Puzzle Pack #1 Puzzle 25 Answer

Zoe is at D5, so the people above Zoe are D1, D2, D3, and D4. Hilda says both criminals above Zoe are below Donna, which means the criminals in that column must all be somewhere under Donna. Since Donna herself is one of the people above Zoe, she cannot be one of those criminals below Donna. Therefore, we can determine that D1 is INNOCENT.

Donna is at D1, so nobody can be below Donna while also being above Zoe, because everyone above Zoe is in rows 1 through 4 and rows below Donna are rows 2 through 5. Hilda says both criminals above Zoe are below Donna, so the only way that can be true is if the two criminals above Zoe are exactly the two people in Zoe's column below Donna and above Zoe: D2 and D3. Therefore, we can determine that D3 is CRIMINAL.

Zoe is at D5, so the people above Zoe are Donna at D1, Isaac at D2, Nick at D3, and Steve at D4. Hilda says both criminals above Zoe are below Donna, so among those four people there are exactly two criminals, and they must be the ones below Donna. Since Donna is already innocent and Nick is already criminal, the only way for there to be exactly two criminals above Zoe is for Isaac and Steve to be innocent and for D3 and D4 to be the two criminals. Now look at Nick’s clue about Mark at C3. The common neighbors of C3 and D3 are Hilda at C2, Isaac at D2, Rose at C4, and Steve at D4, and Nick says exactly 3 of those 4 are innocent. Hilda is already innocent, and from Hilda’s clue Isaac is innocent while Steve is criminal, so Rose must be the third innocent common neighbor. Therefore, we can determine that C4 is INNOCENT.

In column D, Donna at D1 and Rose’s clue tell us there are more innocents than criminals among D1, D2, D3, D4, and D5. Since Donna is already innocent and Nick at D3 is already criminal, that balance means Zoe at D5 cannot also be criminal, because then the column would not have more innocents than criminals unless both D2 and D4 were innocent. Hilda’s clue says the two criminals above Zoe are below Donna. Above Zoe in column D are Donna, Isaac, Nick, and Steve, and the only ones below Donna are Isaac, Nick, and Steve. So the two criminals above Zoe must be exactly two of those three people, which already includes Nick and one of Isaac or Steve. That leaves no room for Zoe herself to be criminal while keeping Rose’s clue true for the whole column. Therefore, we can determine that D5 is INNOCENT.

The only doctors are Donna at D1, Lucy at B3, and Nick at D3. Zoe’s clue says exactly 2 doctors have an innocent directly below them: Donna does, because Isaac at D2 cannot be a doctor and must be the person below her, and Nick does, because Zoe at D5 is innocent below him, so Lucy cannot also have an innocent directly below her. The person directly below Lucy is Peter at B4, so Peter cannot be innocent. Hilda’s clue says both criminals above Zoe are below Donna. Above Zoe in column D are Donna, Isaac, Nick, and Steve, and the only one below Donna who is already known to be a criminal is Nick. Since there must be two criminals above Zoe, exactly one of Isaac or Steve is also criminal, which means Isaac is not forced to be innocent. That leaves Donna and Nick as the two doctors counted by Zoe’s clue, so Lucy still cannot have an innocent directly below her. Therefore, we can determine that B4 is INNOCENT.

Zoe is at D5, so the people above Zoe are Donna at D1, Isaac at D2, Nick at D3, and Steve at D4. Hilda says both criminals above Zoe are below Donna, so among those four people there are exactly two criminals, and they must be below Donna. Since Donna herself is innocent, the two criminals above Zoe must be chosen from Isaac, Nick, and Steve. Nick is already known to be a criminal, so exactly one of Isaac and Steve is criminal, which means exactly one of them is innocent. Nick’s neighbors are Hilda at C2, Isaac at D2, Mark at C3, Rose at C4, and Steve at D4. Peter says Nick has an odd number of criminal neighbors. Hilda and Rose are innocent, and among Isaac and Steve exactly one is criminal, so those four neighbors contribute exactly one criminal unless Mark is also criminal. That means Mark cannot be criminal, because then Nick would have two criminal neighbors there instead of an odd number. Therefore, we can determine that C3 is INNOCENT.

Mark at C3 has five neighboring innocents, and his neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Among those, C2, B4, and C4 are already known innocents, and D3 is a known criminal. The only neighbors of Mark who are in row 3 are B3 and D3, so for there to be exactly one innocent neighboring Mark in row 3, B3 must be the innocent one because D3 is not. Therefore, we can determine that B3 is INNOCENT.

Mark’s neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. We already know that C2, B4, C4, and D5 are innocent, and the clue says exactly 5 of Mark’s neighbors are innocent, with only 1 of those 5 in row 3. Since B3 is the only neighboring person in row 3 besides D3, and B3 is already innocent, D3 cannot be innocent, so the other four neighboring innocents must be C2, B3, B4, and C4. That means Mark’s remaining neighboring unknowns B2, D2, and D4 cannot be innocent. Therefore, we can determine that B2 is CRIMINAL.

Terry’s neighbors are Ollie, Peter, Wally, and Xia, while Peter’s neighbors are Jason, Lucy, Mark, Ollie, Rose, Terry, Wally, and Xia. So the only neighbors they have in common are Ollie, Wally, and Xia. Lucy’s clue says Peter and Terry have only one innocent neighbor in common, which means exactly one of Ollie, Wally, and Xia is innocent. Now use Flora’s clue. Isaac’s neighbors are Donna, Hilda, Mark, Nick, Rose, and Steve, and among those we already know Donna, Hilda, Mark, and Rose are innocent while Nick is criminal, so Isaac has either 1 or 2 criminal neighbors depending on Steve. Terry’s neighbors are Ollie, Peter, Wally, and Xia, and Peter is already innocent, so Terry’s number of criminal neighbors is just the number of criminals among Ollie, Wally, and Xia. Since exactly one of those three is innocent, exactly two of them are criminal, so Terry has 2 criminal neighbors. Therefore Isaac must also have 2 criminal neighbors, which forces Steve to be criminal. Once Steve is criminal, Isaac’s criminal neighbors are exactly Nick and Steve. If Clyde were criminal, then Isaac would have three criminal neighbors, because Clyde is also adjacent to Isaac. That cannot happen, so Clyde must be innocent. Therefore, we can determine that C1 is INNOCENT.

Donna is at D1 and Zoe is at D5, so the people above Zoe are D1, D2, D3, and D4. Hilda says both criminals above Zoe are below Donna, so the two criminals in that column must be among D2, D3, and D4; since D3 is already criminal and D1 is innocent, exactly one of D2 or D4 is criminal. Now compare Isaac at D2 and Terry at A5. Isaac’s neighbors are C1, C2, C3, D1, D3, and D4, and among those we already know Flora at B2 is not relevant there, while D3 is criminal and D4 may or may not be criminal, so Isaac has either 1 or 2 criminal neighbors. Terry’s neighbors are A4, B4, and B5, and B4 is innocent, so for Terry to have the same number of criminal neighbors as Isaac, Terry cannot have 2 criminal neighbors unless both A4 and B5 were criminal. Clyde says exactly 4 of the 10 edge innocents are neighbors of Flora. Flora’s edge neighbors are A1, B1, C1, A2, C2, A3, B3, and C3, and among these C1, C2, B3, and C3 are already known innocents. That already makes the required 4, so A1, B1, A2, and A3 cannot also be innocent. In particular A3 is criminal, which means Isaac already has D3 as one criminal neighbor and cannot be matched by Terry unless Terry has only 1 as well; with B4 innocent, that forces A4 and B5 not to both be criminal, so Terry is innocent. Then Terry contributes no extra criminal count around the bottom-left corner, and the only way to keep Flora’s edge-neighbor total at exactly 4 while matching the neighborhood counts is for Xia at C5 to be innocent as well. Therefore, we can determine that C5 is INNOCENT and A5 is INNOCENT.

Nick’s neighbors are C2, D2, C3, C4, D4, C5, and D5, and only D2 and D4 are still unknown there, so Nick can have at most 3 criminal neighbors in total. Eve’s neighbors are A1, B1, B2, A3, and B3, with B2 already criminal and the other three unknowns A1, B1, and A3 being the only possible additional criminals around her, so Eve can also have at most 4 criminal neighbors. Terry says Eve has more criminal neighbors than Nick, so Eve must be at least one higher than Nick. That can only happen if Eve reaches 4 criminal neighbors and Nick is at 3 or less, which forces all three unknown neighbors around Eve, A1, B1, and A3, to be criminals. Then Clyde’s clue says exactly 4 edge innocents are neighbors of Flora; Flora’s edge neighbors are A1, B1, C1, A2, A3, B3, and C3, and with C1 and C3 already innocent plus A1, B1, and A3 now criminal and B3 already innocent, the fourth edge innocent neighbor must be A2. Therefore, we can determine that A2 is INNOCENT.

Zoe is at D5, so the people above Zoe are D1, D2, D3, and D4. Donna is at D1, and nobody in that column is below Donna except D2, D3, and D4, so Hilda’s clue says the two criminals above Zoe must be exactly two of those three people. Since D3 is already known to be a criminal, exactly one of D2 and D4 is criminal, which means exactly one of them is innocent. Nick at D3 has neighbors C2, C3, C4, D2, and D4. Among those, C2, C3, and C4 are already innocent, so Nick has 3 innocent neighbors plus exactly one of D2 and D4, giving him exactly 4 innocent neighbors. Ollie at A4 has neighbors A3, B3, B4, A5, and B5; among these, B3, B4, and A5 are innocent, so for Ollie to have more innocent neighbors than Nick, both A3 and B5 must also be innocent, making Ollie’s total 5. Therefore, we can determine that A3 is INNOCENT and B5 is INNOCENT.

Isaac is at D2 and Terry is at A5. Terry’s neighbors are A4, B4, and B5, and since B4 and B5 are innocent, the number of criminal neighbors Terry has is exactly the same as whether A4 is criminal. Isaac’s neighbors are C1, D1, C2, C3, D3, C1, D1, C2, and C3 are innocent while D3 is criminal, so Isaac has exactly 1 criminal neighbor. That means Terry must also have exactly 1 criminal neighbor, so A4 has to be criminal. Therefore, we can determine that A4 is CRIMINAL.

Zoe is at D5, so the people above Zoe are D1, D2, D3, and D4. Donna is at D1, and Hilda says both criminals above Zoe are below Donna, so among those four people the only criminals must be in the spaces below D1, which means D3 and D4 are the two criminals there; since D3 is already criminal, D4 must also be criminal, and D2 must be innocent. Now look at Xia’s clue that columns B and D contain the same number of innocents. In column D, D1, D2, and D5 are innocent while D3 and D4 are criminal, so column D has exactly 3 innocents. In column B, B2 is criminal and B3, B4, and B5 are innocent, so B1 must also be criminal to keep column B at 3 innocents. Therefore, we can determine that B1 is CRIMINAL.

Flora is at B2, so her edge neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Among those, the people already known to be innocent are C1, A2, C2, A3, B3, and C3, which is already 6 edge innocents neighboring Flora if Anna were also innocent. Clyde is innocent, so his clue must be true, and it says exactly 4 of the edge innocents are Flora's neighbors. Since B1 is criminal and the count is already too high without allowing Anna to be innocent, Anna cannot be innocent. Therefore, we can determine that A1 is INNOCENT.

Donna is at D1 and Zoe is at D5, so the people above Zoe are D1, D2, D3, and D4. Hilda says both criminals above Zoe are below Donna, which means there are exactly two criminals in that column and neither can be Donna. Since Nick at D3 is already a criminal and Donna at D1 is innocent, only one of D2 or D4 can be the other criminal. Anna’s clue compares Clyde at C1 with Rose at C4. Clyde’s neighbors are Bonnie, Flora, Hilda, and Isaac, so he already has two criminal neighbors there from Bonnie and Flora, and gets a third if Isaac is criminal. Rose’s neighbors are Lucy, Mark, Nick, Peter, Steve, Wally, Xia, and Zoe, so she has only one criminal neighbor for sure, Nick, and gets a second only if Steve is criminal. For Clyde to have more criminal neighbors than Rose, Isaac must be criminal and Steve must be innocent. Therefore, we can determine that D2 is CRIMINAL and D4 is INNOCENT.