Puzzle Pack #1 Puzzle 45 Answer

Carol says that Bobby has exactly 2 innocent neighbors, and only 1 of those 2 is also Carol's neighbor. Among Bobby's neighbors, Carol herself is already a known innocent. The only Bobby-neighbors who are also Carol-neighbors are Freya and Gabe, so Carol is the one innocent neighbor of Bobby who is not Carol's neighbor. That means Bobby's other innocent neighbor must be one of Freya or Gabe, not Amy or Ellie. So Amy and Ellie must be criminals.

Bobby has exactly 2 innocent neighbors in total, and Amy and Ellie are already criminal while Carol is already innocent, so the only way to reach 2 innocents around Bobby is for both Freya and Gabe to be innocent. Ellie says an odd number of innocents in column B neighbor Mark. Among Mark's neighbors, the people in column B are Janet, Nicole, and Terry, so Isaac and Susan are not part of that innocent count. Amy says Bobby has more innocent neighbors than Mark, and Bobby has 2 innocent neighbors, so Mark must have fewer than 2 innocent neighbors. Since Isaac and Susan are outside the column B group mentioned by Ellie's clue, this leaves Isaac and Susan as criminal.

Susan’s clue says there are more innocent builders than innocent sleuths. Among the sleuths, Ellie is already criminal, so Laura is the only sleuth who could possibly be innocent. But all three builders, Janet, Nicole, and Terry, are still available to supply the innocent builders that Susan’s clue requires, so the sleuth side cannot include an innocent here. That makes Laura criminal.

Amy’s clue says Bobby has more innocent neighbors than Mark, and Laura’s clue says Mark has more innocent neighbors than Amy. Right now Bobby’s neighbors already include 1 known innocent while Mark’s have 0, and Mark’s neighbors and Amy’s neighbors are both at 0 known innocents. If Gabe were criminal and Bobby were innocent, then Freya, Janet, Nicole, and Terry would have to make both of those “more innocent neighbors” comparisons true at the same time, but they cannot. That opposite assignment clashes with the two clues together. So Gabe at C2 must be innocent, and Bobby at B1 must be criminal.

Carol’s clue says Bobby has exactly 2 innocent neighbors in total, and exactly 1 of those innocents is also Carol’s neighbor. Bobby’s neighbors already include two known innocents, Carol and Gabe. Among those two, Gabe is Carol’s neighbor and Carol is not her own neighbor, so they already fit the clue’s full innocent count and the split required by the clue. That leaves Bobby’s only unknown neighbor, Freya, unable to be innocent. So Freya must be criminal.

Ellie’s clue applies to the people who are both in column B and neighbors of Mark, and that group is Janet, Nicole, and Terry. It says an odd number of those three are innocent. Freya’s clue applies to the people who are both neighbors of Wally and builders, and that group is Nicole and Terry. It says exactly one of those two is innocent. So among Janet, Nicole, and Terry, the pair Nicole and Terry contributes exactly one innocent, which is already an odd number by itself. For the total for Janet, Nicole, and Terry to still be odd, Janet cannot be innocent. So Janet must be criminal.

Column A already has four known criminals: Amy, Ellie, Isaac, and Susan, with only Mark still unknown there. Janet’s clue says column A has more criminals than any other column, so column A must contain as many criminals as possible here. That means the last person in column A, Mark, has to be a criminal.

Row 5 has to contain an odd number of innocents, and right now row 3 and row 5 both have 0 known innocents. The only unknown person in row 3 is Keith, while the only unknown people these clues can still affect are Keith, Terry, Wally, and Xavi. Since the two rows must end up with the same number of innocents, and row 5 must have an odd number of innocents, row 3 also has to end up with an odd number of innocents. With only Keith still unknown in row 3, Keith must be innocent.

Among the still-unknown edge people who are Olivia's neighbors, there is exactly 1 innocent person: Ronald, Terry, Wally, and Xavi. But in that group, the only people who can still be that innocent person are Terry, Wally, and Xavi. So Ronald cannot be the innocent person in that group, and Ronald must be criminal.

Row 3 has 1 innocent, so row 5 must also have 1 innocent. Right now row 5 has no known innocents, so that one innocent would have to come from Terry, Wally, or Xavi. Also, painters must contain more criminals than pilots. Painters already have 2 known criminals, while pilots have 1, and the only unknown painter is Xavi. If Xavi were innocent, then Terry, Wally, and Hank would have to make both clues true at the same time, but they cannot. So Xavi cannot be innocent. That makes Xavi criminal.

Xavi’s clue says Susan’s neighbors and Hank’s neighbors must contain the same number of criminals. Hank’s neighbors already have 1 known criminal, and the only unresolved person left there is Diane. Freya’s clue fixes the two builder neighbors of Wally, Nicole and Terry, so that exactly 1 of them is a criminal. Those same two people are the unresolved part of Susan’s neighbors, so Susan’s neighbors will add exactly 1 more criminal from Nicole and Terry. So Hank’s neighbors also have to add exactly 1 more criminal, and since Diane is the only unresolved person there, Diane must be criminal.

Janet’s clue says column A has more criminals than any other column. Column A already has all 5 people confirmed as criminals, so no other column is allowed to also reach 5 criminals. Column D already has 4 known criminals, and Hank is its only unresolved person. If Hank were criminal, column D would also have 5 criminals and tie column A, which the clue forbids. So Hank must be innocent.

Wally’s neighboring builders are Nicole and Terry, and Freya’s clue says exactly one innocent is in that group. Diane’s clue says Wally’s neighbors contain an odd number of innocents, and among Wally’s neighbors the only unknown people are Nicole, Olivia, and Terry, with no known innocents there yet. So the innocent count among Nicole, Olivia, and Terry has to be odd, while among Nicole and Terry specifically it has to be exactly one. That makes Olivia criminal.

Bobby’s clue says there is exactly one innocent in the group of neighbors shared by Nicole and Wally. That shared group is only Olivia and Terry. Olivia is already known to be criminal, so the shared group still needs one innocent, and Terry is the only unknown person left in it. So Terry must be innocent.

Ellie’s clue is about the people who are both in column B and neighbors of Mark. That group is Janet, Nicole, and Terry, and the clue says an odd number of them are innocent. Janet is already criminal and Terry is already innocent, so there is already exactly one innocent in that group. To keep the total number of innocents odd, Nicole cannot be innocent. So Nicole must be criminal.

Mark’s clue says row 5 has an odd number of innocents. In row 5, there is already 1 known innocent, and the only unknown person left there is Wally. That means this clue determines Wally’s status, and it makes Wally a criminal.