Clues by Sam May 03, 2026 Answer – Full Solution Explained

01.B2 · Frida → CRIMINAL, D2 · Ike → CRIMINAL

Katie's clue says row 2 has exactly 3 criminals, and exactly 1 of those criminals is a neighbor of Frida. In row 2, the only neighbors of Frida are Eric and Gabe, so the other 2 criminals in row 2 must be people there who are not Frida's neighbors. The only row 2 people who are not neighbors of Frida are Frida and Ike. So Frida and Ike must be criminal.

02.D5 · Zach → INNOCENT

Ike’s clue says Zach is one of the exactly 2 innocents in row 5. That directly includes Zach among the innocents in that row. So Zach must be innocent.

03.C3 · Martin → CRIMINAL

Katie’s clue makes exactly one of the row 2 criminals a neighbor of Frida, and in row 2 the only people who could fill that role are Eric and Gabe. So among Eric and Gabe, exactly one is criminal. Zach’s clue says exactly two of Lisa’s four criminal neighbors also neighbor Frida. In Lisa’s neighbor list, the people who also neighbor Frida are Eric, Gabe, Katie, and Martin, and Katie is innocent. Since exactly one of Eric and Gabe is criminal, that group can reach the required total of two criminals only if Martin is the other one. So Martin must be criminal.

04.D3 · Noah → CRIMINAL

Below Dana are Ike, Noah, Uma, and Zach, and the clue says the two criminals in that group must be connected. Ike is already a criminal and Zach is already innocent, so the only people left to make that clue work are Noah and Uma. If Noah were innocent, then the criminals below Dana could not satisfy the clue's requirement for both criminals there. So Noah must be criminal.

05.D4 · Uma → INNOCENT

Martin’s clue says that below Dana there are exactly two criminals, and those two criminals are connected. Below Dana are Ike, Noah, Uma, and Zach, with Ike and Noah already known to be criminals and Zach already known to be innocent. If Uma were also a criminal, then below Dana there would not be exactly those two criminals required by the clue. So Uma must be innocent.

06.D1 · Dana → INNOCENT

Uma’s clue says column B is the only column with exactly 2 innocents. Column D already has 2 known innocents, Uma and Zach, and Dana is the only person left in that column whose identity is not known. If Dana were criminal, then column D would remain at exactly 2 innocents, which the clue does not allow for any column except B. So Dana must be innocent.

07.C5 · Xena → CRIMINAL

Quita’s neighbors contain exactly 2 criminals, and exactly 1 of those is above Wally. That means among Quita’s neighbors who are not above Wally, there is exactly 1 criminal, and in this step those people are Vince and Wally. Row 5 contains exactly 2 criminals. In row 5, Vince and Wally are the part already accounting for 1 criminal, and the only person in row 5 outside that smaller group is Xena, since Zach is innocent. So the remaining criminal in row 5 has to be Xena. That makes Xena criminal.

08.A4 · Quita → INNOCENT

Lisa has exactly 4 criminal neighbors, and exactly 2 of those also neighbor Frida. Since Frida and Martin are already criminal and both neighbor Lisa, the other 2 criminal neighbors of Lisa must be people who do not neighbor Frida. Among Lisa's row 4 neighbors, the people who do not neighbor Frida are Samira and Terry. So the one criminal in that row 4 part has to be among Samira and Terry, not Quita. So Quita must be innocent.

09.C1 · Celia → CRIMINAL

On the edge, there are already five known innocents, so the remaining edge people can contain only two more innocents. Those two innocents must come from Anna, Bobby, Eric, Vince, and Wally. That leaves Celia outside the only two innocent spots available in that edge group. So Celia must be criminal.

10.A1 · Anna → CRIMINAL

Row 2 has exactly 3 criminals in total, and among the row 2 people who are Frida's neighbors, exactly 1 of those criminals is in that group, namely among Eric and Gabe. Bobby's neighbors must contain an odd number of innocents, and right now the only unknown neighbors there are Anna, Eric, and Gabe. If Anna were innocent, then Anna, Eric, and Gabe would have to satisfy both of those clues at the same time, but they cannot. So Anna cannot be innocent. So Anna must be criminal.

11.C2 · Gabe → CRIMINAL

12.A2 · Eric → INNOCENT

Katie’s clue says row 2 has exactly 3 criminals in total, and exactly 1 of those criminals is Frida’s neighbor. Among the row 2 people who are neighbors of Frida, Gabe is already a known criminal, so that one allowed criminal neighbor is already accounted for. Eric is the only remaining unknown person in that same row-2-and-neighbor group, so Eric cannot also be a criminal. So Eric must be innocent.

13.C4 · Terry → INNOCENT

Dana’s clue says there are exactly 4 criminals in column C. Column C already has 4 known criminals there: Celia, Gabe, Martin, and Xena, and the only person in that column not yet identified is Terry. Since all 4 criminal spots in column C are already filled, Terry cannot be a criminal. So Terry must be innocent.

14.B4 · Samira → CRIMINAL

Zach's clue says Lisa has exactly 4 criminal neighbors, and exactly 2 of those criminal neighbors also neighbor Frida. In the group of Lisa's neighbors who also neighbor Frida, the only criminals are Gabe and Martin, so the 2 criminals required there are already accounted for. That means Lisa's neighbors who do not neighbor Frida must contain exactly 2 criminals. Those people are Frida, Quita, Samira, and Terry, and among them only Frida is already a known criminal while Quita and Terry are innocent. So the remaining criminal spot has to be Samira. So Samira must be criminal.

15.B3 · Lisa → INNOCENT

Frida’s clue says Quita has exactly 2 criminal neighbors, and exactly 1 of those criminal neighbors is above Wally. Among Quita’s neighbors who are above Wally, the only people are Lisa and Samira. Samira is already a known criminal, so that one allowed criminal in this above-Wally group is already accounted for. Therefore Lisa cannot also be a criminal, so Lisa must be innocent.

16.B1 · Bobby → CRIMINAL

Quita’s clue says there are exactly 7 criminals on the edge. On the edge, 5 are already known criminals, and the only unknown edge people are Bobby, Vince, and Wally, so those three must provide the remaining 2 criminals. Frida’s clue fixes Quita’s neighboring criminals so that among Vince and Wally, exactly one is criminal. That accounts for one of the two edge criminals still needed, leaving the remaining edge person outside that pair, Bobby, to supply the last one. So Bobby must be criminal.

17.B5 · Wally → INNOCENT

Uma’s clue says column B is the only column with exactly 2 innocents. In column B, Lisa is already the one known innocent, and Wally is the only unknown person left there, so testing Wally as criminal would leave column B short of the 2 innocents that clue requires. That clashes with the clue, so Wally cannot be criminal. So Wally must be innocent.

18.A5 · Vince → CRIMINAL

Quita’s neighbors contain exactly 2 criminals, and exactly 1 of those criminals is above Wally. Among Quita’s neighbors who are above Wally, the only criminal already present is Samira, so the other criminal neighbor must come from Quita’s neighbors who are not above Wally. Those people are Katie, Vince, and Wally, and Katie and Wally are innocent. So Vince must be criminal.