Clues by Sam Jul 09, 2026 Answer – Full Solution Explained
A1
👩🔧
mech
B1
😬
king
C1
👨⚖️
judge
D1
👨🍳
cook
A2
👩🔧
mech
B2
💂♂️
guard
C2
👩🌾
farmer
D2
💂♀️
guard
A3
👩🔧
mech
B3
👨🌾
farmer
C3
👨⚖️
judge
D3
👩🌾
farmer
A4
👮♂️
cop
B4
👷♀️
builder
C4
👨🍳
cook
D4
👩🏫
teacher
A5
👮♀️
cop
B5
👷♂️
builder
C5
👨🏫
teacher
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 — 5 criminals.
Clues by Sam answer for Jul 09, 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 Erwin (D1), Habiba (C2), Martin (C3), Uma (D4) and Vicky (A5); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.D4 · Uma → CRIMINAL
Carl’s clue says there are no innocents who are both below Erwin and neighboring Zoe. The only person in that shared group is Uma, and there are already 0 innocents in that group. So Uma cannot be innocent. That makes Uma criminal.
02.A2 · Frida → INNOCENT
Uma’s clue directly says that Frida is one of the three innocents below Anna. Since Frida is explicitly included among those innocents, Frida at A2 must be innocent.
03.A4 · Phil → INNOCENT, B4 · Quita → INNOCENT, C4 · Salil → INNOCENT
Frida's clue says every row has at least 3 innocents, so a row of 4 people can have at most 1 criminal. Row 4 already has that 1 known criminal: Uma. That means the other people in row 4 cannot also be criminals. So Phil, Quita, and Salil must be innocent.
04.A1 · Anna → INNOCENT
Uma’s clue says there are exactly 3 innocents below Anna. Below A1, Frida and Phil are already innocent, so among Kiran and Vicky exactly one is innocent and the other is criminal. Salil’s clue says column A has exactly 1 criminal. Since that one criminal is already accounted for among Kiran and Vicky, Anna cannot also be criminal. So Anna must be innocent.
05.B2 · Gary → INNOCENT
Frida’s clue says every row has at least 3 innocents, and Anna’s clue says Erwin’s neighbors contain exactly 2 innocents. Around Erwin, there is already 1 known innocent, Carl, and the only other neighbors involved there are Habiba and Jane, so those same nearby unknowns have to fit Anna’s exact count while the rows also keep meeting Frida’s minimum. If Gary were criminal, then Bobby, Erwin, Habiba, Jane, Kiran, Larry, Martin, Nala, Vicky, Will, Xavi, and Zoe would have to satisfy both of those clues at the same time, and they cannot. So Gary cannot be criminal. That makes Gary innocent.
06.B3 · Larry → INNOCENT
Gary’s clue says exactly 2 builders have an innocent directly above them. One such builder is already known, so there must be 1 more case. The only remaining directly-above person that can affect this builder clue is Larry at B3, so he has to be innocent to provide that second case. So Larry must be innocent.
07.D3 · Nala → INNOCENT, D5 · Zoe → INNOCENT
Frida’s clue says every row has at least 3 innocents, and Larry’s clue says Quita’s neighbors contain exactly 2 criminals. Among Quita’s neighbors there are currently no known criminals, so those 2 criminals would have to come from Kiran, Martin, Vicky, Will, and Xavi. If Nala were a criminal and Zoe were a criminal, then Bobby, Erwin, Habiba, Jane, Kiran, Martin, Vicky, Will, and Xavi would still have to satisfy both of those clues at the same time, but they cannot. That means Nala and Zoe cannot be criminals. So Nala and Zoe must be innocent.
08.C5 · Xavi → INNOCENT
Frida’s clue says every row has at least 3 innocents, so row 5 must reach that minimum. Zoe’s clue also says the innocents to the right of Vicky, namely Will, Xavi, and Zoe’s group on that side, have to form one connected block. If Xavi were a criminal, then to the right of Vicky only Zoe would be a known innocent, with Will separated from Zoe by Xavi. That clashes with the requirement that the innocents there be connected while row 5 still has to satisfy Frida’s minimum. So Xavi must be innocent.
09.D1 · Erwin → CRIMINAL
Jane’s clue says her neighbors contain an odd number of criminals. Among those neighbors, Carl and Nala are already innocent, so the only people who can supply that odd count are Erwin, Habiba, and Martin. At the same time, Salil’s clue fixes column A as the only column with exactly one criminal, so the other columns, including column D, cannot end with exactly one criminal. Since Uma is already a known criminal in column D, that column needs at least one more criminal among Erwin or Jane. If Erwin were innocent, the remaining people involved in these two clues could not satisfy both requirements at once. So Erwin must be criminal.
10.B1 · Bobby → INNOCENT
Frida's clue says every row has at least 3 innocents, so in row 1 there can be at most 1 criminal. Row 1 already has that 1 known criminal, Erwin, and the only person there whose status is not fixed yet is Bobby. So Bobby cannot be the row's second criminal. That makes Bobby innocent.
11.B5 · Will → INNOCENT
Salil’s clue says column A is the only column with exactly one criminal. In column B, there are no known criminals, and Will is the only person there whose identity is not yet fixed. If Will were a criminal, then column B would end up with exactly one criminal too, which would contradict the clue that only column A has that count. So Will must be innocent.
12.A5 · Vicky → CRIMINAL
Quita must have exactly 2 criminal neighbors, and among her eight neighbors five are already known innocents. That means the three unknown neighbors, Kiran, Martin, and Vicky, must contain exactly 1 innocent. Row 5 already has Will, Xavi, and Zoe as innocents, so Frida's clue that each row has at least 3 innocents is already satisfied without needing Vicky to be innocent. The one innocent among Quita's unknown neighbors therefore has to be Kiran or Martin, not Vicky. So Vicky must be criminal.
13.A3 · Kiran → INNOCENT
Salil’s clue says column A is the only column with exactly one criminal, so column B, column C, and column D must not have exactly one criminal. Column A already has one known criminal, Vicky, and Kiran is the only unknown person left in that column. If Kiran were criminal, then column A would no longer have exactly one criminal, which clashes with the clue’s requirement about the columns. So Kiran at A3 must be innocent.
14.C3 · Martin → CRIMINAL
Larry’s clue says Quita has exactly 2 criminal neighbors. Among Quita’s neighbors, 1 is already known to be criminal, and the only neighbor there whose identity is still unknown is Martin. So the second criminal neighbor has to be Martin. That makes Martin criminal.
15.C2 · Habiba → CRIMINAL
Salil’s clue says column A is the only column with exactly one criminal, so no other column can finish with exactly one criminal. Column C already has one known criminal, Martin, and the only person in that column whose identity is not yet fixed is Habiba. If Habiba were innocent, then column C would remain at exactly one criminal, which the clue forbids. So Habiba must be criminal.
16.D2 · Jane → INNOCENT
Frida's clue says every row has at least 3 innocents, so in a 4-person row there can be at most 1 criminal. Row 2 already has one known criminal, Habiba, and the only person there not yet identified is Jane. That means row 2 cannot contain any other criminal. So Jane at D2 must be innocent.