Puzzle Pack #2 Puzzle 1 Answer

In column A, the only unknowns are A1, A2, A3, and A5, because A4 is already known to be innocent. Olivia’s clue says there are exactly two criminals in that column, and those two criminals must be connected by orthogonal adjacency, so they must form one uninterrupted vertical group. Since A4 is innocent, no criminal above A4 can be connected to any criminal below A4, which means both criminals have to be on the same side of A4. That forces the pair above A4 to be A2 and A3, so A2 must be criminal, and A5 cannot be criminal. Therefore, we can determine that A2 is CRIMINAL and A5 is INNOCENT.

Uma’s clue says that Linda is one of the 4 innocents in column C. That directly includes Linda among the innocents in that column. Therefore, we can determine that C3 · Linda is INNOCENT.

Uma says Linda is one of 4 innocents in column C, so in column C exactly four people are innocent. We already know Linda at C3 is innocent, so among C1 Chad, C2 Hal, C4 Rohan, and C5 Wanda, exactly three are innocent. Linda says there is an odd number of innocents below Chad. The people below Chad in column C are C2, C3, C4, and C5, and since C3 is innocent and exactly three of C2, C4, and C5 are innocent, there are 4 innocents below Chad in total if Chad were not innocent. That total is even, so the only way Linda's clue can be true is if Chad is also one of the four innocents in column C. Therefore, we can determine that C1 is INNOCENT.

Alice at A1 and Flora at A2 are neighbors, so their common neighbors are exactly B1 and B2. Chad’s clue says Alice and Flora have no innocent neighbors in common, which means neither of those shared neighbors can be innocent. So both B1 and B2 must be criminal. Therefore, we can determine that B1 is CRIMINAL and B2 is CRIMINAL.

Column C contains Chad at C1, Hal at C2, Linda at C3, Rohan at C4, and Wanda at C5. Uma says Linda is one of 4 innocents in column C, and we already know Chad and Linda are innocent, so among Hal, Rohan, and Wanda exactly two must also be innocent. Gus says exactly 2 innocents below Chad are neighboring Megan; the people below Chad in column C are Hal, Linda, Rohan, and Wanda, and among them the ones neighboring Megan at D3 are Hal, Linda, and Rohan, so exactly two of those three are innocent. Since Linda already is innocent, exactly one of Hal or Rohan can be innocent, which means the second unknown innocent in column C must be Wanda. Therefore, we can determine that C5 is INNOCENT.

Megan is at D3, so the people above her are Isaac at D2 and Denis at D1. Wanda’s clue says there are no innocents above Megan, which means everyone above D3 must be criminal. That applies to both Isaac and Denis. Therefore, we can determine that D1 Denis is CRIMINAL and D2 Isaac is CRIMINAL.

The edge people in row 4 are A4, B4, C4, and D4, and among them A4 is already known to be innocent. Isaac’s clue says that exactly one innocent on the edges is in row 4, so A4 must be the only edge innocent in that row. D4 is also on the edge in row 4, so D4 cannot be innocent. Therefore, we can determine that D4 is CRIMINAL.

Uma’s neighbors are Olivia, Peter, Rohan, Vicky, and Wanda. Since Olivia, Uma, and Wanda are already known innocents among that surrounding group, Uma has at least 3 innocent neighbors, and the only uncertain ones there are Peter, Rohan, and Vicky. Xavi’s only neighbors are Rohan, Salil, and Wanda. Salil is a criminal and Wanda is innocent, so for Xavi to have more innocent neighbors than Uma, Xavi must have 2 innocent neighbors, which forces Rohan to be innocent. Then Xavi has exactly 2 innocent neighbors, so Uma must have fewer than 2; since Uma already has Olivia and Wanda as innocent neighbors, Peter and Vicky cannot also be innocent, and with Rohan now innocent, Uma’s innocent-neighbor total is exactly 3 only if Peter and Vicky are criminals. Therefore, we can determine that C4, Rohan, is INNOCENT, B4, Peter, is CRIMINAL, and B5, Vicky, is CRIMINAL.

Uma says Linda is one of 4 innocents in column C, so column C must contain exactly four innocents in total. In column C, Chad, Linda, Rohan, and Wanda are already known innocents, which already makes four. The only other person in that column is Hal at C2, so Hal cannot also be innocent. Therefore, we can determine that C2 Hal is CRIMINAL.

Joyce at A3 has four neighbors: Flora, Gus, Olivia, and Karen. Flora and Gus are criminals and Olivia is innocent, so Joyce’s number of innocent neighbors depends on Karen: she has 1 innocent neighbor if Karen is criminal, or 2 if Karen is innocent. Xavi at D5 has three neighbors: Salil, Wanda, and Rohan. Salil is criminal while Wanda and Rohan are innocent, so Xavi has exactly 2 innocent neighbors. Peter’s clue says Xavi and Joyce have the same number of innocent neighbors, so Joyce must also have 2 innocent neighbors. That only happens if Karen is innocent. Therefore, we can determine that B3 is INNOCENT.

Karen says column B has more criminals than any other column, so column B must have the unique highest criminal count. Column B already has four criminals: B1, B2, B4, and B5, with only B3 innocent. Column D already has three criminals at D1, D2, and D4, so if either D3 or D5 were criminal then column D would also have four criminals, and column B would no longer have more criminals than every other column. Therefore, we can determine that D3 is INNOCENT and D5 is INNOCENT.

Megan’s clue says every row must contain at least one criminal. In row 3, Karen at B3, Linda at C3, and Megan at D3 are all already known to be innocent, so the only person in that row who could satisfy the clue is Joyce at A3. Therefore, we can determine that A3 is CRIMINAL.

In column A, we already know that Flora at A2 and Joyce at A3 are criminals. The clue says both criminals in column A are connected, so those two must be the only criminals in that column, and they are indeed connected directly. That leaves Alice at A1 unable to be a criminal. Therefore, we can determine that A1 is INNOCENT.