Puzzle Pack #2 Puzzle 49 Answer

Sarah’s clue says Andre is the only person whose neighboring group has no innocents at all. Andre’s neighbors are Bruce, Erwin, and Freya, so that neighboring group must contain exactly 0 innocents. At the same time, every other neighboring group listed in the clue comparison is not allowed to end with 0 innocents. Since Bruce, Erwin, and Freya are all inside Andre’s neighboring group, none of them can be innocent without giving Andre an innocent neighbor. So Bruce, Erwin, and Freya must all be criminals.

Erwin's clue says Will's neighbors contain exactly 3 innocents, and exactly 2 of those innocents are in column C. Among Will's neighbors in column C, Sarah is already known to be innocent, so that column-C part still needs 1 more innocent. The only unknown person left there is Xia. So Xia must be innocent.

Will has exactly 3 innocent neighbors, and 2 of those are already known: Sarah and Xia. So among Will's still-unknown neighbors, A4 Paul, B4 Ruth, and A5 Vicky, exactly 1 is innocent. From the clues already applied here, the only people in that group who can still be that one innocent are Paul and Ruth. That rules Vicky out as the innocent one. So Vicky must be criminal.

Sarah’s neighbors must contain exactly 3 innocents, and exactly 1 of those innocents is in row 4. Since Xia is already an innocent neighbor of Sarah and is not in row 4, the still-unknown neighbors of Sarah who are not in row 4 must contain exactly 1 innocent person: Laura, Nicole, Oscar, Will, and Zach. In that group, the only people who can still be that one innocent are Laura, Will, and Zach. So Nicole and Oscar cannot be innocent, which makes Nicole and Oscar criminals.

Oscar’s clue says Isaac has an odd number of criminal neighbors. Isaac’s neighbors already include exactly 2 known criminals, so among the three unknown neighbors C1 Chuck, D1 Donna, and C2 Hank, exactly one must be a criminal. Sarah’s clue also says Andre is the only person with no innocent neighbors, so Donna’s neighbor group cannot end up with 0 innocents. Donna’s neighbors are C1 Chuck, C2 Hank, and Isaac, so at least one of those three must be innocent. Putting those together, Donna cannot be the one criminal among Isaac’s unknown neighbors, because then Chuck, Hank, and Isaac would all have to be non-criminal to keep Isaac’s total odd, and Donna’s neighbor group would have no innocent at all. So Donna must be innocent, and the one criminal needed among Isaac’s unknown neighbors must be Isaac himself. So Donna must be innocent, and Isaac must be a criminal.

Will’s neighbors must contain exactly 3 innocents, and the two column C neighbors there, Sarah and Xia, are already innocent. So among the other neighbors of Will, which are Paul, Ruth, and Vicky, exactly 1 is innocent; since Vicky is criminal, that means exactly 1 of Paul and Ruth is innocent. Vicky’s neighbors are Paul, Ruth, and Will, and that group must contain an odd number of innocents. Paul and Ruth already contribute exactly 1 innocent, which is odd, so Will cannot also be innocent without making that total even. So Will must be criminal.

Sarah’s clue says Andre is the only person whose neighbor group has no innocents at all, so every other neighbor group must include at least one innocent. Xia’s clue says column D has exactly 2 innocents, and with Donna already innocent, that fixes a very tight limit on D4 Uma and D5 Zach. Vicky’s clue also requires Isaac’s neighbor group to end up with more criminals than Xia’s. If Ruth were criminal, then the remaining open people Andre, Chuck, Hank, Janet, Laura, Paul, Uma, and Zach would have to satisfy all three of those requirements at the same time, and they cannot. So Ruth cannot be criminal. That makes Ruth innocent.

Erwin's clue says Will's neighbors contain exactly 3 innocents in total, and exactly 2 of those innocents are in column C. Those 2 are already Sarah and Xia, so the only innocent among Will's neighbors who is not in column C must be Ruth. That means the remaining unknown neighbor outside column C, Paul, cannot be innocent. So Paul must be criminal.

Bruce’s clue says Sarah’s neighbors contain exactly 3 innocents, and exactly 1 of those innocents is in row 4. Among Sarah’s neighbors who are in row 4, the only people are Ruth and Uma. Ruth is already known to be innocent, so the one row-4 innocent in that group is already accounted for. That means Uma cannot be innocent. So Uma must be criminal.

Xia's clue says there are exactly 2 innocents in column D. In column D, there is already 1 known innocent, and the only unknown person left there is Zach. So Zach must be innocent.

Bruce’s clue says Sarah’s neighbors contain exactly 3 innocents in total, and exactly 1 of those innocents is in row 4. Among Sarah’s row-4 neighbors, Ruth is the only innocent there. That means the neighbors not in row 4 must contain the other 2 innocents, and that group already has Xia and Zach as known innocents. So Laura cannot be innocent, which makes Laura criminal.

Isaac’s clue says his neighbors contain an odd number of criminals, and among Isaac’s neighbors there are already 2 known criminals, with only Chuck and Hank still unknown. Bruce’s clue also says his neighbors contain an odd number of criminals, and among Bruce’s neighbors there are already 2 known criminals, with Andre, Chuck, and Hank still unknown. The only overlap between those two unresolved groups is Chuck and Hank. If Andre were criminal, then the two clue requirements could not both remain true at the same time. So Andre must be innocent.

Zach’s neighbors have 2 innocents, while Freya’s neighbors currently have 1 innocent and the only unknowns there are Chuck, Hank, and Janet. So Laura’s clue says Freya’s neighbors must gain exactly one more innocent. But Sarah’s clue says Andre is the only person whose neighbors have 0 innocents. Donna’s neighbors currently also have 0 known innocents, and the only unknown people there are Chuck and Hank, so at least one of Chuck or Hank must be innocent. If Janet were innocent too, then Freya’s neighbors would need an innocent among Chuck or Hank to match Zach’s 2, and that would also satisfy Donna’s need. That makes Janet impossible as the extra innocent here. So Janet must be criminal.

Janet’s clue says that the two innocents below Chuck must be connected. The people below Chuck are Hank, Nicole, Sarah, and Xia, and among them Nicole is a criminal while Sarah and Xia are already the two innocents. Since Sarah and Xia are the two innocents required by the clue, the only still-unknown person in that group is Hank, and this clue fixes him as not one of those innocents. So Hank must be criminal.

Sarah’s clue says Andre is the only person whose neighbors contain no innocents at all. Andre’s neighbors already have exactly 0 innocents, so every other person’s neighbor set must include at least one innocent. Among all the comparable neighbor sets in this step, the only still-unknown person who can affect that is Chuck, and Donna’s neighbors are the one other set that currently has 0 known innocents and therefore need that unknown to be innocent. So Chuck must be innocent.