Puzzle Pack #1 Puzzle 7 Answer

Carol’s clue is specifically about Rob’s neighbors, and Pam is named as one of the innocents among them. Since everyone tells the truth, that statement directly fixes Pam’s identity. Therefore, we can determine that C4 Pam is INNOCENT.

Betty’s neighbors are Anna, Eric, Flora, Carol, Gary, Isaac, and Jose, and the clue says exactly three of those seven are criminals. “To the right of Eric” means everyone in row 2 to the right of A2, so among Betty’s neighboring criminals, exactly one must be Flora or Gary. Since Carol is already innocent, the other two criminals neighboring Betty must come from Anna, Eric, Isaac, and Jose. That forces Anna to be one of Betty’s three criminal neighbors, and Eric must be one too for the count to work with only one criminal allowed on Eric’s right. Therefore, we can determine that A1 is CRIMINAL and A2 is CRIMINAL.

Will is at B5, so his neighbors are A4 Nicole, B4 Oscar, C4 Pam, A5 Susan, and C5 Xia. Eric’s clue says that exactly two of Will’s neighboring criminals are on the edge, so every criminal among Will’s neighbors must be on the edge. Oscar at B4 is one of Will’s neighbors, but B4 is not an edge position. Therefore, we can determine that B4, Oscar, is INNOCENT.

In column A, Anna at A1 and Eric at A2 are already known criminals. Anna’s clue says all criminals in column A are connected, so any other criminal in that column must form one continuous vertical group with them, with no innocent gap splitting the criminals apart. Since the criminals in column A are determined here to continue downward, the next two spaces in that same connected chain are A3 and A4. Eric’s clue matches that: Will at B5 is neighbored by A4, A5, B4, C4, C5, and those on the edges are A4, A5, B4, and C5. Because B4 and C4 are already innocent, the two criminal neighbors Will has must come from the remaining edge neighbors, which includes A4 and keeps the criminal run in column A going through A3 to A4 rather than breaking at A2. Therefore, we can determine that A3 is CRIMINAL and A4 is CRIMINAL.

Pam is at C4, and her neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. Nicole’s clue says Kevin is one of Pam’s 3 innocent neighbors, so Kevin must be one of those innocent neighbors. Since Kevin is at C3 and is indeed next to Pam, this clue directly fixes Kevin’s identity. Therefore, we can determine that C3 is INNOCENT.

Will is at B5, so the people above him are Betty at B1, Flora at B2, Jose at B3, and Oscar at B4. Kevin says both criminals above Will are connected, so among those four people there are exactly two criminals, and they must be in one unbroken vertical group in column B. Since Oscar at B4 is already innocent, the only way to place exactly two connected criminals above Will is at B2 and B3 or at B1 and B2. But B1 and B2 cannot be the pair, because that would leave Jose at B3 innocent, and then the known innocents at B3 and B4 would separate the criminals from the rest in a way that does not fit the clue’s required connected pair above Will. Therefore, Flora at B2 has to be one of the two connected criminals above Will, so we can determine that B2 is CRIMINAL.

Betty is at B1, so her three neighboring criminals are exactly Anna at A1, Eric at A2, and Flora at B2. Pam’s clue says that only one of those three criminals is to the right of Eric. Of those three, only Flora at B2 is to the right of Eric at A2, so the clue fits only if Betty really does have exactly those three neighboring criminals and no fourth criminal neighbor. Gary at C2 is also Betty’s neighbor, so Gary cannot be a criminal. Therefore, we can determine that C2, Gary, is INNOCENT.

Oscar’s clue compares how many innocents each profession has, and it says the teachers have uniquely the most. We already know two innocent teachers, Betty at B1 and Kevin at C3, while Carol at C1 is also an innocent teacher, so the teacher group has three innocents in total. No other profession can reach three innocent people, and among builders we already have Gary at C2 and Oscar at B4 as innocents, so Henry at D2 cannot also be innocent or builders would tie teachers. That fixes Henry as criminal, and since the teachers must be the profession with the most innocent people, Betty must be the missing innocent teacher. Therefore, we can determine that B1 is INNOCENT and D2 is CRIMINAL.

Will is at B5, so the people above him in column B are Betty at B1, Flora at B2, Jose at B3, and Oscar at B4. Among them, Betty and Oscar are innocent while Flora is already known to be criminal, so the two criminals above Will must be Flora and Jose. Kevin’s clue says both criminals above Will are connected, and in one column that means they must be adjacent with no innocent between them. Since Flora at B2 is one of those criminals, Jose at B3 has to be the other. Therefore, we can determine that B3, Jose, is CRIMINAL.

Row 3 contains Isaac at A3, Jose at B3, Kevin at C3, and Megan at D3. Betty’s clue says there is only one innocent in that row, and Kevin at C3 is already known to be innocent. That means the rest of row 3 must all be criminals, including Megan at D3. Therefore, we can determine that D3 is CRIMINAL.

Rob is at D4 and Xia is at C5. Their common neighbors are C4, C5, D5, and D4, excluding Rob and Xia themselves, so the only common neighbors that count are Pam at C4 and Zed at D5. Megan says Rob and Xia have only one innocent neighbor in common. Pam is already known to be innocent, so among those two shared neighbors, the other one cannot also be innocent. Therefore, we can determine that D5, Zed, is CRIMINAL.

Henry is at D2, and the people below him in the same column are D3, D4, and D5. We already know D3 is criminal and D5 is criminal, so there are 2 criminals below Henry so far. Zed’s clue says the total number of criminals below Henry is odd, so D4 must also be criminal to make that total 3. Therefore, we can determine that D4 is CRIMINAL.

Will is at B5. His neighbors are A4, A5, B4, C4, and C5, and Eric’s clue says exactly two of those neighbors are criminals, with both of those criminals on the edge. Since A4 is already a criminal and is on the edge, while B4 and C4 are already innocent, the second criminal neighbor of Will must be either A5 or C5, so C5 cannot be innocent. Henry is at D2, and his innocent neighbors are C1, C2, C3, and D1 except that D1 is still unknown. Rob’s clue says Will and Henry have the same number of innocent neighbors, and Will already has exactly two innocent neighbors next to him, namely B4 and C4, because his other three neighboring spots contain one criminal at A4, one more criminal among A5 and C5, and the remaining one is not innocent. So Henry must also have exactly two innocent neighbors, which forces D1 to be criminal since C1, C2, and C3 are already three innocent neighbors otherwise. Therefore, we can determine that D1 is CRIMINAL.

Diane’s clue says there are exactly 13 criminals in total. On the board, 12 people are already known to be criminals, so among the three unknowns at A5, B5, and C5, exactly one must be a criminal. Eric’s clue is about Will at B5. Will’s neighbors are A4, A5, B4, C4, and C5, and the edge positions among those are A4, A5, and C5. Since A4 is already a known criminal, the clue says the only criminals neighboring Will are on edges, so B4 and C4 cannot be criminals, which fits, and any second neighboring criminal must also be on an edge. Because exactly one of A5, B5, and C5 is still criminal, and both A5 and C5 would be neighboring criminals of Will if either of them were that one criminal, Eric’s clue already accounts for Will having two criminal neighbors on edge squares. That leaves Will himself as the remaining criminal needed to make the total 13. Therefore, we can determine that B5 is CRIMINAL.

Pam is at C4, so her neighbors are B3, C3, D3, B4, D4, B5, C5, and D5. From the board, among those eight people, Jose, Megan, Rob, Will, and Zed are criminal, while Kevin and Oscar are innocent, leaving Xia as the only undecided neighbor. Nicole’s clue says Kevin is one of Pam’s 3 innocent neighbors, so Pam must have exactly three innocent neighbors in total. Since Kevin and Oscar already account for two of them, Xia has to be the third innocent neighbor. Therefore, we can determine that C5 is INNOCENT.

Will is at B5, so his neighbors are A4, B4, C4, A5, C5, and there are no cells below row 5. Among these, the known criminals are A4, and the clue says both criminals neighboring Will are on the edges, so Will has exactly two criminal neighbors and both must be edge positions. Since B4 is not on the edge and C4 and C5 are already innocent, the only possible second criminal neighbor is A5, which is on the edge. Therefore, we can determine that A5 is CRIMINAL.