Clues by Sam Jun 07, 2026 Answer – Full Solution Explained
A1
💂♂️
guard
B1
🕵️♂️
sleuth
C1
👨⚕️
doctor
D1
👩💼
clerk
A2
🕵️♀️
sleuth
B2
👨🏫
teacher
C2
👩🌾
farmer
D2
👨🔧
mech
A3
👨⚖️
judge
B3
👩⚖️
judge
C3
👩🏫
teacher
D3
👩🎤
singer
A4
👨⚕️
doctor
B4
👩⚕️
doctor
C4
👩🔧
mech
D4
👩🎤
singer
A5
👩💼
clerk
B5
💂♂️
guard
C5
👨🌾
farmer
D5
👨🎤
singer
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 07, 2026 — a Brutal solved in 15 steps
Today's Clues by Sam puzzle is rated Brutal and resolves with 5 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Andre (A1), Ghani (B2), Isaac (D2), Nala (B3) and Tina (C4); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.B1 · Brian → INNOCENT
Uma’s clue says that Brian is one of Denis’s three innocent neighbors. That directly places Brian among the innocent people in Denis’s neighborhood. So Brian must be innocent.
02.A4 · Ryan → INNOCENT, B4 · Samira → INNOCENT
Brian’s clue says exactly 2 innocents are both to the left of Uma and neighboring Vicky. That shared group contains only Ryan and Samira. Since the clue needs 2 innocents there, and Ryan and Samira are the only people in that group, both of them have to fill those 2 innocent spots. So Ryan and Samira must be innocent.
03.C3 · Pam → INNOCENT
Samira’s clue says that Pam is one of Quita’s three innocent neighbors. That directly identifies Pam as innocent. So Pam must be innocent.
04.C4 · Tina → CRIMINAL
Quita has exactly 3 innocent neighbors. Two of them are already known: Pam and Uma. That leaves exactly 1 more innocent among Helen, Isaac, and Tina. In Quita's neighbors, the people who fit the needed remaining innocent slot are Helen and Isaac, so Tina cannot be that innocent person. That makes Tina criminal.
05.B3 · Nala → CRIMINAL
Helen’s neighbors below Ellie are just Isaac and Quita, and exactly 1 of them is criminal. Pam’s neighbors who are not neighbors of Nala are Isaac, Nala, Quita, and Uma, and exactly 2 of that group are criminals. Since Uma is innocent, those 2 criminals in Pam’s group must come from Isaac, Nala, and Quita. But Isaac and Quita together account for only 1 criminal, so the remaining person needed to make that total 2 is Nala. So Nala must be criminal.
06.C1 · Denis → INNOCENT
Helen has exactly 3 criminal neighbors, and exactly 1 of those criminals is below Ellie. Among Helen's neighbors who are not below Ellie, Nala is already one known criminal, so there is room for only 1 more criminal there, and that extra criminal would have to be among Ellie and Ghani. That leaves Denis unable to be the criminal in that not-below-Ellie group. So Denis must be innocent.
07.D5 · Zach → INNOCENT
Below Ellie there is already 1 known innocent, Uma, and the other people there are Isaac, Quita, and Zach. Denis's clue says the total number of innocents below Ellie must be odd. If Zach were criminal, then Isaac and Quita would be the only remaining people below Ellie who could change that total, while those same unknowns also still have to fit Ryan's requirement about Pam's neighbors. That combination cannot be made to satisfy both clues at once. So Zach must be innocent.
08.B5 · Will → INNOCENT
Samira’s clue fixes Quita’s neighbors at exactly three innocents. Pam and Uma are already two of them, so Helen and Isaac cannot both be innocent. Zach’s clue says exactly one clerk has exactly two innocent neighbors. Vicky already has two known innocent neighbors, and if Will were criminal, Vicky would stay at exactly two innocent neighbors. Then Ellie would also have to avoid ending up with exactly two innocent neighbors, which means among Helen and Isaac there could not be exactly one innocent. But those two requirements clash: Helen and Isaac cannot both be innocent, and they also cannot have exactly one innocent between them. So Will cannot be criminal. That makes Will innocent.
09.A5 · Vicky → INNOCENT, C5 · Xavi → INNOCENT
Will’s clue says row 1 has more criminals than row 5. Right now both rows have 0 known criminals, so any criminals placed in row 5 would have to be outweighed by criminals among the remaining people in row 1, Andre and Ellie. If Vicky and Xavi were both criminals, row 5 would already have 2 criminals. Then Andre and Ellie would have to make row 1 satisfy that clue at the same time, but that cannot be done. So Vicky and Xavi cannot both be criminals. That makes Vicky and Xavi innocent.
10.A1 · Andre → CRIMINAL, D1 · Ellie → INNOCENT
If Andre were innocent and Ellie were criminal, the remaining people named in these clues would have to make all four statements true at once: Quita would still need exactly 3 innocent neighbors including Pam, Pam would still need exactly 4 criminal neighbors with exactly 2 of those also neighboring Nala, row 2 would have to be the only row with exactly 2 innocents, and Frida’s neighbors would have to contain an odd number of criminals. But with Andre innocent and Ellie criminal, A2 Frida’s neighbor set starts from 1 known criminal and 1 known innocent, with only Andre, Ghani, and Kumar left to adjust that odd criminal count, while the same group of unknown people is also tied up by the row 2, Quita, and Pam requirements. Those requirements conflict, so Andre and Ellie cannot have those opposite identities. So Andre must be criminal, and Ellie must be innocent.
11.B2 · Ghani → CRIMINAL
Helen’s neighbors contain exactly 3 criminals, and exactly 1 of those criminals is below Ellie. That means 2 of Helen’s neighboring criminals must be not below Ellie. Among Helen’s neighbors who are not below Ellie, the people are Brian, Denis, Ellie, Ghani, Nala, and Pam. In that group, Nala is already a known criminal, while Brian, Denis, Ellie, and Pam are innocent, so there is room for exactly one more criminal there. The only unknown person left in that not-below-Ellie group is Ghani, so Ghani must be criminal.
12.C2 · Helen → INNOCENT
Ryan’s clue says Pam has exactly 4 criminal neighbors, and exactly 2 of those criminals also neighbor Nala. Among the people who are both Pam’s neighbors and Nala’s neighbors, Ghani and Tina are already known criminals. That already fills the clue’s full total of 2 criminals in that shared group, so no other person there can be criminal. Helen is in that same shared group, so Helen must be innocent.
13.D2 · Isaac → CRIMINAL
Samira’s clue says Pam is one of Quita’s exactly 3 innocent neighbors. Among Quita’s neighbors, Helen, Pam, and Uma are already known to be innocent, while Tina is criminal and Isaac is the only unknown. If Isaac were innocent too, Quita would have 4 innocent neighbors, which contradicts the clue that there are exactly 3. So Isaac must be criminal.
14.D3 · Quita → INNOCENT
Ryan’s clue says Pam has exactly 4 criminal neighbors in total, and exactly 2 of those criminals also neighbor Nala. Among Pam’s neighbors who also neighbor Nala, the two criminals are already Ghani and Tina. That means the other part of Pam’s neighborhood, the people who do not neighbor Nala, can contain only the other 2 criminals. In that remaining group, Isaac and Nala are already those 2 known criminals, so the only unknown person left there cannot be a criminal. So Quita must be innocent.
15.A2 · Frida → INNOCENT, A3 · Kumar → INNOCENT
Nala’s clue says row 2 is the only row with exactly 2 innocents, so row 2 must end with exactly 2 innocents and row 3 must not end with exactly 2 innocents. In row 2, Helen is the only known innocent and Frida is the only unknown there, so if Frida were criminal, row 2 would stay at just 1 innocent. In row 3, there are already 2 known innocents, so if Kumar were criminal, row 3 would stay at exactly 2 innocents. Those two results clash with the clue at the same time. So Frida and Kumar must be innocent.