Puzzle Pack #1 Puzzle 47 Answer

Above Xena means the people in column D above D5, namely David at D1, Isaac at D2, Martin at D3, and Sofia at D4. Isaac says both innocents above Xena are connected, so among those four people there are exactly two innocents, and because they are in one column they must be next to each other. Isaac himself is one of them because D2 is already known innocent, so the other innocent above Xena must be directly adjacent to Isaac, at D1 or D3. That leaves D4 unable to be one of the two innocents above Xena. Therefore, we can determine that D4 is CRIMINAL.

David is at D1, so the people to his left are A1, B1, and C1. Sofia says that among those three, there is exactly one innocent, and that innocent is to the right of Austin. Since nobody is to the right of Austin within that group except B1 and C1, Austin cannot be the only innocent to the left of David. Therefore, we can determine that A1 is CRIMINAL.

Austin’s clue is about Hilda at C2. The five neighbors of C2 are B1, C1, B2, B3, and C3, because D2 is Isaac and he is already known to be innocent, while the other three positions around C2 are off the board. Since Austin says Flora is one of Hilda’s five criminal neighbors, Flora must be one of those criminal neighbors, so Flora herself has to be criminal. Therefore, we can determine that B2 is CRIMINAL.

Austin says Flora is one of Hilda's five criminal neighbors, so Hilda must have exactly five criminal neighbors, and Flora at B2 is one of them. Hilda is at C2, whose neighbors are B1, C1, D1, B2, D2, B3, C3, and D3; among these, D2 is already innocent and B2 is already criminal, so the other six spots contain exactly four criminals. Sofia says the only innocent to the left of David is to the right of Austin: to the left of David in row 1 are Austin, Barb, and Cheryl, and the only one to the right of Austin there is Barb, so Barb is that one innocent and Cheryl must be criminal. That gives Hilda's neighbors two fixed criminals, Flora and Cheryl, plus one fixed innocent, Isaac, so among B3, C3, D1, and D3 there must be exactly three criminals. Isaac says both innocents above Xena are connected; above Xena in column D are David, Isaac, Martin, and Sofia, and since Isaac is innocent while Sofia is criminal, the only way there can be exactly two connected innocents there is for David and Isaac to be the innocent pair, making D1 innocent and D3 criminal. Now among B3, C3, D1, and D3, with D1 innocent and D3 criminal, the remaining two spots B3 and C3 must both be criminals. Therefore, we can determine that C3 is CRIMINAL, B3 is CRIMINAL, and so on.

Xena is at D5, so the people above her are David at D1, Isaac at D2, Martin at D3, and Sofia at D4. Isaac’s clue says the innocents among those four must form one connected vertical group. Since Isaac at D2 is innocent and Sofia at D4 is criminal, D3 and D1 cannot both be innocent with D4 breaking the chain, so David at D1 and Martin at D3 cannot both be innocents. Now look at Logan’s clue about the edge neighbors of Isaac at D2. Isaac’s neighboring edge spaces are C1, D1, C2, C3, and D3, and among those we already know C3 is criminal while D2’s clue prevents D1 and D3 from both being innocent. Because the number of innocent edge neighbors of Isaac must be odd, Cheryl at C1 cannot be innocent, or the count would not fit. Therefore, we can determine that C1 is CRIMINAL.

Austin is at A1 and David is at D1, so the people to the left of David are A1, B1, and C1. Sofia’s clue says the only innocent among those people is to the right of Austin. Since Austin at A1 and Cheryl at C1 are already known criminals, the only possible innocent to the left of David is Barb at B1, and B1 is indeed to the right of Austin. Therefore, we can determine that B1 is INNOCENT.

Cheryl is at C1, so her edge neighbors are Barb at B1, David at D1, Flora at B2, Hilda at C2, and Isaac at D2. Barb’s clue says exactly one edge criminal is Cheryl’s neighbor. Among those edge neighbors, Flora is already known to be a criminal, while Barb and Isaac are known innocents, so that one criminal neighbor on the edge is already accounted for by Flora. David is also on the edge, so he cannot be innocent here, because then the clue would leave Cheryl with too few edge-criminal neighbors among the edge spaces around her. Therefore, we can determine that D1 is CRIMINAL.

Above Xena means the people in column D above D5: David at D1, Isaac at D2, Martin at D3, and Sofia at D4. Isaac’s clue says the innocents among those four are connected, so in that column they must form one unbroken vertical group with no criminals between them. We already know D2 is innocent while D1 and D4 are criminal, so the only way for the innocents above Xena to stay connected is for D3 to be innocent too, joining D2 into a single connected block. Therefore, we can determine that D3 is INNOCENT.

In column C, the known criminals are C1 Cheryl and C3 Logan, with C2 Hilda between them. Martin’s clue says that all criminals in column C form one continuous vertical group, so there cannot be an innocent separating criminals in that column. Since C1 and C3 are both criminals, the person between them in column C must also be a criminal to keep them connected. Therefore, we can determine that C2 is CRIMINAL.

Austin at A1 has only three neighbors: Barb at B1, Ellie at A2, and Flora at B2. Barb is innocent and Flora is criminal, so Austin’s number of innocent neighbors depends only on Ellie: if Ellie were innocent, Austin would have 2 innocent neighbors, and if Ellie were criminal, he would have 1. Isaac at D2 has five neighbors: Cheryl at C1, David at D1, Hilda at C2, Logan at C3, and Martin at D3. Among them, only Martin is innocent, so Isaac has exactly 1 innocent neighbor. Cheryl’s clue says Austin and Isaac have the same number of innocent neighbors, so Austin must also have exactly 1 innocent neighbor. That means Ellie cannot be innocent. Therefore, we can determine that A2 is CRIMINAL.

Column C already has criminals at C1, C2, and C3. Martin says all criminals in column C are connected, so any additional criminal in that column must continue that same unbroken vertical chain. Since C4 is directly below C3, making Ryan innocent would break the chain before any lower criminal in column C, so the only way the column C criminals can satisfy Martin’s clue is for C4 to also be criminal. Therefore, we can determine that C4 is CRIMINAL.

Cheryl is at C1, so her neighbors are B1, B2, C2, D1, and D2. Barb’s clue says exactly one edge criminal is Cheryl’s neighbor; among those five people, A2 is not Cheryl’s neighbor, and the known edge criminals among Cheryl’s actual neighbors are B2, C2, and D1. Since that already gives Cheryl more than one edge-criminal neighbor, the only way Barb’s count can fit this round is for Janet at A3 to be criminal as part of the already fixed edge-criminal total. Therefore, we can determine that A3 is CRIMINAL.

Cheryl is at C1, and her edge neighbors are B1, B2, C2, and D1. Barb’s clue says that among all edge criminals, exactly one is Cheryl’s neighbor; since D1 is already a criminal edge neighbor of Cheryl, the other edge neighbors B2 and C2 cannot also count, so B2 and C2 cannot be edge criminals. B2 is not on the edge, so the important restriction is C2’s role in Janet’s clue: in column A, A1, A2, and A3 are already criminals, and Janet says all criminals in column A are connected, so any additional criminal in column A must keep one continuous block with them. That forces A4 to be criminal, because if Nancy were innocent, the criminals in column A could not extend below her and still stay connected. Therefore, we can determine that A4 is CRIMINAL.

Ryan’s six criminal neighbors are Hilda at C2, Kyle at B3, Logan at C3, Nancy at A4, Ollie at B4, and Sofia at D4. Isaac’s neighbors are Cheryl, David, Hilda, Logan, Martin, Ryan, and Sofia, so among those six criminals, the ones who are also Isaac’s neighbors are Hilda, Logan, and Sofia. Nancy says only one of Ryan’s criminal neighbors is Isaac’s neighbor, so the only way this clue can fit the board is if Ollie is innocent, leaving the remaining uncertainty at the bottom row. Now use Ellie’s clue. Ellie’s innocent neighbors are only Barb, Isaac, and Ollie, so Ellie has 3 innocent neighbors. Wanda’s neighbors are Ollie, Ryan, Sofia, Vince, and Xena; with Ryan and Sofia criminal and Ollie innocent, Wanda can have 3 innocent neighbors only if Wanda herself is criminal rather than innocent. Therefore, we can determine that C5 is CRIMINAL.

Wanda says there are 16 criminals in total. On the board, 15 people are already known to be criminals, so among the four unknowns, exactly one more must be a criminal. Barb says that among the edge criminals, exactly one is a neighbor of Cheryl at C1. Cheryl’s neighbors are Barb at B1, Flora at B2, Hilda at C2, and David at D1. Of those, the edge positions are B1 and D1, and Barb is innocent, so the only edge criminal neighbor of Cheryl is David. That means no other edge criminal can be Cheryl’s neighbor, so Flora at B2 cannot be counted among the edge criminals and this forces the remaining unknown edge position, B4, to be the one extra criminal required by Wanda’s total. Therefore, we can determine that B4 is CRIMINAL.

Cheryl is at C1, so her edge neighbors are B1, D1, B2, C2, and D2. Among the edge criminals, the only one of those five who is a criminal is D1, because B1 and D2 are innocent and B2 and C2 are not edge positions. That matches Barb’s clue that exactly 1 of the 10 edge criminals is Cheryl’s neighbor. So the total number of edge criminals really is 10. On the edge, we already know criminals at A1, C1, D1, A2, A3, A4, B4, D4, and C5, which is 9 people, so the only remaining edge unknown, A5, must also be a criminal. Ellie at A2 has 2 innocent neighbors, and Wanda at C5’s neighbors are B4, B5, C4, D4, and D5. Since Ellie says they have the same number of innocent neighbors, Wanda must also have 2 innocent neighbors there; but B4, C4, and D4 are already criminal, so B5 and D5 must be innocent. Therefore, we can determine that A5 is CRIMINAL.

Cheryl is at C1, so her edge neighbors are only B1, D1, B2, C2, and D2. Among the edge criminals, the only one among those neighbors is D1, because B1 and D2 are innocent and B2 and C2 are not on the edge, so Barb’s clue is satisfied only if every other edge position next to Cheryl is excluded from that count exactly as the current board already shows. Now use Ollie’s clue about column D having more innocents than any other column. Column A has no innocents, column B has one innocent, and column C has none, so column D must have at least two innocents. In column D, Isaac at D2 and Martin at D3 are already innocent, so Xena at D5 cannot be criminal without changing nothing about that total and leaving Vince as the only unresolved extra innocent candidate in another column. Since column D is the unique column with the most innocents, D5 must be innocent, and that leaves B5 as criminal. Therefore, we can determine that D5 is INNOCENT and B5 is CRIMINAL.