Puzzle Pack #1 Puzzle 22 Answer

Sam’s clue is specifically about Brian’s neighbors, and it says Ellie is one of two or more criminals neighboring Brian. Ellie is at A2, which does neighbor Brian at B1. Since the clue directly says Ellie is among the criminals next to Brian, Ellie must be a criminal. Therefore, we can determine that A2 is CRIMINAL.

Vicky is at A5, so her neighbors are A4, B4, and B5. Ellie’s clue says Vicky has exactly 2 innocent neighbors, and only 1 of those 2 is in column B. Among Vicky’s neighbors, the only person not in column B is A4, so one of the two innocent neighbors must be A4, while the other is either B4 or B5. Therefore, we can determine that A4 is INNOCENT.

Vicky at A5 has exactly two neighbors, A4 and B4, and the clue says both of Vicky’s innocent neighbors are those two people, with only one of them in column B. Since Oscar at A4 is already known to be innocent, Phil at B4 must be the other innocent neighbor of Vicky, so B4 is also innocent. Sam at C4 is innocent, and the people in column B who neighbor Sam are B3, B4, and B5. Oscar’s clue says an odd number of those column B neighbors are innocent. Since B4 is innocent, the total can only stay odd if B3 and B5 are both criminal or both innocent; but B4 is the only one forced innocent here, so the clue rules out having exactly one of B3 and B5 innocent. In this step, that leaves Lucy at B3 as criminal. Therefore, we can determine that B3 is CRIMINAL.

Vicky’s neighbors are A4, B4, and B5. Since A4 is already innocent, Ellie’s clue says there are exactly two innocent neighbors of Vicky in total, and only one of those two is in column B. That means exactly one of B4 and B5 is innocent, so the other one is criminal. Below Brian are Freya at B2, Lucy at B3, Phil at B4, and Will at B5. Lucy is already criminal, and among B4 and B5 exactly one is innocent. Lucy’s clue says there is only one innocent below Brian, so that one innocent below Brian must be whichever of B4 or B5 is innocent. That leaves Freya unable to be innocent. Therefore, we can determine that B2 is CRIMINAL.

Oscar is at A4, and his neighbors are Katie at A3, Lucy at B3, Phil at B4, Vicky at A5, and Will at B5. Freya says Oscar has only one innocent neighbor, and we already know Sam at C4 is innocent, so this clue is specifically about Oscar’s own neighbors; among them, Lucy is criminal and the only confirmed innocent there is not yet fixed. Ellie says Vicky has exactly two innocent neighbors, and exactly one of those two is in column B; Vicky’s neighbors are Oscar at A4, Phil at B4, and Will at B5, so her only possible innocent neighbors must come from those three, with exactly one of Phil and Will being innocent. Since Oscar has only one innocent neighbor total, Vicky cannot be innocent along with Oscar, because then Oscar would already have Vicky as an innocent neighbor and still need to fit Ellie’s statement about Vicky having two innocent neighbors among Oscar, Phil, and Will. The only way both clues fit together is that Oscar’s single innocent neighbor is Vicky’s one non-column-B innocent neighbor, which makes Katie the remaining non-innocent neighbor of Oscar above him and forces Vicky herself not to be innocent. Therefore, we can determine that A3 is CRIMINAL and A5 is CRIMINAL.

Katie says columns A and D contain the same number of innocents, and Vicky says column D has exactly one innocent. So column A must also have exactly one innocent. In column A, Ellie, Katie, and Vicky are already known criminals, and Oscar is already known innocent, so the only innocent in column A is A4. That leaves A1 unable to be innocent. Therefore, we can determine that A1 is CRIMINAL.

Column D contains Diane, Isaac, Noah, Tyler, and Zara, and Vicky says there is only one innocent in that whole column. Austin says the number of criminals above Zara is odd, and the people above Zara are exactly Diane, Isaac, Noah, and Tyler. So among those four people above Zara, there must be either 1 or 3 criminals, which means there are either 3 or 1 innocents above Zara. Since column D has only one innocent in total, the group above Zara cannot contain 3 innocents, so it must contain exactly 1 innocent, leaving Zara as a criminal. Therefore, we can determine that D5 is CRIMINAL.

In column D, Zara at D5 is criminal, so her clue is true: there is an odd number of innocents below Diane. The people below Diane are Isaac at D2, Noah at D3, Tyler at D4, and Zara at D5, and Vicky’s true clue says column D has only one innocent in total. Since Zara is already criminal, that one innocent must be among D2, D3, and D4, so below Diane there is exactly one innocent, which is odd. That means Diane cannot also be innocent, because then column D would have two innocents. Therefore, we can determine that D1 is CRIMINAL.

Isaac is in D2 and Zara is in D5, so the people in between them are Noah at D3 and Tyler at D4. Diane’s clue says there is an odd number of criminals among those two people, so exactly one of Noah and Tyler is criminal. Vicky’s clue says column D has only one innocent, and since Diane at D1 and Zara at D5 are already criminal, that single innocent in column D must be either Isaac, Noah, or Tyler. Because Noah and Tyler are split with exactly one criminal and one innocent, the only way column D can still have just one innocent in total is for Isaac to be criminal. Therefore, we can determine that D2 is CRIMINAL.

Vicky’s neighbors are A4, B4, and B5, and Ellie says exactly two of those three are innocent, with only one of those innocents in column B. Since A4 is already known innocent and is not in column B, exactly one of B4 and B5 is innocent. Vicky also says column D has exactly one innocent, and in that column D1, D2, and D5 are already criminal, so exactly one of D3 and D4 is innocent. Isaac says columns B and D have the same number of innocents, so column B also has exactly one innocent in total. But B4 and B5 already contain exactly one innocent, so there cannot be any other innocent in column B. Therefore, we can determine that B1 is CRIMINAL.

Vicky’s neighbors are A4 Oscar, B4 Phil, and B5 Will. Ellie says exactly two of those three are innocent, and only one of those innocents is in column B, so among Phil and Will exactly one is innocent and Oscar must be the other innocent. Brian says Cheryl has more criminal neighbors than Oscar. Oscar’s neighbors are Katie, Lucy, Phil, and Vicky, and since Katie, Lucy, and Vicky are already criminal, Oscar has either 3 or 4 criminal neighbors depending on Phil. Cheryl’s neighbors are Brian, Hilda, Diane, Freya, Isaac, Lucy, Martin, and Noah, and among those Brian, Diane, Freya, Isaac, and Lucy are already criminal, so Cheryl already has 5 criminal neighbors no matter what the unknown neighbors are. That means Cheryl has more criminal neighbors than Oscar regardless of Phil, and for Cheryl to already have those 5 criminal neighbors, Hilda must be one of them. Therefore, we can determine that C2 is CRIMINAL.

Look at column D: D1, D2, D3, D4, and D5. Vicky says there is only one innocent in that whole column, so among Diane, Isaac, Noah, Tyler, and Zara exactly one is innocent. Hilda says all criminals above Zara are connected, and the people above Zara in column D are D1, D2, D3, and D4. Since D1 and D2 are already criminals, every criminal above Zara has to form one continuous vertical block with them, so D3 must also be criminal; then D4 cannot also be criminal, because that would leave no innocent at all above Zara and in column D there must be exactly one innocent. Therefore, we can determine that D3 is CRIMINAL and D4 is INNOCENT.

Vicky’s neighbors are A4, B4, and B5, and Ellie’s clue says exactly one of Vicky’s two innocent neighbors is in column B. Since A4 is already known innocent and is not in column B, Vicky must have exactly one innocent among B4 and B5, so one of them is innocent and the other is criminal. Noah’s clue says Vicky and Zara have the same number of criminal neighbors. Vicky already has A4 innocent, A5 herself not counted, and among B4 and B5 exactly one criminal, while Zara’s neighbors are C4, C5, and D4, with C4 and D4 already innocent. So Zara can have the same number of criminal neighbors as Vicky only if C5 is the one criminal neighbor on Zara’s side. Therefore, we can determine that C5 is CRIMINAL.

Vicky’s neighbors are A4, B4, and B5. Since A4 is innocent and Vicky has exactly two innocent neighbors, Ellie’s clue says that among those two innocents, only one is in column B. That means exactly one of B4 and B5 is innocent. Tyler’s neighbors are C3, C4, C5, D3, and D5, and among those we already know C4 is innocent while C5, D3, and D5 are criminal, so Tyler has at most two innocent neighbors: C4 and possibly C3. Xia says Vicky has more innocent neighbors than Tyler, and Vicky has two innocent neighbors, so Tyler cannot also have two. Therefore C3 cannot be innocent. Therefore, we can determine that C3 is CRIMINAL.

Vicky at A5 has exactly two neighbors: Oscar at A4 and Will at B5, so those are the two innocents neighboring Vicky mentioned in Ellie’s clue. Ellie says only one of those two innocents is in column B, and since Oscar is in column A, that means both Oscar and Will must be innocents, so B5 is innocent. Martin says there are 16 criminals in total, so with B5 now fixed as innocent, every remaining unknown must be criminal to reach that total. Therefore, we can determine that C1 is CRIMINAL.

Katie is at A3, so her neighboring criminals are A2 Ellie, B2 Freya, and B3 Lucy. Oscar is at A4, whose neighbors are Katie at A3, Phil at B4, and Vicky at A5. Cheryl’s clue says that only 1 of Katie’s 3 criminal neighbors is also Oscar’s neighbor, and among Ellie, Freya, and Lucy, only Lucy at B3 is next to Oscar. That means Oscar cannot have any other criminal neighbors, so Phil at B4 cannot be criminal. Therefore, we can determine that B4 is INNOCENT.

Vicky is at A5, so her neighbors are A4, B4, and B5. We already know A4 and B4 are innocent, and Ellie says that of Vicky’s 2 innocent neighbors, only 1 is in column B. Since B4 is already one innocent neighbor in column B, the other innocent neighbor cannot also be in column B, so B5 cannot be innocent. Therefore, we can determine that B5 is CRIMINAL.