Puzzle Pack #2 Puzzle 33 Answer

Zoe’s clue says that Logan is one of the exactly 2 innocents above Xavi. That directly identifies Logan as innocent. So Logan must be innocent.

Logan’s clue says row 5 is the only row with exactly one innocent. Row 5 already has one known innocent, Zoe. That means the other three people in row 5 cannot be innocent, because that would give row 5 more than one innocent instead of exactly one. So Tina, Uma, and Xavi must be criminals.

Xavi’s clue says exactly 2 innocents above Xavi are neighboring Oscar. In the shared group identified for this step, there is currently 1 known innocent, Logan, and no unknown people left in that shared group. Since the clue requires 2 innocents there, the remaining target from Oscar’s neighbors for this step, Phil, has to supply the second innocent. So Phil must be innocent.

Zoe’s clue says Logan is one of exactly 2 innocents above Xavi. The people above Xavi are Celia, Gus, Logan, and Phil, and Logan and Phil are already the 2 known innocents there. If Celia and Gus were innocent as well, that same group above Xavi would have more than 2 innocents, which clashes with the clue. So Celia and Gus must be criminal.

Phil is already one of Logan's innocent neighbors who also neighbors Uma. The clue says there must be exactly 2 such people, and among Logan's neighbors who also neighbor Uma, the only other possible person is Oscar. So Oscar has to be the second one, which makes Oscar innocent.

Phil's clue fixes Logan's neighbors at exactly 3 innocents in total. Oscar and Phil are already two known innocents there, so among Logan's five still-unknown neighbors, exactly one more is innocent. In that unknown group, the only people who can still be that one innocent are Frank, Hilda, Joyce, and Ruby. Mark is not one of them, so he cannot be the remaining innocent person there. So Mark must be criminal.

Mark’s clue says that the two criminals above Uma are connected. Above Uma are Bobby, Frank, Joyce, and Oscar, and Oscar is already innocent, so the clue can only affect Bobby, Frank, and Joyce. For the two criminals in that group to be connected, Frank has to be one of them. So Frank must be criminal.

Frank’s clue says there is exactly one innocent to the left of Phil. To the left of Phil, there is already one known innocent, Oscar, and the only unknown person left there is Nick. Since that group already has its full one innocent, Nick cannot be innocent. So Nick must be criminal.

In column D, Mark is already a known criminal and Zoe is a known innocent. Oscar’s clue says all criminals in column D have to form one connected block, and Nick’s clue says column D must contain at least 3 criminals in total. With only one known criminal there now, the column needs more criminals, and to keep the criminals in one connected block above Mark while Zoe remains innocent at the bottom, Hilda has to be one of them. So Hilda must be criminal.

Logan’s clue says row 5 is the only row with exactly one innocent, so no other row can finish with exactly one innocent. Row 2 currently has no known innocents, and the only unknown person in that row is Evie. If Evie were innocent, row 2 would end with exactly one innocent, which the clue forbids. So Evie must be criminal.

Row 5 already has exactly one innocent, so Logan’s clue means no other row can finish with exactly one innocent. Row 3 already contains Logan as one known innocent, and the only unknown people left in that row are Isaac and Joyce. So at least one of Isaac or Joyce must also be innocent so row 3 does not end with exactly one innocent. This leaves Isaac as innocent.

Gus’s listed neighbors are Bobby, Celia, Diane, Frank, Hilda, Joyce, Logan, and Mark, and among them there are already 4 known criminals, 1 known innocent, and 3 unknowns. Evie’s clue says Gus has the most criminal neighbors, so Gus’s neighbor count of criminals has to be as high as needed to stay uniquely highest. With Celia, Frank, Hilda, and Mark already criminal, Diane is also required to be criminal among Gus’s neighbors. So Diane must be criminal.

Row 5 is the only row with exactly one innocent, so row 1 cannot end with exactly one innocent. Row 3 and row 4 already each have two known innocents, and Logan's clue says exactly one of his still-unknown neighbors, Joyce or Ruby, is innocent, so exactly one of those two rows will stay at exactly two innocents. Since Isaac says only one row has exactly two innocents, row 1 cannot have exactly two innocents either. Row 1's only unknown people are Amy and Bobby, so its possible innocent totals were 0, 1, or 2; with 1 and 2 ruled out, row 1 must have 0 innocents. So Amy and Bobby must be criminal.

Mark’s clue says the criminals above Uma are both connected, so among the people above Uma there must be exactly two criminals forming one connected group. Above Uma we already have Bobby and Frank as criminals, Oscar as innocent, and Joyce is the only unknown person there. Since Bobby and Frank already make the required connected pair, the clue leaves no room for Joyce to be a criminal. So Joyce must be innocent.

Logan's neighbors contain exactly 3 innocents in total. The clue says exactly 2 of those innocents are also neighbors of Uma, and those two are already Oscar and Phil. That leaves only 1 innocent among Logan's neighbors who are not neighbors of Uma, and that 1 is already Joyce. So Ruby cannot be innocent, which makes Ruby criminal.