Puzzle Pack #1 Puzzle 19 Answer

Ollie is at D3, so his neighbors are C2, C3, C4, D2, and D4. Sue says exactly two of those neighbors are criminals, and neither of those two is a neighbor of Ruth at B4. Ruth’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5, so among Ollie’s neighbors the only one who is also Ruth’s neighbor is C3, Nicole. Since none of Ollie’s two criminal neighbors can be Ruth’s neighbor, Nicole cannot be one of those criminals. Therefore, we can determine that C3 is INNOCENT.

Nicole’s clue says there are exactly 2 innocents to the right of Flora, and the people to Flora’s right are Gus at B2, Hilda at C2, and Isaac at D2. So among B2, C2, and D2, exactly two are innocent. Sue’s clue says Ollie has exactly 2 criminal neighbors, and neither of those criminals is a neighbor of Ruth. Ollie at D3 is next to Hilda, Isaac, Nicole, Sue, and Terry, while Ruth’s neighbors are Logan, Martin, Nicole, Paula, Sue, Will, Xia, and Terry. The only neighbors of Ollie who are not also neighbors of Ruth are Hilda at C2 and Isaac at D2, so those must be Ollie’s 2 criminal neighbors. That makes Hilda and Isaac the two criminals among Flora’s right-side group, so the remaining person there, Gus at B2, must be innocent. Ollie’s clue also leaves Terry at D4 out of the “2 criminals neighboring Ollie,” but Terry’s status is already forced by the exact neighbor count around Ollie together with Nicole and Sue being innocent, so Terry is the remaining criminal neighbor needed in that local arrangement. Therefore, we can determine that B2 is INNOCENT and D4 is CRIMINAL.

Row 3 contains Logan at A3, Martin at B3, Nicole at C3, and Ollie at D3. Gus says that Martin is one of 3 innocents in that row, which directly tells us Martin is innocent. Therefore, we can determine that B3 Martin is INNOCENT.

Ollie is at D3, so his neighbors are C2, D2, C3, C4, C5, D4, and D5. Sue says exactly two of those neighbors are criminals, and neither of those two is a neighbor of Ruth at B4. Ruth’s neighbors are A3, B3, C3, A4, C4, A5, B5, and C5, so among Ollie’s neighbors the ones who are not Ruth’s neighbors are only C2, D2, D4, and D5. Since C3 and C4 are innocent and D4 is already known to be criminal, Ollie’s two criminal neighbors must be D4 and exactly one of C2, D2, or D5. Martin says row 2 has an odd number of innocents. In row 2, B2 is already innocent, so among A2, C2, and D2 there must be an even number of innocents. Because at most one of C2 and D2 can be criminal from Sue’s clue, at least one of C2 or D2 is innocent, so A2 cannot be criminal; otherwise A2, C2, and D2 would contain either one or three innocents, making the total in row 2 even instead of odd. Therefore, we can determine that A2 is INNOCENT.

The corners are A1, D1, A5, and D5. Flora says there is exactly one innocent among those four corners, and that innocent is in row 1, so the only possible innocent corners are A1 or D1. That means the two corners in row 5, A5 and D5, cannot be innocent. Therefore, we can determine that A5 is CRIMINAL and D5 is CRIMINAL.

Isaac is at D2, so his neighbors are C1, C2, C3, D1, and D3. Among those, C3 is already known innocent, so the two criminal neighbors mentioned in Vince’s clue must come from C1, C2, D1, and D3. Hilda is at C2, and her neighbors are B1, B2, B3, C1, C3, D1, D2, and D3, which means C2 itself is not one of Hilda’s neighbors. So if Isaac had been a criminal, then C2 would be one of Isaac’s criminal neighbors, but it could not be one of the two criminals that are Hilda’s neighbors, since Hilda cannot neighbor herself. That would make Vince’s statement impossible. Therefore, we can determine that C2, Hilda, is INNOCENT.

Ollie is at D3, and his neighbors are C2, C3, D2, C4, and D4. Among those, C2, C3, and C4 are innocent, and D4 is criminal, so the clue’s “2 criminals neighboring Ollie” must be D4 and D2. Ruth is at B4, and her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5, so D4 is not one of Ruth’s neighbors, which matches the clue. Since neither of Ollie’s two criminal neighbors can be Ruth’s neighbor, the other one also cannot be Ruth’s neighbor, and among Ollie’s neighbors only D2 fits that as well. Therefore, we can determine that D2 is CRIMINAL.

Below Ben are Martin at B3, Ruth at B4, and Will at B5. The ones among them who neighbor Logan at A3 are Martin and Ruth, because Logan’s neighbors include B3 and B4 but not B5. Hilda says an odd number of innocents in that group neighbor Logan, and Martin is already known to be innocent, so Ruth must also be innocent to make that count odd. Therefore, we can determine that B4 is INNOCENT.

Ruth’s clue is about Anna and row 1. It says Anna is one of two or more innocents in that row, which directly includes Anna among the innocents there. Therefore, we can determine that A1 is INNOCENT.

The corner people are Anna at A1, Eve at D1, Vince at A5, and Zed at D5. Flora’s clue says there is exactly one innocent among all four corners, and that innocent is in row 1. We already know Anna at A1 is innocent, while Vince at A5 and Zed at D5 are criminal, so Anna is the one innocent corner Flora is talking about. That leaves Eve at D1 as not innocent. Therefore, we can determine that D1 is CRIMINAL.

Isaac’s neighbors are C1, D1, C2, C3, D3, C1 Debra, D1 Eve, Hilda, Nicole, and Ollie, and the criminals among those neighbors must both also be neighbors of Hilda. Since Eve is already a criminal neighbor of Isaac and Eve is not a neighbor of Hilda, Eve cannot be one of those two criminals, so Isaac’s criminal neighbors must be the two who are also next to Hilda: Debra at C1 and Ollie at D3. That means Hilda’s neighbors contain exactly those two criminals, and her other neighbors A1 Anna, B1 Ben, B2 Gus, B3 Martin, C1 Debra, C3 Nicole, D1 Eve, D2 Isaac, and D3 Ollie reduce to five innocents plus Ben besides the two criminals. Eve’s clue says Hilda has more innocent than criminal neighbors, so with two criminal neighbors, Hilda must have at least three innocent neighbors; among her unknown neighbors, that forces Ben to be innocent. Therefore, we can determine that B1 is INNOCENT.

Debra at C1 has exactly two criminal neighbors already visible: Eve at D1 and Isaac at D2. So Xia at C5 must also have exactly two criminal neighbors. Xia’s neighbors are Ruth at B4, Sue at C4, Terry at D4, Will at B5, and Zed at D5; among these, Terry and Zed are criminal while Ruth and Sue are innocent, so the only way for Xia to have the same total of two criminal neighbors is for Will not to be criminal. Therefore, we can determine that B5 is INNOCENT.

In row 5, the two innocents must be connected by orthogonal adjacency, so they have to form one continuous block within that row. We already know B5 is innocent, while A5 and D5 are criminals. If C5 were also criminal, then B5 would be the only innocent in row 5, so there would not be both innocents there. The only way for row 5 to contain exactly two connected innocents is for C5 to be the second one next to B5. Therefore, we can determine that C5 is INNOCENT.

Zed’s clue compares the total number of innocents in row 4 and row 5. Row 5 already has two innocents, Will at B5 and Xia at C5, while row 4 already has two innocents, Ruth at B4 and Sue at C4. For row 4 to have more innocents than row 5, row 4 must gain at least one additional innocent, and the only unknown in that row who can provide that is Paula at A4, since Terry at D4 is already criminal. Therefore, we can determine that A4 is INNOCENT.

Paula says row 5 has more criminals than any other row, so row 5 must have uniquely the highest criminal count. Row 5 already has exactly two criminals, Vince at A5 and Zed at D5, because Will and Xia are innocent. That means every other row must have fewer than two criminals, so at most one each. Row 1 already contains Eve at D1 as a criminal, so C1 cannot also be a criminal. Therefore, we can determine that C1 is INNOCENT.

Isaac is at D2, so his neighbors are C1, C2, C3, D1, and D3. Among those, D1 is already criminal, while C1, C2, and C3 are innocent, so the only way Vince’s clue can be true is if the other criminal neighboring Isaac is D3, Ollie. Hilda is at C2, and both D1 and D3 are indeed her neighbors, matching the clue exactly. Therefore, we can determine that D3 is CRIMINAL.

Row 3 contains Logan at A3, Martin at B3, Nicole at C3, and Ollie at D3. Gus says Martin is one of 3 innocents in that row, so row 3 must contain exactly 3 innocents in total. Martin and Nicole are already known to be innocent, and Ollie is already known to be criminal, so the only way for row 3 to have 3 innocents is for Logan to be innocent as well. Therefore, we can determine that A3 is INNOCENT.