Puzzle Pack #2 Puzzle 29 Answer

Terry’s clue says Sarah and Vince have exactly 2 innocent neighbors in common. The shared neighbor group is just Ruby and Wanda, and that group still needs 2 innocents. Since Ruby and Wanda are the only people in that shared group, both of them have to supply those 2 innocents. So Ruby and Wanda must be innocent.

Ruby’s clue says an odd number of Isaac’s neighbors in column B are innocent, and that group is exactly Frank, Katie, and Peter. Wanda’s clue says Katie has exactly 6 innocent neighbors in total, and exactly 1 of those innocent neighbors is in column B, which means among the column B people relevant to Katie’s clue, only Frank or Peter can be that innocent one. So Katie cannot be one of the innocent people in that Isaac-and-column-B group. That makes Katie criminal.

Katie’s clue says there are more criminals than innocents below Bruce. The people below Bruce in column B are Frank, Katie, Peter, and Vince, and Katie is already a known criminal with no known innocents among those people. Ruby’s clue says an odd number of the people in column B who neighbor Isaac are innocents; those people are Frank, Katie, and Peter, and since Katie is criminal, exactly one of Frank or Peter is innocent. That gives at most one innocent among the people below Bruce before Vince is counted, so Vince has to be a criminal for the group below Bruce to still have more criminals than innocents. So Vince must be criminal.

Ruby’s clue says the number of innocents in the overlap of column B and Isaac’s neighbors is odd. That overlap is exactly Frank, Katie, and Peter, and Katie is already criminal, so among Frank and Peter there must be an odd number of innocents. Vince’s clue says every column has at least 2 innocents, so column B must contain at least 2 innocents among Bruce, Frank, and Peter, since Katie and Vince are already criminals. If Bruce were criminal, then Frank and Peter would have to provide those innocents for column B, but that clashes with Ruby’s odd-number condition on Frank and Peter. So Bruce at B1 must be innocent.

Bruce's clue says Nicole is one of Isaac's exactly 3 innocent neighbors. That directly identifies Nicole as innocent. So Nicole must be innocent.

Nicole's clue says row 2 has exactly 3 innocents, and exactly 2 of those innocents are Bruce's neighbors: Erwin, Frank, and Gabe. So the remaining 1 innocent in row 2 must be someone in row 2 who is not Bruce's neighbor. The only person in row 2 who is not Bruce's neighbor is Hope, and that group still needs that 1 innocent. So Hope must be innocent.

Isaac’s still-unknown neighbors are A2 Erwin, B2 Frank, and B4 Peter, and among those three there is exactly 1 criminal. In that group, the only people who can still be that criminal are B2 Frank and B4 Peter. So Erwin cannot be the criminal one there. That makes Erwin innocent.

Katie has exactly 6 innocent neighbors, and Wanda says only 1 of those innocents is in column B. Among Katie's neighbors, that means the five neighbors not in column B contain exactly 5 innocents, so among the three unknown people there, C2 Gabe, A3 Isaac, and C3 Luigi, exactly 1 is criminal. Erwin says Katie and Ruby have only one innocent neighbor in common, and the only shared neighbors are C3 Luigi and B4 Peter. That fixes the single criminal among C2 Gabe, A3 Isaac, and C3 Luigi as coming from Gabe or Luigi, not Isaac. So Isaac must be innocent.

Isaac’s clue says the mechs and the teachers have the same number of innocents. The teachers already have Hope as one innocent, and Hope’s clue says exactly one of Carol and Gabe is innocent, so the teachers will end up with one more innocent there. The mechs already have Terry as one innocent, and Zed is the only unresolved mech left. So Zed must be innocent.

Row 2 has exactly 3 innocents in total, and Erwin and Hope are already innocent, so exactly one of Frank and Gabe is innocent. Carol’s neighbors must contain exactly 3 innocents, and Bruce and Hope already account for 2 of them. The only other neighbors Carol has besides Frank and Gabe are Debra, so the single remaining innocent among Carol’s neighbors has to come from Frank or Gabe, not Debra. So Debra must be criminal.

Debra’s clue says Zed’s neighbors and Isaac’s neighbors must contain the same number of innocents. Right now Zed’s neighbors already have 2 known innocents, and Isaac’s neighbors also already have 2 known innocents. The only unknown person still on Zed’s side is Sarah, so this step fixes her as innocent. So Sarah must be innocent.

Katie and Ruby’s common neighbors contain exactly one innocent, and the only people in that shared group are Luigi and Peter. Ruby’s neighbors also must contain an odd number of innocents; there are already 3 known innocents there, so among the unknown neighbors Luigi, Maria, and Peter, the total number of innocents has to stay even. If Maria were criminal, then only Luigi and Peter could supply the needed innocents. But Luigi and Peter must also make the shared pair contain exactly one innocent, which means they contribute an odd number of innocents there, not an even one. That clashes with Ruby’s odd-neighbor requirement, so Maria cannot be criminal. So Maria must be innocent.

Maria’s clue says column A has more innocents than any other column. In column A, Erwin, Isaac, Nicole, and Terry are already known innocents, so Anna is the only undecided person there. For column A to stand as the column with the most innocents, Anna has to be innocent too. So Anna must be innocent.

Isaac’s clue says the innocent mechs and innocent teachers must match, and the counts are already 2 innocent mechs against only 1 innocent teacher. Anna’s clue says Bruce must have more innocent neighbors than Nicole, but both currently have 2 known innocent neighbors. If Frank were criminal and Peter were innocent, the only other people left to make both clues work would be Carol and Gabe. But with Frank criminal and Peter innocent, Carol and Gabe cannot satisfy both the teacher count and the neighbor comparison at the same time. So Frank must be innocent and Peter must be criminal.

Row 2 has exactly 3 innocents in total. The clue says exactly 2 of those innocents are Bruce's neighbors, and among the row 2 neighbors of Bruce we already have 2 known innocents: Erwin and Frank. That means the remaining row 2 neighbor of Bruce, Gabe, cannot be innocent. So Gabe must be criminal.

Wanda’s clue says Katie has exactly 6 innocent neighbors, and exactly 1 of those innocents is in column B. That means the other 5 innocent neighbors of Katie must be not in column B. Among Katie’s neighbors who are not in column B, Erwin, Isaac, Nicole, and Ruby are already known innocents, so there is still 1 more innocent needed there. The only unknown person left in that group is Luigi, so Luigi must be innocent.

Isaac's clue says the mechs and the teachers must have the same number of innocents. The mechs already have 2 known innocents, while the teachers currently have only 1 known innocent. The only unknown person left in those groups is Carol among the teachers, so she has to supply that missing second innocent. So Carol must be innocent.