Clues by Sam Sep 23, 2026 Answer – Full Solution Explained
A1
👨🎤
singer
B1
👨💼
clerk
C1
👩💻
coder
D1
👨💻
coder
A2
👷♀️
builder
B2
👩💼
clerk
C2
👨🎤
singer
D2
👨🍳
cook
A3
👷♀️
builder
B3
👷♂️
builder
C3
👩⚖️
judge
D3
👨🍳
cook
A4
💂♀️
guard
B4
💂♀️
guard
C4
👮♀️
cop
D4
👮♂️
cop
A5
👨⚖️
judge
B5
👨🌾
farmer
C5
👩🌾
farmer
D5
👩⚖️
judge
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 — 10 criminals.
Clues by Sam answer for Sep 23, 2026 — a Medium solved in 17 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 10 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Hope (B2), Isaac (C2), John (D2), Katie (A3), Luigi (B3), Peter (D3), Quita (A4), Umar (A5), Wanda (C5) and Xia (D5); the remaining 10 suspects are innocent.
The deduction chain, in plain English
01.D3 · Peter → CRIMINAL
Eric’s clue says that Peter is one of John’s 2 criminal neighbors. That directly identifies Peter as a criminal. So Peter must be criminal.
02.C3 · Nicole → INNOCENT
If Nicole were criminal, then with Peter already being one of John's exactly 2 criminal neighbors, John's neighbor list would already use up those 2 criminals with Nicole and Peter. That would make Carol and Isaac innocent. Peter's clue says exactly 1 of the 2 clerks has a criminal directly to the right. The clerks are Brian and Hope, and with Carol innocent and Isaac innocent, neither clerk would have a criminal directly to the right. That clashes with the clue. So Nicole must be innocent.
03.C2 · Isaac → CRIMINAL
Nicole’s clue says exactly 1 innocent is both below Carol and neighboring Hope. The only people in that shared group are Isaac and Nicole, and Nicole is already known to be innocent. That means the one innocent allowed there is already accounted for, so Isaac cannot also be innocent. So Isaac must be criminal.
04.C1 · Carol → INNOCENT
Eric’s clue says Peter is one of John’s exactly 2 criminal neighbors. John’s neighbors are Carol, Eric, Isaac, Nicole, and Peter, and among them Isaac and Peter are already the 2 known criminals. That leaves no room for Carol to be a criminal neighbor of John. So Carol must be innocent.
05.D2 · John → CRIMINAL
Isaac’s clue says column D contains exactly 3 criminals, and exactly 1 of those criminals is a neighbor of Wanda. Since the column-D neighbors of Wanda are Thor and Xia, that means exactly 2 criminals in column D are not Wanda’s neighbors. The people in column D who are not Wanda’s neighbors are Eric, John, and Peter. Eric is innocent and Peter is already a criminal, so that group has 1 criminal so far and needs 1 more. The only unknown person left in that group is John, so John at D2 must be criminal.
06.C5 · Wanda → CRIMINAL
Column D must contain exactly three criminals, and John and Peter are already two of them, so among Thor and Xia exactly one is criminal. Among Stella's neighbors, the people who are not neighbors of Quita are Nicole, Peter, Thor, Wanda, and Xia, and that group must contain exactly three criminals. Nicole is innocent, Peter is already one criminal, and Thor and Xia together contribute exactly one more criminal, so Wanda has to be the remaining criminal needed to make that total three. So Wanda must be criminal.
07.B4 · Ruby → INNOCENT
Among the people below Brian who matter here, Luigi and Vince contain exactly 1 criminal. Among Quita's neighbors who matter here, Luigi, Ruby, and Vince also contain exactly 1 criminal. The only extra person in that second group is Ruby, so the 1 criminal allowed there is already fully accounted for by Luigi and Vince, and Ruby cannot be a criminal. So Ruby must be innocent.
08.A4 · Quita → CRIMINAL
Ruby’s clue says Vince has exactly 3 criminal neighbors, and exactly 1 of those criminals is also a neighbor of Umar. Among the people who are both Vince’s neighbors and Umar’s neighbors, Ruby is innocent and the only other person there is Quita. So that group still needs 1 criminal, and Quita is the only person who can fill it. That makes Quita criminal.
09.B2 · Hope → CRIMINAL
Below Brian there are exactly 2 innocents, and Ruby is already one of them, so among Hope, Luigi, and Vince there can be only one more innocent. That means the people below Brian contain exactly 2 criminals. Among the people below Brian who are also Stella's neighbors and Quita's neighbors, Luigi, Ruby, and Vince, John's clue says exactly 1 is criminal. Since Ruby is innocent, that one criminal is among Luigi and Vince. So in the larger below-Brian group, the only person outside that smaller set is Hope, and the below-Brian group still needs one more criminal. That makes Hope criminal.
10.B1 · Brian → INNOCENT
Column B must contain an odd number of innocents. Right now column B has 1 known innocent, Ruby, and the unknown people there are Brian, Luigi, and Vince. If Brian were criminal, then only Luigi and Vince could change the innocent count in column B. At the same time, Stella's neighbors must contain exactly 4 criminals, and exactly 1 of those criminal neighbors of Stella is also a neighbor of Quita, which is the group containing Luigi, Ruby, and Vince. Making Brian criminal leaves the same remaining unknown people to satisfy both of those clue requirements, and they cannot do so. So Brian cannot be criminal. That makes Brian innocent.
11.A3 · Katie → CRIMINAL
Vince has exactly 3 criminal neighbors, and Quita and Wanda are already two of them, so among Stella and Umar there is exactly 1 more criminal. Ruby has exactly 5 criminal neighbors, and exactly 1 of those is below Brian, so 4 of Ruby's criminal neighbors are not below Brian. In Ruby's neighbors who are not below Brian, Quita and Wanda are already criminals, Nicole is innocent, and Stella and Umar contribute exactly 1 more criminal, leaving only Katie to supply the last needed criminal in that group. So Katie must be criminal.
12.A5 · Umar → CRIMINAL
Stella’s neighbors must contain exactly 4 criminals, and exactly 1 of those criminals is also a neighbor of Quita. Among the people who are neighbors of both Stella and Quita, Ruby is already innocent, so that one criminal in the shared group has to come from Luigi or Vince. Quita’s neighbors must contain an odd number of criminals. Quita already has 1 known criminal neighbor, and the remaining unknown neighbors are Luigi, Umar, and Vince. If Umar were innocent, then Luigi, Vince, Thor, and Xia would have to satisfy both clues at once, but they cannot do that while keeping Stella’s total at 4 criminals, keeping exactly 1 shared criminal for John’s clue, and keeping Quita’s criminal-neighbor count odd for Brian’s clue. So Umar must be criminal.
13.C4 · Stella → INNOCENT
Vince's neighbors contain exactly 3 criminals in total. Ruby's clue also says exactly 1 of those criminals is a neighbor of Umar, and among the people who are both Vince's neighbors and Umar's neighbors, the only criminal is Quita. That leaves the rest of Vince's neighbors outside Umar's neighborhood as Stella, Umar, and Wanda, and Umar and Wanda are already known criminals. Since Vince's neighborhood already has its full total of 3 criminals, Stella cannot also be a criminal. So Stella must be innocent.
14.A2 · Frida → INNOCENT
If Frida were criminal, then the edge clue would force Thor to be innocent and also Xia to be innocent, because exactly 2 edge people must have an innocent directly below them. Then Stella’s neighbors would have Peter and Wanda as the known criminals, with Luigi and Vince as the only possible criminals among Quita’s neighbors. But John’s clue says Stella has exactly 4 criminal neighbors in total, and exactly 1 of those criminals is also Quita’s neighbor. With Thor innocent and Xia innocent, the only way to reach 4 criminal neighbors around Stella is to make both Luigi and Vince criminal, and that would give 2 criminal neighbors of Stella who are also Quita’s neighbors, not 1. So Frida must be innocent.
15.A1 · Alex → INNOCENT
If Alex were criminal, then column A would have 4 criminals. Frida’s clue says columns A and D must have the same number of criminals, so column D would also have to reach 4 criminals. Column D already has John and Peter as criminals, so both Thor and Xia would have to be criminals to bring column D up to 4. But Stella’s clue says she has exactly 4 criminal neighbors in total, and exactly 1 of those is also Quita’s neighbor among Luigi, Ruby, and Vince. With Peter and Wanda already criminal neighbors of Stella, making both Thor and Xia criminals gives Stella 4 criminal neighbors, and then Peter, Thor, and Xia are all criminal neighbors of Stella who are not Quita’s neighbors, leaving only 1 criminal slot among Luigi, Ruby, and Vince. That clashes with the required balance from the clues, so Alex cannot be criminal. So Alex must be innocent.
16.B5 · Vince → INNOCENT, B3 · Luigi → CRIMINAL
Stella’s clue says she has exactly 4 criminal neighbors, and exactly 1 of those criminals is also a neighbor of Quita. Among the people who are both Stella’s neighbors and Quita’s neighbors, only Luigi, Ruby, and Vince fit that description, and Ruby is already innocent. If Vince were criminal and Luigi were innocent, then Vince would be the one criminal in that shared group. But Umar’s clue says the edge has an odd number of criminals, while the edge already has 6 known criminals and the only unknown edge people here are Thor, Vince, and Xia. With Vince criminal, Thor and Xia would have to satisfy that edge clue and Stella’s clue at the same time, and they cannot. So Vince must be innocent and Luigi must be criminal.
17.D4 · Thor → INNOCENT, D5 · Xia → CRIMINAL
Luigi says row 4 has more innocents than row 5. Right now row 4 already has 2 known innocents, while row 5 has 1 known innocent, and the only unknowns in those rows are Thor in row 4 and Xia in row 5. If Thor were criminal and Xia were innocent, then row 4 would stay at 2 innocents and row 5 would also rise to 2 innocents. That would make the two rows equal, which clashes with the clue that row 4 must have more innocents than row 5. So Thor must be innocent and Xia must be criminal.