Clues by Sam Jun 26, 2026 Answer – Full Solution Explained
A1
💂♂️
guard
B1
👩🍳
cook
C1
💂♂️
guard
D1
💂♂️
guard
A2
👨🌾
farmer
B2
👩🎤
singer
C2
👨🎤
singer
D2
👨🎤
singer
A3
👩🌾
farmer
B3
👩🎨
painter
C3
🧛♂️
vampire
D3
🧛♀️
vampire
A4
👩⚕️
doctor
B4
👩🍳
cook
C4
👮♀️
cop
D4
👮♂️
cop
A5
👨🍳
cook
B5
👩⚕️
doctor
C5
👮♂️
cop
D5
👩⚕️
doctor
Final Board State
This puzzle is fully solved.
All characters have been identified as innocent or criminal based on today's clues.
See how each clue leads to the final result
Skip the reasoning — 5 criminals.
Clues by Sam answer for Jun 26, 2026 — a Tricky solved in 16 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 5 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Bonnie (B1), Flora (B2), Isaac (D2), Joy (A3) and Martin (C3); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.A3 · Joy → CRIMINAL
Andre’s clue says row 3 has exactly 2 criminals, and exactly 1 of those criminals is a neighbor of Gabe. That means the other 1 criminal in row 3 must be someone in row 3 who is not a neighbor of Gabe. Among the people in row 3, Gabe’s neighbors there are Kay, Martin, and Petra, so the only person in row 3 who is not a neighbor of Gabe is Joy. Since that non-neighbor part of row 3 still needs 1 criminal, Joy must fill it. So Joy must be criminal.
02.B3 · Kay → INNOCENT
Joy’s clue says that Kay is one of the two innocents above Wanda. That identifies Kay directly as an innocent. So Kay must be innocent.
03.C5 · Xavi → INNOCENT, D5 · Zoe → INNOCENT
Kay’s clue says exactly 2 of the 3 cops have an innocent directly below them. Sofia has Xavi directly below her, and Thor has Zoe directly below him. Those are the only remaining cop-and-person-below cases available to satisfy the clue, so the people below those two cops have to be innocent. So Xavi and Zoe must be innocent.
04.B5 · Wanda → INNOCENT
Joy’s clue says Kay is one of exactly 2 innocents above Wanda. Among the people above Wanda, Kay is already the only known innocent, so that group must end up with exactly one more innocent among Bonnie, Flora, and Rose. Zoe’s clue also requires every column to have at least 3 innocents. If Wanda were a criminal, then Bonnie, Flora, Rose, and the other people involved would have to satisfy both of those requirements at the same time, but they cannot. So Wanda must be innocent.
05.A5 · Umar → INNOCENT
Xavi’s clue says row 1 is the only row with exactly 3 innocents. Row 5 already has 3 known innocents: Wanda, Xavi, and Zoe, and Umar is the only person left in that row whose status is not known. If Umar were criminal, row 5 would remain at exactly 3 innocents, which the clue does not allow for any row except row 1. So Umar must be innocent.
06.B2 · Flora → CRIMINAL
Xavi’s clue fixes row 1 as the only row with exactly 3 innocents, and Wanda’s clue fixes that among the guards Andre, Carl, and Derek, exactly one has a criminal directly below. If Flora were innocent, then Bonnie, Carl, Derek, Eli, Gabe, Isaac, Martin, Petra, Quita, Rose, Sofia, and Thor would have to satisfy both of those requirements at the same time. That cannot be done without breaking either the row 1 requirement or the guard requirement. So Flora at B2 must be criminal.
07.D1 · Derek → INNOCENT
The people to the left of Derek contain exactly 2 innocents, and row 1 contains exactly 3 innocents. In row 1, the only person who is not among those people to the left of Derek is Derek himself. So once the 2 innocent spots to Derek's left are accounted for, the remaining innocent required for row 1 has to be Derek. That makes Derek innocent.
08.A2 · Eli → INNOCENT
Row 1 must have exactly 3 innocents, and Flora’s edge neighbors must contain an odd number of innocents. The relevant edge neighbors of Flora are Andre, Bonnie, Carl, Eli, and Joy, with Andre already innocent and Joy already criminal. If Eli were criminal, then Bonnie and Carl would be the only row 1 unknowns available to bring row 1 up from its current 2 innocents to exactly 3, while also keeping Flora’s edge-neighbor count odd. But with Eli set to criminal, the remaining people involved in these clues cannot satisfy all of those requirements at the same time. So Eli at A2 must be innocent.
09.A4 · Quita → INNOCENT
Joy’s clue fixes the people above Wanda so that Kay is one of exactly two innocents in that group of Bonnie, Flora, Kay, and Rose. Eli’s clue also says that only one column can have exactly two criminals. If Quita were a criminal, then the remaining unknown people involved here, including Bonnie and Rose in Wanda’s column and the unknowns in columns C and D, could not satisfy both of those restrictions at the same time. That opposite choice makes the clues conflict. So Quita must be innocent.
10.B4 · Rose → INNOCENT, C4 · Sofia → INNOCENT, D4 · Thor → INNOCENT
Row 1 has to be the only row with exactly 3 innocents, and every column has to have at least 3 innocents. The rows that must fit around that are row 2, row 3, row 4, and row 5, and the columns that must fit around that are columns A, B, C, and D. If Rose, Sofia, and Thor were all criminals, then Bonnie, Carl, Gabe, Isaac, Martin, and Petra would have to make all of those row and column requirements come out correctly at the same time. But with Rose, Sofia, and Thor all criminal, those same remaining people cannot satisfy both the “only row 1 has exactly 3 innocents” condition and the “each column has at least 3 innocents” condition. So Rose, Sofia, and Thor cannot be criminals. That makes Rose, Sofia, and Thor innocent.
11.B1 · Bonnie → CRIMINAL
Joy’s clue says Kay is one of exactly 2 innocents above Wanda. Above Wanda are Bonnie, Flora, Kay, and Rose, and among them Kay and Rose are already innocent while Flora is criminal. So the 2 innocent places above Wanda are already filled by Kay and Rose. If Bonnie were innocent too, there would be 3 innocents above Wanda, which conflicts with the clue. So Bonnie must be criminal.
12.C1 · Carl → INNOCENT
Xavi’s clue says row 1 is the only row with exactly 3 innocents. In row 1, Andre and Derek are already innocent and Bonnie is criminal, so Carl is the only person who could supply the third innocent there. If Carl were criminal, then row 1 would not have exactly 3 innocents, and the remaining unknown people tied to this clue, Gabe, Isaac, Martin, and Petra, could not make all of the row facts fit at the same time. So Carl cannot be criminal. That makes Carl innocent.
13.C3 · Martin → CRIMINAL
Petra’s neighbors must contain an odd number of innocents. Right now Sofia and Thor are already innocent there, so Petra’s neighbor group has 2 known innocents, and the unknown neighbors are Gabe, Isaac, and Martin. If Martin were innocent, then Gabe and Isaac would be the remaining people that have to make both clues fit. But with Martin counted as an innocent neighbor of Petra, those remaining placements for Gabe and Isaac cannot satisfy the clues at the same time. So Martin must be criminal.
14.D3 · Petra → INNOCENT
Andre's clue says row 3 has exactly 2 criminals, and exactly 1 of those criminals is Gabe's neighbor. Among the people in row 3 who are Gabe's neighbors, C3 Martin is already a known criminal. That means the one criminal neighbor from row 3 is already accounted for, so the remaining unknown person in that same group cannot also be a criminal. So Petra must be innocent.
15.C2 · Gabe → INNOCENT
Eli says only one column has exactly 2 criminals. Column B already has exactly 2 criminals, Bonnie and Flora, with Kay, Rose, and Wanda innocent. Column C currently has 1 known criminal, Martin, and Gabe is the only person there not yet identified. If Gabe were a criminal, then column C would also have exactly 2 criminals, which would clash with Eli's clue that only one column does. So Gabe must be innocent.
16.D2 · Isaac → CRIMINAL
Xavi’s clue says row 1 is the only row with exactly 3 innocents. Row 2 already has 2 known innocents, and Isaac is the only person left in that row whose identity is not yet known. If Isaac were innocent, then row 2 would also have exactly 3 innocents, which would break the clue. So Isaac must be criminal.