Puzzle Pack #1 Puzzle 45 Answer

Carol is at C1, so her neighbors are Bobby at B1, Freya at B2, and Gabe at C2. Bobby’s neighbors are Amy at A1, Ellie at A2, Freya at B2, Carol at C1, and Gabe at C2. Carol says that of Bobby’s 2 innocent neighbors, only 1 is also her neighbor. Since Carol herself is innocent and is one of Bobby’s neighbors, she must be one of those 2 innocent neighbors, so exactly one of Bobby’s other innocent neighbors is among Freya and Gabe, and none of Bobby’s innocent neighbors can be Amy or Ellie because those two are not neighbors of Carol. Therefore, we can determine that A1 is CRIMINAL and A2 is CRIMINAL.

Mark’s neighbors are A3, B3, A5, B5, and B4, so among the people in column B who neighbor Mark, only B3, B4, and B5 count. Ellie says an odd number of innocents in column B neighbor Mark, so among B3, B4, and B5 there must be 1 or 3 innocents. Carol says Bobby has exactly two innocent neighbors, and only one of those two is also Carol’s neighbor. Bobby’s neighbors are A1, C1, A2, B2, and C2, and among those Carol’s neighbors are B1, B2, and C2, so the only way exactly one of Bobby’s two innocent neighbors is also Carol’s neighbor is that B2 and C2 cannot both be innocent. Since A1 and A2 are criminals and C1 is innocent, Bobby cannot reach two innocent neighbors unless at least one of B2 or C2 is innocent, so exactly one of B2 and C2 is innocent and Bobby has exactly two innocent neighbors in total. Amy says Bobby has more innocent neighbors than Mark, so Mark must have fewer than two innocent neighbors. Combining that with Ellie’s clue, Mark cannot have 1 innocent among B3, B4, and B5 and also any innocent at A3 or A5, so A3 and A5 cannot be innocent. Therefore, we can determine that A3 is CRIMINAL and A5 is CRIMINAL.

Carol’s clue says Bobby has exactly 2 innocent neighbors, and exactly 1 of those 2 is also Carol’s neighbor. Bobby’s neighbors are Amy, Carol, Ellie, Freya, Gabe, Isaac, and Janet; among these, Amy, Ellie, and Isaac are already criminals, and Carol is an innocent, so Bobby’s second innocent neighbor must be either Freya, Gabe, or Janet. Since Carol’s neighbors among Bobby’s neighbors are only Amy, B1, B2, C2, and D2, that means the one innocent neighbor of Bobby who is also Carol’s neighbor can only be Freya or Gabe, so Janet must be the other innocent neighbor and is innocent, and exactly one of Freya and Gabe is innocent. Amy’s clue says Bobby has more innocent neighbors than Mark, so Mark must have fewer than 2 innocent neighbors. Mark’s neighbors are Isaac, Janet, Nicole, Susan, and Terry; with Isaac and Susan already criminals and Janet now innocent, Mark cannot have 2 or more innocent neighbors, so Nicole and Terry cannot both be innocent. Susan’s clue says there are more innocent builders than innocent sleuths. Janet is now an innocent builder, and among sleuths, Ellie is criminal and Laura is the only remaining possible innocent sleuth, so to avoid having the innocent builders outnumber the innocent sleuths, Laura cannot be innocent. Therefore, we can determine that D3 is CRIMINAL.

Amy is at A1, so she has exactly three neighbors: Bobby at B1, Ellie at A2, and Freya at B2. Since Amy is known to have fewer innocent neighbors than Mark, and Ellie is already known criminal, Amy’s innocent-neighbor count can only be raised by Bobby and Freya. Amy also says Bobby has more innocent neighbors than Mark, so Bobby must have more innocent neighbors than Amy. Bobby is next to Amy, Carol, Ellie, Freya, Gabe, and Janet. Among those, Amy and Ellie are known criminals and Carol is known innocent, so for Bobby to have more innocent neighbors than Mark and Mark to have more innocent neighbors than Amy, Bobby needs additional innocent neighbors beyond Carol. That forces Gabe at C2 to be innocent, and Bobby himself cannot be innocent here, because Amy’s count and Bobby’s comparison only work if Bobby is one of the non-innocent neighbors affecting Mark and Amy differently. Therefore, we can determine that C2 is INNOCENT and B1 is CRIMINAL.