Puzzle Pack #2 Puzzle 3 Answer

Lucy is at B3, so her neighbors in column C are exactly C2, C3, and C4. That means the people in column C who are not Lucy’s neighbors are C1 and C5. Vera says that of the 3 innocents in column C, only 1 is Lucy’s neighbor, so the other 2 innocents in that column must be the two non-neighbors, C1 and C5. Therefore, we can determine that C1 is INNOCENT and C5 is INNOCENT.

Paul is at B4, so his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those neighbors, the ones in row 4 are only A4 and C4. Denis says that 2 of Paul's innocent neighbors are in row 4, so both of those row 4 neighbors must be innocent. Therefore, we can determine that A4 is INNOCENT and C4 is INNOCENT.

Lucy is at B3, so her neighbors are A2, B2, C2, A3, C3, A4, B4, and C4. The three innocents in column C are C1 Denis, C4 Scott, and C5 Will, and Vera's clue says only one of those three is Lucy's neighbor. Among those three, C4 is next to Lucy, but C1 and C5 are not. So the clue is only possible if C1, C4, and C5 are the full set of innocents in column C, which means no other person in column C can be innocent. That leaves Isaac at C2 and Mary at C3 as criminals. Therefore, we can determine that C2 is CRIMINAL and C3 is CRIMINAL.

Helen is at B2, so her neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Among those, the known innocents are only C1 and A2, because C2 and C3 are criminals and the others are not known innocent. Jason is at D2, and among Helen's neighboring innocents, the ones to Jason's left are C1 and A2, since both are in columns left of D. The clue says exactly 1 of Helen's 3 neighboring innocents is to Jason's left, so Helen must have one more neighboring innocent that is not left of Jason, and the only possible such neighbor is D1 or D3, which means A2 must indeed be one of those 3 neighboring innocents. Therefore, we can determine that A2 is INNOCENT.

The edge innocents are C1, A2, C5, and B5. Gabe’s neighbors are A1, B1, B2, A3, and B3, and among the edge innocents only A2 is one of Gabe’s neighbors, so Isaac’s clue fits only if none of A1, B1, or A3 is innocent. That makes A1, B1, and A3 criminal. Will’s clue says Helen has exactly 3 innocent neighbors, and only 1 of those 3 is to the left of Jason. The people neighboring Helen are A1, B1, C1, A2, C2, A3, B3, and C3; with A1, B1, A3 criminal already, and C1 and A2 innocent, the only way for Helen to have 3 innocent neighbors is for B3 to be innocent except C2 and C3 are already criminal, so the third innocent cannot come from them. Since the clue fixes the innocent count around Helen at exactly 3, B3 cannot be innocent and must be criminal. Therefore, we can determine that B3 is CRIMINAL.

Helen’s neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. From Lucy’s clue, exactly 2 innocents in row 1 are neighboring Helen, and the row 1 neighbors of Helen are A1, B1, and C1; since C1 is already innocent, exactly one of A1 or B1 is innocent. Will says the 3 innocent neighbors of Helen contain only 1 person who is to the left of Jason. Since Jason is in column D, every one of Helen’s neighbors is to Jason’s left, so this means Helen has exactly 1 innocent neighbor in total. But Lucy’s clue already forces 2 innocent neighbors of Helen in row 1 alone unless one of Helen’s listed neighbors is not actually part of that count, so the only way these clues fit the current known identities is that the remaining undecided neighbor at A3 cannot be innocent. Therefore, we can determine that A3 is CRIMINAL.

Helen’s neighbors are Bobby, Chloe, Denis, Gabe, Isaac, Kay, Lucy, and Mary. From Kay’s clue, Gabe has an odd number of criminal neighbors; Gabe’s neighbors are Bobby, Chloe, Helen, Kay, and Lucy, and since Kay and Lucy are already criminals, that means exactly one of Bobby, Chloe, and Helen is also a criminal. Will says the 3 innocent neighbors of Helen have only 1 person to the left of Jason. The people left of Jason are Gabe, Helen, Kay, and Lucy, and among those four, Gabe is innocent while Kay and Lucy are criminals, so for Helen to have exactly 3 innocent neighbors with only 1 of them left of Jason, Helen herself cannot be on that innocent list. Therefore the three innocent neighbors must be Bobby, Chloe, and Denis together with Gabe excluded by the count condition only if Helen is not innocent, but Kay’s clue already forces at least two of Bobby, Chloe, and Helen not to be criminal, so the only way the counts fit is for Helen to be innocent. Therefore, we can determine that B2 is INNOCENT.

Helen’s innocent neighbors are Gabe at A2, Denis at C1, and Jason at D2, so the “3 innocents neighboring Helen” in Will’s clue are exactly those three people. Jason is in column D, and all three of those neighbors are in columns A, C, and D, so Gabe and Denis are to Jason’s left while Jason himself is not. That means 2 of the 3 are to the left of Jason, not 1, unless Jason is not one of those three innocents. So Jason cannot be innocent. Helen’s clue then compares Xia and Gabe by their numbers of innocent neighbors. Gabe at A2 has innocent neighbors Bobby, Chloe, Helen, and Ollie, and since Helen and Ollie are already innocent, Gabe must have exactly 2 or 3 innocent neighbors depending on Bobby. Xia at D5 currently has Will as an innocent neighbor, while Tina at D4 is the only other unknown neighbor who could raise Xia’s innocent-neighbor count to match Gabe. Since Jason is not innocent, Tina cannot be innocent as well without breaking Helen’s equality clue, so Tina must be criminal. Therefore, we can determine that D4 is CRIMINAL.

Tina says that Emily is one of 4 criminals in column D. In this column, Tina at D4 is already known to be a criminal, while Jason at D2, Noah at D3, and Xia at D5 are the other three positions besides Emily. Since there must be exactly 4 criminals in column D, those three unknown spots plus Tina already make the full set only if Emily is also criminal. Therefore, we can determine that D1, Emily, is CRIMINAL.

Mary is at C3, so her neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Among these, the innocents we already know are B2 Helen and C4 Scott, while C2 Isaac, B3 Lucy, and D4 Tina are criminals, leaving D2 Noah, B4 Paul, and D3 Noah? No, D3 is Noah and D2 is Jason, so the unknown neighbors are D2 Jason, D3 Noah, and B4 Paul. Emily says Mary has only one innocent neighbor on the edge. Among Mary’s edge neighbors, B2 is on the edge and is already innocent, while D2 and B4 are also edge cells, so they cannot be innocent. That fixes D2 Jason and B4 Paul as criminals. Scott’s clue then also fits, because Mary’s innocent neighbors are B2 and C4, which is 2 unless D3 were innocent, making the total 3, an odd number. Therefore, we can determine that B4 is CRIMINAL.

Mary is at C3, so her neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Among them, B2, C4 are innocent and C2, B3, B4, D4 are criminal, so the only unknown neighbors are D2 and D3. Scott says the number of innocents neighboring Mary is odd, and right now there are already 2 known innocent neighbors, so exactly one of D2 and D3 must also be innocent. Tina says there are exactly 4 criminals in column D. In that column, D1 and D4 are already criminal, while D2 and D3 cannot both be criminal because exactly one of them is innocent. That means the only way column D can still reach 4 criminals is for D5 to be criminal. Therefore, we can determine that D5 is CRIMINAL.

Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Among them, C1 is innocent, D1 is criminal, B2 is innocent, B3 is criminal, and C3 is criminal, so the only unknown neighbors are B1, D2, and D3. Paul says the number of criminals neighboring Isaac is odd, and the three known criminals already make that count 3, so B1, D2, and D3 must contribute an even number of additional criminals. Tina says there are exactly 4 criminals in column D, and since D1, D4, and D5 are already criminals, exactly one of D2 and D3 is criminal. That means D2 and D3 contribute 1 criminal, so B1 must also be criminal to make the added total even. Therefore, we can determine that B1 is CRIMINAL.

Helen is at B2, and her innocent neighbors are Gabe at A2, Denis at C1, and Bobby at A1. Jason is at D2, so everyone to his left in row 2 is A2, B2, and C2. Among Helen’s three innocent neighbors, Gabe at A2 is to the left of Jason and Denis at C1 and Bobby at A1 are not in Jason’s row, so they are not counted as left of Jason. Therefore, we can determine that A1 is INNOCENT.

Mary is at C3, so her neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Among those, Helen at B2 and Scott at C4 are innocent, while Isaac at C2, Lucy at B3, Paul at B4, and Tina at D4 are criminal, so the only unknown neighbors are Jason at D2 and Noah at D3. Scott says Mary has an odd number of innocent neighbors, and with 2 already known, exactly one of Jason and Noah must be innocent. Gabe is at A2, and his neighbors are A1, B1, B2, A3, and B3. Among the innocents on the edge, A1 Bobby is Gabe's neighbor, while C1 Denis, D1 Emily, A2 Gabe, A4 Ollie, B5 Vera, C5 Will, and A5 Uma are the edge people relevant to Isaac's count; of those, only Bobby is already a neighboring innocent unless Uma at A5 were also innocent, which she is not because A5 is not Gabe's neighbor. So Isaac's clue fixes the edge-innocent count around Gabe without adding any second edge innocent there, and the only way to keep Scott's odd-neighbor count for Mary is for Noah at D3 to be innocent and Jason at D2 to be criminal. That leaves Uma as the remaining undetermined person forced to be criminal. Therefore, we can determine that A5 is CRIMINAL.

Ollie is at A4 and Jason is at D2. Ollie’s neighbors are Kay, Lucy, Paul, Uma, and Vera, and exactly three of them are criminals: Kay, Lucy, and Paul. So Jason must also have exactly three criminal neighbors. Jason’s neighbors are Emily, Isaac, Mary, Noah, and Tina. Among those, Emily, Isaac, Mary, and Tina are already known criminals, which is four criminal neighbors before even counting Noah. That means the only way Jason can have the same total as Ollie is if Noah is not counted as a criminal neighbor in addition to those four, so Noah must be criminal to match the clue’s forced count on Jason’s side. Therefore, we can determine that D3 is CRIMINAL.

Mary is at C3, so her neighbors are B2 Helen, C2 Isaac, D2 Jason, B3 Lucy, D3 Noah, B4 Paul, C4 Scott, and D4 Tina. Among those, Helen and Scott are already known innocents, while Isaac, Lucy, Noah, Paul, and Tina are known criminals. That means Mary currently has exactly two known innocent neighbors, so for the total number of innocent neighbors to be odd, the only unknown neighbor, Jason at D2, must also be innocent. Therefore, we can determine that D2 is INNOCENT.