Clues by Sam May 02, 2026 Answer – Full Solution Explained
A1
💂♀️
guard
B1
👩🎤
singer
C1
👮♀️
cop
D1
👨🎤
singer
A2
💂♂️
guard
B2
👮♀️
cop
C2
👮♀️
cop
D2
👨🎤
singer
A3
👩🔧
mech
B3
👨🔧
mech
C3
🕵️♂️
sleuth
D3
👩🏫
teacher
A4
👩🔧
mech
B4
🕵️♀️
sleuth
C4
👨🏫
teacher
D4
💂♂️
guard
A5
🕵️♀️
sleuth
B5
👨✈️
pilot
C5
👩✈️
pilot
D5
👨🏫
teacher
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 — 11 criminals.
Clues by Sam answer for May 02, 2026 — a Hard solved in 16 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 11 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chloe (C1), Gladys (B2), Kay (C2), Mark (D2), Olsi (B3), Peter (C3), Quita (D3), Ruby (A4), Stella (B4), Xia (C5) and Ziad (D5); the remaining 9 suspects are innocent.
The deduction chain, in plain English
01.A5 · Vera → INNOCENT, C5 · Xia → CRIMINAL, D5 · Ziad → CRIMINAL
Wally’s clue says that the two criminals in row 5 must be connected. Row 5 is Vera, Wally, Xia, and Ziad, and Wally is innocent. If Vera were criminal while Xia and Ziad were innocent, then row 5 would not have the connected pair of criminals that the clue requires. So the opposite pattern cannot be right. That makes Vera innocent, Xia criminal, and Ziad criminal.
02.A1 · Amy → INNOCENT
Ziad says Frank has exactly 2 criminal neighbors, and only 1 of those 2 is also Chloe's neighbor. So among Frank's neighbors who are not Chloe's neighbors, there is exactly 1 criminal. That group is A1 Amy, A3 Nicole, and B3 Olsi, and that 1 criminal has to be Nicole or Olsi, not Amy. So Amy must be innocent.
03.C3 · Peter → CRIMINAL
Frank has exactly 2 criminal neighbors, and only 1 of them is Chloe's neighbor. Among Frank's neighbors, the ones who are not Chloe's neighbors are exactly Nicole and Olsi, so exactly 1 of Nicole and Olsi is criminal. Stella has exactly 4 criminal neighbors, and exactly 2 of those are in row 3. The row 3 neighbors of Stella are Nicole, Olsi, and Peter, so exactly 2 of Nicole, Olsi, and Peter are criminal. Since Nicole and Olsi contribute exactly 1 criminal there, the remaining criminal in Stella's row 3 neighbor group has to be Peter. So Peter is criminal.
04.D1 · Denis → INNOCENT
Vera's clue says row 1 has exactly one criminal, so row 1 needs exactly one criminal among Bunty, Chloe, and Denis. Peter's clue says the row 1 neighbors of Gladys contain exactly one criminal, and in that group the only unknowns are Bunty and Chloe, so exactly one of Bunty and Chloe is the criminal in row 1. Denis is in row 1 but not in that group, so the one criminal row 1 needs is already accounted for without him. So Denis must be innocent.
05.D3 · Quita → CRIMINAL
Row 3 already has only 1 known criminal, while row 5 has 2 criminals, and Denis says row 3 has more criminals than row 5. So row 3 must end up with at least 3 criminals. Xia also says Stella has exactly 4 criminal neighbors, and exactly 2 of those criminal neighbors are in row 3. In Stella's neighboring row 3 spaces, Peter is already one of those criminals, so the remaining row 3 criminal neighbors are limited by that clue. If Quita were innocent, then Nicole, Olsi, Ruby, and Tyler would have to make both clues true at the same time, but they cannot. So Quita cannot be innocent. That makes Quita criminal.
06.A2 · Frank → INNOCENT
Bunty and Gladys are the two people who are both neighbors of Frank and neighbors of Chloe, and Ziad's clue says exactly one of those two is a criminal. Amy's neighbors are Bunty, Frank, and Gladys, and Quita's clue says Amy has an odd number of criminal neighbors. Since Bunty and Gladys already contribute exactly one criminal among Amy's neighbors, Frank cannot be a criminal too. So Frank is innocent.
07.C2 · Kay → CRIMINAL
Stella’s clue fixes the row 3 neighbors of Stella at exactly 2 criminals, and those people are Nicole, Olsi, and Peter. Since Peter is already a criminal, that means Nicole and Olsi together supply exactly 1 more criminal there. Gladys’s clue fixes her neighbors outside row 1 at exactly 3 criminals, and that group is Frank, Kay, Nicole, Olsi, and Peter. Frank is innocent and Peter is already a criminal, so besides Peter this group needs 2 more criminals. But Nicole and Olsi together account for only 1 of those, leaving the remaining person in that group, Kay, to supply the last one. So Kay must be criminal.
08.D2 · Mark → CRIMINAL
Frank’s clue says he has exactly 2 criminal neighbors, and exactly 1 of those 2 is in the shared pair with Chloe, Bunty and Gladys. So among Frank’s other neighbors, A3 Nicole and B3 Olsi, exactly 1 must also be criminal. Chloe’s clue says she has an odd number of criminal neighbors. She already has 1 known criminal neighbor, Kay, and the only other people involved here are Bunty, Gladys, and Mark. If Mark were innocent, then Chloe’s remaining criminal neighbors would have to come only from Bunty and Gladys, but Frank’s clue allows exactly 1 criminal in that shared pair, making Chloe’s total 2 criminals, which is even, not odd. So Mark must be criminal.
09.A4 · Ruby → CRIMINAL
Vera’s clue says row 1 is the only row with exactly one criminal, and Mark’s clue says exactly two innocents are shared by Peter’s neighbors and Xia’s neighbors, namely Stella, Tyler, and Umar. If Ruby were innocent, then Bunty, Chloe, Gladys, Nicole, Olsi, Stella, Tyler, and Umar would have to make both of those clues true at the same time. But that combination cannot satisfy both requirements together. So Ruby at A4 must be criminal.
10.C4 · Tyler → INNOCENT
Xia's clue says Stella has exactly 4 criminal neighbors, and exactly 2 of those criminals are in row 3. Among Stella's neighbors outside row 3, the people are Ruby, Tyler, Vera, Wally, and Xia, and Ruby and Xia are already known criminals. That already fills the 2 criminal-neighbor spots outside row 3, so the only unknown person there cannot also be a criminal. So Tyler must be innocent.
11.D4 · Umar → INNOCENT
Tyler says exactly 2 criminals in row 3 have an innocent directly below them. In row 3, the known criminals are Peter and Quita. If Umar were criminal, then Quita would not have an innocent directly below her, so the two required row 3 criminals would both have to come from Peter together with Nicole or Olsi. But Nicole and Olsi are the only other people in row 3 whose identities are not fixed here, and that cannot satisfy the clue. So Umar must be innocent.
12.B4 · Stella → CRIMINAL
Vera’s clue says row 1 is the only row with exactly one criminal, so no other row can finish with exactly one criminal. Row 4 already has 1 known criminal, and Stella is the only person in that row not yet identified. If Stella were innocent, row 4 would remain at exactly 1 criminal, which the clue forbids. So Stella must be criminal.
13.A3 · Nicole → INNOCENT, B3 · Olsi → CRIMINAL
Frank’s clue says his neighbors contain exactly two criminals, and exactly one of those criminal neighbors is also Chloe’s neighbor, so among Bunty and Gladys exactly one must be criminal. Amy’s clue says column B has an odd number of criminals, and column B already has one known criminal, Stella. If Nicole were criminal and Olsi were innocent, then Bunty and Gladys would have to make both clues work, but that cannot be done: Frank’s clue requires exactly one criminal among Bunty and Gladys, while column B would also need an even number of criminals from Bunty, Gladys, and Olsi, and with Olsi innocent that means Bunty and Gladys together would also have to be even. Those requirements clash, so Nicole cannot be criminal while Olsi is innocent. So Nicole must be innocent and Olsi must be criminal.
14.C1 · Chloe → CRIMINAL
Frank’s clue fixes his neighborhood very tightly: Frank has exactly 2 criminal neighbors in total, and exactly 1 of those criminals is also Chloe’s neighbor. The only unknown people involved in that part are Bunty and Gladys, because they are the shared neighbors of Frank and Chloe. Now test Chloe as innocent. Then Bunty and Gladys would have to make Frank’s clue come out right while also fitting Kay’s clue that only one column has exactly one innocent, but those same people cannot satisfy both requirements at once. So Chloe must be criminal.
15.B1 · Bunty → INNOCENT
Vera’s clue says row 1 is the only row with exactly one criminal. In row 1, Chloe is already the one known criminal, and Bunty is the only person there not yet identified. If Bunty were criminal, then row 1 would no longer have exactly one criminal, which clashes with Vera’s clue. So Bunty cannot be criminal. That makes Bunty innocent.
16.B2 · Gladys → CRIMINAL
Frank has exactly 2 criminal neighbors. Among Frank's neighbors, exactly 1 of the criminals must also be a neighbor of Chloe, and in that shared group the people are Bunty and Gladys. Bunty is innocent, so that shared group still needs 1 criminal, and Gladys is the only person there who could fill it. So Gladys must be criminal.