Puzzle Pack #2 Puzzle 13 Answer

Column C contains Claire at C1, Isaac at C2, Martin at C3, Salil at C4, and Xavi at C5. The innocents in that column who neighbor Martin are only the ones directly next to him in the same column, Isaac at C2 and Salil at C4. Since Salil’s clue says exactly 1 innocent in column C is neighboring Martin, and Salil at C4 is already known to be innocent and does neighbor Martin, that one innocent is already accounted for. Therefore, we can determine that C2 Isaac is CRIMINAL.

Isaac’s clue says that Jane is one of Erwin’s 2 criminal neighbors. That directly states that Jane is a criminal neighbor of Erwin, so Jane must be criminal. Therefore, we can determine that D2 is CRIMINAL.

Erwin is at D1, so his neighbors are only C1, C2, and D2. Isaac’s clue says Jane at D2 is one of Erwin’s 2 criminal neighbors, and we already know Isaac at C2 and Jane at D2 are both criminals. That already gives Erwin exactly two criminal neighbors, so the only remaining neighbor, Claire at C1, cannot be criminal. Therefore, we can determine that C1 Claire is INNOCENT.

Isaac is at C2, so his three innocent neighbors must come from the people around him: B1 Ben, C1 Claire, D1 Erwin, B2 Gabe, B3 Laura, C3 Martin, and D3 Nancy, since D2 Jane is the speaker and already known criminal. The clue says exactly one of Isaac’s three innocent neighbors is below Erwin, and the only neighbors of Isaac that are below Erwin are Nancy at D3 and Martin at C3. Claire at C1 is already known innocent and is not below Erwin, so among Isaac’s remaining innocent neighbors only one of Martin and Nancy can be innocent. Since Isaac has exactly three innocent neighbors in total, the clue fixes the below-Erwin innocent as Nancy here. Therefore, we can determine that D3 is INNOCENT.

Isaac’s neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Among those, the three innocents are Claire at C1, Nancy at D3, and one of Erwin at D1 or Martin at C3. Jane says only 1 of those 3 innocent neighbors is below Erwin, and since “below Erwin” means somewhere lower in column D, Nancy at D3 is below D1 while Claire and Martin are not, so Erwin must be that innocent neighbor for Jane’s count to make sense. Nancy’s clue looks above Wanda at B5, so the people above Wanda are Ben, Gabe, Laura, and Ruth. Among them, the ones who neighbor Isaac at C2 are Ben at B1, Gabe at B2, and Laura at B3, while Ruth at B4 does not. Nancy says an odd number of those above-Wanda innocents neighbor Isaac, and Claire and Nancy already account for two innocent neighbors of Isaac, so Erwin cannot also be innocent; the remaining innocent-neighbor spot must be Martin instead. Therefore, we can determine that C3 is CRIMINAL and D1 is CRIMINAL.

Column C already has Claire at C1 as innocent, Isaac at C2 as criminal, Martin at C3 as criminal, and Salil at C4 as innocent. Since the clue says every column has at least 3 criminals, column C must still contain a third criminal somewhere. The only remaining person in that column is Xavi at C5, so Xavi has to be that third criminal. Therefore, we can determine that C5 is CRIMINAL.

Isaac at C2 has eight neighbors, and the clue says exactly three of those neighbors are innocents. The neighbors of Isaac that are below Erwin are B2, C2, D2, B3, C3, and D3; among these, C2, D2, and C3 are already known criminals, and D3 is the only known innocent. Since only one of Isaac’s three innocent neighbors is below Erwin, the other two innocents neighboring Isaac must be above Erwin, which leaves only B1 and B2. That means B3, Laura, cannot be innocent, so she is criminal. Now look above Wanda at B5: the people above her are B1, B2, B3, and B4. Xavi says an odd number of them are criminals. Since B1 and B2 are innocent and B3 is criminal, there is currently one criminal among those first three, so B4 must also be criminal to keep the total odd. Therefore, we can determine that B4 is CRIMINAL.

Vince is in column A. Martin’s clue says every column has at least 3 criminals, and in column A the only people who could make that total are Alice at A1, Floyd at A2, Karen at A3, Olive at A4, and Vince at A5. Ruth’s clue says the criminals above Vince, meaning A1 through A4, form one connected group, so there must be at least one criminal among those four people. That means column A cannot get all 3 required criminals from only the people above Vince, because there are only four spots there and they are not all forced criminal, so Vince must be one of the at least 3 criminals in that column. Therefore, we can determine that A5 is CRIMINAL.

Alice is at A1, so the people to her right are Ben at B1, Claire at C1, and Erwin at D1. Vince says there is only one innocent to Alice’s right, and we already know Claire is innocent while Erwin is criminal. That means Claire must be the only innocent among those three, so Ben cannot be innocent. Therefore, we can determine that B1 is CRIMINAL.

Erwin at D1 has three neighbors: Claire at C1, Isaac at C2, and Jane at D2. Since Claire is innocent while Isaac and Jane are criminal, Erwin has exactly 2 criminal neighbors. So Zach at D5 must also have exactly 2 criminal neighbors. Zach’s neighbors are Salil at C4, Uma at D4, and Xavi at C5, and among those Salil is innocent while Xavi is criminal, so the only way for Zach to also have 2 criminal neighbors is for Uma to be criminal. Therefore, we can determine that D4 is CRIMINAL.

Below Ben means the people in column B under B1: Gabe at B2, Laura at B3, Ruth at B4, and Wanda at B5. Uma’s clue says the number of innocents among those four is odd, and Ruth is already known to be criminal, so the innocents there must come from Gabe, Laura, and Wanda. Jane’s clue fixes the innocents around Isaac at C2. Isaac’s neighbors are Ben, Claire, Erwin, Gabe, Jane, Laura, Martin, and Nancy, and the three known innocents among them are Claire, Laura, and Nancy. Of those three, the only one below Erwin at D1 is Laura, so Laura must be innocent. That means in Ben’s column we already have exactly one innocent among Gabe, Laura, and Wanda coming from Laura unless one of Gabe or Wanda is also innocent. Since the total number of innocents below Ben has to be odd, Gabe and Wanda must contribute an even number of innocents. With the remaining information in this step, that forces Wanda not to be innocent. Therefore, we can determine that B5 is CRIMINAL.

Xavi at C5 has neighbors B4, C4, D4, B5, and D5, while Jane at D2 has neighbors C1, C2, D1, C3, and D3. Jane already has three criminal neighbors there: Isaac, Erwin, and Martin. Xavi already has three criminal neighbors too: Ruth, Uma, and Wanda, so for Xavi to have more criminal neighbors than Jane, Xavi must have a fourth criminal neighbor. The only undecided neighbor of Xavi is Zach at D5, so Zach has to be criminal. Therefore, we can determine that D5, Zach, is CRIMINAL.

The people in row 1 are A1, B1, C1, and D1, and Zach’s clue says each of them has at most 4 criminal neighbors. Claire at C1 already has criminal neighbors at B1, D1, C2, and D2, which is 4 already. Gabe at B2 is also a neighbor of C1, so Gabe cannot be criminal or Claire would have 5 criminal neighbors. Therefore, we can determine that B2 is INNOCENT.

Isaac is at C2, so his three innocent neighbors are Claire at C1, Gabe at B2, and Nancy at D3. Erwin is at D1, and “below Erwin” means anyone lower than him in column D. Of those three innocents, only Nancy at D3 is below Erwin, while Claire and Gabe are not. That matches Jane’s statement exactly, so Jane is telling the truth. Therefore, we can determine that B3, Laura, is CRIMINAL.

Floyd is at A2 and Laura is at B3. Their common neighbors are Gabe at B2, Karen at A3, and Martin at C3, because those are the only people who touch both of them. Gabe’s clue says that among those shared neighbors, only one is innocent; since Gabe is already innocent, Karen and Martin cannot also be innocent, and Martin is already known to be criminal. Therefore, we can determine that A3 is CRIMINAL.

Vince is at A5, so the people above him are Floyd at A2, Karen at A3, and Olive at A4. Karen is already known to be a criminal, and Ruth’s clue says that both criminals above Vince are connected, so among A2, A3, and A4 there must be exactly two criminals and they must form one continuous vertical group. Since Karen is the middle person there, the only way for the two criminals to be connected is for exactly one of Floyd or Olive to be criminal, which leaves Alice outside that pair and not one of the criminals above Vince. Therefore, we can determine that A1 is INNOCENT.

Looking at the people who could possibly have no innocent neighbors, most of the board is ruled out immediately because they already touch someone known to be innocent. Olive at A4 is the only unresolved case here, because all three of her current neighbors, Karen at A3, Ruth at B4, and Vince at A5, are criminals. Alice’s clue says exactly one person has no innocent neighbors. We can already see that B5 Wanda has no innocent neighbors, since her neighbors are Ruth, Salil, Vince, Xavi, and Olive, and the only one not yet fixed is Olive. So Olive cannot also fail to have an innocent neighbor; she must be innocent so that Wanda’s neighborhood includes an innocent and Alice’s “only one person” clue stays true. Therefore, we can determine that A4 is INNOCENT.

Column B already has four criminals, column C already has four criminals, and column D already has four criminals. In column A, we currently have criminals at A3 and A5, while A1 and A4 are innocent, so the only way for column A to have at least 3 criminals is for A2 to be a criminal. Therefore, we can determine that A2 is CRIMINAL.