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.

Among Olof's neighbors, exactly 2 innocent people are also neighbors of Xena. In that shared group, Sue and Tina are already known to be innocent, and the only other person there is Uma. Since the clue's total of 2 is already filled by Sue and Tina, Uma cannot be innocent. So Uma must be criminal.

Isaac’s clue says that among the people strictly between Eve and Vera, there is exactly 1 innocent, and that group is just Mark and Ryan. Uma’s clue says that among Sue’s neighbors in column A, there is exactly 1 innocent, and that larger group is Mark, Ryan, and Vera. Since the only extra person in that larger group is Vera, the one allowed innocent in the larger group must already be the same one coming from Mark and Ryan. That means Vera cannot be innocent. So Vera must be criminal.

Isaac’s clue says the people strictly between Eve and Vera contain exactly one innocent, and those two people are Mark and Ryan. Since neither of them is already known innocent, that clue uses up exactly one innocent somewhere between Mark and Ryan. Will’s clue also requires every column, including column A, to have at least 3 innocents. In column A, Eve is already innocent and Vera is already criminal, so Austin must be innocent to satisfy that column requirement. So Austin must be innocent.

To the left of Phil there are exactly 2 innocents among Mark, Nicole, and Olof, and Olof is already one of them. Gary’s neighbors contain exactly 7 innocents, and in that larger group the only unknown person who is not also in Phil’s-left group is Betty. So once the needed innocents in Mark and Nicole account for the difference between the two groups, Betty also has to be innocent. That makes Betty innocent.

Olof’s neighbors already have 2 known criminals, Gary and Uma, and that group must contain exactly 5 innocents in total, so among Olof’s three still-unknown neighbors, D2 Logan, B3 Nicole, and D3 Phil, exactly 1 is criminal. In this round, the only people in that unknown trio who can still be that criminal are Logan and Nicole. That means Phil cannot be the criminal in Olof’s neighbor group. So D3 Phil must be innocent.

David’s neighbors are Celia and Isaac, who are both innocent, plus Logan. So David currently has 0 known criminal neighbors, and only Logan is still unknown there. Mark’s neighbors already include Gary, who is a criminal, so Mark already has at least 1 criminal neighbor. Betty’s clue says Mark has more criminal neighbors than David. Since Mark already has more criminal neighbors than David’s current 0, David cannot gain a criminal neighbor in his last unknown spot. So Logan must be innocent.

Olof already has exactly 5 innocent neighbors in total, and those 5 are already known except for Nicole. Among Olof's neighbors, exactly 2 of the innocents also neighbor Xena, and the people there are Sue, Tina, and Uma, with Uma criminal, so Sue and Tina are the 2 innocents for that count. That means Nicole cannot be innocent, so Nicole must be criminal. Uma's neighbors must have fewer criminals than Eve's neighbors, and Eve already has 1 known criminal neighbor while Uma has none known and only Xena still unknown there. So Xena cannot be criminal. That makes Xena innocent.

Tina’s clue says Olof is one of the exactly 7 innocent neighbors of Gary. Among Gary’s neighbors, Austin, Betty, Celia, Eve, Isaac, and Olof are already known innocents, Nicole is known criminal, and Mark is the only neighbor there whose status is still unknown. That means the seventh innocent neighbor has to be Mark. So Mark must be innocent.

Isaac's clue says the people between Eve and Vera contain exactly one innocent. That group already has its one known innocent. The only unknown person left in that same group is Ryan, so Ryan cannot be innocent. So Ryan must be criminal.

Above Zoe, the people are David, Logan, Phil, and Uma. Nicole’s clue says the two innocents in that group must be connected, and Logan and Phil are already the two known innocents there. The only still-unknown person involved in this clue is David, so the clue fixes his status. So David must be criminal.