Puzzle Pack #2 Puzzle 50 Answer

Lucy’s clue says there are no innocents among the people who are both to the right of Alex and neighbors of Isaac. That shared group is exactly Daniel and Eric, and there are already 0 known innocents in it. So Daniel and Eric cannot be innocent. That makes Daniel and Eric criminals.

Eric’s clue says there is exactly one innocent to the left of Mark. That group already contains one known innocent, Lucy. The only unknown people left in that group are Janet and Kay, so neither of them can be innocent. So Janet and Kay must be criminals.

Rose’s neighbors must contain an odd number of innocents. Right now those neighbors already contain 1 known innocent, Lucy, so the unknown neighbors must contribute an even number of additional innocents. Kay’s clue says there are exactly 4 innocents on the edge, and exactly 2 of those edge innocents are neighbors of Rose. So among Rose’s unknown neighbors on the edge, the number of innocents is fixed by that edge condition. That leaves Susan, who is the only unknown neighbor of Rose here who is not on the edge, unable to be one of the innocents needed to change the parity of Rose’s neighbor total. So Susan must be criminal.

Kay’s clue fixes the edge people who are not neighboring Rose: among the still-unknown members of that group, exactly 2 are innocent. In that group, the only people who can still be innocent are Alex, Chad, Freya, Mark, Terry, and Ziad. Isaac is in that same edge-and-not-neighboring-Rose group, but he is not one of the people who can fill those 2 innocent spots. So Isaac at D2 must be criminal.

Janet says Xia and Chad have the same number of innocent neighbors, and Susan says Eric and Freya have the same number of criminal neighbors. Right now both sides of each comparison are tied on the known counts, so those clues leave the remaining unknowns around Xia, Chad, Eric, and Freya having to keep both balances true. At the same time, Kay fixes the edge total at exactly 4 innocents, with exactly 2 of those edge innocents neighboring Rose. If Hank were innocent, the remaining open people touched by these three clues would have to make all of those requirements hold together at once, but they cannot. That makes Hank at C2 criminal.

Hank’s and Susan’s clues fix two important counts. Freya’s neighbors must contain the same number of criminals as Eric’s neighbors, and Eric already has 3 criminal neighbors, so among Alex, Chad, and Gus there is exactly 1 criminal and therefore exactly 2 innocents. Kay’s clue also fixes the 4 edge innocents into two groups of 2: exactly 2 among Nicole, Uma, Vera, and Xia, and exactly 2 among Alex, Chad, Freya, Mark, Terry, and Ziad. Now use Hank’s clue that exactly one row has exactly 2 innocents. If Rose were innocent, then row 4 would be counting Rose together with Nicole and Terry, and with the required splits above, the innocents among Alex, Chad, and Gus and among the edge groups cannot be arranged so that there is still exactly one row with exactly 2 innocents; they always produce either a second such row or none at all. So Rose cannot be innocent. That makes Rose criminal.

Kay’s clue says there are exactly 4 innocents on the edge, and exactly 2 of those edge innocents are not neighbors of Rose. In the still-unknown part of that non-neighbor group, the 2 innocent spots are already limited to Alex, Chad, Mark, Terry, and Ziad. Freya is not one of the people who can still fill those innocent spots, so she cannot be innocent there. So Freya must be criminal.

Among the edge people who are not neighbors of Rose, there are exactly 2 innocents, and in the still-unknown part of that group those 2 innocents must come from Alex, Chad, Mark, Terry, and Ziad. But the only people in that group who can still be innocent are Alex, Chad, Terry, and Ziad. So Mark cannot be one of the 2 innocents there. That makes Mark criminal.

Isaac's neighbors must have the same number of criminals as Xia's neighbors. Isaac's neighbors already have 4 criminals, while Xia's neighbors have 2 known criminals, so among Terry, Vera, and Ziad there must be exactly 2 more criminals, which means exactly 1 of those three is innocent. Janet's clue says Xia's neighbors and Chad's neighbors have the same number of innocents, so Chad's neighbors also need exactly 1 innocent among Alex and Gus. Susan's clue says Eric's neighbors and Freya's neighbors have the same number of criminals. Eric's neighbors already have 3 criminals, while Freya's neighbors have 2 known criminals, so among Alex, Chad, and Gus there must be exactly 1 more criminal, meaning those three contain exactly 2 innocents. Since Alex and Gus together contribute exactly 1 innocent, Chad must be the other innocent. So Chad must be innocent.

Kay’s clue says exactly 2 of the 4 innocent edge people are Rose’s neighbors. In the edge cells that are Rose’s neighbors, the possible innocents are A4 Nicole, A5 Uma, B5 Vera, and C5 Xia, so exactly 2 innocents must come from that set. From this step’s facts, those 2 innocent spots in that group are accounted for by Nicole, Uma, and Vera, which means Xia is not one of the 2 innocents there. So Xia must be criminal.

Kay’s clue says there are exactly 4 innocents on the edge, and exactly 2 of those edge innocents are Rose’s neighbors. Among the edge people who are Rose’s neighbors, none is already known innocent, and the only unknown ones there are Nicole, Uma, and Vera. So the needed innocent edge-neighbors of Rose must come from that small group, which makes Nicole innocent. So Nicole must be innocent.

Among the still-unknown edge cells that are not neighboring Rose, there is exactly 1 innocent person, and that group is Alex, Terry, and Ziad. From the current board and these clues, the only people in that group who can still be that innocent person are Alex and Ziad. So Terry cannot be the innocent one there. That makes Terry criminal.

Kay says exactly 2 of the 4 innocents on the edges are Rose's neighbors. Among edge cells that neighbor Rose, Nicole is already one innocent, so there is room for exactly 1 more innocent there, and the only unknown people in that group are Uma and Vera. The remaining innocent in that group has to be Uma, so Vera cannot be innocent. That makes Vera criminal.

Eric’s neighbors already have 3 criminals, so Freya’s neighbors must also total 3 criminals. Freya’s neighbors currently have 2 known criminals, so among Alex and Gus, exactly one must be criminal. Chad’s neighbors must have more criminals than Vera’s neighbors. Right now both Chad and Vera have 3 known criminal neighbors, and the only unknown in Vera’s neighbors is Uma. If Uma were criminal, Vera’s total would become 4, but Chad can only gain criminals from Alex and Gus, and the first clue allows only one of Alex and Gus to be criminal, so Chad could reach only 4, not more than Vera. So Uma must be innocent.

Isaac's neighbors and Xia's neighbors must contain the same number of criminals. Isaac's neighbors already have 4 criminals, and Xia's neighbors also already have 4 criminals. The only unknown person still involved here is Ziad among Xia's neighbors, so he cannot also be a criminal without making Xia's total higher than Isaac's. So Ziad must be innocent.

Kay’s clue says there are exactly 4 innocents on the edge, and exactly 2 of those edge innocents are neighbors of Rose. Among the edge people who are neighbors of Rose, Nicole and Uma are already the 2 known innocents. That means the edge people who are not neighbors of Rose must contain the other 2 edge innocents, and that group already has Chad and Ziad as 2 known innocents. Alex is the only unknown person left in that non-neighbor edge group, so Alex cannot be innocent. So Alex must be criminal.

Susan’s clue says Eric’s neighbors and Freya’s neighbors must contain the same number of criminals. Eric’s neighbors already have 3 criminals, and Freya’s neighbors also already have 3 criminals. The only unknown person still involved in Freya’s neighbor group is Gus, so he cannot be a criminal without making Freya’s total higher than Eric’s. So Gus must be innocent.