Puzzle Pack #1 Puzzle 48 Answer

Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Since Nancy is at D3, the neighbors of Isaac that are to Nancy’s left are B1, C1, B2, B3, and C3, while D1, D2, and D3 are not to her left. Logan’s clue says exactly one of Isaac’s neighboring innocents is to the left of Nancy, and Logan himself at B3 is already known to be innocent, so every other Isaac-neighbor to Nancy’s left must not be innocent. Maria is C3, one of those other left-of-Nancy neighbors. Therefore, we can determine that C3 is CRIMINAL.

Peter is at B4, so the people above him are B1 Debra, B2 Hilda, and B3 Logan. Evie is at C1, and among those three, Debra at B1 and Hilda at B2 are neighbors of Evie, while Logan at B3 is not. Maria says exactly 0 innocents above Peter are neighboring Evie, so none of those neighboring people above Peter can be innocent. Therefore, we can determine that B1 is CRIMINAL and B2 is CRIMINAL.

Nancy is at D3, so everyone to the left of Nancy is every person in row 3 except Nancy herself. Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3, and among those, the ones to the left of Nancy are B1, C1, B2, D2, B3, and C3. Since Logan says exactly one of Isaac’s four innocent neighbors is to the left of Nancy, and we already know B3 is innocent while B2 and C3 are criminal, the other three innocents neighboring Isaac must be D1, D2, and D3, so Flora at D1 is innocent. Hilda then says both innocents below Flora are connected. The people below Flora are D2, D3, D4, and D5, and with D2 already innocent there must be exactly one more innocent among D3, D4, and D5 that is connected to D2. D3 is already innocent, so D4 and D5 cannot also be innocent, and in particular Zach at D5 is not innocent. Therefore, we can determine that D5 is CRIMINAL.

Flora is at D1, so the people below Flora are Janet at D2, Nancy at D3, Sarah at D4, and Zach at D5. Hilda’s clue says there are exactly two innocents among those four, and those two innocents must be connected by orthogonal adjacency in that same column. Since Zach at D5 is already known to be criminal, the only way to have two connected innocents below Flora is for Nancy at D3 and Sarah at D4 to be the innocent pair. Therefore, we can determine that D3 is INNOCENT.

Maria is at C3, so her row 4 neighbors are B4 Peter, C4 Rob, and D4 Sarah. Zach says there is an odd number of innocents among those three people, so row 4 must contain either 1 innocent or 3 innocents in those positions. Debra says row 4 has more innocents than any other row. Row 3 already has two known innocents, Logan at B3 and Nancy at D3, so for row 4 to have more innocents than row 3, row 4 must have at least 3 innocents. That rules out having only 1 innocent among Peter, Rob, and Sarah, so all three of them must be innocent. Therefore, we can determine that B4 is INNOCENT, C4 is INNOCENT, and D4 is INNOCENT.

Flora is at D1, so the people below Flora are Janet at D2, Nancy at D3, Sarah at D4, and Zach at D5. Hilda’s clue says the innocents among those people are exactly two, and those two innocents must be connected in one continuous vertical group. We already know Nancy and Sarah are innocent, and they are directly next to each other, so they are the connected pair of innocents below Flora. That leaves Janet and Zach as the non-innocents below Flora, and Zach is already known to be criminal. Therefore, we can determine that D2 is CRIMINAL.

Isaac is at C2, so his four innocent neighbors must come from the five unknown-or-innocent neighbors around him: Evie at C1, Flora at D1, Logan at B3, and Nancy at D3 are already innocent or could be, while the known criminals B1, B2, D2, and C3 are not counted. “To the left of Nancy” in Nancy’s row means only the people in row 3 to her left, so among Isaac’s neighboring innocents the only possible one there is Logan at B3. Since the clue says exactly 1 of Isaac’s 4 innocent neighbors is to the left of Nancy, Logan must be that one, and the other three innocent neighbors of Isaac must be the remaining possible neighbors: Evie, Flora, and Gus is not in Nancy’s row so this left-of-Nancy condition does not apply, but Isaac needs four innocent neighbors total and the two top neighbors C1 and D1 are forced to be included. Therefore, we can determine that C1 is INNOCENT, D1 is INNOCENT, and so on.

Row 4 contains Peter, Rob, Sarah, and Olof, and we already know Peter, Rob, and Sarah are innocent. That means row 4 already has at least 3 innocents. Row 1 also already has 3 innocents, because Debra is criminal while Evie and Flora are innocent, so Bruce would have to be innocent to make row 1 also reach 3. Since Debra says row 4 has more innocents than any other row, row 4 must have more innocents than row 1, so row 4 cannot stay at just 3 innocents and Olof must be innocent. Therefore, we can determine that A4 is INNOCENT.

Nancy says there are 9 criminals in total. On the board, 5 people are already known criminals and 7 are already known innocents, so the 8 unknown people must contain exactly 4 criminals. Flora says there are more innocents than criminals on the edges. The 14 edge spaces currently contain 4 known criminals, 7 known innocents, and 3 unknowns at A1, A5, and B5, so those 3 unknown edge spaces can contain at most 2 criminals if the edge is to still have more innocents than criminals. That means the 4 criminals still needed among all unknowns cannot all be on the edge, so the remaining interior unknown, C2 Isaac, must be one of them. Therefore, we can determine that C2 is CRIMINAL.

The edge squares are A1, B1, C1, D1, A2, D2, A3, D3, A4, D4, A5, B5, C5, and D5. Among those, the known edge innocents are C1, D1, D3, A4, and D4, while the known edge criminals are B1, D2, and D5, so the three unknown edge people A1, A5, and B5 together must include at least one more innocent for Flora’s statement to be true. Rob’s clue compares Sarah at D4 and Olof at A4. Olof’s neighbors are A3, B3, B4, A5, and B5, so he currently has 2 known innocent neighbors there, while Sarah’s neighbors are C3, D3, C4, C5, and D5, so she also has 2 known innocent neighbors there plus Xavi at C5 still unknown. Because their numbers of innocent neighbors are equal, Xavi cannot be criminal, since that would leave Sarah with 2 innocent neighbors while Olof could not drop below his already known 2. Therefore, we can determine that C5 is INNOCENT.

Olof is at A4 and Sarah is at D4. From the people already known around them, Sarah’s innocent neighbors are C3, C4, and D3, so she has 3 innocent neighbors; Olof’s known innocent neighbors are A5, B4, and B3 if A5 is innocent, plus A3 and B5 are still undecided. Rob says Olof and Sarah have the same number of innocent neighbors, so Olof also must have exactly 3 innocent neighbors, which means A5 must be innocent and A3 and B5 must not be. Now use Xavi’s clue about Olof and Bruce having 4 innocent neighbors in total. Since Olof already has 3 innocent neighbors, Bruce must have 1 innocent neighbor. Bruce’s neighbors are Gus at A2, Debra at B1, and Hilda at B2, and Debra and Hilda are already criminal, so the only way Bruce can have 1 innocent neighbor is if Gus is innocent. Therefore, we can determine that A2 is INNOCENT.

Nancy says there are 9 criminals in total. On the board, 7 people are already confirmed criminals, so the three unknown people A1, A3, A5, and B5 must contain exactly 2 criminals altogether. Rob says Sarah and Olof have the same number of innocent neighbors. Sarah's neighbors are Maria, Nancy, Rob, Xavi, and Zach, and 3 of those are innocent: Nancy, Rob, and Xavi. Olof's neighbors are Kay, Logan, Peter, Vicky, and Will, and Logan and Peter are already innocent, so for Olof to also have 3 innocent neighbors, exactly one of Kay, Vicky, and Will must be innocent. That means among A3, A5, and B5, exactly 2 are criminals. Since the four unknown people contain exactly 2 criminals total, A1 cannot be one of the innocents and must be the remaining criminal. Therefore, we can determine that A1 is CRIMINAL.

We already know 8 criminals: A1, B1, B2, C2, D2, C3, and D5, so Nancy’s clue that there are 9 criminals in total means there is exactly 1 criminal left among the three unknown people A3, A5, and B5. Bruce’s clue says Xavi and Hilda have 7 innocent neighbors in total. Xavi’s neighbors are B4, B5, C4, D4, and D5, and among those B4, C4, and D4 are innocent while D5 is criminal, so Xavi contributes 3 innocent neighbors if B5 is criminal or 4 if B5 is innocent. Hilda’s neighbors are A1, B1, C1, A2, C2, A3, B3, and C3, and among those C1, A2, and B3 are innocent while A1, B1, C2, and C3 are criminal, so Hilda contributes 3 innocent neighbors if A3 is criminal or 4 if A3 is innocent. To make their total exactly 7, one of A3 and B5 must be innocent and the other criminal. Since the only remaining criminal among A3, A5, and B5 is exactly one person, that leaves A5 as the remaining criminal. Therefore, we can determine that A5 is CRIMINAL.

Gus is at A2 and Vicky is at A5, so the people in between them are exactly A3 and A4. Vicky says there are as many criminals as innocents among those two people, which means one of them must be criminal and the other innocent. Since A4, Olof, is already known to be innocent, A3, Kay, must be the criminal one. Therefore, we can determine that A3 is CRIMINAL.

Sarah is at D4 and Olof is at A4, so we compare the innocent neighbors around those two positions. Sarah’s neighbors are C3, D3, C4, C5, and D5, and exactly three of them are innocent: D3, C4, and C5. Olof’s neighbors are A3, B3, A4, B4, A5, and B5, and among those we already know B3, A4, and B4 are innocent, so Olof can have the same total of three innocent neighbors only if B5 is also innocent. Therefore, we can determine that B5 is INNOCENT.