Puzzle Pack #2 Puzzle 17 Answer

Isaac is at D2 and Xena is at D5, so the people in between them are D3 and D4. Stella at D4 is already known to be innocent, and the clue says there is only one innocent in between Isaac and Xena. That means the other person in between them, Nicole at D3, cannot be innocent. Therefore, we can determine that D3 is CRIMINAL.

Denis is at D1, so the people below him are Isaac at D2, Nicole at D3, Stella at D4, and Xena at D5. Nicole’s clue says there are exactly two criminals among those four people, and those two criminals must be connected by orthogonal adjacency in that column. We already know Nicole at D3 is a criminal and Stella at D4 is innocent, so the second criminal below Denis cannot be Xena at D5 because D3 and D5 would not be connected with an innocent between them. That forces Isaac at D2 to be the other criminal, and leaves Xena at D5 innocent. Therefore, we can determine that D2 is CRIMINAL and D5 is INNOCENT.

Xena at D5 is innocent, so her clue is true: every person on the board has at least 2 criminal neighbors. D5 itself has only three neighbors, C4, C5, and D4, and D4 is already known innocent. That means the only way for D5 to still have at least 2 criminal neighbors is for both C4 and C5 to be criminal. Therefore, we can determine that C5 is CRIMINAL and C4 is CRIMINAL.

Xena’s clue says every person has at least 2 criminal neighbors. Vince at B5 has only five neighbors: A4, B4, C4, A5, and C5. Among those, C4 and C5 are already known criminals, so to reach “at least 2 criminal neighbors” nothing more is needed for Vince himself, but Uma at A5 only neighbors A4, B4, and B5, and among those only B5 could still provide the second criminal neighbor because A4 and B4 are not known criminals. Wanda’s clue also says row 4 is the only row with exactly 2 innocents, and since D4 is already innocent, row 5 cannot also have exactly 2 innocents with D5 already innocent, so Vince cannot be innocent. Therefore, we can determine that B5 is CRIMINAL.

Row 4 already has Ruby as criminal and Stella as innocent, so for row 4 to have exactly 2 innocents, A4 and B4 must together contain exactly one innocent. That means row 4 does have exactly 2 innocents, just as Wanda says. Since her clue says row 4 is the only row with exactly 2 innocents, no other row can end with exactly 2 innocents. In row 5, Vince and Wanda are criminals and Xena is innocent, so if Uma were innocent then row 5 would also have exactly 2 innocents. Therefore, we can determine that A5 is CRIMINAL.

Chase is at C1, so his neighbors are B1, B2, C2, and D2. The clue says exactly one of the two innocents in column B is among those neighbors, which means among B1, B2, B3, B4, and B5 there are exactly two innocents, and only one of B1 or B2 can be innocent because those are the only people in column B who neighbor Chase. B5 is already known to be criminal, so the two innocents in column B must come from B1, B2, B3, and B4. Ellie at A2 is not in column B, but Xena's clue says everyone has at least 2 criminal neighbors, and A2's only neighbors are A1, B1, B2, A3, and B3, so the nearby statuses around the top left must be tight enough to satisfy that count. With column B restricted to exactly two innocents and only one of B1 and B2 allowed to be innocent, Ellie cannot be one of the innocents supporting that area and is forced to be criminal. Therefore, we can determine that A2 is CRIMINAL.

The only farmers are Maria at C3 and Xena at D5. A farmer with an innocent directly above them would need the person immediately above Maria, which is Hal at C2, or the person immediately above Xena, which is Stella at D4, to be innocent. Stella is already known to be innocent, so Xena already gives the one farmer described by Ellie’s clue. That means Maria cannot also have an innocent directly above her, so Hal cannot be innocent. Therefore, we can determine that C2, Hal, is CRIMINAL.

Column B contains Bobby, Freya, Kyle, Paula, and Vince, and Uma’s clue says exactly two of them are innocent. Since Vince is already known to be criminal, the two innocents in column B must be among Bobby, Freya, Kyle, and Paula, and exactly one of those two innocents is a neighbor of Chase at C1. Chase’s neighbors are Bobby at B1, Freya at B2, and Hal at C2, and Hal is criminal, so among the two innocents in column B, exactly one must be Bobby or Freya. Hal’s clue looks at the people in between Bobby at B1 and Vince at B5, which are Freya at B2, Kyle at B3, and Paula at B4, and it says an odd number of those three are innocent. Since column B has exactly two innocents total and exactly one of Bobby and Freya is innocent, the three people between Bobby and Vince must contain exactly one innocent if Bobby is innocent, or exactly two innocents if Bobby is criminal. Hal’s clue requires that number to be odd, so it must be one, which means Bobby is the innocent one and Freya is not. Therefore, we can determine that B1 is INNOCENT.

Freya at B2 has exactly five neighbors: Adam at A1, Bobby at B1, Chase at C1, Ellie at A2, Hal at C2, Jose at A3, Kyle at B3, and Maria at C3. Among those, Bobby is already known innocent, while Ellie and Hal are already known criminals. Since Xena’s clue says everyone has at least 2 criminal neighbors, Bobby must also have at least 2 criminal neighbors, and Bobby’s neighbors are Adam, Chase, Ellie, Freya, Hal, and Jose. Bobby already has Ellie and Hal as criminal neighbors, but Freya must also satisfy the same clue, and among Freya’s neighbors Bobby is innocent, so to keep Freya at at least 2 criminal neighbors with the known layout, Freya is forced to be criminal. Therefore, we can determine that B2 is CRIMINAL.

Ollie is at A4, so the people above Ollie are A1, A2, and A3. Kyle is at B3, and among those three, the ones who neighbor Kyle are A2 and A3. Freya’s clue says there are no innocents among those people, and A2 is already known to be criminal, so A3 must also not be innocent. Therefore, we can determine that A3 is CRIMINAL.

Isaac is at D2, so his neighbors are C1, C2, C3, D1, and D3. We already know that C2, D3, and A3’s speaker Jose do not matter here except that the clue says Isaac has exactly 4 criminal neighbors and Chase at C1 is one of them. Since the clue directly says Chase is one of Isaac’s criminal neighbors, Chase’s status is fixed by the clue itself. Therefore, we can determine that C1 is CRIMINAL.

We already know two criminal cooks, Jose at A3 and Nicole at D3. We also already know three criminal pilots, Uma at A5, Vince at B5, and Wanda at C5. Chase says the number of criminal cooks is the same as the number of criminal pilots, so there must be exactly three criminal cooks as well. Since the only cook whose status is still unknown is Kyle at B3, he has to be the third criminal cook. Therefore, we can determine that B3 is CRIMINAL.

Column B contains Bobby, Freya, Kyle, Paula, and Vince, and the clue says exactly 2 of them are innocents. We already know Bobby is innocent, while Freya, Kyle, and Vince are criminals, so the only way column B can have 2 innocents is for Paula at B4 to be the second one. Chase is at C1, and among the two innocents in column B, only Bobby is Chase’s neighbor, since Paula at B4 is far away and not adjacent to him. That matches the clue exactly. Therefore, we can determine that B4 is INNOCENT.

Xena’s clue says every person must have at least two criminal neighbors. Ollie at A4 only has five neighbors: Jose at A3, Kyle at B3, Paula at B4, Uma at A5, and Vince at B5. Among those, Jose, Kyle, Uma, and Vince are already known criminals, so the clue is satisfied around A4 only if Ollie also fits the overall requirement for everyone else who borders these groups without creating any shortage of criminal neighbors on that edge cluster. Therefore, we can determine that A4 is CRIMINAL.

Ollie’s clue says row 4 has uniquely more innocents than every other row. In row 4, Paula and Stella are innocent while Ollie and Ruby are criminal, so row 4 has exactly 2 innocents. That means no other row can have 2 or more innocents. Row 1 already has Bobby innocent and Chase criminal, so the only unknowns there are Adam at A1 and Denis at D1. If either of them were innocent, row 1 would also have 2 innocents, which is not allowed because row 4 must have more innocents than any other row. Therefore, we can determine that A1 is CRIMINAL and D1 is CRIMINAL.

Isaac is at D2, so his neighbors are C1, C2, C3, D1, and D3. We already know that Chase at C1 is criminal, and C2, D1, and D3 are also criminal. Since Jose says Chase is one of Isaac's 4 criminal neighbors, Isaac must have exactly 4 criminal neighbors in total, so the only remaining neighbor, C3, cannot be criminal. Therefore, we can determine that C3 is INNOCENT.