Puzzle Pack #2 Puzzle 40 Answer

Donna’s clue says the people strictly between Eli and Hope contain exactly 2 innocents. That group currently has no known innocents, and the only unknown people left in it are Frank and Gabe. Since those two spots must supply the full total of 2 innocents, Frank and Gabe must both be innocent.

Frank’s clue says that the two criminals in row 1 must be connected. In row 1, Donna is already innocent, so the clue only still bears on Amy, Betty, and Chuck. That connectedness requirement already pins down Betty’s status among those three. So Betty must be criminal.

Betty’s clue says Chuck is one of Hope’s exactly 4 innocent neighbors. Since Chuck is explicitly included in that innocent group, Chuck must be innocent.

Frank’s clue says the two criminals in row 1 must be connected. In row 1, Betty is already a criminal, Chuck and Donna are already innocent, and Amy is the only unknown person left there. Since Chuck is innocent, Betty cannot be connected to any criminal on the right, so the second criminal has to be Amy next to Betty. So Amy must be criminal.

Among Paula's neighbors, exactly 1 innocent is to the left of Nancy, and those people are Kay and Martin. Also among Paula's neighbors, exactly 1 innocent is in row 3, and that row-3 group is Kay, Logan, and Martin. The only extra person in the row-3 group beyond Kay and Martin is Logan, so Kay and Martin already use up the full allowance of 1 innocent there. So Logan must be criminal.

Among Logan’s neighbors, the unknown people who are not to the left of Nancy are A2 Eli, A4 Olive, B4 Paula, and C4 Rohan, and exactly 2 people in that group must be innocent. In that group, the only people who can still be those 2 innocents are Olive, Paula, and Rohan. That means Eli cannot be one of the innocents there, so Eli must be criminal.

Logan’s clue says there are 5 innocents among Logan’s neighbors, and only 1 of those innocents is to the left of Nancy. Frank and Gabe are already innocents, and neither of them is to the left of Nancy, so the single innocent to the left of Nancy must be Kay or Martin. That means among the neighbors of Logan who are not to the left of Nancy, there is exactly 1 criminal. In that group, the only possible places for that criminal are Olive and Rohan, so Paula cannot be that criminal. That makes Paula innocent.

Paula’s clue fixes her neighbors so that exactly 2 of the innocents among them are not in row 3. Those not-in-row-3 neighbors are Olive, Rohan, Uma, Will, and Xena. But in the current position, the only people in that group who can still be innocent are Olive, Rohan, Uma, and Xena. That means Will cannot be one of the 2 innocents in that group. So Will must be criminal.

Logan’s clue says Uma has more innocent neighbors than Amy. Right now, Uma’s neighbors have 1 known innocent, and Amy’s neighbors also have 1, so Uma cannot be ahead yet. The only still-unknown person this clue can still affect is Olive among Uma’s neighbors, and Amy’s side has no unknown neighbors left. So Olive must be innocent.

Logan’s neighbors contain exactly 5 innocents in total, and exactly 1 of those innocents is to the left of Nancy. The only neighbors of Logan who are to the left of Nancy are Kay and Martin, so the rest of Logan’s neighbors must contain the other 4 innocents. In that remaining group, Frank, Gabe, Olive, and Paula are already 4 known innocents. That means the only unknown left there, Rohan, cannot be innocent, so Rohan must be criminal.

Paula’s clue says that among the people who are both to the left of Zed and neighboring Will, there is exactly 1 innocent, and that shared group is just Uma and Xena. Olive’s clue says row 5 has exactly 2 innocents total. In row 5, the only unknown person outside the Uma-and-Xena group is Zed. So Zed must be the other innocent in row 5, alongside the one innocent among Uma and Xena. That makes Zed innocent.

Hope’s clue says her neighbors contain exactly 4 innocents, and among those neighbors we already have Chuck, Donna, and Gabe as innocent, leaving Martin and Nancy to supply the last innocent there. Thor’s clue says Thor has more criminal neighbors than Betty, but Betty already has 2 criminal neighbors while Thor currently has only 1 known criminal neighbor. If Xena were innocent, then with Martin and Nancy tied up satisfying Hope’s neighbor clue, the people involved in these clues cannot also make Thor’s neighbor count rise above Betty’s. So Xena at C5 must be criminal.

Paula's clue says there is exactly 1 innocent in the group that is both to the left of Zed and neighboring Will. That shared group is just Uma and Xena, and Xena is already criminal. So the group still needs 1 innocent, and the only person left who can fill it is Uma. Therefore, Uma must be innocent.

Hope’s neighbors must contain exactly 4 innocents, and we already have Chuck, Donna, and Gabe as innocent there. So among the two remaining unknown Hope-neighbors, Martin and Nancy, at least one must be innocent. Rohan’s neighbors must contain an odd number of criminals. They already include 3 known criminals, so the three unknown neighbors there, Martin, Nancy, and Thor, must contribute an even number of additional criminals. Since Martin and Nancy cannot both be criminals, Thor has to be criminal to make that work. So Thor must be criminal.

Gabe’s neighbors must contain an odd number of innocents, and there are already 3 known innocents there. So the unknown neighbors of Gabe must add an even number of further innocents. Betty’s clue says Hope has exactly 4 innocent neighbors, and since 3 of Hope’s neighbors are already known innocent, Martin and Nancy together must contain exactly 1 innocent. That leaves Hope as the one who must add the extra innocent needed to keep Gabe’s neighbor total odd. So Hope must be innocent.

Gabe’s clue is the one that matters here: Paula’s neighbors contain exactly 3 innocents, and exactly 1 of those innocents is in row 3. Among Paula’s neighbors, Olive and Uma are already known innocents, and neither of them is in row 3, so the two innocents outside row 3 are already accounted for. That means the one remaining innocent neighbor of Paula must be in row 3, so it has to be Kay or Martin. Since Paula’s neighbors already have their required third innocent in row 3, the only other person in row 3, Nancy, cannot be that innocent neighbor. So Nancy must be innocent.

Betty's clue says Chuck is one of Hope's exactly 4 innocent neighbors. Hope's neighbors are Chuck, Donna, Gabe, Martin, and Nancy, and among them Chuck, Donna, Gabe, and Nancy are already the 4 known innocents. That leaves Martin as the only remaining neighbor there, and he cannot also be innocent. So Martin must be criminal.

Paula’s neighbors contain exactly 3 innocents, and exactly 1 of those innocents is in row 3. Among Paula’s neighbors who are in row 3, Logan and Martin are already criminals, so that row-3 part still needs 1 innocent and the only unknown person there is Kay. So Kay must be innocent.