3 innocent neighbors, so Zara must have at most 3 innocent neighbors. Zara’s neighbors are Thor at D4, Stella at C4, and Xena at C5, and among those only Xena is innocent, so Zara cannot be innocent herself, because then she would have 4 innocent neighbors in total when counting the neighboring innocents around her. Therefore, we can determine that D5 is CRIMINAL.

In row 5, the two innocents are connected, so the innocent people in that row must form one unbroken horizontal group. We already know C5 is innocent, while A5 and D5 are criminals. Since A5 and D5 cannot be part of the innocent group, the only way for the two innocents in row 5 to be connected is for them to be B5 and C5, next to each other. Therefore, we can determine that B5 is INNOCENT.

Noah is at C3, and his four innocent neighbors are the already known innocents B3 Kay, D3 Olof, B4 Ryan, and C4 Stella is not innocent, so the four innocents neighboring Noah must be A2 Freya, B3 Kay, D3 Olof, and B4 Ryan only if we count actual neighbors; among Noah’s actual neighboring innocents we have B3, D3, B4, and D2 is unknown, so Gary’s clue fixes how many of Noah’s innocent neighbors are above Wally at B5. Everyone above row 5 is above Wally, so Gary’s clue means exactly two of Noah’s neighboring positions are innocent, which matches the already known innocents B3 and D3 and leaves C2 unable to be innocent. Since C2 is not innocent, Zara’s clue about Emily at D1 matters: Emily’s neighbors are C1, C2, and D2, and Zara says an odd number of those are innocent. With C2 not innocent and D2 still not fixed here, the parity forces C1 not to be innocent. Therefore, we can determine that C1 is CRIMINAL.

Denis’s clue is about row 1: among A1, B1, C1, and D1, only one of them has exactly one innocent neighbor. In that row, C1 already has exactly one innocent neighbor because its neighbors are B1, B2, C2, and D2, and only B1 could be innocent there since B2 is criminal. D1 cannot be the unique one, because D1’s neighbors include C2 and D2 and it would need both of them not to be innocent to stay at exactly one, while Gary’s clue about Noah’s four innocent neighbors forces the count above Wally to come out correctly only if B1 is innocent among the row-1 positions above Wally. That makes B1 the needed innocent and fixes Claire’s status. Therefore, we can determine that B1 is INNOCENT.

Noah’s neighbors are Gary, Helen, Isaac, Kay, Stella, Ryan, Thor, and Wally. Among them, the innocents are Kay, Ryan, and Wally for sure, and Gary’s clue says exactly 2 of Noah’s 4 innocent neighbors are above Wally; since Wally is on row 5, every neighbor of Noah is above Wally, so Noah must have exactly 4 innocent neighbors. That means Helen and Isaac are the remaining two innocents among Noah’s neighbors, so D2 is innocent. Now compare Denis and Thor’s criminal neighbors. Denis’s neighbors are Claire, Emily, Gary, Helen, Isaac, and Noah, and with Claire and Noah innocent, Gary criminal, Helen and Isaac innocent, Denis can have only 2 criminal neighbors if Emily is criminal but only 1 if Emily is innocent. Thor’s neighbors are Noah, Olof, Stella, Xena, and Zara, of whom Stella and Zara are criminal and the others innocent, so Thor has exactly 2 criminal neighbors. Claire says Denis has more criminal neighbors than Thor, so Denis must have more than 2, which forces Emily to be criminal. Therefore, we can determine that D1 is CRIMINAL.

Paul at A4 has five neighbors: Julie, Kay, Ryan, Vince, and Wally. Since Kay, Ryan, Wally, and Paul’s other known neighbors there are innocent while Vince is criminal, Paul’s number of innocent neighbors depends only on Julie, so Paul has either 4 innocent neighbors if Julie is innocent or 3 if Julie is criminal. Thor at D4 has five neighbors too: Noah, Olof, Stella, Xena, and Zara. Among them, Noah, Olof, and Xena are innocent, while Stella and Zara are criminal, so Thor has exactly 3 innocent neighbors. Emily’s clue says Thor and Paul have the same number of innocent neighbors, so Paul must also have 3 innocent neighbors. That only happens if Julie is criminal. Therefore, we can determine that A3 is CRIMINAL.

Kay is at B3, so her neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. Among those eight people, the criminals already known are Gary at B2, Julie at A3, and Stella at C4, which makes exactly 3 criminal neighbors. Since Olof says Stella is one of Kay's 3 criminal neighbors, Kay cannot have any other criminal neighbor besides those three, and the only still-unknown neighbor in that group is Helen at C2. Therefore, we can determine that C2 is INNOCENT.

Noah is at C3, so his four innocent neighbors are Helen at C2, Olof at D3, Ryan at B4, and Isaac at D2. Wally is at B5, and each of C2, D3, and B4 is above Wally, so those three already count as innocents neighboring Noah who are above Wally. Gary’s clue says exactly 2 of Noah’s 4 innocent neighbors are above Wally, so Isaac cannot be one of those four innocent neighbors. Since Isaac is Noah’s neighbor, Isaac cannot be innocent. Therefore, we can determine that D2 is CRIMINAL.

Helen says there are 10 innocents in total. On the board, 10 people are already known to be innocent: B1, A2, C2, B3, C3, D3, A4, B4, B5, and C5. Since that already reaches the total of 10, no unknown person can also be innocent, so A1 must be criminal. Therefore, we can determine that A1 is CRIMINAL.