Puzzle Pack #2 Puzzle 15 Answer

Row 5 contains Uma, Wanda, Xena, and Zane, and Rohan is at C4. Every person in row 5 neighbors Rohan except Uma, because Wanda at B5, Xena at C5, and Zane at D5 all touch C4 diagonally or directly. Xena’s clue says exactly one innocent in row 5 is neighboring Rohan, and since Xena herself is in row 5, does neighbor Rohan, and is already known to be innocent, she must be that one innocent neighbor. That leaves Wanda and Zane, the other row 5 neighbors of Rohan, as criminals. Therefore, we can determine that D5 is CRIMINAL and B5 is CRIMINAL.

The only sleuths on the board are Megan at C3, Rohan at C4, and Xena at C5. Wanda’s clue says exactly one sleuth has an innocent directly below them. Xena has nobody below her, and Rohan has Xena directly below him, who is already known to be innocent, so Rohan is one sleuth who fits the clue. That means Megan cannot also have an innocent directly below her, so Isaac’s status must not be innocent. Therefore, we can determine that C4, Rohan, is CRIMINAL.

The corners are A1, D1, A5, and D5. We already know D5 is a criminal, so the two innocent corners must be among A1, D1, and A5, and Rohan’s clue says exactly one of those innocent corners is in column A. Since both A1 and A5 are in column A, they cannot both be innocent, so the other innocent corner has to be D1. Therefore, we can determine that D1 is INNOCENT.

Lucy is at B3, so her neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. Among those neighbors, the sleuths are Megan at C3 and Rohan at C4, and we already know Rohan is criminal. Zane says exactly 1 of Lucy's innocent neighbors is a sleuth, so the only way for that to be true is for the other neighboring sleuth, Megan, to be innocent. Therefore, we can determine that C3 is INNOCENT.

The corners are A1, D1, A5, and D5. We already know D1 is innocent and D5 is criminal, so the two innocent corners must be D1 and exactly one of the two corners in column A, meaning exactly one of A1 and A5 is innocent. Debra’s clue says both innocents in column A are connected, so column A must contain exactly two innocents that form one continuous vertical group. Since only one of A1 and A5 is innocent, A3 cannot be one of those two connected innocents in column A, because then the second innocent would have to be separated from it by at least one non-innocent. Therefore, we can determine that A3 is CRIMINAL.

Kumar’s clue says that Janet is one of Chloe’s three criminal neighbors, so Janet must be a neighbor of Chloe and must be criminal. Chloe is at C1, and Janet at D2 is indeed one of Chloe’s neighboring spaces. Therefore, we can determine that D2, Janet, is CRIMINAL.

Lucy’s neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. Zane’s clue says Lucy has exactly 4 innocent neighbors, and only 1 of those 4 is a sleuth. Among those neighbors, Megan at C3 is already a known innocent sleuth, and Rohan at C4 is a known criminal, so Megan must be the only innocent sleuth around Lucy. Kumar’s clue says Janet is one of Chloe’s 3 criminal neighbors. Chloe’s neighbors are B1, B2, C2, D2, and D1, and we already know Janet at D2 is criminal while Debra at D1 is innocent, so exactly two of B1, B2, and C2 are criminal. That means at most one of A2, B2, and A4 can be innocent, because B2 cannot be both one of Chloe’s noncriminal neighbors and also leave Chloe with three criminal neighbors. So around Lucy, besides Megan, the innocents must include Phil at B4, since the other side of Lucy cannot supply enough innocents without breaking Chloe’s clue. Then Debra’s clue says the two innocents in column A are connected; with Kumar at A3 criminal, the only way column A can contain two connected innocents is for A4 to be innocent and for A1 or A2 to match it. That leaves Bonnie at B1 as one of Chloe’s required criminals. Therefore, we can determine that B1 is CRIMINAL and B4 is INNOCENT.

Chloe’s neighbors are Bonnie, Debra, Gus, Janet, Lucy, Megan, and Nick, and Kumar’s clue says exactly 3 of them are criminals, with Janet definitely being one of those 3. Among those neighbors, Bonnie and Janet are already known criminals, while Debra and Megan are already known innocents, so exactly one of Gus, Lucy, and Nick is criminal. Xena’s neighbors are Phil, Rohan, Wanda, and Zane, and only Phil is innocent there, so Xena has 1 innocent neighbor. Megan’s clue says Chloe has more innocent neighbors than Xena, so Chloe must have at least 2 innocent neighbors. Since Debra and Megan already give Chloe exactly 2 known innocent neighbors, Chloe cannot have only those 2; this forces Nick to be the one criminal among Gus, Lucy, and Nick, so Gus and Lucy are innocent. Now look at Steve’s neighbors: Megan, Nick, Rohan, Xena, and Zane. With Megan and Xena innocent and Nick, Rohan, and Zane criminal, Steve has 3 criminal neighbors. Therefore, we can determine that D4 is CRIMINAL.

Bonnie’s neighbors are Andre, Chloe, Freya, Gus, Kumar, and Lucy. We already know Bonnie and Kumar are criminals, and Steve says Bonnie has an odd number of criminal neighbors, so among those six neighbors there must be exactly one more criminal besides Kumar. Kumar’s clue says Janet is one of Chloe’s three criminal neighbors, and Chloe’s neighbors are Bonnie, Debra, Gus, Isaac, and Janet; since Bonnie and Janet are already criminal and Debra is innocent, Chloe must also border exactly one of Gus or Isaac as a criminal. That means Gus cannot be the extra criminal needed next to Bonnie, because if Gus were criminal then Chloe would already have Bonnie, Janet, and Gus as her three criminal neighbors, leaving no room for Isaac, and Bonnie would then have two criminal neighbors already with Kumar and Gus, which would make Steve’s odd-count clue fail unless yet another neighbor were criminal. So Gus is innocent, and Chloe’s third criminal neighbor must be Isaac. Then Bonnie’s criminal neighbors are just Kumar and Isaac unless Chloe were criminal too, but Steve says Bonnie has an odd number of criminal neighbors, so Chloe must be the third one? No: Isaac is not a neighbor of Bonnie, so Bonnie’s only possible extra criminal among her neighbors is Chloe, Freya, Andre, or Lucy, and Debra’s column-A clue keeps the column-A innocents connected, which prevents Andre and Freya from both being arranged to satisfy that while Chloe is criminal. Therefore, Chloe cannot be criminal and must be innocent. Therefore, we can determine that C1, Chloe, is INNOCENT.

Megan at C3 has three innocent neighbors. The neighbors of Megan are B2, C2, D2, B3, D3, B4, C4, and D4, and among those we already know D2, C4, and D4 are not innocent, while B4 is innocent. So the other two innocent neighbors must be chosen from B2, C2, B3, and D3. “Below Debra” means in column D under D1, so among Megan’s neighbors only D2 and D3 are below Debra. Chloe says exactly one of Megan’s three innocent neighbors is below Debra. Since D2 is criminal, that one must be D3, so D3 has to be innocent. Therefore, we can determine that D3 is INNOCENT.

Chloe’s three neighbors are Bonnie, Gus, and Isaac, and Kumar says Janet is one of Chloe’s three criminal neighbors. Since Janet is not next to Chloe, that tells us Chloe has exactly three criminal neighbors among those actual neighbors, so Bonnie, Gus, and Isaac are all criminal. Megan’s neighbors are Gus, Isaac, Janet, Lucy, Phil, Rohan, Nick, and Steve. The innocents among them already include Phil and Nick, while Janet, Rohan, and Steve are criminal, and Gus and Isaac have just been forced to be criminal, so the only possible third innocent neighboring Megan is Lucy. Chloe says only one of Megan’s three innocent neighbors is below Debra, and below Debra in column D are Janet, Nick, Steve, and Zane, so among Megan’s neighboring innocents only Nick is below Debra. That means Lucy cannot be innocent, so Lucy must be criminal. Therefore, we can determine that B3 is CRIMINAL.

Lucy’s neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. Among them, the known innocents are C3, B4, and C4, and two of those, C3 and C4, are sleuths. Zane’s clue says that of Lucy’s four innocent neighbors, only one is a sleuth, so the fourth innocent neighbor cannot be A2 or B2, because that would still leave both C3 and C4 as sleuth innocents among Lucy’s innocent neighbors. The only remaining possibility is C2, and since Lucy must have four innocent neighbors in total, C2 has to be innocent. Therefore, we can determine that C2 is INNOCENT.

Chloe is at C1, so her neighbors are B1, B2, C2, D2, and D1. We already know B1 and D2 are criminals, while C2 and D1 are innocents. Kumar’s clue says Janet is one of Chloe’s 3 criminal neighbors, so Chloe must have exactly 3 criminal neighbors in total. The only remaining neighbor who can supply that third criminal neighbor is B2, Gus. Therefore, we can determine that B2 is CRIMINAL.

Explanation text is not available for this step.

The corner positions are A1, D1, A5, and D5. We already know A1 and D1 are innocent, D5 is criminal, so the two innocent corners must be A1 and exactly one of A5 or D1, which means the clue is really deciding whether A5 is innocent. Since the clue says only one of the two innocent corners is in column A, and A1 is already one innocent corner in column A, the other innocent corner cannot also be in column A. Therefore, we can determine that A5 is CRIMINAL.

In column A, the known innocents are A1 and no one else yet, while A3 and A5 are criminals, so the only way for there to be exactly two innocents in that column is for A2 or A4 to be the second innocent. Debra says both innocents in column A are connected, and in one column that means the two innocent cells must touch through up-and-down adjacency. A1 is already innocent, and A4 cannot be connected to A1 because A2 and A3 lie between them, with A3 already criminal. So the second innocent has to be A2, which leaves A4 as criminal. Therefore, we can determine that A2 is INNOCENT and A4 is CRIMINAL.