Clues by Sam Jun 02, 2026 Answer – Full Solution Explained
A1
👮♂️
cop
B1
👨🎨
painter
C1
👩🏫
teacher
D1
👮♂️
cop
A2
👮♂️
cop
B2
🕵️♀️
sleuth
C2
👨🌾
farmer
D2
👨🎨
painter
A3
👷♀️
builder
B3
💂♂️
guard
C3
👩🌾
farmer
D3
👨💻
coder
A4
👷♀️
builder
B4
👩🎤
singer
C4
👩🎤
singer
D4
💂♀️
guard
A5
👨🏫
teacher
B5
👨🎤
singer
C5
👩💻
coder
D5
🕵️♀️
sleuth
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 — 13 criminals.
Clues by Sam answer for Jun 02, 2026 — a Medium solved in 17 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 13 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Berat (B1), Celia (C1), Daniel (D1), Habiba (B2), Jason (D2), Martin (B3), Quita (A4), Samira (B4), Tina (C4), Uma (D4), Wally (B5), Xena (C5) and Zoe (D5); the remaining 7 suspects are innocent.
The deduction chain, in plain English
01.C5 · Xena → CRIMINAL
Karen’s clue says that Xena is one of the exactly 3 criminals in column C. Since Xena is explicitly included among those criminals, her identity is fixed by the clue itself. So Xena must be criminal.
02.B1 · Berat → CRIMINAL, B2 · Habiba → CRIMINAL
Xena’s clue says Celia has exactly 4 criminal neighbors, and exactly 2 of those criminals are also neighbors of Alex. The only people who are both neighbors of Celia and neighbors of Alex are Berat and Habiba. Since that shared group still needs 2 criminals, both of those people have to fill those 2 spots. So Berat and Habiba must be criminal.
03.C3 · Nicole → INNOCENT
Below Celia are Ike, Nicole, Tina, and Xena, and Habiba’s clue says there are exactly two innocents there and those two must be connected. Xena is already a criminal, so the only candidates for those two innocents are Ike, Nicole, and Tina. If Nicole were a criminal, then Ike and Tina would have to be the two innocents, but that cannot satisfy the clue. So Nicole at C3 must be innocent.
04.B5 · Wally → CRIMINAL, D5 · Zoe → CRIMINAL
Nicole’s clue says there are no innocents to the right of Vince, so that group must contain exactly 0 innocents. That group already has 0 known innocents, and the only people there not yet identified are Wally and Zoe. Since no innocents are allowed in that group, neither Wally nor Zoe can be innocent. So Wally and Zoe must be criminal.
05.C1 · Celia → CRIMINAL
Karen’s clue says column C has exactly 3 criminals. In that column, Nicole is already innocent and Xena is already criminal, so among Celia, Ike, and Tina there can be only 1 innocent. Habiba’s clue says both innocents below Celia are connected. The people below Celia are Ike, Nicole, Tina, and Xena, and Nicole is already one innocent there while Xena is criminal. That means the other innocent below Celia must be either Ike or Tina, so the single innocent among Celia, Ike, and Tina has to come from Ike or Tina. So Celia cannot be that innocent. That makes Celia criminal.
06.A1 · Alex → INNOCENT
Wally’s clue says there is exactly one innocent to the left of Daniel. Among the people there, there are currently no known innocents, and the only person there whose identity is still unknown is Alex. So the one innocent in that group has to be Alex. That makes Alex innocent.
07.A2 · Floyd → INNOCENT
Alex's clue says the people between Alex and Quita contain exactly 2 innocents. In that group, there is already 1 known innocent. The only person there whose identity is still unknown is Floyd, so the group needs Floyd to supply the second innocent. So Floyd must be innocent.
08.D1 · Daniel → CRIMINAL
Xena's clue says Celia's neighbors contain exactly 4 criminals, so among Celia's five neighbors there is exactly 1 innocent. The two neighbors who also neighbor Alex are Berat and Habiba, and both are already criminals, so that one innocent cannot be one of them. That leaves the single innocent among Celia's neighbors to be either Ike or Jason, not Daniel. So Daniel must be criminal.
09.A5 · Vince → INNOCENT
Daniel’s clue says rows 1 and 5 must have the same number of criminals. Row 1 already has exactly 3 criminals, and row 5 already also has 3 criminals: Wally, Xena, and Zoe. If Vince were a criminal too, then row 5 would have 4 criminals, which would no longer match row 1. So Vince must be innocent.
10.A4 · Quita → CRIMINAL
Vince’s clue says column A contains exactly one criminal. In column A, there are currently no known criminals, and Alex, Floyd, Karen, and Vince are all already innocent. That leaves Quita as the only person in column A not yet identified, so the one criminal required in that column has to be Quita. So Quita must be criminal.
11.B3 · Martin → CRIMINAL
Quita’s clue says Martin is one of Tina’s 6 criminal neighbors. Since the clue directly places Martin among Tina’s criminal neighbors, Martin must be criminal. So Martin must be criminal.
12.B4 · Samira → CRIMINAL
Martin says he has exactly 4 criminal neighbors, so among his three unknown neighbors C2 Ike, B4 Samira, and C4 Tina, exactly one must be innocent. Habiba’s clue places that one innocent below Celia, and within Martin’s unknown neighbors the people below Celia are only Ike and Tina. So the single innocent among those three must be Ike or Tina, not Samira. That makes Samira criminal.
13.D3 · Oscar → INNOCENT
Celia’s clue fixes her neighborhood very tightly: she has exactly 4 criminal neighbors, and exactly 2 of those criminals are neighbors of Alex, namely Berat and Habiba. So among Celia’s neighbors, the other 2 criminals must come from Daniel, Ike, and Jason. At the same time, Samira’s clue says exactly one row has exactly 2 criminals across rows 1 through 5. If Oscar were criminal, then the remaining unknown people involved here, Ike, Jason, Tina, and Uma, would have to fit both Celia’s neighbor count and that row-count condition at once, and they cannot do that. So Oscar cannot be criminal. That makes Oscar innocent.
14.D4 · Uma → CRIMINAL
Quita’s clue says Martin is one of Tina’s exactly 6 criminal neighbors. Among Tina’s neighbors, Martin, Samira, Wally, Xena, and Zoe are already known criminals, while Nicole and Oscar are innocent, so the only neighbor not yet identified is Uma. If Uma were innocent, Tina would have only 5 criminal neighbors, not 6. So Uma must be criminal.
15.D2 · Jason → CRIMINAL
Below Celia, the two innocents have to be connected, and among C2 Ike, C3 Nicole, C4 Tina, and C5 Xena we already know Nicole is innocent and Xena is criminal. Oscar’s clue says his neighbors contain an odd number of criminals, and among those neighbors there is already 1 known criminal, with only Ike, Jason, and Tina undecided. If Jason were innocent, then Ike and Tina would have to satisfy both clues together, but they cannot do that. So Jason cannot be innocent. That makes Jason criminal.
16.C2 · Ike → INNOCENT
Celia’s neighbors contain exactly 4 criminals in total. The clue says exactly 2 of those criminal neighbors also neighbor Alex, and those two are already Berat and Habiba. That leaves Daniel, Ike, and Jason as the neighbors of Celia who do not neighbor Alex, and Daniel and Jason already account for the 2 criminals allowed in that group. So Ike must be innocent.
17.C4 · Tina → CRIMINAL
Habiba’s clue says there are exactly two innocents below Celia, and those two innocents are connected. Below Celia we have Ike and Nicole, who are already innocent, Tina, whose status is unknown, and Xena, who is criminal. If Tina were innocent, there would be three innocents below Celia instead of both innocents being Ike and Nicole. So Tina at C4 must be criminal.