Clues by Sam Oct 27, 2026 Answer – Full Solution Explained

01.A1 · Anna → CRIMINAL

Daniel’s clue directly says that Anna is one of the exactly 2 criminals in column A. That identifies Anna herself as criminal. So Anna must be criminal.

02.A2 · Eli → INNOCENT

If Eli were criminal, then column A would already have its two criminals: Anna and Eli, because Daniel says Anna is one of exactly 2 criminals in column A. That means Joy, Quita, and Uday would all have to be innocent. But Anna also says there is an odd number of criminals below Eli, and the people below Eli are exactly Joy, Quita, and Uday. If all three of them are innocent, that gives 0 criminals below Eli, which is not odd. So Eli must be innocent.

03.B4 · Rob → CRIMINAL

Eli's clue says Sue has exactly 2 criminal neighbors, and exactly 1 of those criminals is to Sue's left. Among Sue's neighbors, the only person who is to Sue's left is Rob. So the one criminal neighbor to Sue's left has to be Rob. That makes Rob criminal.

04.B3 · Kim → INNOCENT

Rob’s clue says every profession has at least one innocent. In the judge profession, Rob is already a known criminal, so the judges can have at most one criminal if there still has to be at least one innocent judge. The only other judge is Kim. So Kim at B3 must be innocent.

05.B1 · Bella → INNOCENT, B5 · Xavi → INNOCENT, B2 · Gary → INNOCENT

Kim’s clue says column B is the only column with exactly 4 innocents. In column B, Kim is already innocent and Rob is already criminal, so the other three people in that column have to supply the remaining three innocents needed to make column B reach exactly 4 innocents. That means none of Bella, Gary, or Xavi can be criminal. So Bella, Gary, and Xavi must be innocent.

06.C1 · Chad → CRIMINAL

Gary’s clue says there are exactly 2 innocents to the right of Anna. That group already has 2 known innocents, Bella and Daniel. The only person there whose identity is still unknown is Chad, so Chad cannot also be innocent. So Chad must be criminal.

07.C4 · Sue → INNOCENT

Chad’s clue says Xavi and Yasmin have exactly one innocent neighbor in common. Their shared neighbors are only Rob and Sue, and Rob is already criminal. That means the shared pair still needs one innocent, and Sue is the only person there who can fill it. So Sue must be innocent.

08.D4 · Taylor → INNOCENT

Sue’s clue says she has exactly 2 criminal neighbors, and exactly 1 of those criminals is to Sue’s left. The only neighbor of Sue who is to her left is Rob, and Rob is already that one criminal. So if Taylor were criminal, then the other unknown neighbors named here cannot also be criminals, or Sue would have more than 2 criminal neighbors. That makes Nancy, Petra, Yasmin, and Zulu innocent. But then below Daniel the innocents would be Petra at D3 and Zulu at D5, with Taylor at D4 criminal between them, so those innocents would not be one connected group. That contradicts Sue’s clue about all innocents below Daniel being connected. So Taylor at D4 must be innocent.

09.A5 · Uday → INNOCENT

Assume Uday were criminal. Then the edge clue about exactly 3 edge people having a criminal directly to their left, together with the edge cells listed here, leads to Helen, Nancy, and Yasmin being innocent. But then column C would have exactly 4 innocents: Chad is criminal, while Helen, Nancy, Sue, and Yasmin would be innocent. Kim’s clue says column B is the only column with exactly 4 innocents, so column C cannot also have 4 innocents. That contradiction means Uday cannot be criminal. So Uday must be innocent.

10.D5 · Zulu → INNOCENT

Uday’s clue says there are more criminal judges than criminal teachers. The judges already have 1 criminal, while the teachers currently have 0 criminals. Since Zulu is the only teacher not yet identified, Zulu cannot be a criminal, because that would make the number of criminal teachers 1 and no longer less than the number of criminal judges. So Zulu must be innocent.

11.A4 · Quita → INNOCENT

Zulu’s clue says there are exactly 2 innocents below Joy. Below Joy there is already 1 known innocent, Uday, and the only person there whose status is still unknown is Quita. That group still needs exactly 1 more innocent, so Quita has to supply it. So Quita must be innocent.

12.A3 · Joy → CRIMINAL

Daniel says Anna is one of exactly 2 criminals in column A. In that column, Anna is already a criminal, while Eli, Quita, and Uday are innocent, so the only other person there who can fill the second criminal spot is Joy. So Joy must be criminal.

13.D3 · Petra → INNOCENT, D2 · Ivan → INNOCENT

14.C5 · Yasmin → CRIMINAL

Petra’s clue says the number of criminal cops and criminal vampires is the same. The cops already have 1 criminal, while the vampires currently have 0 known criminals. Since Yasmin is the only vampire whose identity is not yet known, she has to supply that one criminal vampire. So Yasmin must be criminal.

15.C3 · Nancy → INNOCENT

Eli's clue says Sue has exactly 2 criminal neighbors, and exactly 1 of those criminals is to Sue's left. The criminal neighbor to Sue's left is already identified as Rob. That means the other criminal neighbor must be among Sue's neighbors who are not to her left, and that group already contains Yasmin as a known criminal. Since Nancy is the only unknown person among Sue's neighbors, Nancy cannot also be a criminal. So Nancy must be innocent.

16.C2 · Helen → INNOCENT

Taylor’s clue says exactly 3 edge people have a criminal directly to their left. Those 3 are already Bella at B1, Daniel at D1, and Zulu at D5. The only remaining person who could change that count is Helen at C2, because if Helen were criminal then Ivan at D2 would also be an edge person with a criminal directly to the left, raising the total above 3. So Helen must be innocent.