Puzzle Pack #2 Puzzle 46 Answer

Nicole’s clue says there are exactly 3 innocents in the corners, and exactly 1 of those corner innocents is in row 1. That means the other 2 corner innocents must be corners that are not in row 1. The only corner cells not in row 1 are A5 Uma and D5 Zoe, and there are not already any known innocents there. So Uma and Zoe must be innocent.

Uma’s clue says the number of innocents among Gus’s neighbors who are in row 3 is odd. In that group, the only people are Jane, Kumar, and Laura. Zoe’s clue says Henry has exactly 3 innocent neighbors, and exactly 1 of those is Ruby’s neighbor. Among Henry’s neighbors, the ones who are also Ruby’s neighbors are only Kumar and Laura, so exactly 1 of Kumar and Laura is innocent. That means among Jane, Kumar, and Laura, the pair Kumar and Laura contributes exactly 1 innocent. To make the total for Jane, Kumar, and Laura odd, Jane cannot be innocent. So Jane must be criminal.

Nicole's clue says the corners contain exactly 3 innocents, and only 1 of those corner innocents is in row 1. Uma and Zoe are already known innocents in the corners, and both are not in row 1, so the one row-1 corner innocent has to be Anna or Denis, which also means the other row-1 corner person is criminal. Jane's clue says Bonnie has exactly 4 innocent neighbors. Among Bonnie's neighbors, that means 4 of Anna, Chris, Ethan, Floyd, and Gus are innocent. Since one of the two row-1 corners Anna and Denis is criminal, the row-1 neighbor of Bonnie that can still help make up those 4 innocents is Anna, so Anna is the row-1 corner innocent and Denis is the row-1 corner criminal. Then Bonnie still needs three more innocent neighbors besides Anna, and the targets among Bonnie's remaining neighbors are Chris and Gus, so they are innocent. So Gus is innocent, Anna is innocent, Denis is criminal, and Chris is innocent.

Anna’s clue says Nicole’s neighbors include exactly 5 innocents, and exactly 1 of those innocents is in column B. Among Nicole’s neighbors who are in column B, Jane is already known to be criminal, so the one innocent in that column-B neighbor group still has to be found. The only unknown person left in that group is Vince. So Vince must be innocent.

Nicole’s unknown neighbors are Isaac, Kumar, Megan, Ruby, and Wanda, and exactly 2 of those 5 are criminals. The only people in that group who can still be those 2 criminals are Isaac, Kumar, Megan, and Ruby. That leaves Wanda outside the criminal spots, so Wanda must be innocent.

Wanda’s clue says that Henry is one of the exactly 10 innocents on the edges. Since Henry is explicitly included in that innocent edge group, Henry must be innocent.

Jane’s clue says Bonnie has exactly 4 innocent neighbors. Bonnie’s listed neighbors are Anna, Chris, Ethan, Floyd, and Gus, and among them Anna, Chris, and Gus are already known to be innocent. That means exactly one of Ethan and Floyd must also be innocent, so Bonnie cannot be innocent as well under this clue’s requirement. So Bonnie must be criminal.

Chris’s clue says Jane has exactly 5 innocent neighbors, and exactly 2 of those innocents are on edge cells. Jane already has 2 known innocent neighbors, Gus and Nicole, and neither of them is on an edge, so the 3 other innocent neighbors must all come from Jane’s 6 unknown neighbors, with exactly 2 of those 3 on edge cells. Among Jane’s unknown neighbors, the edge-cell ones are Ethan, Isaac, and Megan, so two of those three must be innocent. That makes Isaac innocent.

Anna’s clue says Nicole’s neighbors contain exactly 5 innocents, and exactly 1 of those innocents is in column B. Among Nicole’s neighbors, the only person in column B who is already innocent is B5 Vince, so that one innocent in column B is already accounted for. Nicole already has 4 known innocent neighbors total: Isaac, Uma, Vince, and Wanda. That means the remaining unknown neighbors of Nicole cannot be innocent, so Megan must be criminal.

Chris's clue says that among Jane's neighbors, exactly 2 innocents are on the edge. In that edge-neighbor group, Isaac is already known to be innocent, Megan is known to be criminal, and the only unknown person there is Ethan. So that edge-neighbor group still needs 1 more innocent, and Ethan must be that person. That makes Ethan innocent.

Jane’s clue says Bonnie’s neighbors contain exactly 4 innocents. Bonnie’s neighbors already include 4 known innocents, and the only unknown neighbor left there is Floyd. Since all 4 innocent spots are already accounted for, Floyd cannot be innocent. So Floyd must be criminal.

Nicole has exactly 5 innocent neighbors, and 4 of those are already known, so among her two unknown neighbors, Kumar and Ruby, exactly 1 is innocent. Laura has exactly 4 innocent neighbors; Gus and Henry already account for 2 of them, and Kumar and Ruby can contribute only 1 more innocent between them. That leaves Sam to provide the last needed innocent neighbor for Laura. So Sam must be innocent.

Row 1 already has two criminals, row 2 already has one, row 4 already has one, and row 5 has none. Denis’s clue says only one row has exactly one criminal, so rows 2 and 4 cannot both stay at one criminal. Since the only unknown in row 4 is Ruby, row 4 needs a second criminal instead of staying at exactly one. So Ruby must be criminal.

Anna’s clue says Nicole’s neighbors contain exactly 5 innocents, and exactly 1 of those innocents is in column B. That means the other 4 innocent neighbors of Nicole must be outside column B. Among Nicole’s neighbors who are not in column B, Isaac, Uma, and Wanda are already known innocents, while Kumar is the only unknown person left there. So Kumar must be innocent.

Uma's clue is about the people who are both neighboring Gus and in row 3. That group is Jane, Kumar, and Laura, and the clue says an odd number of them are innocents. In that group, Jane is already known to be criminal and Kumar is already known to be innocent, and Laura is the only unknown person left in the group. So Laura must be criminal.