Puzzle Pack #2 Puzzle 7 Answer

Logan is at C3, so his row 3 neighbors are Karen at B3 and Mark at D3. Wanda’s clue says exactly 2 innocents in row 3 are neighboring Logan, and those are the only people in row 3 who neighbor him. That means both Karen and Mark must be the two innocents described. Therefore, we can determine that B3 is INNOCENT and D3 is INNOCENT.

The innocents above Xavi are the innocent people in column D above D5, so they can only be Diane at D1, Helen at D2, and Mark at D3. Mark is already known innocent, and the clue says both innocents above Xavi are connected, so there must be exactly two innocents among those three positions and they must form one orthogonally connected group. Since D3 is one of them, the only way for the two innocents to be connected is for D2 to be the other innocent. Therefore, D1 cannot be innocent. Therefore, we can determine that D1 is CRIMINAL.

Diane’s clue says that Bonnie is one of Evie’s four innocent neighbors. Bonnie is next to Evie at B1, so Bonnie is one of the neighbors this clue is talking about, and Diane explicitly says that neighbor is innocent. Therefore, we can determine that B1 is INNOCENT.

In Xavi’s column, Bonnie’s clue means column D must contain at least three innocents. We already know Mark at D3 is innocent, and the only people above Xavi are Diane at D1, Helen at D2, and Mark at D3. Mark’s clue says that both innocents above Xavi are connected, so there are exactly two innocents among D1, D2, and D3, and those two must touch orthogonally. Since D1 is criminal and D3 is innocent, the two innocents above Xavi can only be D2 and D3, so Helen is innocent. That gives column D exactly two known innocents above Xavi, namely D2 and D3. Since the column must have at least three innocents, Xavi must be the third. Therefore, we can determine that D5 is INNOCENT.

Column A already has no confirmed criminals, so Bonnie’s clue means it must contain at least three innocents among A1, A2, A3, A4, and A5. Xavi’s clue says all innocents in column A form one connected vertical group. Since A5 is the only possible innocent at the bottom of that column with no one below it, having three or more innocents in one connected group forces the group to extend upward through A4 and A3. Therefore, we can determine that A3 is INNOCENT.

Mark is at D3, so the people above him in the same column are Diane at D1 and Helen at D2. Isaac says there are no innocents above Mark, which means everyone above Mark is criminal. Diane already fits that, so Helen must be criminal as well. Therefore, we can determine that D2 Helen is CRIMINAL.

Xavi is at D5, so the people above him are Diane at D1, Helen at D2, Mark at D3, and Rohan at D4. The clue says both innocents above Xavi are connected, so there must be exactly two innocents in that column above Xavi, and those two must form one continuous vertical group. We already know Diane at D1 and Helen at D2 are criminals, while Mark at D3 is innocent. That means the second innocent above Xavi must be Rohan at D4, and D3 and D4 are directly adjacent, so the two innocents are connected exactly as the clue requires. Therefore, we can determine that D4 is INNOCENT.

Paula is at C4, so her neighbors in column C are exactly Logan at C3 and Wanda at C5. Rohan says exactly 2 innocents in column C are neighboring Paula. Since those are the only two column C neighbors Paula has, both of them must be innocent, and Wanda is already known innocent. Therefore, we can determine that C3 is INNOCENT.

In column C, Logan at C3 and Wanda at C5 are already innocent, so Bonnie’s clue that every column has at least 3 innocents means column C needs at least one more innocent among Chloe at C1, Gary at C2, and Paula at C4. Logan’s clue says Helen at D2 and Xavi at D5 have the same number of innocent neighbors. Xavi already has three innocent neighbors, namely Paula at C4, Wanda at C5, and Rohan at D4, so Helen must also have three innocent neighbors. Around Helen, Bonnie at B1, Isaac at A3, Karen at B3, Logan at C3, and Mark at D3 are already innocent, which means the count can only match Xavi’s if Paula at C4 is not adding to Xavi’s total. Therefore, we can determine that C4 is INNOCENT.

Frank is at B2, so the innocents below him are the people in column B beneath him: Karen at B3, Olof at B4, and Terry at B5. Paula says exactly one of those innocents is neighboring Isaac at A3. Isaac’s neighbors among those three are only Karen at B3 and Olof at B4, since Terry at B5 is too far away. Karen is already known to be innocent, so she already accounts for the one innocent below Frank who neighbors Isaac. That means Olof cannot also be innocent. Therefore, we can determine that B4 is CRIMINAL.

Helen’s neighbors are Chloe, Diane, Gary, Logan, Mark, Paula, Rohan, and Xavi. Among those, Diane is criminal and Logan, Mark, Paula, Rohan, and Xavi are innocent, so Helen already has 5 innocent neighbors, which means Chloe and Gary must contribute equally to Helen and Xavi’s innocent-neighbor counts from Logan’s clue. Since Xavi’s neighbors are Paula, Rohan, Terry, and Wanda, and Paula, Rohan, and Wanda are already innocent, Xavi has 3 innocent neighbors, so Helen must also end up with 3 innocent neighbors among the neighbors still relevant to that comparison, forcing both Chloe and Gary to be criminal. Then Olof’s clue says Helen has more criminal neighbors than Wanda. With Chloe, Diane, and Gary criminal, Helen has 3 criminal neighbors. Wanda’s neighbors are Paula, Terry, Olof, and Xavi, and Paula and Xavi are innocent while Olof is criminal, so for Helen to have more criminal neighbors than Wanda, Terry cannot be criminal as well. Therefore, we can determine that B5 is INNOCENT.

Column A already has Isaac at A3 as innocent, and the clue about Alice says there is an odd number of innocents below A1. The people below Alice are A2, A3, A4, and A5, so among those four spots the total number of innocents must be odd; since A3 is already innocent, A2, A4, and A5 must contain an even number of innocents. Bonnie says every column has at least 3 innocents, so column A must have at least three innocents in total. Since A1 is still unknown and A2, A4, and A5 need to contribute an even number of innocents, the only way column A can reach at least three innocents is for A2, A4, and A5 all to be innocent. That also fits Xavi's clue, because then all innocents in column A form one connected vertical group. Therefore, we can determine that A4 is INNOCENT.

Nancy says row 3 has more innocents than any other row, so row 3 must have uniquely the highest number of innocents. Row 3 already has 4 innocents. That means every other row must have fewer than 4 innocents. Row 5 already has three known innocents at B5, C5, and D5, so A5 cannot also be innocent, because then row 5 would also have 4 innocents and would tie row 3. Therefore, we can determine that A5 is CRIMINAL.

Alice is at A1, so the people below her are Evie at A2, Isaac at A3, Nancy at A4, and Scott at A5. We already know Isaac and Nancy are innocent, and Scott is criminal, so among those four, the only unknown is Evie. That means there are currently 2 known innocents below Alice, and Terry’s clue says the total number of innocents below Alice is odd. So Evie must also be innocent to make that total 3. Therefore, we can determine that A2 is INNOCENT.

Evie’s neighbors are A1, B1, B2, A3, and B3. Diane says Bonnie is one of Evie’s 4 innocent neighbors, so Evie has exactly 4 innocent neighbors among those 5 people; since Bonnie, Isaac, and Karen are already known innocent, exactly one of A1 and B2 is not innocent. Mark’s neighbors are C2, D2, C3, C4, and D4. Scott says Evie and Mark have an equal number of innocent neighbors, so Mark also has exactly 4 innocent neighbors among those 5 people. Since Logan, Paula, and Rohan are already known innocent and Helen is criminal, the only way for Mark to have 4 innocent neighbors is for Gary at C2 to be innocent. Therefore, we can determine that C2 is INNOCENT.

Helen at D2 and Xavi at D5 must have the same number of innocent neighbors. Xavi’s neighbors are C4, C5, and D4, and all three are innocent, so Xavi has 3 innocent neighbors. Helen’s neighbors are C1, D1, C2, C3, D3, C1 is Chloe, D1 is Diane, and the other three are all innocent. Since Diane is criminal and Helen must also have exactly 3 innocent neighbors, Chloe cannot be innocent. Therefore, we can determine that C1 is CRIMINAL.

Row 1 contains Alice at A1, Bonnie at B1, Chloe at C1, and Diane at D1. Gary’s clue says there is only one innocent in that row, and Bonnie is already known to be innocent. That means the other three people in row 1 must all be criminals, including Alice. Therefore, we can determine that A1 is CRIMINAL.

Evie is at A2, so her neighbors are A1, B1, B2, A3, and B3. Diane’s clue says Bonnie is one of Evie’s 4 innocent neighbors, which means exactly four of those five neighbors are innocent. Among them, Bonnie at B1, Isaac at A3, and Karen at B3 are already known to be innocent, while Alice at A1 is known to be criminal, so the only way Evie can have four innocent neighbors is if Frank at B2 is also innocent. Therefore, we can determine that B2 Frank is INNOCENT.