Clues by Sam Sep 18, 2026 Answer – Full Solution Explained
A1
👨🍳
cook
B1
👩💼
clerk
C1
👩💼
clerk
D1
👨💼
clerk
A2
👩🍳
cook
B2
👨✈️
pilot
C2
🕵️♂️
sleuth
D2
🕵️♂️
sleuth
A3
👨🍳
cook
B3
👩🔧
mech
C3
👩🔧
mech
D3
🕵️♀️
sleuth
A4
👮♀️
cop
B4
👨✈️
pilot
C4
👩🌾
farmer
D4
👩🏫
teacher
A5
👮♂️
cop
B5
👨🌾
farmer
C5
👩🌾
farmer
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 — 8 criminals.
Clues by Sam answer for Sep 18, 2026 — a Hard solved in 19 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Akira (A1), Betty (B1), Freya (A2), Jose (C2), Olivia (C3), Quita (A4), Vince (B5) and Xia (C5); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.B1 · Betty → CRIMINAL
Derek’s clue says the two criminals in row 1 must be connected. If Betty were innocent, then with Derek already innocent, the two criminals in that row would have to be Akira and Carol. But then the two criminals in row 1 would be separated by Betty, so they would not be connected. So Betty at B1 must be criminal.
02.A2 · Freya → CRIMINAL
Betty’s clue says column A contains exactly 3 criminals, and exactly 1 of those criminals is to the left of Kumar. In column A, the only person who is to the left of Kumar is Freya. That means the one criminal in that part of column A has to be Freya. So Freya must be criminal.
03.A1 · Akira → CRIMINAL
Freya’s clue says Akira is one of the 6 criminals on the edge. That directly places Akira among the edge criminals. So Akira must be criminal.
04.C1 · Carol → INNOCENT
Derek’s clue says the two criminals in row 1 are connected. In row 1, the two criminals are already Akira and Betty, and row 1 contains Akira, Betty, Carol, and Derek. Since those two criminals are the pair the clue is talking about, Carol cannot also be a criminal there. So Carol must be innocent.
05.D2 · Kumar → INNOCENT
Freya’s clue says there are exactly 6 criminals on the edge. We already have 3 known edge criminals there, so the remaining 3 edge criminals have to come from A3 Luigi, D3 Paula, A4 Quita, D4 Tina, A5 Umar, B5 Vince, C5 Xia, and D5 Zach. That leaves Kumar at D2 outside that needed group of 3 edge criminals, so Kumar cannot be a criminal. So Kumar must be innocent.
06.D3 · Paula → INNOCENT
Freya’s clue says there are exactly 6 criminals on the edge. The edge already has 3 known criminals, so the other 3 edge criminals have to come from the remaining edge people. That leaves the 3 edge criminals to be chosen from A3 Luigi, A4 Quita, D4 Tina, A5 Umar, B5 Vince, C5 Xia, and D5 Zach. Paula is not in that group, so she cannot be one of those 3 edge criminals. So Paula must be innocent.
07.C5 · Xia → CRIMINAL
Umar and Vince are the only people to the left of Xia, and that group contains exactly 1 innocent. Ronald's row 5 neighbors are Umar, Vince, and Xia, and that larger group also contains exactly 1 innocent. Since Umar and Vince already account for the full innocent count allowed in the larger group, the only remaining person there, Xia, cannot be innocent. So Xia must be criminal.
08.D4 · Tina → INNOCENT
The edge clue says there are exactly 6 criminals on the edge. There are already 4 known edge criminals, so the remaining 2 edge criminals have to come from the unknown edge people. Those possible remaining edge criminals are A3 Luigi, A4 Quita, A5 Umar, B5 Vince, and D5 Zach, which leaves D4 Tina out of that pair. So Tina cannot be one of the 2 remaining edge criminals. That makes Tina innocent.
09.B3 · Megan → INNOCENT
Ronald has exactly 4 criminal neighbors, and exactly 2 of those criminals are in row 5. Since Xia is already a criminal in row 5 among Ronald's neighbors, the other 2 criminals among Ronald's neighbors must be outside row 5, in the group A3 Luigi, B3 Megan, C3 Olivia, A4 Quita, and C4 Stella. Those 2 criminals must come from A3 Luigi, C3 Olivia, A4 Quita, and C4 Stella, so Megan is not one of them. So Megan must be innocent.
10.D5 · Zach → INNOCENT
Freya’s clue says there are exactly 6 criminals on the edge. There are already 4 known edge criminals, so exactly 2 more edge criminals must come from the 5 unknown edge people: Luigi, Quita, Umar, Vince, and Zach. Those 2 edge criminals are accounted for by Luigi, Quita, Umar, and Vince, so Zach cannot be one of them. That makes Zach innocent.
11.B5 · Vince → CRIMINAL
Betty’s clue says column A has exactly 3 criminals. Since Akira and Freya are already criminals there, exactly one of A3 Luigi, A4 Quita, and A5 Umar can also be a criminal. Freya’s clue says there are exactly 6 criminals on the edge. The edge already has 4 known criminals, and among the edge people in column A only one more criminal can appear, so the edge can get only 1 more criminal from A3, A4, and A5. That still leaves one more edge criminal needed, and the only other edge person available is B5 Vince. So Vince must be criminal.
12.A5 · Umar → INNOCENT
Kumar’s clue says there is exactly one innocent to the left of Xia. Among the people to the left of Xia, there are currently no known innocents, and the only person there whose identity is not yet known is Umar. So the one innocent required by the clue has to be Umar. That makes Umar innocent.
13.C3 · Olivia → CRIMINAL
Freya’s clue says there are exactly 6 criminals on the edge, and 5 edge criminals are already known, so among the edge people A3 Luigi and A4 Quita, exactly 1 is a criminal. Zach’s clue says Megan has exactly 4 criminal neighbors, and exactly 1 of those is also Olivia’s neighbor. That means Megan’s neighbors who are not neighbors of Olivia must account for the other 3 criminals. In that group, A2 Freya is already a criminal, and among A3 Luigi and A4 Quita exactly 1 is a criminal, so those three people contribute exactly 2 criminals total. The only other person in that group is Olivia, so Olivia has to be the third criminal. So Olivia must be criminal.
14.C4 · Stella → INNOCENT
Akira’s clue says Tina’s neighbors contain an odd number of innocents. Among Tina’s neighbors, there are already 2 known innocents, and the only neighbor there whose identity is still unknown is Stella. Since 2 is even, Stella has to be innocent to make the total number of innocents odd. So Stella must be innocent.
15.B2 · Ivan → INNOCENT
Olivia’s clue says exactly one criminal in row 2 has a criminal directly above them. In row 2, Freya is already a known criminal, while Ivan and Jose are the only other people there not yet identified, and Kumar is innocent. If Ivan were criminal, then with Freya already filling the row’s criminal requirement from this clue, the remaining row 2 people would have to fit that same condition in a way that is impossible. So Ivan cannot be criminal. That makes Ivan innocent.
16.C2 · Jose → CRIMINAL
If Jose were innocent, then Ivan’s clue says exactly 3 edge people have a criminal directly below them. With Jose innocent, that count makes Luigi a criminal. Freya’s clue also says there are exactly 6 criminals on the edges, and the edge already has 5 known criminals, so the two unknown edge people would have to be Luigi and Quita, making Quita a criminal too. But then the edge people with a criminal directly below them would be Akira, Betty, Freya, and Luigi, which is 4, not 3. So Jose at C2 must be criminal.
17.B4 · Ronald → INNOCENT
Zach's clue says that Megan has exactly four criminal neighbors, and exactly one of those criminals is also a neighbor of Olivia. Among the people who are neighbors of both Megan and Olivia, Jose is already a known criminal, while Ivan and Stella are innocent and Ronald is the only unknown. Since that shared group already contains the one criminal the clue allows, Ronald cannot also be a criminal. So Ronald must be innocent.
18.A3 · Luigi → INNOCENT
In row 4, exactly one person can have exactly 3 innocent neighbors. Tina already has exactly 3 innocent neighbors, and Stella cannot be that person because she has 5. Quita also currently has 3 known innocent neighbors, but her total would go above 3 if Luigi were not innocent. Ronald currently has 3 known innocent neighbors too, and with his unknown neighbors, Luigi being innocent is what keeps the clue from conflicting with Tina’s count. So Luigi must be innocent.
19.A4 · Quita → CRIMINAL
Megan’s neighbors must contain exactly 4 criminals, and exactly 1 of those criminals is also a neighbor of Olivia. Among the people who are neighbors of both Megan and Olivia, the only criminal is Jose, so the other 3 criminals among Megan’s neighbors must be people who are not neighbors of Olivia. Those Megan-neighbors who are not Olivia-neighbors are Freya, Luigi, Olivia, and Quita. Freya and Olivia are already criminals there, so that group still needs 1 more criminal. The only unknown person left in that group is Quita, so Quita at A4 must be criminal.