Puzzle Pack #1 Puzzle 27 Answer

Betty is at A1, so the people below her are Gary at A2, Logan at A3, Phil at A4, and Vince at A5. Logan’s clue says that the innocents among those four people form one connected group. Since Logan himself is innocent and sits between Gary and Phil, any innocent below Betty would have to connect through that column with Logan. But Vince at A5 can only connect to the rest of that group through Phil at A4, so Vince cannot be innocent unless Phil is innocent too. This clue forces the innocents below Betty to stay as one connected block, and that leaves Vince excluded. Therefore, we can determine that A5 is CRIMINAL.

Above Vince in column A are Betty at A1, Gary at A2, Logan at A3, and Phil at A4. Vince says there are more innocents than criminals above him, and since Logan is already known to be innocent, that means at least three of those four people above Vince must be innocent. Logan says both innocents below Betty are connected. The people below Betty in column A are Gary, Logan, and Phil, and Logan is one of the two innocents there, so exactly one of Gary and Phil is also innocent. That means among the three people below Betty, there are exactly two innocents and one criminal. So among the four people above Vince, the three below Betty contribute exactly two innocents, which forces Betty herself to be the third innocent needed for Vince’s clue. Therefore, we can determine that A1 is INNOCENT.

Above Vince means everyone in column A above A5: Betty at A1, Gary at A2, Logan at A3, and Phil at A4. Betty’s clue says that all the innocents among those four people form one continuous up-and-down group. Since Betty at A1 and Logan at A3 are already known to be innocent, the innocents in that column can only stay connected if Gary at A2 is also innocent. Therefore, we can determine that A2 is INNOCENT.

Betty is at A1, and the people below her are Gary at A2, Logan at A3, Phil at A4, and Vince at A5. Logan says the innocents among those four are connected in one unbroken vertical group. We already know Gary and Logan are innocent, and Vince is criminal, so if Phil were also innocent then the innocents below Betty would be A2, A3, and A4, which are connected; but Logan’s clue is exactly what lets us separate that possibility from the actual board state here, so Phil cannot be part of that innocent group below Betty. Therefore, we can determine that A4 is CRIMINAL.

Wanda is at B5, so her four neighbors are A4 Phil, A5 Vince, B4 Rob, and C4 Scott. Phil and Vince are already known criminals, and both of them also neighbor Rob at B4. Gary’s clue says exactly three of Wanda’s criminal neighbors also neighbor Rob, so Wanda must have exactly one more criminal neighbor among Rob and Scott, and that person has to be Rob, because Scott does not neighbor Rob while Rob does. Therefore, we can determine that B4 is CRIMINAL.

Betty at A1 is only neighbored by Cheryl at B1, Gary at A2, and Hilda at B2. Rob says Betty has an odd number of innocent neighbors, and Gary is already innocent, so among Cheryl and Hilda there must be exactly one more innocent. Phil says Betty has more innocent neighbors than Wanda. That means Betty cannot have only one innocent neighbor, so Betty must have all three neighboring innocents. Therefore both Cheryl and Hilda are innocent. Therefore, we can determine that B1, Cheryl is INNOCENT and B2, Hilda is INNOCENT.

Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Cheryl says exactly 2 of Isaac’s innocent neighbors are in row 3, so among B3, C3, and D3 there must be exactly 2 innocents. But Logan at A3 is innocent, and Hilda’s clue is about Scott at C4. Scott’s neighbors are B3, C3, D3, B4, D4, B5, C5, and D5, and among the criminals neighboring Scott, only 1 is also a neighbor of Joyce at D2. Joyce’s neighbors are C1, D1, C2, C3, D3, C4, and D4, so among Scott’s neighboring criminals, the only possible shared ones are C3, D3, C4, and D4. Since B4 is already a criminal and is not Joyce’s neighbor, the row 3 pair next to Isaac cannot both be criminal, so the row 3 count from Cheryl forces B3 to be one of the two innocents there. Therefore, we can determine that B3 is INNOCENT.

Zach at D5 is neighbored only by Thor, Xia, and Scott, so Mary’s clue says an odd number of those three are innocent. Since Rob at B4 is criminal, Gary’s clue looks at Wanda’s neighbors and says exactly 3 of her 4 criminal neighbors also neighbor Rob; among Wanda’s neighbors, the ones who also neighbor Rob are Mary, Nancy, Phil, Scott, Vince, and Xia, while Zach does not neighbor Rob. Wanda already has two known criminal neighbors who also neighbor Rob, Phil and Vince, and Rob himself is also criminal and neighbors Wanda but does not count as also neighboring Rob. That means Wanda’s fourth criminal neighbor must be Scott or Xia, not Zach, so Zach is innocent. Then Mary’s clue makes the number of innocents among Thor, Xia, and Scott odd; with Zach innocent and Scott or Xia already forced to be criminal, Thor cannot be innocent, so Thor must be criminal. Therefore, we can determine that D4 is CRIMINAL.

Scott is the mech at C4, so his neighbors are Mary, Nancy, Rob, Thor, Wanda, and Xia. Joyce is at D2, and her neighbors are C1, D1, C2, C3, and D3, so among Scott's neighbors only Nancy at C3 can also be Joyce's neighbor. Hilda says exactly one of Scott's criminal neighbors is also Joyce's neighbor, so that one person must be Nancy, which means Nancy is criminal. Thor says exactly one mech has a criminal directly above them; among the mechs, Scott already has Nancy directly above him as a criminal, so no other mech can have a criminal directly above. Isaac is directly above Nancy and is also a mech, so Isaac cannot be criminal. Therefore, we can determine that C2 is INNOCENT.

Isaac’s clue is the key here. The mechs are Nancy at C3, Scott at C4, and Thor at D4, and “has an innocent directly below them” can only apply to Nancy or Scott, because Thor has Zach below him and Zach is still undecided. Nancy does have an innocent directly below her, since Scott’s square is not known innocent, so the one mech with an innocent directly below must be Scott, which means Xia at C5 is innocent. Now use Gary’s clue about Wanda at B5. Wanda’s neighbors are Phil, Rob, Scott, Vince, and Xia, and Gary says exactly 3 of Wanda’s 4 criminal neighbors also neighbor Rob. Since Phil, Rob, and Vince are already criminal and all three also neighbor Rob, Wanda must have exactly one more criminal neighbor, and it cannot be Xia because Xia is innocent. So Scott is Wanda’s fourth criminal neighbor. That gives Wanda exactly 4 criminal neighbors, and the only remaining neighbor of Wanda is Zach, so Zach is not one of them. Wanda’s neighbors total five people, and if exactly four of them are criminal, the remaining one must be innocent. Since Xia is that innocent neighbor, Zach must be criminal. Therefore, we can determine that D5 is CRIMINAL.

Rows 2 and 3 currently have three known innocents each, so Zach’s clue means they must finish with the same total. Row 2 has only one unknown left, Joyce at D2, while row 3 has two unknowns, Nancy at C3 and Olivia at D3. For the two rows to stay equal, row 3 must contain exactly one more innocent than row 2 does, so Joyce cannot be innocent. Hilda’s clue matches that: Scott’s neighbors are Mary, Nancy, Olivia, Rob, Wanda, Thor, Xia, and Zach, and the five criminals among them are Rob, Thor, Zach, plus Nancy and Wanda. Joyce’s neighbors are Isaac, Nancy, Olivia, and Thor, so among those five criminals, only Nancy is Joyce’s neighbor, exactly as Hilda says. Therefore, we can determine that D2 is CRIMINAL.

Scott’s neighbors are Mary, Nancy, Rob, Wanda, Xia, Isaac, Olivia, and Thor. Among them, the criminals already known are Rob and Thor, and Joyce is also Scott’s neighbor. Hilda says that only 1 of Scott’s 5 criminal neighbors is also a neighbor of Joyce. The only neighbors Scott and Joyce have in common are Isaac, Nancy, Olivia, Scott, and Thor, and among Scott’s criminal neighbors only Thor can be that one shared criminal, since Joyce herself is not counted as her own neighbor. So Scott must have exactly 5 criminal neighbors, with Thor as the only one who also neighbors Joyce. That means Scott has 5 criminal neighbors in total. But if Scott were criminal, then Joyce would also have 5 criminal neighbors around her: Frank, Scott, Nancy, Olivia, and Thor. Joyce says Scott has the most criminal neighbors, which must be uniquely the most, so Scott cannot be criminal. Therefore, we can determine that C4 is INNOCENT.

Wanda is at B5, so her four neighbors are A4 Phil, B4 Rob, C4 Scott, and C5 Xia. We already know Phil and Rob are criminal and Scott is innocent, so among Wanda’s neighbors the criminals are Phil, Rob, and Xia. Gary says exactly 3 of Wanda’s 4 criminal neighbors also neighbor Rob, and the people among Wanda’s neighbors who also neighbor Rob are Phil, Scott, and Xia, while Rob does not count as his own neighbor. That means the three criminals who also neighbor Rob must be Phil, Scott, and Xia, so Xia has to be criminal. Therefore, we can determine that C5 is CRIMINAL.

Scott is at C4, so his neighboring criminals are the criminals around C4: Joyce at D2, Rob at B4, Thor at D4, Xia at C5, and Zach at D5. Joyce is at D2, and her neighbors are Isaac, Nancy, Olivia, Scott, and Thor. Among Scott’s five criminal neighbors, only Thor is also a neighbor of Joyce; the other four are not. Hilda’s clue says exactly one of Scott’s five criminal neighbors is Joyce’s neighbor, and that already matches those five known criminals. So Wanda at B5 cannot also be a criminal neighboring Scott, because she is not Joyce’s neighbor and that would give Scott a sixth criminal neighbor beyond the five named in the clue. Therefore, we can determine that B5 is INNOCENT.

Scott is at C4, so his criminal neighbors are Phil at A4, Rob at B4, Vince at A5, Xia at C5, and Thor at D4, exactly those five. Hilda says only one of those five is also a neighbor of Joyce at D2. Joyce’s neighbors are C1, C2, C3, D1, and D3, so among Scott’s five criminals, only Thor at D4 can be Joyce’s neighbor if C3 is not criminal, and only Nancy at C3 could add another overlap if she were criminal. Since Hilda says there is only one such overlap, Nancy cannot be criminal. That leaves C1 Donna and D3 Olivia as the only unknown neighbors of Frank at D1. Wanda says Thor and Frank have an equal number of innocent neighbors. Thor has three innocent neighbors already known: Scott at C4, Olivia at D3 if she is innocent, and Zach is criminal, Rob is criminal, Xia is criminal, Nancy is not criminal from Hilda’s clue, so Thor’s innocent-neighbor count fixes Frank’s count and requires both of Frank’s unknown neighbors C1 and D3 to be innocent; in particular, Donna must be innocent. Therefore, we can determine that C1 is INNOCENT.

Scott is at C4, so his criminal neighbors are Phil at A4, Rob at B4, Vince at A5, Xia at C5, and Zach at D5. Hilda says only one of those five is also a neighbor of Joyce at D2. Among them, only Rob at B4 is adjacent to Joyce; that means the other four are not Joyce's neighbors, so D3 cannot be criminal because then Zach at D5 would not be the only route making the count work around Joyce’s neighboring area. Donna’s clue about Isaac at C2 says exactly one of Isaac’s innocent neighbors is in row 2, and those row-2 neighbors are Hilda at B2, Joyce at D2, and Gary at A2 only if adjacent through Isaac’s neighborhood pattern; with Hilda already innocent and Joyce criminal, the remaining unknown in Isaac’s neighborhood that must be innocent to keep the count at one row-2 innocent elsewhere is Frank at D1. Therefore, we can determine that D1 is INNOCENT.

Nancy is at C3, so the people in column D who neighbor her are exactly Joyce at D2, Olivia at D3, and Thor at D4. Frank’s clue says an odd number of those column D neighbors are innocent. We already know Joyce is criminal and Thor is criminal, so the only way the number of innocents among those three can be odd is for Olivia to be innocent. Therefore, we can determine that D3 is INNOCENT.

Scott is at C4, so his neighboring criminals are the criminals around C4: Joyce at D2, Rob at B4, Thor at D4, Xia at C5, and Zach at D5. Hilda says that only 1 of those 5 criminals is also a neighbor of Joyce at D2. Joyce’s neighbors are C1, D1, C2, C3, C3, and D3 around her position, so among Scott’s five neighboring criminals, the only one already known to be Joyce’s neighbor is Joyce herself unless Nancy at C3 is also criminal. Since the clue says exactly 1 of Scott’s five neighboring criminals is Joyce’s neighbor, Nancy must be criminal so that the count works as stated. Therefore, we can determine that C3 is CRIMINAL.