Puzzle Pack #1 Puzzle 18 Answer

Chad is at B1, so the people below him are B2 Gus, B3 Kay, B4 Olivia, and B5 Will. Nancy is at A4, and her neighbors are A3, B3, B4, A5, and B5. Olivia’s clue says that both innocents below Chad are Nancy’s neighbors, so every innocent among those four people below Chad must be one of B3, B4, or B5. That leaves B2 Gus as the only person below Chad who is not Nancy’s neighbor, so he cannot be one of the innocents below Chad. Therefore, we can determine that B2 is CRIMINAL.

Gus directly says that Bobby is one of the three criminals in the corners. Since everyone tells the truth, that fixes Bobby’s identity immediately. Therefore, we can determine that A1 is CRIMINAL.

Will is at B5, so his neighbors are A4 Nancy, B4 Olivia, C4 Peter, A5 Vince, and C5 Xena. Bobby is at A1, and everyone except A1 is below him, so among Will’s neighboring innocents, exactly one is below Bobby. Olivia is already known to be innocent, and she is below Bobby, so she is that one innocent neighbor of Will. That means none of Will’s other neighbors can be innocent, so Peter at C4 and Xena at C5 must both be criminal. Therefore, we can determine that C4 is CRIMINAL, C5 is CRIMINAL, and so on.

Vince is at A5, so the people above Vince are A1 Bobby, A2 Flora, A3 Janet, and A4 Nancy. The clue says both innocents above Vince are connected, so among those four people there are exactly two innocents, and they must form one continuous vertical group in column A. We already know A1 Bobby is a criminal, so the two innocents cannot include A1. That means the only possible connected pair of innocents above Vince is A3 and A4. Therefore, we can determine that A3 Janet is INNOCENT.

Janet says there is exactly one innocent in row 5, and that person is Peter’s neighbor. Peter is at C4, so his neighbors in row 5 are B5, C5, and D5. Since C5 is already known to be criminal, the only possible innocent in row 5 must be either B5 or D5, not A5. Therefore, we can determine that A5 is CRIMINAL.

Will is at B5, so his neighbors are A4 Nancy, B4 Olivia, C4 Peter, A5 Vince, C5 Xena, and the people on row 5 do not count as below Bobby because Bobby is at A1 and “below” means lower in the same column or, in this clue’s plain placement sense, on a lower row than him. Since Bobby says exactly 2 of Will’s neighbors are innocents, and exactly 1 of those 2 innocents is below Bobby, the two innocents among Will’s neighbors must be on rows below row 1. We already know Olivia at B4 is innocent, while Peter, Vince, and Xena are criminals, so the only possible second innocent neighboring Will is Nancy at A4. Therefore, we can determine that A4 is INNOCENT.

The people above Vince at A5 are Flora at A2, Janet at A3, and Nancy at A4. Peter’s clue says that both innocents above Vince are connected, so among those three people exactly two are innocent, and those two must touch orthogonally as one continuous group. Janet and Nancy are already known to be innocent, and they are directly above and below each other, so they are the connected pair of innocents above Vince. That leaves Flora as the only one above Vince who is not innocent. Therefore, we can determine that A2 is CRIMINAL.

Eric is at D1, and his neighbors are C1, C2, and D2. Flora’s clue says Isaac is one of two or more criminals neighboring Eric, so Isaac must be a criminal, and Eric must have at least one other criminal neighbor as well. The extra criminal is not needed for this step; the clue already fixes Isaac’s identity. Therefore, we can determine that D2, Isaac, is CRIMINAL.

Isaac says row 2 has more criminals than any other row, so row 2 must have uniquely the highest number of criminals. We already know row 2 has three criminals there now: A2, B2, and D2, while row 5 already has at least two criminals at A5 and C5. Janet says the only innocent in row 5 is Peter's neighbor. Peter is at C4, and his neighbors in row 5 are B5, C5, and D5. Since C5 is already criminal, the single innocent in row 5 must be either B5 or D5, which means row 5 can have at most one innocent and therefore at least three criminals. So for row 2 to have more criminals than row 5, row 2 cannot stay at only three criminals; it must have four. The only unknown in row 2 is Hank at C2, so Hank must be the fourth criminal. Therefore, we can determine that C2 is CRIMINAL.

In column B, the only people who could be innocents are Chad at B1, Kay at B3, Olivia at B4, and Will at B5, since Gus at B2 is already criminal. The clue says all innocents in that column must form one continuous vertical group with no criminals between them. But Olivia at B4 is already innocent, and if Chad at B1 were also innocent, then the criminal at B2 would split the innocents in column B into separate groups. Therefore, we can determine that B1 is CRIMINAL.

Janet’s clue is about row 5, which contains Vince, Will, Xena, and Zara. Peter is at C4, so his neighbors in row 5 are B5, C5, and D5, meaning the only innocent in row 5 must be Will, Xena, or Zara. But Xena is already known to be criminal, so the only possible innocent in row 5 is either Will at B5 or Zara at D5. Now use Xena’s clue. Bobby at A1 has exactly 3 criminal neighbors: Chad at B1, Flora at A2, and Gus at B2. Xena at C5 currently has 2 known criminal neighbors, Peter at C4 and Vince at A5? Wait, A5 is not Xena’s neighbor; Xena’s actual neighbors are B4, C4, D4, B5, and D5, so the known criminals among them are only Peter at C4, while Olivia at B4 is innocent. To match Bobby’s total of 3 criminal neighbors, Xena must have exactly two more criminal neighbors among Ronald at D4, Will at B5, and Zara at D5. Since Janet’s clue says row 5 has exactly one innocent, at least one of Will and Zara is innocent, so they cannot both be the two extra criminals Xena needs. That forces Ronald at D4 to be one of those extra criminal neighbors. Therefore, we can determine that D4, Ronald, is CRIMINAL.

Kay is at B3, so her neighbors are A2 Flora, B2 Gus, C2 Hank, A3 Janet, C3 Lisa, A4 Nancy, B4 Olivia, and C4 Peter. Among those, Flora, Gus, Hank, and Peter are already known criminals, while Janet, Nancy, and Olivia are already known innocents. That gives Kay exactly 3 known innocent neighbors so far, and Ronald's clue says she has exactly 4 innocent neighbors in total, so the only remaining undecided neighbor, Lisa at C3, must be innocent. Therefore, we can determine that C3 is INNOCENT.

In row 3, the innocents are Janet at A3 and Lisa at C3, while Megan at D3 is the only unknown there. Ronald is at D4, so his row 3 neighbors are C3 and D3. Nancy’s clue says exactly one innocent in row 3 is neighboring Ronald, and Lisa at C3 already is one such innocent. That leaves no room for Megan at D3 to be innocent. Therefore, we can determine that D3 is CRIMINAL.

Will is at B5, so the people above him are Chad at B1, Gus at B2, Kay at B3, and Olivia at B4. Among those, Chad and Gus are already known criminals and Olivia is already known innocent, so the only undecided person in that group is Kay. For there to be more criminals than innocents above Will, that column must contain at least 3 criminals and at most 1 innocent. That only works if Kay is a criminal. Therefore, we can determine that B3 is CRIMINAL.

Chad is at B1, so the people below him are B2, B3, B4, and B5. Among those, the two innocents must be Nancy's neighbors. Nancy is at A4, and her neighbors are A3, B3, B4, A5, and B5. Intersecting those two groups leaves only B3, B4, and B5, so the two innocents below Chad must be B3 and B5. Since B3 is already known to be innocent, B5 has to be the other one. Therefore, we can determine that B5 is INNOCENT.

Xena is at C5 and Bobby is at A1. Bobby’s neighbors are B1, A2, and B2, and all three of them are criminals, so Bobby has 3 criminal neighbors. Xena’s neighbors are B4, C4, D4, B5, and D5; among these, C4 and D4 are already criminals while B4 and B5 are innocents, so for Xena to also have 3 criminal neighbors, D5 must be criminal. Therefore, we can determine that D5 is CRIMINAL.

The corners are A1, D1, A5, and D5. We already know A1 Bobby, A5 Vince, and D5 Zara are criminals, so those are the 3 criminals in the corners that Gus is talking about. That leaves D1 Eric as the only corner who is not a criminal. Therefore, we can determine that D1 Eric is INNOCENT.

Row 1 contains Bobby at A1, Chad at B1, Diane at C1, and Eric at D1. We already know Bobby and Chad are criminals and Eric is innocent, so the only unknown person in that row is Diane. That means row 1 currently has 1 innocent for sure, and it stays an odd total only if Diane is also a criminal. Therefore, we can determine that C1 is CRIMINAL.