Puzzle Pack #2 Puzzle 5 Answer

Ethan at A2 has exactly three neighbors: Adam at A1, Bonnie at B1, Frank at B2, Janet at A3, and Karen at B3 are not all neighbors; correcting that, his row 3 neighbors are only Janet at A3 and Karen at B3. His clue says that 2 of his innocent neighbors are in row 3, and those are the only neighbors he has in row 3. So both Janet and Karen must be innocent. Therefore, we can determine that B3 is INNOCENT and A3 is INNOCENT.

Vicky is at B5, so the people above her are Paul at B4, Karen at B3, Frank at B2, and Bonnie at B1. Karen is innocent, so among those four, the criminals above Vicky must be Bonnie and Frank, or Paul and Frank, or Paul and Bonnie, because the clue says exactly two of them are criminals and those two must be connected in one continuous vertical group. Since Karen sits between Paul and Frank, and also between Paul and Bonnie, any pair involving Paul would be split by an innocent and would not be connected. That leaves only Bonnie and Frank as the connected criminals above Vicky, which makes Paul not one of them. Therefore, we can determine that B2 is CRIMINAL, B1 is CRIMINAL, and B4 is INNOCENT.

Row 2 contains Ethan at A2, Frank at B2, Henry at C2, and Isaac at D2. Frank’s clue says there is only one innocent in that row, and Ethan is already known to be innocent. That means Ethan is the only innocent in row 2, so the other two unknown people in that row, Henry and Isaac, cannot be innocent. Therefore, we can determine that C2 is CRIMINAL and D2 is CRIMINAL.

The cops on the board are Adam at A1, Denis at D1, Ethan at A2, and Isaac at D2. Isaac’s clue says exactly one cop has exactly 2 criminal neighbors. Ethan already has exactly 2 criminal neighbors, because his neighbors are Adam, Bonnie, Frank, Janet, and Karen, and among those Bonnie and Frank are criminal while Janet and Karen are innocent. So no other cop can also have exactly 2 criminal neighbors. Adam’s neighbors are Bonnie, Ethan, and Frank; with Bonnie and Frank criminal and Ethan innocent, Adam has exactly 2 criminal neighbors. Denis’s neighbors are Celia, Henry, and Isaac, and Henry and Isaac are criminal, so Denis has exactly 2 criminal neighbors only if Celia is innocent. Since Ethan is the only cop allowed to have exactly 2 criminal neighbors, Adam cannot be a cop and Celia cannot be innocent. Therefore, we can determine that A1 is CRIMINAL and C1 is CRIMINAL.

The pilots are Frank at B2, Olive at A4, and Rose at C4. Adam’s clue says exactly one pilot has an innocent directly above them. Frank has Bonnie above him, who is criminal, and Rose has Martin above her at C3, so for there to be exactly one such pilot, Olive must be the one with an innocent above her and Rose cannot have an innocent above her. Since the person directly above Rose is Martin, Martin cannot be innocent. Therefore, we can determine that C3, Martin, is CRIMINAL.

Martin says that row 5 is the only row with exactly 3 innocents, so no other row can have exactly 3 innocents. In row 3, Janet and Karen are already innocent and Martin is criminal, leaving only Nick undetermined in that row. If Nick were innocent, then row 3 would have exactly 3 innocents, which Martin says cannot happen because only row 5 has that total. Therefore, we can determine that D3 is CRIMINAL.

Row 1 currently has Adam, Bonnie, and Celia already confirmed as criminals, so the only person left in that row who could satisfy Nick’s clue is Denis at D1. Since the clue says every row has at least one innocent, row 1 must contain an innocent, and Denis is the only possible one there. Therefore, we can determine that D1 is INNOCENT.

Martin’s clue says row 5 has exactly 3 innocents, so in that row there are exactly 1 criminal and 3 people who are not criminals. Denis says there is an odd number of criminals to the right of Uma, and to Uma’s right are exactly B5, C5, and D5. Since those three people contain only 0 or 1 criminal, the only odd possibility is 1 criminal, which means Uma herself cannot be the criminal in row 5. Therefore, we can determine that A5 is INNOCENT.

Uma’s clue is specifically about Xena’s neighbors and says that Vicky is one of Xena’s 4 innocent neighbors. Vicky at B5 is indeed a neighbor of Xena at C5, so the clue is directly telling us that Vicky is innocent. Therefore, we can determine that B5 is INNOCENT.

Row 5 already has two known innocents, Uma at A5 and Vicky at B5, so Martin’s clue means there must be exactly one more innocent somewhere in that row. That makes C5 and D5 split: one is innocent and the other is criminal. Vicky says Sue at D4 has exactly two innocent neighbors. Sue’s neighbors are Martin at C3, Nick at D3, Rose at C4, Xena at C5, and Zane at D5. Martin and Nick are criminal, so Sue’s only possible innocent neighbors are Rose, Xena, and Zane. Since exactly one of Xena and Zane is innocent, those two contribute exactly one innocent neighbor in total. To reach Sue’s total of exactly two innocent neighbors, Rose must be the other innocent neighbor. Therefore, we can determine that C4 is INNOCENT.

Ethan at A2 has neighbors A1, B1, B2, A3, and B3, and among them only Janet and Karen are innocent, so Ethan has 2 innocent neighbors. Uma at A5 has neighbors A4, B4, and B5, and Rose says Uma must have the same number of innocent neighbors as Ethan, so Uma also must have 2 innocent neighbors. Since Paul at B4 and Vicky at B5 are already innocent, Olive at A4 cannot be innocent as well. Therefore, we can determine that A4 is CRIMINAL.

Row 5 currently has Uma and Vicky as innocents, while Xena and Zane are still undecided. Martin’s clue says row 5 is the only row with exactly 3 innocents, so row 5 must end up with one more innocent and one criminal. Row 4 already has Paul and Rose as innocents, Olive as criminal, and only Sue undecided. If Sue were innocent, then row 4 would also have exactly 3 innocents, which is not allowed because row 5 is supposed to be the only such row. Therefore, we can determine that D4 is CRIMINAL.

Uma’s clue says that Xena has exactly 4 innocent neighbors, and Vicky is one of them. Xena is at C5, so her neighbors are B4, C4, D4, B5, and D5. Among those, Paul at B4, Rose at C4, and Vicky at B5 are already innocent, while Sue at D4 is criminal, so to reach exactly 4 innocent neighbors the only remaining neighbor, Zane at D5, must also be innocent. Therefore, we can determine that D5 is INNOCENT.

Row 5 contains Uma, Vicky, Xena, and Zane, and we already know Uma, Vicky, and Zane are innocent. Martin says row 5 is the only row with exactly 3 innocents, so row 5 must have exactly 3 innocents in total. Since those three innocents are already accounted for, Xena cannot also be innocent. Therefore, we can determine that C5 is CRIMINAL.