Clues by Sam Jul 16, 2026 Answer – Full Solution Explained

01.B2 · Isaac → CRIMINAL, B3 · Maria → CRIMINAL, B1 · Claire → CRIMINAL

Tyler’s clue says Claire is one of exactly three criminals above Ruth. The people above Ruth are Claire, Isaac, and Maria. So those three people have to be the exact group that satisfies the clue, and Claire must be included in it. If Claire, Isaac, and Maria were innocent, that would make the clue impossible to satisfy in that group. That makes Isaac, Maria, and Claire criminals.

02.B5 · Wanda → CRIMINAL

Claire’s clue says there are exactly 5 innocents on the edge, and exactly 1 of those edge innocents is a neighbor of Ruth. Among the edge neighbors of Ruth, the possible people are A3 Lisa, A4 Peter, A5 Vince, B5 Wanda, and C5 Xavi. That one innocent in this group has to be one of A3 Lisa, A4 Peter, A5 Vince, or C5 Xavi, so it cannot be Wanda. That makes Wanda at B5 criminal.

03.C5 · Xavi → CRIMINAL

Claire’s clue says there are exactly 5 innocents on the edge, and only 1 of those edge innocents is Ruth’s neighbor. Within the edge cells that are Ruth’s neighbors, the possible innocent spots being used here are A3 Lisa, A4 Peter, A5 Vince, and C5 Xavi. That one innocent person must come from Lisa, Peter, or Vince, so C5 cannot be the innocent one in that group. That makes Xavi criminal.

04.D1 · Frank → INNOCENT

Row 5 already has 2 criminals, Wanda and Xavi, so Maria's clue means the two unknown people there, Vince and Zach, must contribute an odd number of criminals. That means exactly one of Vince and Zach is a criminal. Wanda's clue says the number of criminal clerks equals the number of criminal cops. There is already 1 criminal clerk, Maria, and if Frank were also a criminal then the clerks would have 2 criminals. But the only cops are Vince and Zach, and row 5 allows exactly one of them to be a criminal, not 2. So Frank at D1 must be innocent.

05.A2 · Hank → CRIMINAL

Lisa, Peter, and Vince are the edge neighbors of Ruth, and Claire's clue says exactly one edge neighbor of Ruth is innocent. So that three-person group contains an odd number of innocents: exactly 1. Below Aaron are Hank, Lisa, Peter, and Vince, and Frank's clue says that group must contain an odd number of innocents. Since Lisa, Peter, and Vince already account for an odd number of innocents, adding Hank as another innocent would make the total below Aaron even instead of odd. So Hank must be criminal.

06.A1 · Aaron → CRIMINAL

Claire’s clue says there are exactly 5 innocents on the edge, and only 1 of them is Ruth’s neighbor, so exactly 4 edge innocents are not Ruth’s neighbors. Frank and Tyler are already known edge innocents and neither is Ruth’s neighbor, which means the remaining non-neighbor edge innocents must be the last 2 people from A1 Aaron, C1 David, D2 Karen, D3 Olive, and D5 Zach. From Xavi’s clue, column A has an odd number of innocents. Since Hank is already a criminal and there are currently 0 known innocents in column A, the innocents in column A must come from A1 Aaron, A3 Lisa, A4 Peter, and A5 Vince. But the 2 non-neighbor edge innocents are taken from C1 David, D2 Karen, D3 Olive, and D5 Zach, so Aaron cannot be one of those innocents. That makes Aaron criminal.

07.C1 · David → CRIMINAL

Row 5 already has 2 known criminals, and Maria’s clue says row 5 must contain an odd number of criminals. Aaron’s clue says row 1 and row 5 must contain the same number of criminals, and row 1 also currently has 2 known criminals. If David were innocent, then row 1 would stay at 2 criminals. That would force row 5 to stay at 2 criminals as well to match row 1, but 2 is not odd, so that conflicts with Maria’s clue. So David must be criminal.

08.B4 · Ruth → CRIMINAL

09.A3 · Lisa → CRIMINAL

Claire’s clue says that among the edge people, exactly one innocent is a neighbor of Ruth. In the shared group of edge neighbors of Ruth, the only unknown people are Lisa, Peter, and Vince. That one innocent person in this group has to be Peter or Vince, so Lisa cannot be the innocent one there. So Lisa must be criminal.

10.C4 · Sarah → CRIMINAL

Claire’s clue says there are exactly 5 innocents on the edge, and exactly 1 of those edge innocents is Ruth’s neighbor. Hank’s clue says Wanda’s neighbors contain an odd number of innocents, and right now the only unknown people among Wanda’s neighbors are Peter, Sarah, and Vince. If Sarah were innocent, then Karen, Olive, Peter, Vince, and Zach would still have to fill the remaining edge-innocent requirement from Claire’s clue while also making Wanda’s neighbor count come out odd. Those same people cannot satisfy both conditions at once. So Sarah must be criminal.

11.C2 · Joyce → INNOCENT

12.C3 · Nancy → CRIMINAL

Isaac's neighbors and Wanda's neighbors must contain the same number of innocents. Isaac's neighbors already include 1 known innocent, Joyce, and the only person there not yet identified is Nancy. Among the edge people who are Ruth's neighbors, exactly 1 is innocent, and that 1 must come from Peter and Vince. So Peter and Vince cannot add any innocent neighbors to Wanda's neighbor group. That means Wanda's neighbors gain no additional innocents, so Isaac's neighbors cannot gain one either. Therefore Nancy at C3 must be criminal.

13.D3 · Olive → CRIMINAL

Maria’s clue says row 5 has an odd number of criminals. Row 5 already has 2 known criminals, so Vince and Zach must be the people who make that row fit the clue. Nancy’s clue says there are more innocent cops than innocent sleuths. Right now the only people this can depend on are Vince and Zach among the cops, and Olive among the sleuths. If Olive were innocent, then Vince and Zach would have to satisfy both clues at once, but they cannot do that. So Olive must be criminal.

14.D2 · Karen → INNOCENT, D5 · Zach → INNOCENT

Claire’s clue says there are exactly 5 innocents on the edge, and exactly 1 of those edge innocents is Ruth’s neighbor. That means 4 edge innocents must be people on the edge who are not neighbors of Ruth. Among those edge non-neighbors, Frank and Tyler are already known innocents, so 2 more innocents are still needed there. The only unknown edge non-neighbors left are Karen and Zach, so Karen and Zach must be innocent.

15.A5 · Vince → CRIMINAL

Maria’s clue says row 5 contains an odd number of criminals. In row 5, Wanda and Xavi are already known criminals, so the current criminal count there is 2, and Vince is the only unknown person left in that row. If Vince were innocent, row 5 would stay at 2 criminals, which is even, not odd. So Vince at A5 must be criminal.

16.A4 · Peter → INNOCENT

Claire’s clue says there are exactly 5 innocents on the edge, and exactly 1 of those edge innocents is a neighbor of Ruth. Among the edge people who are neighbors of Ruth, Lisa, Vince, Wanda, and Xavi are already known criminals, so that group still needs its 1 innocent. The only unknown person left in that edge-and-neighbor group is Peter. So Peter must be innocent.