Clues by Sam May 09, 2026 Answer – Full Solution Explained

01.C5 · Xena → CRIMINAL, C1 · Chad → CRIMINAL

Habiba’s clue says there are no innocents in the people who are both in column C and on the edge. That shared group is exactly Chad at C1 and Xena at C5, and it already contains 0 known innocents. So Chad and Xena cannot be innocent. That makes Chad and Xena criminal.

02.A1 · Austin → CRIMINAL, A3 · Jose → CRIMINAL

Frank’s neighbors contain exactly 4 criminals, and exactly 2 of those criminals are above Vera. That means the other 2 criminals among Frank’s neighbors must be people who are not above Vera. In Frank’s neighborhood, the only people not above Vera are Austin and Jose, and that group currently has no known criminals in it. So Austin and Jose must be criminal.

03.B4 · Rohan → CRIMINAL

Frank’s neighbors contain exactly 1 innocent, and among the people above Vera those same three people are Barnie, Gabe, and Mary. So Barnie, Gabe, and Mary already account for an odd number of innocents above Vera: exactly 1. The only person above Vera who is not in that exact-count group is Rohan, so if Rohan were innocent, the number of innocents above Vera would become 2, which is not odd. So Rohan must be criminal.

04.A2 · Frank → INNOCENT, B3 · Mary → CRIMINAL

Rohan’s clue says exactly 2 of the 3 doctors have an innocent directly above them. The doctors are Gabe, Jose, and Mary. If Frank were criminal and Mary were innocent, then among the people this clue still depends on, only Gabe is left to make that doctor clue work. But that cannot satisfy the requirement about the doctors. So Frank and Mary cannot have those opposite identities. Therefore Frank must be innocent and Mary must be criminal.

05.B5 · Vera → INNOCENT

Frank’s clue says there are exactly 4 criminals among Frank’s neighbors, and exactly 2 of those neighboring criminals are above Vera. In the overlap relevant to that clue, the people above Vera who are also Frank’s neighbors are Barnie, Gabe, and Mary, with Mary already a criminal. Frank’s other clue says column B has an odd number of criminals. Column B already has 2 known criminals, and the unknown people there are Barnie, Gabe, and Vera. If Vera were a criminal, then Barnie and Gabe would have to make both clues work at the same time, but they cannot. So Vera must be innocent.

06.A5 · Uma → INNOCENT

07.C3 · Nala → CRIMINAL, C4 · Susan → CRIMINAL

08.D5 · Zach → INNOCENT

The two clues both point at Zach and Olga. Terry must have at least one innocent neighbor, but Terry’s neighbors currently have no known innocent, so that innocent would have to be either Olga or Zach. At the same time, the coder group must contain more criminals than the farmer group, with coders currently at 1 known criminal and farmers at 0, so Olga and Zach cannot be assigned in a way that makes Zach criminal and still satisfy that comparison. If Zach were criminal, Olga would be the only possible innocent neighbor for Terry, and those same two people would also have to keep the coder total higher than the farmer total. That cannot be done, so Zach cannot be criminal. That makes Zach innocent.

09.D1 · Diane → INNOCENT

Zach’s clue says exactly 4 edge people have an innocent directly to their right. Among the edge people involved here, if Diane were criminal, then Barnie, Ike, Olga, Quita, and Terry would have to supply the rest of that clue’s required pattern along with the edge people already known. But making Diane criminal leaves those same people unable to satisfy that exact total of 4 on the edge. So Diane at D1 must be innocent.

10.D4 · Terry → INNOCENT

Row 1 and row 4 must have the same number of innocents. Row 1 already has 1 known innocent, while row 4 has none yet, so row 4 needs an innocent among Quita and Terry. At the same time, Frank's neighbors must contain exactly 4 criminals with exactly 2 of those criminals above Vera, and Mary's neighbors must contain exactly 5 criminals. The unknown people involved across those requirements are Barnie, Gabe, and Quita. If Terry were criminal, then Barnie, Gabe, and Quita would have to satisfy those clue counts while row 4 still matched row 1's innocent count, and those requirements cannot all be met together. So Terry at D4 must be innocent.

11.D2 · Ike → CRIMINAL

Frank has exactly 4 criminal neighbors, and among Frank's neighbors exactly 2 criminals are above Vera. In that above-Vera part, Mary is already one criminal, so Barnie and Gabe cannot both be criminals, because that would make 3 criminals there instead of 2. Chad's neighbors must contain an odd number of innocents. Chad already has 2 known innocents there, so among Barnie, Gabe, and Ike the number of innocents must be odd. If Ike were innocent, then Barnie and Gabe would also have to contribute an odd number of innocents between them, but from Frank's clue they cannot both be criminals, so that combination cannot satisfy both clues at once. So Ike must be criminal.

12.D3 · Olga → CRIMINAL

Frank’s clue fixes Frank’s neighbors at exactly 4 criminals. Since 3 of Frank’s neighbors are already known criminals, Barnie and Gabe account for the remaining 1 criminal among those two spots. Ike’s clue says Habiba has exactly 2 innocent neighbors, so Habiba’s neighbors contain exactly 6 criminals. In Habiba’s neighborhood, the people shared with Frank’s neighborhood are Barnie, Gabe, and Mary, and the only person in Habiba’s neighborhood beyond that shared part who matters here is Olga. Since Barnie and Gabe provide only 1 more criminal there, the remaining criminal needed in Habiba’s neighborhood has to be Olga. So Olga must be criminal.

13.A4 · Quita → CRIMINAL, B2 · Gabe → INNOCENT

14.B1 · Barnie → CRIMINAL

Frank’s neighbors contain exactly 4 criminals, and among Frank’s neighbors who are above Vera, exactly 2 must be criminals. In that above-Vera part of Frank’s neighborhood, Mary is already a known criminal and Gabe is innocent, so that group still needs 1 more criminal. The only unknown person left in that group is Barnie. So Barnie must be criminal.