Puzzle Pack #2 Puzzle 36 Answer

Tina’s clue says that Olof is one of Gary’s exactly 7 innocent neighbors. That directly identifies Olof as innocent. So Olof must be innocent.

Tina’s clue says Gary has exactly 7 innocent neighbors, and Olof is one of them. Olof’s clue says Eve has exactly 3 innocent neighbors. If Eve, Celia, and Isaac were criminal while Gary were innocent, then the only other people involved in these two clues would be Austin, Betty, Mark, and Nicole, and they would have to make both counts come out correctly at the same time. That cannot be done, so Eve, Celia, Isaac, and Gary cannot have those opposite identities. So Eve and Celia and Isaac must be innocent, and Gary must be criminal.

Xena’s neighbors must have more innocents than Eve’s neighbors, and right now Xena’s side has 1 known innocent while Eve’s side has 0. If Will and Zoe were both criminals, then only Sue and Uma would be left among Xena’s unknown neighbors to raise Xena’s innocent-neighbor count, while Austin, Betty, Mark, Nicole, Sue, and Uma would also have to fit the other clue requirements about Gary’s and Olof’s neighbor counts at the same time. That combination cannot satisfy all of those facts together. So Will and Zoe cannot both be criminals. That makes Will and Zoe innocent.

Eve’s clue says there is exactly 1 innocent between Eve and Vera, and the only people between them are Mark and Ryan, so one of those two must be innocent. Isaac’s clue says Olof has exactly 5 innocent neighbors, and exactly 2 of those innocent neighbors also neighbor Xena. Among the people who both neighbor Olof and neighbor Xena, Tina is already known innocent, so exactly one of Sue and Uma must also be innocent. Since Sue is one of those required shared innocent neighbors, Sue must be innocent.