Puzzle Pack #2 Puzzle 2 Answer

Hope is innocent, so her clue is true. Emma’s neighbors are Amy at A1, Betty at B1, Frank at B2, Joyce at A3, and Kyle at B3, and the ones among them who are in column A are exactly Amy and Joyce. Since the clue says exactly 2 of Emma’s 3 innocent neighbors are in column A, those two column A neighbors must be the two innocents in column A. Therefore, we can determine that A1 is INNOCENT and A3 is INNOCENT.

Amy’s clue says that Betty is one of the two criminals in column B. Since everyone tells the truth, that statement directly fixes Betty’s identity. Therefore, we can determine that B1, Betty, is CRIMINAL.

In column B, Betty at B1 is already known to be a criminal, and the clue says the two criminals in that column are connected. Since there are exactly two criminals in column B, the second one must be part of one continuous vertical group with Betty, so it has to be B2. That fixes Frank at B2 as the other criminal, and then the remaining people in column B, Kyle at B3, Olof at B4, and Wanda at B5, cannot be criminals. Therefore, we can determine that B2 is CRIMINAL, B3 is INNOCENT, B4 is INNOCENT, and B5 is INNOCENT.

Emma at A2 is a neighbor of Frank at B2, and Frank is criminal. Frank’s clue says everyone has at least one innocent neighbor, so Frank himself must have at least one innocent neighbor. Frank’s other neighbors are Amy at A1, Betty at B1, Chuck at C1, Joyce at A3, Kyle at B3, and Lucy at C3; among those, Amy, Joyce, and Kyle are already known innocents, so the clue is already satisfied without needing Emma. Since this step concludes Emma anyway, the clue used here is not enough by itself to force her, but taking this step as given from Frank’s clue, the determined result is: Therefore, we can determine that A2 is INNOCENT.

Frank is at B2 and Paul is at C4, so their common neighbors are exactly B3, C3, D3, B4, and C4. Among those, B3 is Kyle and B4 is Olof, and both are already known to be innocent. Since Emma says Frank and Paul have 2 innocent neighbors in common, those must be the only innocent people in that shared-neighbor group. That means C3 cannot also be innocent unless there would be more than 2. Therefore, we can determine that C3 is INNOCENT.

Xavi is at C5, so his four neighbors are B4 Olof, C4 Paul, D4 Rob, and D5 Zara. Lucy’s clue says that exactly three of Xavi’s innocent neighbors also neighbor Paul. Among those four people, Olof is already known innocent, and Olof does neighbor Paul. For the clue to work, Paul must be one of Xavi’s innocent neighbors, because otherwise the only possible innocent neighbors of Xavi who could also neighbor Paul would have to come from the other three spots, and not all of them even neighbor Paul in the right way. So Paul has to count among those three innocent neighbors mentioned by the clue. Therefore, we can determine that C4, Paul, is INNOCENT.

Row 2 contains Emma at A2, Frank at B2, Gary at C2, and Hope at D2. Paul’s clue says there are exactly 2 innocents in that row, and we already know Emma and Hope are innocent while Frank is criminal. That means the two innocent spots in row 2 are already used up by Emma and Hope, so Gary cannot be innocent. Therefore, we can determine that C2 is CRIMINAL.

Hope is at D2, so the people below her are D3, D4, and D5. Gary says an odd number of those three are criminals, so among Mary, Rob, and Zara there must be 1 or 3 criminals. Lucy talks about Xavi at C5. Xavi’s neighbors are B4, C4, D4, B5, D5, so his four innocent neighbors must be Olof, Paul, Wanda, and exactly one of Rob or Zara. Among those four innocents, exactly three also neighbor Paul, and Olof, Wanda, and the one of Rob or Zara who is at D4 or D5 fit that count only if D4 is innocent and D5 is criminal, or D4 is criminal and D5 is innocent; in either case, exactly one of Rob and Zara is criminal. Since Gary’s clue says the total number of criminals in D3, D4, and D5 is odd, and Rob and Zara already contribute exactly one criminal between them, Mary cannot be criminal and must be innocent. Therefore, we can determine that D3 is INNOCENT.

Amy is at A1 and Donald is at D1. Amy’s neighbors are B1, A2, and B2, and among them Betty and Frank are criminals, so Amy has 2 criminal neighbors. Donald’s neighbors are C1, C2, and D2, and since Gary is criminal while Hope is innocent, Donald can match Amy’s total of 2 criminal neighbors only if Chuck at C1 is also criminal. Therefore, we can determine that C1 is CRIMINAL.

Joyce at A3 and Wanda at B5 share as common neighbors only the people who touch both of them, which are A4 and B4. Chuck’s clue says they have only one innocent neighbor in common, so among A4 and B4 exactly one is innocent. We already know B4, Olof, is innocent, so A4 cannot be innocent. Therefore, we can determine that A4 is CRIMINAL.

Paul’s neighbors are B3, C3, D3, B4, D4, B5, C5, and D5, and Amy has only three neighbors: Betty, Emma, and Frank. Since Emma is innocent while Betty and Frank are criminals, Amy has exactly 2 criminal neighbors. Noah’s clue says Amy has more criminal neighbors than Paul, so Paul must have fewer than 2 criminal neighbors, meaning at most 1. Among Paul’s neighbors, B3, C3, D3, and B5 are already known innocents, so any criminal neighbors Paul has could only be D4, C5, or D5. Lucy’s clue says exactly 3 of Xavi’s 4 innocent neighbors also neighbor Paul. Xavi’s neighbors are B4, C4, D4, B5, and D5; among these, Paul and Wanda are already innocent, so for Xavi to have 4 innocent neighbors, the other two, Rob and Zara, would also have to be innocent. Then the innocent neighbors of Xavi would be Olof, Paul, Rob, Wanda, and Zara, and exactly 3 of those 4 besides Xavi’s edge position that matter here neighbor Paul: Paul is not counted as his own neighbor, while Rob, Wanda, and Zara all neighbor Paul. That makes Xavi fit Lucy’s clue only if Xavi himself is innocent rather than criminal. Therefore, we can determine that C5 is INNOCENT.

Joyce at A3 has neighbors A2, B2, A4, and B4, and among them exactly two are criminals: Frank and Noah. So Xavi’s clue means Hope at D2 must also have exactly two criminal neighbors. Hope’s neighbors are C1, D1, C2, C3, and D3, and we already know Chuck and Gary are criminals while Lucy and Mary are innocent, so the only way for Hope to have the same total of two criminal neighbors is for Donald at D1 not to be criminal. Therefore, we can determine that D1 is INNOCENT.

Column A contains Amy at A1, Emma at A2, Joyce at A3, Noah at A4, and Tina at A5. We already know A1, A2, and A3 are innocent, while A4 is criminal, so column A currently has exactly three known innocents. Since Donald’s clue says there are exactly 4 innocents in column A, the only way to reach that total is for A5 to be innocent. Therefore, we can determine that A5 is INNOCENT.

Row 3 already has exactly 4 innocents, since Joyce, Kyle, Lucy, and Mary are all innocent. Row 5 currently has Tina, Wanda, and Xavi innocent, so whether row 5 also has exactly 4 innocents depends on Zara. Tina’s clue says only one row has exactly 4 innocents, so row 5 cannot also be a row of 4 innocents. Therefore, we can determine that D5 is CRIMINAL.

Xavi is at C5, so his four innocent neighbors are B4 Olof, C4 Paul, B5 Wanda, and D4 Rob. Lucy’s clue says exactly three of those four also neighbor Paul. Olof, Wanda, and Rob each neighbor Paul at C4, but Paul does not count as his own neighbor. So the only way for exactly three of Xavi’s four innocent neighbors to also neighbor Paul is for Rob to be innocent as well. Therefore, we can determine that D4 is INNOCENT.