Clues by Sam Sep 11, 2026 Answer – Full Solution Explained
A1
👩🔬
scientist
B1
👨🔬
scientist
C1
👨🌾
farmer
D1
👩🌾
farmer
A2
👨🔬
scientist
B2
👨⚕️
doctor
C2
😬
streamer
D2
👨🎨
painter
A3
🕵️♂️
sleuth
B3
👨⚕️
doctor
C3
👷♀️
builder
D3
👷♂️
builder
A4
🕵️♀️
sleuth
B4
👨💻
coder
C4
👨💼
clerk
D4
👷♂️
builder
A5
🕵️♀️
sleuth
B5
👨💻
coder
C5
👨💼
clerk
D5
👩💼
clerk
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 Sep 11, 2026 — a Tricky solved in 19 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 Emily (D1), Fred (A2), Jason (B2), Mathew (A3) and Randy (B4); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.A3 · Mathew → CRIMINAL
Zara’s clue says there are exactly 2 criminals in column A, and exactly 1 of those column A criminals is in row 3. In column A, the only person who is in row 3 is Mathew at A3. So the one criminal required in the row 3 part of column A has to be Mathew. That makes Mathew criminal.
02.A4 · Quita → INNOCENT
Mathew’s clue says that Quita is one of Uma’s exactly two innocent neighbors. That identifies Quita herself as innocent, not just one of the possible neighbors. So Quita must be innocent.
03.A5 · Uma → INNOCENT
Quita’s clue says there is exactly 1 innocent among the people who are both above Uma and neighboring Brian, and that group is just Alice and Fred. In column A, that leaves the rest of the needed innocents to come from the people in column A but not in that smaller group. Among those remaining column A people, Mathew is already criminal and Quita is already innocent, so the remaining person there is Uma. Therefore Uma at A5 must be innocent.
04.B3 · Noah → INNOCENT
Assume Noah were criminal. Then, from Uma's clue, everyone in row 4 can have at most 2 criminal neighbors, but Quita already has 2 known criminal neighbors and Randy already has 2 known criminal neighbors. Since Mathew also says Quita is one of Uma's exactly 2 innocent neighbors, the other innocent neighbor of Uma must be Randy. That leaves Uma's three neighbors as Quita innocent, Randy innocent, and Vincent innocent. But then Uma would have 3 innocent neighbors, contradicting the statement that Quita is one of Uma's 2 innocent neighbors. So Noah at B3 must be innocent.
05.B2 · Jason → CRIMINAL
Uma’s neighbors need exactly one more innocent besides Quita, so among Randy and Vincent there is exactly one innocent. Brian’s below group needs exactly two innocents in total, and Noah is already one of them. That means the other innocent below Brian must come from Randy or Vincent, which are the shared people between the two clues. Jason is the only person below Brian who is outside that shared pair, so he cannot be innocent. So Jason must be criminal.
06.D2 · Luigi → INNOCENT
Jason’s clue says there are exactly two criminals in row 2, and those two are connected. Jason is already one of those criminals, so the other criminal in row 2 has to be directly next to him in that row. That limits the second criminal to Fred at A2 or Kathy at C2, and Luigi at D2 is not next to Jason in that line. So Luigi at D2 must be innocent.
07.B1 · Brian → INNOCENT
Uma has exactly 2 innocent neighbors, and Quita is one of them. Among Uma's neighbors, Quita is already innocent, so the other innocent neighbor must be either Randy or Vincent; with the listed neighbor statuses here, Randy and Vincent are both the innocent neighbors in that group. Now suppose Brian were criminal. Luigi's clue says every column has at least 3 innocents, and that assumption leads to an impossible situation with those column requirements. So the assumption fails, and Brian at B1 must be innocent.
08.C3 · Olga → INNOCENT
Column C has Cj, Kathy, Olga, Scott, and Will, and this column must contain at least 3 innocents. Brian’s clue also says that all innocents in column C have to form one connected block. If Olga were criminal, the remaining people named in these clues would have to provide at least 3 innocents in column C while also keeping all the innocents there in one connected block, and that cannot be done. So Olga cannot be criminal. Therefore Olga at C3 must be innocent.
09.C4 · Scott → INNOCENT
Assume Scott were criminal. Then row 4 would still have to satisfy Uma's clue that no one there has more than 2 criminal neighbors. With Scott criminal, Randy already has 2 criminal neighbors, so Vincent cannot also be criminal and Vincent must be innocent. Randy is then still capped at 2, so Will cannot be criminal either and Will must be innocent. But Brian's clue says all innocents in column C must be connected, and in column C that would make Olga at C3 and Will at C5 both innocent while Scott at C4 is criminal between them. That breaks the required connected block of innocents in column C, so the assumption fails. So Scott must be innocent.
10.D3 · Poe → INNOCENT
Scott’s clue says row 2 is the only row with exactly 2 criminals. Row 3 already has 1 known criminal, and Poe is the only person in that row not yet identified. If Poe were a criminal, then row 3 would also have exactly 2 criminals, which is not allowed because only row 2 can have that total. So Poe must be innocent.
11.A1 · Alice → INNOCENT
If Alice were criminal, the remaining people named here would still have to satisfy all three clue restrictions at once. Uma’s neighbors would still need exactly 2 innocents among Quita, Randy, and Vincent, row 2 would have to be the only row with exactly 2 criminals, and the coders would have to contain the same number of criminals as the farmers. But with Alice set to criminal, those requirements clash for the people involved in rows 1, 2, 4, and 5 and for the coder and farmer groups. So Alice cannot be criminal. That makes Alice innocent.
12.A2 · Fred → CRIMINAL
Zara's clue says column A has exactly 2 criminals, and exactly 1 of them is in row 3. Mathew at A3 is that one criminal in row 3, so the other criminal in column A must be someone in column A who is not in row 3. Those people are Alice, Fred, Quita, and Uma. Alice, Quita, and Uma are already innocent, leaving Fred as the only person there who can fill that remaining criminal spot. So Fred must be criminal.
13.C2 · Kathy → INNOCENT
Scott’s clue says row 2 is the only row with exactly 2 criminals. In row 2, Fred and Jason are already criminals, Luigi is innocent, and Kathy is the only person there not yet identified. That means row 2 already has its exact 2 criminals, so Kathy cannot also be a criminal. So Kathy must be innocent.
14.D4 · Ted → INNOCENT
Uma’s clue says her neighbors contain exactly 2 innocents, and one of them is already Quita. So among Uma’s other two neighbors, Randy and Vincent, exactly one must be innocent and the other must be criminal. Will’s clue says his neighbors contain an odd number of criminals. Right now, among Will’s unknown neighboring people, those are Randy, Ted, and Vincent. If Ted were criminal, then Randy and Vincent would still have to be one innocent and one criminal from Uma’s clue, which would make Will’s criminal count among these three equal to 2, not an odd number. So Ted must be innocent.
15.C5 · Will → INNOCENT
Ted’s clue fixes the edge count: exactly 3 edge people have a criminal directly to their right. If Will at C5 were criminal, then the remaining people involved here, Cj, Emily, Randy, and Vincent, would also have to fit Olga’s clue that the number of criminal coders equals the number of criminal farmers. But with Will set to criminal, those same people cannot satisfy both clues at the same time. That means Will cannot be criminal. So Will must be innocent.
16.B4 · Randy → CRIMINAL
The two doctors are Jason and Noah, and the clue says exactly one of them has a criminal directly below them. Jason does not satisfy that, because the person directly below Jason is Noah, who is innocent. So the one doctor who does satisfy the clue must be Noah, and the person directly below Noah is Randy. That makes Randy criminal.
17.B5 · Vincent → INNOCENT
Mathew's clue says Quita is one of Uma's exactly 2 innocent neighbors. Among Uma's neighbors, Quita is already innocent, Randy is criminal, and Vincent is the only neighbor not yet identified. Since Uma must have 2 innocent neighbors, the second innocent neighbor has to be Vincent. So Vincent must be innocent.
18.C1 · Cj → INNOCENT
Fred’s clue says exactly one edge person has exactly 3 criminal neighbors. Mathew already has 3 known criminal neighbors, and no unknown neighbors, so he is that one person. Brian is the only other edge person who could possibly reach 3, because he already has 2 known criminal neighbors and his only unknown neighbor is Cj. Everyone else on the edge either is already below 3 with no unknown neighbors, or still cannot reach 3 from the unknown neighbors they have. So if Cj were criminal, Brian would also have exactly 3 criminal neighbors, giving a second edge person with 3. That would break Fred’s clue. So Cj at C1 must be innocent.
19.D1 · Emily → CRIMINAL
Olga’s clue says the number of criminal coders and criminal farmers is the same. The coders already have 1 criminal, while the farmers currently have 0 known criminals. Since Emily is the only farmer not yet identified, she has to supply that one criminal farmer. So Emily must be criminal.