Clues by Sam Jun 03, 2026 Answer – Full Solution Explained
A1
👩🎤
singer
B1
👮♀️
cop
C1
🕵️♀️
sleuth
D1
👩⚕️
doctor
A2
👨🎤
singer
B2
👮♂️
cop
C2
🕵️♀️
sleuth
D2
😬
android
A3
👮♂️
cop
B3
🕵️♂️
sleuth
C3
👩🏫
teacher
D3
👨🏫
teacher
A4
💂♀️
guard
B4
👩💼
clerk
C4
👩💼
clerk
D4
👨💼
clerk
A5
👩🎤
singer
B5
💂♂️
guard
C5
💂♂️
guard
D5
👨💻
coder
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 03, 2026 — a Medium solved in 18 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 Celia (B1), Dana (C1), Ivan (B2), Janet (C2), Kumar (D2), Martin (A3), Quita (A4), Sue (B4), Tina (C4), Umar (D4), Vicky (A5), Will (B5) and Zulu (D5); the remaining 7 suspects are innocent.
The deduction chain, in plain English
01.C4 · Tina → CRIMINAL, C2 · Janet → CRIMINAL
Olivia’s clue says exactly 1 innocent is in the overlap of the people above Xavi and the people neighboring Nick. That overlap is Janet, Olivia, and Tina, and Olivia is already known to be innocent. Since the shared group already contains its one innocent, the other two people in that group cannot be innocent. So Tina and Janet must be criminal.
02.C5 · Xavi → INNOCENT
Tina’s clue says there are exactly 2 innocents below Janet. In that group, there is already 1 known innocent, and the only person there whose status is still unknown is Xavi. That group still needs exactly 1 more innocent, so Xavi must be innocent.
03.C1 · Dana → CRIMINAL
Xavi's clue says every column has at least 3 criminals, so in column C there can be at most 2 innocents. Column C already has its 2 known innocents, and the only person there whose status is not yet fixed is Dana at C1. That means no remaining person in column C can be innocent. So Dana must be criminal.
04.A1 · Betsy → INNOCENT
Dana’s clue says the two criminals in row 1 are connected, and Dana is already one of those criminals. That means the other criminal in row 1 has to be next to Dana, so it can only be Celia or Eve. Betsy is not next to Dana in that row, so Betsy cannot be the second criminal in the connected pair. So Betsy must be innocent.
05.D4 · Umar → CRIMINAL
Betsy’s clue says every corner person can have at most one innocent neighbor. For D5 Zulu, there is already one known innocent neighbor, Xavi. That means Zulu cannot have any other innocent neighbor. Umar is one of Zulu’s unknown neighbors, so if Umar were innocent, Zulu would have more than one innocent neighbor, which breaks Betsy’s clue. So Umar must be criminal.
06.A5 · Vicky → CRIMINAL
Column A already contains Betsy, who is innocent, and that column must still end up with at least 3 criminals. So in column A there is room for at most one more innocent. Umar’s clue says an odd number of the people who are both in column A and neighbors of Nick are innocent. Those people are Fantine, Martin, and Quita, and the known innocent count there is currently even, so at least one of Fantine, Martin, and Quita must be innocent. That uses the only remaining innocent place available in column A, so Vicky must be criminal.
07.A3 · Martin → CRIMINAL
Fantine’s neighbors contain exactly 3 criminals, and exactly 2 of those criminals are also neighbors of Janet. That means exactly 1 of Fantine’s criminal neighbors is not a neighbor of Janet. Among Fantine’s neighbors who are not neighbors of Janet, the only people are Betsy and Martin. Betsy is already innocent, and that group currently has no known criminal, so the one criminal required there has to be Martin. So Martin must be criminal.
08.B4 · Sue → CRIMINAL
Celia, Ivan, and Nick are the exact-count group here: Fantine's neighbors contain exactly 2 criminals among those three people. Those same three people are also above Will, and Martin's clue says the people above Will must contain an odd number of criminals. So within the group above Will, Celia, Ivan, and Nick contribute 2 criminals, which is even, and Sue is the only person above Will outside that group. That makes Sue criminal.
09.B5 · Will → CRIMINAL
Fantine's neighbors must contain exactly three criminals, and Martin is already one of them, so the other two criminals there have to be among Celia, Ivan, and Nick. Those same three people are the column B people that overlap with Fantine's neighbors. Sue says column B has only one innocent, so column B has four criminals in total. In column B, Celia, Ivan, Nick, and Sue account for only three of those four criminals, and the only remaining person in column B is Will. So Will must be criminal.
10.B1 · Celia → CRIMINAL
Will’s clue says exactly 3 edge people have an innocent directly to the left of them. That total is already accounted for by B1 Celia, D3 Pip, and D5 Zulu, and the only unresolved direct-left neighbor still affecting this clue is Celia. If Celia were innocent, this clue would create too many counted edge people. So Celia must be criminal.
11.D1 · Eve → INNOCENT
Dana’s clue says that row 1 has exactly two criminals, and those two criminals are connected. In row 1, Celia and Dana are already the two known criminals, Betsy is innocent, and Eve is the only unknown. If Eve were criminal, row 1 would no longer have exactly two criminals, which breaks the clue. So Eve cannot be criminal. That makes Eve innocent.
12.D5 · Zulu → CRIMINAL
Row 1 has 2 innocents, and row 5 currently has 1 innocent. Eve’s clue says row 1 must have more innocents than row 5. If Zulu were innocent, then row 5 would also have 2 innocents. That would make the number of innocents in row 5 equal to row 1, not fewer, so Eve’s clue would be false. So Zulu must be criminal.
13.B3 · Nick → INNOCENT
Zulu’s clue says the people between Ivan and Will contain exactly one innocent. In that group, there are currently no known innocents, and the only person there whose identity is still unknown is Nick. Since that group still needs exactly one innocent, Nick has to be that innocent. So Nick must be innocent.
14.B2 · Ivan → CRIMINAL
Fantine’s neighbors must contain exactly 3 criminals in total. Among the Fantine-neighbors who also neighbor Janet, the relevant people are Celia, Ivan, and Nick, and exactly 2 of those are criminals. Celia is already a criminal and Nick is innocent, so that shared group still needs 1 more criminal, and Ivan is the only person there who could fill it. So Ivan must be criminal.
15.D3 · Pip → INNOCENT
Ivan’s clue says only one row can have exactly 2 innocents. Row 1 already has exactly 2 innocents, Betsy and Eve. Row 3 also already has 2 known innocents, Nick and Olivia, and Pip is the only person in that row not yet identified. If Pip were criminal, row 3 would also have exactly 2 innocents, which would clash with Ivan’s clue. So Pip must be innocent.
16.D2 · Kumar → CRIMINAL
Xavi's clue says every column has at least 3 criminals, so in column D there can be at most 2 innocents. Column D already has 2 known innocents, Eve at D1 and Pip at D3. The only person in that column whose status was not yet fixed is Kumar at D2, so Kumar cannot also be innocent. So Kumar must be criminal.
17.A2 · Fantine → INNOCENT
Pip's clue says rows 2 and 5 have the same number of innocents. Row 5 already has 1 innocent, while row 2 currently has none. The only person in row 2 whose identity is not fixed yet is Fantine, so if Fantine were criminal, row 2 would stay at 0 innocents and could not match row 5's 1. So Fantine must be innocent.
18.A4 · Quita → CRIMINAL
Xavi’s clue says every column has at least 3 criminals, so in column A there can be at most 2 innocents. Column A already has 2 known innocents, Betsy and Fantine. The only person in column A whose status is not yet fixed is Quita, so she cannot also be innocent. So Quita must be criminal.