Clues by Sam Sep 16, 2026 Answer – Full Solution Explained
A1
👨🔧
mech
B1
💂♂️
guard
C1
💂♀️
guard
D1
👩🔧
mech
A2
👩🎨
painter
B2
💂♂️
guard
C2
👩🍳
cook
D2
👨🔧
mech
A3
👩🎨
painter
B3
👩🎨
painter
C3
👨🍳
cook
D3
👨🍳
cook
A4
👨🏫
teacher
B4
👩🎤
singer
C4
👨🎤
singer
D4
🕵️♀️
sleuth
A5
👩🏫
teacher
B5
👨🏫
teacher
C5
🕵️♂️
sleuth
D5
🕵️♂️
sleuth
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 — 12 criminals.
Clues by Sam answer for Sep 16, 2026 — a Tricky solved in 14 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Andre (A1), Brian (B1), Cheryl (C1), Frida (A2), Ghani (B2), Karen (A3), Lucy (B3), Nick (D3), Penny (B4), Susan (D4), Xavi (C5) and Zach (D5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.B3 · Lucy → CRIMINAL
Will’s clue says there are exactly 2 criminals to the right of Karen, and exactly 1 of those criminals is a neighbor of Olsi. Among the people to the right of Karen, the only person who is also a neighbor of Olsi is Lucy. That means the one criminal who must satisfy both parts of the clue has to be Lucy. So Lucy must be criminal.
02.B1 · Brian → CRIMINAL
Lucy’s clue says that Brian is one of the exactly 3 criminals in row 1. That directly identifies Brian’s status without needing anything else. So Brian must be criminal.
03.A3 · Karen → CRIMINAL
To Karen’s right, there are exactly 2 criminals among Lucy, Martin, and Nick. In row 3, there are exactly 3 criminals among Karen, Lucy, Martin, and Nick. The only person in row 3 who is not in the group to Karen’s right is Karen herself. So when row 3 needs one more criminal than the right-of-Karen group has, that extra criminal has to be Karen. So Karen must be criminal.
04.A1 · Andre → CRIMINAL, C2 · Hazel → INNOCENT
Row 1 must contain exactly 3 criminals, and Brian is already one of them. Jason’s neighbors must contain exactly 3 innocents, but none of Jason’s neighbors is known innocent yet. If Andre were innocent and Hazel were criminal, then Cheryl, Diane, Martin, and Nick would have to cover both clues at the same time: row 1 would still need enough criminals among Cheryl and Diane, while Jason’s neighborhood would still need enough innocents among Cheryl, Diane, Martin, and Nick. That combination cannot be made to fit both exact counts together. So Andre cannot be innocent and Hazel cannot be criminal. That makes Andre criminal and Hazel innocent.
05.C1 · Cheryl → CRIMINAL
Assume Cheryl were innocent. Hazel says she has exactly 5 criminal neighbors, with exactly 1 of those criminals in column D, and Will’s clue makes the two criminals to Karen’s right be Lucy and Martin, so Nick is innocent. The shared neighbors of Cheryl and Hazel must contain exactly 2 innocents; in that shared group Brian is criminal, so Ghani must be criminal and Diane and Jason must be the 2 innocents. But then among Hazel’s neighbors, the column D people are Diane, Jason, and Nick, and all three would be innocent. That leaves Hazel with no criminal neighbor in column D, which contradicts Hazel’s clue that exactly 1 of her criminal neighbors is in column D. So Cheryl at C1 must be criminal.
06.D1 · Diane → INNOCENT
Lucy’s clue says Brian is one of exactly 3 criminals in row 1. In row 1, the people are Andre, Brian, Cheryl, and Diane, and Andre, Brian, and Cheryl are already the 3 known criminals there. That leaves no room for a fourth criminal in that row, so Diane at D1 must be innocent.
07.C4 · Ryan → INNOCENT
Diane’s clue says that Ryan is one of Nick’s 4 innocent neighbors. That directly places Ryan among the innocent neighbors named by the clue. So Ryan must be innocent.
08.C3 · Martin → INNOCENT, D3 · Nick → CRIMINAL, D4 · Susan → CRIMINAL, B2 · Ghani → CRIMINAL, D2 · Jason → INNOCENT
Nick’s neighbors must include exactly four innocents, and Ryan is one of them. Those neighbors are Hazel, Jason, Martin, Ryan, and Susan, with Hazel and Ryan already innocent. If you try Martin as criminal, Nick as innocent, Susan as innocent, Ghani as innocent, and Jason as criminal, those same people cannot satisfy the clues together. That tested combination conflicts with the required clue facts, so each of those tested opposite statuses is ruled out. So Martin must be innocent, Nick must be criminal, Susan must be criminal, Ghani must be criminal, and Jason must be innocent.
09.A2 · Frida → CRIMINAL
Martin’s clue says every row has at least 2 criminals, so in a 4-person row there can be at most 2 innocents. Row 2 already has 2 known innocents, Hazel and Jason. The only person in row 2 whose status was not yet fixed is Frida, so Frida cannot also be innocent. So Frida must be criminal.
10.D5 · Zach → CRIMINAL
Martin’s clue says every row has at least 2 criminals, and Cheryl’s clue says exactly 2 innocents on edge cells have an innocent directly to their left. Among the edge cells, the known innocents are Diane, Jason, and Will, and the other people involved here are Olsi, Penny, Uma, Xavi, and Zach. If Zach were innocent, those remaining people would have to satisfy both of those clues at the same time, but they cannot. So Zach must be criminal.
11.B4 · Penny → CRIMINAL
Will has exactly 2 criminal neighbors. Of those criminal neighbors, exactly 1 is also a neighbor of Martin, and in the overlap between Will's neighbors and Martin's neighbors the only people are Penny and Ryan. Ryan is already innocent, so that overlap still needs 1 criminal and Penny is the only unknown person left there. That makes Penny criminal.
12.A4 · Olsi → INNOCENT
Will’s neighbors contain exactly 2 criminals in total, and exactly 1 of those criminals is also Martin’s neighbor. Among the neighbors shared by Will and Martin, Penny is the only criminal there, so if Olsi were a criminal, that would already give Will’s neighborhood its second criminal. That would make Uma and Xavi innocent. But then row 5 would have only Zach as a criminal, which breaks the rule that each row has at least 2 criminals. So Olsi at A4 must be innocent.
13.C5 · Xavi → CRIMINAL
Olsi’s clue says exactly one of the three teachers has an innocent directly to the right. Olsi does not, because Penny is to Olsi’s right, and Uma does, because Will is to Uma’s right and Will is innocent. That means the clue’s single matching teacher is already Uma. Will is also a teacher, so if Xavi were innocent, then Will would also have an innocent directly to the right, which would make a second teacher fit the clue. So Xavi at C5 must be criminal.
14.A5 · Uma → INNOCENT
Will’s neighbors must contain exactly 2 criminals in total. Among Will’s neighbors, the only people who are also Martin’s neighbors are Penny and Ryan, and exactly 1 of Will’s criminal neighbors has to be in that overlap; since Ryan is innocent, that one is Penny. That leaves the other criminal neighbor to come from the remaining Will-neighbors A4 Olsi, A5 Uma, and C5 Xavi, and Xavi is already a known criminal there. So Uma must be innocent.