Clues by Sam May 28, 2026 Answer – Full Solution Explained

01.A2 · Freya → CRIMINAL, B2 · Gary → CRIMINAL

Donna’s clue says there are no innocents among the people who are both Laura’s neighbors and in row 2. That shared group is exactly Freya and Gary, and it already contains 0 known innocents. So neither of the unknown people in that group can be innocent. Therefore Freya and Gary must be criminal.

02.D5 · Ziad → CRIMINAL

Gary's clue says row 5 has exactly 3 criminals, and exactly 2 of those row-5 criminals are neighbors of Sam. So exactly 1 criminal in row 5 must be someone who is not a neighbor of Sam. In row 5, the only person who is not a neighbor of Sam is Ziad, and that group currently has no known criminal in it. That makes Ziad criminal.

03.B3 · Megan → CRIMINAL

Ziad’s clue says Megan is one of Freya’s 4 criminal neighbors. That directly places Megan among the criminal neighbors named by the clue. So Megan must be criminal.

04.C1 · Chris → CRIMINAL

Freya’s clue says an odd number of the edge cells that neighbor Gary are innocent. That group is Adam, Barnie, Chris, Freya, and Laura, and Freya is already criminal. Ziad’s clue says Freya has exactly 4 criminal neighbors, and among Freya’s neighbors we already know Gary and Megan are criminal, leaving only Adam, Barnie, and Laura to finish that requirement. If Chris were innocent, then Adam, Barnie, and Laura would have to satisfy both clues at the same time, but they cannot do that. So Chris cannot be innocent. So Chris must be criminal.

05.C2 · Ike → CRIMINAL, D2 · Kay → CRIMINAL

Row 5 has exactly 3 criminals in total, and exactly 2 of those row 5 criminals are Sam's neighbors. Since Ziad is already a criminal in row 5 and is not one of those neighboring row 5 people, the three row 5 people who can fill the remaining row 5 criminal spots are Vicky, Wanda, and Xavi. Chris's clue says row 2 has more criminals than row 5, and row 5 has 3 criminals, so row 2 must have more than 3 criminals. Row 2 already has Freya and Gary as criminals, so if Ike and Kay were both innocent, row 2 would stay at only 2 criminals, which clashes with that clue. So Ike and Kay must be criminal.

06.A4 · Rob → INNOCENT

Ike’s clue says there are exactly two innocents below Adam, and those two innocents must be connected. Below Adam are Freya, Laura, Rob, and Vicky, and Freya is already a criminal. If Rob were a criminal, then Laura and Vicky would be the only possible innocents below Adam, but that would not satisfy the clue’s requirement for the two innocents below Adam. So Rob must be innocent.

07.A1 · Adam → CRIMINAL

Megan’s clue says column A is the only column with exactly 2 innocents, so column A itself must have exactly 2 innocents. In column A, Rob is already known innocent, while Freya is known criminal and Adam, Laura, and Vicky are the unknowns. Ike’s clue says the innocents below Adam are exactly two people and those two must be connected. The people below Adam are Freya, Laura, Rob, and Vicky, with Freya criminal and Rob innocent. If Adam were innocent, then Laura and Vicky would still have to fit those two clue requirements together with the other unknown people named here, and that cannot be done. So Adam cannot be innocent. That makes Adam criminal.

08.D3 · Penny → CRIMINAL, D4 · Uma → CRIMINAL

Megan’s clue says column A is the only column with exactly 2 innocents, and Adam’s clue says every column has at least 3 criminals. The relevant columns here are A, B, C, and D. If Penny and Uma were both innocents, then the remaining people named in these columns, Barnie, Laura, Olivia, Sam, Terry, Vicky, Wanda, and Xavi, would have to make all of those column requirements true at the same time, but they cannot. That means Penny and Uma cannot both be innocents. So Penny and Uma must be criminals.

09.C3 · Olivia → CRIMINAL

Megan’s clue says column A is the only column with exactly 2 innocents, and Kay’s clue says Ziad’s neighbors contain exactly 1 innocent. In Ziad’s neighbors, there are no known innocents yet, so that one innocent has to come from Terry or Xavi. If Olivia were innocent, then the remaining people named here, including Laura and Vicky in column A and Barnie, Sam, Wanda, Terry, and Xavi, would have to satisfy both of those clue requirements at the same time, but they cannot. So Olivia must be criminal.

10.B5 · Wanda → CRIMINAL

Megan’s clue fixes column A at exactly 2 innocents, Kay’s clue fixes Ziad’s neighbors at exactly 1 innocent, and Rob’s clue says Sam’s neighbors contain an odd number of criminals. In the groups touched by those clues, the remaining people involved are Barnie, Laura, Sam, Terry, Vicky, and Xavi, together with Wanda. If Wanda were innocent, those same remaining people would have to make all three of those facts true at once, but they cannot. So Wanda must be criminal.

11.B4 · Sam → CRIMINAL

Rob's neighbors must contain an odd number of innocents. Among those neighbors, the only people in column A are Laura and Vicky, and column A has exactly 2 innocents total; since Rob is already the one known innocent in column A, Laura and Vicky contribute exactly 1 more innocent. That means the unknown neighbors already contribute the required odd number of innocents through Laura and Vicky. If Sam were innocent too, the unknown neighbors would contain 2 innocents instead of an odd number, which clashes with Rob's clue. So Sam must be criminal.

12.A3 · Laura → INNOCENT

13.A5 · Vicky → CRIMINAL

Megan’s clue says column A is the only column with exactly 2 innocents, so column A must have exactly 2 innocents and columns B, C, and D must not. In column A, Laura and Rob are already the 2 known innocents. If Vicky were innocent too, then that would clash with the requirement that column A have exactly 2 innocents and be the only column of its kind with that count. So Vicky must be criminal.

14.C5 · Xavi → INNOCENT

Row 5 has exactly 3 criminals in total. Among the row 5 people who are Sam's neighbors, the clue says exactly 2 of those criminals are there, and that group is A5 Vicky, B5 Wanda, and C5 Xavi. Vicky and Wanda are already the 2 known criminals in that neighboring group, so Xavi cannot also be a criminal. So Xavi must be innocent.

15.B1 · Barnie → CRIMINAL

Ziad’s clue says Megan is one of Freya’s exactly 4 criminal neighbors. Freya’s neighbors are Adam, Barnie, Gary, Laura, and Megan, and among them Adam, Gary, and Megan are already criminal while Laura is innocent. If Barnie were innocent too, then Freya’s neighbors would contain only those 3 criminals, which clashes with the clue that Freya has exactly 4 criminal neighbors. So Barnie must be criminal.

16.C4 · Terry → CRIMINAL

Kay’s clue says Ziad has exactly one innocent neighbor. Ziad’s neighbors already include one known innocent, Xavi. The only neighbor of Ziad whose identity is still unknown is Terry, so Terry cannot also be innocent without giving Ziad more than one innocent neighbor. So Terry must be criminal.