Clues by Sam Oct 06, 2026 Answer – Full Solution Explained

01.B4 · Nancy → CRIMINAL, B3 · John → CRIMINAL

Emma’s clue says the people strictly between Frank and Vicky contain exactly 2 criminals. That group currently has no known criminals, and the only people left in it are John and Nancy. Since those two spots must supply all 2 criminals, both of them have to be criminals. So Nancy and John must be criminals.

02.C3 · Kumar → INNOCENT, D3 · Lucy → INNOCENT, A3 · Ike → CRIMINAL

Nancy’s clue says the two innocents in row 3 must be connected. Row 3 contains Ike, John, Kumar, and Lucy, and John is already a criminal. If Kumar were a criminal, Lucy were a criminal, and Ike were innocent, then row 3 could not satisfy that clue: there would not be the required connected pair of innocents in that row. So those opposite identities are impossible. That makes Kumar and Lucy innocent, and Ike criminal.

03.C4 · Penny → CRIMINAL

Lucy’s clue says there are exactly 2 criminals above Wally, and exactly 1 of those criminals is Bobby’s neighbor. So among the people above Wally who are not Bobby’s neighbors, there must be exactly 1 criminal. Those people are Kumar and Penny. Kumar is already innocent, so that group currently has no known criminal and still needs 1 criminal. Penny is the only unknown person left in that group, so Penny at C4 must be criminal.

04.C5 · Wally → INNOCENT

Lucy’s clue fixes the people above Wally at exactly two criminals, and exactly one of those two must be Bobby’s neighbor. Penny is already one criminal above Wally, so the second criminal above Wally has to be either Chloe or Gus, because those are the only unknown people there who are also Bobby’s neighbors. Now test Wally as a criminal. With Penny’s clue that column A is the only column with exactly three criminals, the remaining unknown people named in these clues cannot be assigned in a way that keeps column A as the unique column with exactly three criminals while also satisfying Lucy’s requirement about the two criminals above Wally. So Wally must be innocent.

05.A1 · Adam → CRIMINAL

If Adam were innocent, then in column A the only criminals would have to come from Ike, Maria, and Uma. Penny says column A has exactly 3 criminals, so Maria and Uma would both have to be criminals to bring column A up to three. But Wally says there is exactly one innocent below Emma, and the people below Emma are Ike, Maria, and Uma. With Ike already criminal, making both Maria and Uma criminals would leave no innocent below Emma, which clashes with that clue. So Adam must be criminal.

06.B5 · Vicky → CRIMINAL

Adam’s clue says the people above Vicky contain exactly one innocent. Right now, the people above B5 are Bobby, Frank, John, and Nancy, and among them John and Nancy are already criminals, so that one innocent would have to come from Bobby or Frank. At the same time, Penny’s clue says column A is the only column with exactly 3 criminals, so column B cannot end up with exactly 3 criminals. But if Vicky were innocent, then column B’s total would be controlled entirely by Bobby and Frank together with the two criminals already there, and that clashes with the requirement about the one innocent above Vicky and column A being the only column with exactly 3 criminals. So Vicky at B5 must be criminal.

07.B2 · Frank → CRIMINAL

If Frank were innocent, then Bobby’s neighbors would have to get to exactly 3 criminals without Frank. Since Adam is already the 1 known criminal there, Chloe and Gus would both have to be criminals to make that total 3. But then column B would contain Bobby, John, Nancy, and Vicky as criminals. That gives column B more than 3 criminals, so column A can still be the only column with exactly 3 criminals only if Bobby is criminal. This leaves no way for Frank to be innocent without forcing that clash, so Frank at B2 must be criminal.

08.B1 · Bobby → INNOCENT

Adam’s clue says the people above Vicky contain exactly one innocent. In that group, there are currently no known innocents, and the only person there whose identity is still unknown is Bobby. Since that group still needs exactly one innocent, Bobby has to be the one. So Bobby must be innocent.

09.D1 · Debra → INNOCENT

Above Wally there are exactly 2 criminals in total, and exactly 1 of those criminals is Bobby's neighbor. Penny is already one of the people above Wally and is already a criminal, so among Chloe and Gus exactly one must be a criminal. Hal's neighbors must contain an odd number of criminals. Right now Hal has no known criminal neighbors, and the only unknown neighbors are Chloe, Debra, and Gus. Since Chloe and Gus contribute exactly one criminal between them, making Debra a criminal too would give Hal exactly 2 criminal neighbors there, which is even, not odd. So Debra must be innocent.

10.C1 · Chloe → CRIMINAL

Debra’s clue says there is exactly 1 innocent among the people who are both in column C and on the edge. That group is just Chloe and Wally, and Wally is already known to be innocent. Since the one innocent in that group is already accounted for, Chloe cannot also be innocent. So Chloe must be criminal.

11.C2 · Gus → INNOCENT

Penny’s clue says column A is the only column with exactly 3 criminals. In column C, Chloe and Penny are already criminals, and Gus is the only person there not yet identified. If Gus were a criminal, then column C would also have exactly 3 criminals, which is not allowed. So Gus must be innocent.

12.D2 · Hal → INNOCENT

If Hal were a criminal, then row 2 would stay at 2 innocents, while column A already has 1 innocent and two unknown people, Maria and Uma. Gus's clue says row 2 has more innocents than column A, so column A could not gain enough innocents to catch up. That means Maria and Uma would both have to be criminals. But then column A would contain Adam, Ike, Maria, and Uma as criminals, for 4 criminals in that column. Penny's clue says column A is the only column with exactly 3 criminals, so column A cannot have 4. So Hal must be innocent.

13.D4 · Tina → INNOCENT

14.A4 · Maria → INNOCENT, A5 · Uma → CRIMINAL, D5 · Xavi → INNOCENT

Rows 4 and 5 have to contain the same number of innocents. Right now each row already has 1 known innocent, with Maria the only unknown in row 4 and Uma and Xavi the unknowns in row 5. If Maria were criminal, Uma were innocent, and Xavi were criminal, then row 4 would stay at 1 innocent while row 5 would become 2 innocents. That conflicts with the requirement that rows 4 and 5 have equal numbers of innocents. So Maria cannot be criminal, Uma cannot be innocent, and Xavi cannot be criminal. Maria must be innocent, Uma must be criminal, and Xavi must be innocent.