Puzzle Pack #1 Puzzle 34 Answer

Ruth is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Zoe says both innocents in row 3 are Ruth's neighbors, so the two innocents in row 3 must be among A3, B3, and C3, because those are the only row 3 neighbors Ruth has. The people in row 3 are A3 Lucy, B3 Mary, C3 Nancy, and D3 Olive, so Olive is the only person in that row who is not a neighbor of Ruth and therefore cannot be one of the two innocents. Therefore, we can determine that D3 is CRIMINAL.

Row 3 is Lucy at A3, Mary at B3, Nancy at C3, and Olive at D3. Olive says there is an odd number of innocents between Lucy and Olive, and the only people between them are Mary and Nancy, so exactly one of Mary and Nancy is innocent. Zoe says both innocents in row 3 are neighbors of Ruth at B4, and Ruth’s neighbors in row 3 are exactly Lucy at A3, Mary at B3, and Nancy at C3, not Olive at D3. Since Olive is already criminal and exactly one of Mary or Nancy is innocent, the other innocent in row 3 must be Lucy. Therefore, we can determine that A3 is INNOCENT.

Lucy’s clue is about Sofia at C4. Sofia has exactly two criminal neighbors, and among Sofia’s neighbors the only people in row 5 are Wally at B5, Xavi at C5, and Zoe at D5. The clue says neither of Sofia’s two criminal neighbors is in row 5, so none of those row 5 neighbors can be criminal. Since Zoe is already known to be innocent, the row 5 neighbors still undetermined there, Wally and Xavi, must also be innocent. Therefore, we can determine that B5 is INNOCENT and C5 is INNOCENT.

Row 3 already has Lucy at A3 as an innocent, and Zoe says that both innocents in row 3 are neighbors of Ruth at B4. The only people in row 3 who are neighbors of B4 are Lucy at A3, Mary at B3, Nancy at C3, and Olive at D3, and Olive is already known to be criminal, so the second innocent in row 3 must be either Mary or Nancy. That means both criminals in row 3 are not next to Ruth, so Ruth cannot be one of the two criminals neighboring Sofia at C4. Lucy says Sofia has exactly two criminal neighbors, and neither of them is in row 5. Sofia's neighbors are Mary, Nancy, Olive, Ruth, Thor, Wally, Xavi, and Zoe; among the row 5 neighbors, Wally, Xavi, and Zoe are all innocent, and Olive at D3 is already one criminal neighbor. Since Ruth cannot be the other criminal neighbor, the only possible second criminal neighbor is Nancy at C3, which leaves both Ruth at B4 and Thor at D4 as non-criminal neighbors of Sofia. Therefore, we can determine that B4 is INNOCENT and D4 is INNOCENT.

Daniel is at C1, so his neighbors are B1, D1, B2, C2, and D2. In row 2, the only people who are Daniel's neighbors are B2, C2, and D2, and Ruth's clue says the two innocents in row 2 are both among those neighbors. That means the only person in row 2 who cannot be innocent is A2. Therefore, we can determine that A2 is CRIMINAL.

Row 2 is A2 Gabe, B2 Helen, C2 Jane, and D2 Kevin, and we already know A2 is criminal. Ruth says the two innocents in row 2 are both neighbors of Daniel at C1, so in row 2 the innocents must come from B2, C2, and D2, but only B2 and C2 neighbor C1. That means the two innocents in row 2 are Helen and Jane. Wally’s clue fits this as well, because among those row 2 innocents, only Jane is neighboring Jane. Therefore, we can determine that C2 is INNOCENT.

Vince is in A5, and the people above him in column A are Anna at A1, Gabe at A2, Lucy at A3, and Paul at A4. Jane says there are exactly 2 innocents above Vince, and Xavi says Lucy is one of 3 innocents in column A. Since Lucy is already one innocent in that column and Gabe is already a criminal, the only way to have exactly 2 innocents above Vince is for exactly one of Anna and Paul to be innocent, making the three innocents in column A be Lucy, Vince, and that one person. Therefore, we can determine that A5 is INNOCENT.

Daniel is at C1, so his four neighbors are B1 Carl, B2 Helen, C2 Jane, and D2 Kevin. Among those, the ones who also neighbor Kevin at D2 are C1 Daniel, C2 Jane, C3 Nancy, D1 Floyd, and D3 Olive, so from Daniel's neighbors only C2 Jane and D2 Kevin also neighbor Kevin. Since the clue says exactly 2 of Daniel's 4 innocent neighbors also neighbor Kevin, Daniel must have exactly 2 innocent neighbors among those overlapping spots, which forces D2 Kevin to be innocent along with C2 Jane. Then Daniel's other two neighbors, B1 Carl and B2 Helen, cannot also be innocent, so Daniel still needs a fourth innocent neighbor elsewhere, and the only remaining neighbor of Kevin who can fit that role next to Daniel's area is D1 Floyd. Therefore, we can determine that D1 is INNOCENT.

Daniel is at C1, so his neighbors are B1 Carl, B2 Helen, C2 Jane, and D2 Kevin. Ruth says both innocents in row 2 are Daniel’s neighbors, and in row 2 Jane is already innocent, so the other innocent in that row must be either Gabe or Kevin; since Gabe is criminal, it has to be Kevin. Vince then says exactly 2 of Daniel’s 4 innocent neighbors also neighbor Kevin. Among Daniel’s neighbors, Jane and Carl both neighbor Kevin, while Daniel himself does not and Helen is not known innocent; with Kevin already one of Daniel’s innocent neighbors, the count reaches exactly 2 only if Carl is also innocent. Therefore, we can determine that B1 is INNOCENT.

Ruth is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. In row 3, the two innocents must both be among Ruth’s neighbors, and the only row 3 people who are not Ruth’s neighbors are Olive at D3 and Lucy at A3 is already innocent, so the second innocent in row 3 has to be Mary at B3 or Nancy at C3, which means Olive cannot be one of the two innocents in that row. Since Olive is already known to be criminal, row 3’s two innocents are Lucy and exactly one of Mary or Nancy. Now use Gabe’s clue. Anna at A1 has criminal neighbors Gabe at A2 and Helen at B2, so Anna has 1 criminal neighbor if Helen is innocent, or 2 if Helen is criminal. Sofia at C4 neighbors Mary, Nancy, Olive, Ruth, Thor, Wally, Xavi, and Zoe; among these, Olive is criminal, Ruth, Thor, Wally, Xavi, and Zoe are innocent, and exactly one of Mary or Nancy is criminal because row 3 has exactly two innocents. So Sofia has exactly 2 criminal neighbors. To match Sofia’s total, Anna must also have 2 criminal neighbors, so Helen must be criminal. Therefore, we can determine that B2 is CRIMINAL.

Daniel is at C1, so his neighbors are B1, B2, C2, D1, and D2. The two innocents in row 2 are Jane at C2 and one other person, because Gabe and Helen are already criminals. Ruth says both row 2 innocents are Daniel's neighbors, and among the row 2 spaces only C2 and D2 neighbor Daniel. So the second innocent in row 2 must be Kevin at D2. Therefore, we can determine that D2 is INNOCENT.

Ruth is at B4, so her neighbors in row 3 are A3, B3, and C3. Zoe says both innocents in row 3 are Ruth's neighbors, which means the only possible innocents in that row are Lucy at A3, Mary at B3, and Nancy at C3; Olive at D3 is already criminal, so row 3 contains exactly two innocents among Lucy, Mary, and Nancy. Jane is at C2, and her five innocent neighbors are B1 Carl, D1 Floyd, D2 Kevin, A3 Lucy, and D3 Olive is not innocent, so the five innocent neighbors are actually B1, D1, D2, and then exactly one of B3 Mary or C3 Nancy must also be innocent together with Lucy to make the total fit row 3. Among Jane's innocent neighbors, the ones who also neighbor Daniel at C1 are only B1, D1, and either B3 or C3 does not count because neither is adjacent to C1 except B2, B1, C2, D2, and the top row neighbors. Kevin says exactly three of Jane's five innocent neighbors also neighbor Daniel, and that already forces Daniel himself to be innocent so the count works with the known innocent neighbors around him. Therefore, we can determine that C1 is INNOCENT.

In row 3, the known people are Lucy at A3 as innocent and Olive at D3 as criminal, so the two unknowns there are Mary at B3 and Nancy at C3. Zoe says both innocents in row 3 are neighbors of Ruth at B4, and Ruth’s neighbors in row 3 are only A3, B3, and C3. Since A3 is already one innocent, the other row 3 innocent must be either Mary or Nancy, which means at least one of Jane’s neighbors at B4, C4, and C3 is innocent. Thor says Jane at C2 and Xavi at C5 have the same number of innocent neighbors. Xavi already has four innocent neighbors: Ruth at B4, Sofia at C4, Thor at D4, and Wally at B5. Jane’s neighbors are Carl at B1, Daniel at C1, Floyd at D1, Helen at B2, Kevin at D2, Mary at B3, Nancy at C3, and Olive at D3, and among these she already has four known innocent neighbors: Carl, Daniel, Floyd, and Kevin. Because at least one of Mary or Nancy is also innocent, Jane would have at least five innocent neighbors unless Sofia is innocent and counted for Xavi as well. So the only way for Jane and Xavi to have equal numbers of innocent neighbors is for Sofia to be innocent. Therefore, we can determine that C4 is INNOCENT.

Ruth is at B4, so her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Sofia says exactly 7 of those 8 people are innocent, and we already know A3, C4, A5, B5, and C5 are innocent, so among B3, C3, and A4 exactly two are innocent. Zoe says both innocents in row 3 are Ruth's neighbors, so the only innocents in row 3 must be among A3, B3, and C3; since A3 is already innocent, exactly one of B3 and C3 is innocent, which means the second innocent among B3, C3, and A4 has to be A4. Therefore, we can determine that A4 is INNOCENT.

Column A contains Anna at A1, Gabe at A2, Lucy at A3, Paul at A4, and Vince at A5. Xavi says Lucy is one of exactly 3 innocents in that column, and we already know Gabe is criminal while Lucy, Paul, and Vince are innocent. That already gives column A its 3 innocents, so Anna cannot also be innocent. Therefore, we can determine that A1 Anna is CRIMINAL.

The only unknown people in row 3 are Mary at B3 and Nancy at C3, because Lucy at A3 is already innocent and Olive at D3 is already criminal. Zoe says both innocents in row 3 are Ruth’s neighbors, and Ruth is at B4, whose neighbors in row 3 are A3, B3, and C3. Since Lucy at A3 is one of the innocents in row 3, the other innocent in row 3 must be either Mary or Nancy, so row 3 contains exactly one more innocent and one more criminal. Floyd says Lucy has more criminal neighbors than Thor. Lucy’s neighbors are A2, B2, B3, and A4, so she already has two criminal neighbors there from Gabe and Helen. Thor’s neighbors are C3, D3, C4, C5, and D5, so he already has one criminal neighbor there from Olive. For Lucy to have more criminal neighbors than Thor, Mary at B3 must be criminal and Nancy at C3 must be innocent; otherwise Thor would have at least as many criminal neighbors as Lucy. Therefore, we can determine that B3 is CRIMINAL and C3 is INNOCENT.