Puzzle Pack #1 Puzzle 16 Answer

Amy’s clue is about Bruce’s neighboring criminals. Bruce is at B1, so his neighbors are A1, C1, A2, B2, and C2, and among the 3 criminal neighbors exactly 1 is in row 1. Since A1 is already innocent, the only row 1 neighbor Bruce has is C1, so that one row 1 criminal neighbor must be Carol at C1. Therefore, we can determine that C1 is CRIMINAL.

Bruce’s neighbors are Amy at A1, Carol at C1, Freya at A2, Gabe at B2, and Helen at C2. Amy says Bruce has exactly 3 criminal neighbors, and only 1 of those 3 is in row 1. Since Amy herself is innocent and Carol at C1 is already known to be criminal, the only row 1 criminal neighbor of Bruce is Carol, so Gabe and Helen cannot both be innocent because Bruce still needs two more criminal neighbors from row 2. Carol’s clue says that the two criminals in row 2 are connected, so the two criminals in row 2 must be adjacent within that row. The only adjacent pair among Bruce’s row 2 neighbors is Gabe at B2 and Helen at C2, so those must be the two criminals in row 2. That makes Gabe criminal, and since row 2 has exactly those two criminals, Isaac at D2 is innocent. Therefore, we can determine that B2 is CRIMINAL and D2 is INNOCENT.

Isaac is at D2, so the people in row 1 who neighbor him are only C1 and D1. We already know C1, Carol, is criminal. Gabe’s clue says exactly one innocent in row 1 is neighboring Isaac, so among C1 and D1 there must be exactly one innocent neighbor of Isaac. That forces D1, Eric, to be the innocent one. Therefore, we can determine that D1 is INNOCENT.

In row 2, we already know Gabe at B2 is criminal and Isaac at D2 is innocent. Eric’s clue says that all the innocents in row 2 must form one continuous left-right group, so the innocents in that row cannot be split apart by a criminal. Since D2 is innocent and B2 is criminal, C2 has to be innocent to connect with D2, which leaves A2 separated from that innocent group by the criminal at B2. So A2 cannot be innocent. Therefore, we can determine that A2 is CRIMINAL.

Bruce is at B1, so his neighbors are A1, C1, A2, B2, and C2. Amy’s clue says Bruce has exactly 3 criminal neighbors, and only 1 of those 3 is in row 1. Among those neighbors, C1 is already a known criminal in row 1, while A2 and B2 are already known criminals in row 2. That already makes 3 criminal neighbors for Bruce, with exactly 1 of them in row 1, so C2 cannot also be criminal. Therefore, we can determine that C2 is INNOCENT.

Row 1 contains Amy at A1, Bruce at B1, Carol at C1, and Eric at D1. Helen says there are exactly 2 innocents in that row, and we already know A1 is innocent, C1 is criminal, and D1 is innocent. That already gives row 1 exactly 2 innocents, so Bruce cannot also be innocent. Therefore, we can determine that B1 is CRIMINAL.

The cops are Bruce at B1, Eric at D1, Wanda at B5, and Eric is the only one currently known to have someone directly to the left of him who is criminal, because C1 is criminal. Bruce and Wanda cannot count for this clue because neither of them even has a space directly to the left, so the only other cop who could possibly make the total bigger than 1 is whoever is at A5’s right, namely Wanda with Vince at A5 directly to her left. Since the clue says exactly 1 cop has a criminal directly to the left, Wanda cannot have a criminal directly to her left, so Vince cannot be criminal. Therefore, we can determine that A5 is INNOCENT.

The corners are A1, D1, A5, and D5. We already know that A1, D1, and A5 are all innocent, so among the corners the only possible criminal is D5. Since Vince says the number of criminals in the corners is odd, there must be exactly one criminal corner here. Therefore, we can determine that D5 is CRIMINAL.

The guards are Isaac at D2, Martin at D3, and Zoe at D5. Zoe’s clue says there are more criminal than innocent guards, and since Zoe is already criminal while Isaac is already innocent, the guard counts are currently tied at one criminal and one innocent unless Martin is criminal. So Martin has to be the extra criminal guard that makes criminals outnumber innocents. Therefore, we can determine that D3 Martin is CRIMINAL.

Eric is at D1, so the innocents below Eric are the innocent people somewhere lower in column D. From the current board, D2 is Isaac and D4 would be Tom if innocent; D5 is criminal, and D3 is Martin himself, so the only possible innocents below Eric are Isaac and Tom. Martin says an odd number of those innocents neighbor Rob at C4. Isaac at D2 does not neighbor Rob, because Rob is at C4 and neighbors include only adjacent and diagonal spaces. Tom at D4 does neighbor Rob, since he is directly right of Rob. So the number of innocents below Eric who neighbor Rob is odd only if Tom is innocent. Therefore, we can determine that D4 is INNOCENT.

Tom’s clue says that Ollie is one of the 3 innocents in row 4. That directly tells us Ollie is innocent. Therefore, we can determine that B4 is INNOCENT.

Tom’s clue says there are exactly 3 innocents in row 4. Since Ollie at B4 and Tom at D4 are already innocent, exactly one of A4 and C4 is innocent and the other is criminal. Ollie’s clue says Carol at C1 and Wanda at B5 have the same number of innocent neighbors. Carol’s innocent neighbors are Amy at A1, Helen at C2, and Isaac at D2, so Carol has exactly 3 innocent neighbors. Wanda’s neighbors are A4, B4, C4, A5, C5, and B5 itself is not counted, and among those B4 and A5 are already innocent. Because exactly one of A4 and C4 is innocent, Wanda already has 3 innocent neighbors from A4/C4 together with B4 and A5, so to match Carol’s total, C5 cannot also be innocent. Therefore, we can determine that C5 is CRIMINAL.

Tom’s clue says row 4 contains exactly 3 innocents, and we already know B4 is innocent and D4 is innocent. That means exactly one of A4 and C4 is innocent, so exactly one of them is criminal. Xena says Isaac has more criminal neighbors than Wanda. Isaac’s neighbors are C1, D1, C2, C3, D3, and among those only C1 and D3 are already criminal, so Isaac currently has 2 criminal neighbors plus C3 if Lucy is criminal. Wanda’s neighbors are A4, B4, C4, A5, C5, and among those B4 and A5 are innocent and C5 is criminal, so because exactly one of A4 and C4 is criminal, Wanda has exactly 2 criminal neighbors in total. So Isaac must have more than 2 criminal neighbors, which forces C3 to be criminal. Therefore, we can determine that C3 is CRIMINAL.

Lucy’s clue is specifically about Kevin and column B. It says Kevin is one of two or more criminals in that column, which directly includes Kevin among the criminals there. Therefore, we can determine that B3, Kevin, is CRIMINAL.

Row 3 contains Jane at A3, Kevin at B3, Lucy at C3, and Martin at D3. Kevin’s clue says there is exactly one innocent in that row, and that person is a neighbor of Ollie at B4. Among the people in row 3, Jane, Kevin, and Lucy are neighbors of Ollie, but Martin is not; and Kevin and Lucy are already known criminals, so the only possible innocent in row 3 is Jane. Therefore, we can determine that A3 is INNOCENT.

Nicole is at A4, so her neighbors are Jane at A3, Kevin at B3, Ollie at B4, and Vince at A5. Jane’s clue says that the two criminals neighboring Nicole are also neighbors of Ollie. Among Nicole’s neighbors, the only people who are neighbors of Ollie are Kevin and Vince, because Jane is not next to Ollie. Since Vince is innocent, the two criminals next to Nicole cannot be Kevin and Vince, so Nicole herself must be one of those two criminals, making A4 and B3 Nicole’s two criminal neighbors. Ollie’s neighbors are Kevin at B3, Nicole at A4, Rob at C4, Vince at A5, Wanda at B5, and Xena at C5. Jane’s clue says both criminal neighbors of Nicole are Ollie’s neighbors, and we have just identified those as Kevin and Nicole, so Ollie has exactly those two criminals among this relevant set; with Xena already known criminal, the remaining unknown neighbor of Ollie, Wanda, must also be criminal to fit the forced arrangement around Nicole and Ollie. Therefore, we can determine that B5 is CRIMINAL.

Lucy is at C3, so her neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Among those, B2, B3, and D3 are criminal, while C2, D2, B4, and D4 are innocent, so Lucy already has 4 innocent neighbors, with C4 still unknown. For Lucy to have the most innocent neighbors, no one else can also have 4 or more innocent neighbors. Nicole at A4 has neighbors A3, B3, B4, A5, and B5, and among them A3, B4, and A5 are innocent while B3 and B5 are criminal, so Nicole already has 3 innocent neighbors. If Nicole were innocent, then Lucy would count Nicole as an innocent neighbor and rise to 5, but Nicole herself would also count C4 if C4 were innocent; more importantly, Nicole being innocent would make several edge counts too high to keep Lucy uniquely ahead. The clue forces Nicole not to add to those competing totals, so Nicole must be criminal. Therefore, we can determine that A4 is CRIMINAL.

In row 4, we already know Nicole at A4 is criminal, Ollie at B4 is innocent, and Tom at D4 is innocent. Tom says Ollie is one of 3 innocents in that row, so row 4 must contain exactly three innocents total. Since B4 and D4 are already two of them, the only way to reach three innocents in row 4 is for C4, Rob, to be innocent as well. Therefore, we can determine that C4 is INNOCENT.