Clues by Sam Jan 14, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👮♀️
cop
C1
💂♀️
guard
D1
👨⚕️
doctor
A2
🕵️♂️
sleuth
B2
👩🍳
cook
C2
💂♂️
guard
D2
🕵️♀️
sleuth
A3
👩✈️
pilot
B3
👩✈️
pilot
C3
👩⚕️
doctor
D3
👨🍳
cook
A4
👨🎤
singer
B4
👩🍳
cook
C4
👩⚖️
judge
D4
👨🔧
mech
A5
👨🎤
singer
B5
👨🔧
mech
C5
👨⚖️
judge
D5
👨🔧
mech
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 Jan 14, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 10 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Barb (B1), Chloe (C1), Eric (D1), Hazel (B2), Maria (B3), Nancy (C3), Phil (A4), Ruby (B4), Wally (B5) and Ziad (D5); the remaining 10 suspects are innocent.
The deduction chain, in plain English
01.B1 · Barb → CRIMINAL, C1 · Chloe → CRIMINAL
Alice is at A1 and Eric is at D1, so the only people strictly between them in row 1 are Barb at B1 and Chloe at C1. Gabe’s clue says there are exactly two criminals in between Alice and Eric. Since exactly two people lie between them—Barb and Chloe—both of those in-between people must be criminals. Therefore, we can determine that Barb is CRIMINAL, Chloe is CRIMINAL, and so on.
02.C2 · Isaac → INNOCENT
Barb (B1) has neighbors A1, A2 (Gabe), B2, C2 (Isaac), and C1 (Chloe); Gabe is INNOCENT and Chloe is CRIMINAL. Her clue says she has exactly three innocent neighbors, and among those three, exactly one is also a neighbor of Gabe. Among Barb’s neighbors, the ones who are also neighbors of Gabe are A1 and B2; C2 (Isaac) is not, and Gabe does not count as his own neighbor. Since Gabe already accounts for one of the three innocents and only one of the remaining innocents may also be a neighbor of Gabe, the other remaining innocent must be the non-Gabe neighbor C2. Therefore, we can determine that C2 Isaac is INNOCENT.
03.A3 · Karen → INNOCENT
The clue explicitly says “Karen is one of Maria’s 4 innocent neighbors,” which directly states that Karen is an innocent neighbor of Maria. Therefore, we can determine that A3 Karen is INNOCENT.
04.B2 · Hazel → CRIMINAL
Around Maria (B3), we already know three neighbors are innocent: Gabe (A2), Isaac (C2), and Karen (A3). Isaac’s clue says Karen is one of Maria’s four innocent neighbors, so among the remaining five neighbors—Hazel (B2), Nancy (C3), Phil (A4), Ruby (B4), and Sofia (C4)—exactly one is innocent and the other four are criminals; in particular, at least one of Hazel or Ruby (the column B neighbors) must be a criminal. Karen’s clue says all criminals in column B are connected. With Barb (B1) already a criminal, any other criminal in column B can only be connected to her if B2 is also criminal, since B2 is the required first link below B1. Putting these together, because at least one of Hazel or Ruby is a criminal in column B, B2 must be criminal to keep column B’s criminals connected. Therefore, we can determine that B2 · Hazel is CRIMINAL.
05.A1 · Alice → INNOCENT
Barb (B1) has five neighbors: Alice (A1), Gabe (A2), Hazel (B2), Chloe (C1), and Isaac (C2). Her clue says she has exactly three innocent neighbors. Among those five, Gabe and Isaac are already known innocents, while Hazel and Chloe are criminals, leaving only Alice undecided. To make the total of three innocent neighbors, Alice must be the third innocent. Therefore, we can determine that A1 Alice is INNOCENT.
06.A4 · Phil → CRIMINAL
Karen is at A3, and her neighbors in column A are A2 (Gabe) and A4 (Phil). Alice’s clue says that exactly one innocent in column A is neighboring Karen, so among A2 and A4, only one can be innocent. Since A2 Gabe is already known to be innocent, A4 cannot also be innocent. Therefore, we can determine that A4 Phil is CRIMINAL.
07.D2 · Joy → INNOCENT
Row 2 contains Gabe (A2), Hazel (B2), Isaac (C2), and Joy (D2). Phil’s clue says there is only one criminal in row 2. We already know Hazel is a criminal while Gabe and Isaac are innocents, so the “one criminal” quota is filled by Hazel. If Joy were a criminal, row 2 would have two criminals, which contradicts the clue, so Joy must be innocent. Therefore, we can determine that D2 Joy is INNOCENT.
08.B3 · Maria → CRIMINAL, B4 · Ruby → CRIMINAL
We focus on column B (B1–B5) and the corners. Joy’s clue says only one corner has exactly one criminal neighbor; D1 already does (C1 is criminal, while C2 and D2 are innocent), so no other corner can. A5’s neighbors are A4 (criminal), B4, and B5; if both B4 and B5 were innocent, A5 would also have exactly one criminal neighbor, which is forbidden, so at least one of B4 or B5 is criminal. Karen’s clue says all criminals in column B are connected, so with B1 and B2 already criminal and at least one of B4 or B5 criminal, every position between must be criminal as well, forcing B3 and B4 to be criminal. Therefore, we can determine that B3 Maria is CRIMINAL and B4 Ruby is CRIMINAL.
09.C5 · Xavi → INNOCENT
Maria at B3 has neighbors A2, B2, C2, A3, C3, A4, B4, and C4; among these, A2 (Gabe), C2 (Isaac), and A3 (Karen) are known innocents, while B2 (Hazel), A4 (Phil), and B4 (Ruby) are criminals, leaving only C3 (Nancy) and C4 (Sofia) unknown. Isaac’s clue that “Karen is one of Maria’s 4 innocent neighbors” fixes the total at four, so exactly one of Nancy or Sofia is innocent and the other is criminal. Maria’s clue says column C has an odd number of innocents; with C2 already innocent and C1 criminal, the number of innocents among C3, C4, and C5 must be even. Since C3 and C4 contribute exactly one innocent (odd), C5 must also be innocent to make that even. Therefore, we can determine that Xavi (C5) is INNOCENT.
10.C4 · Sofia → INNOCENT
Row 4 has Phil at A4 and Ruby at B4 already known as criminals, leaving only Sofia at C4 and Thor at D4. Xavi’s clue says there are exactly 2 innocents in row 4. Since the only two open spots are C4 and D4, those two must be the two innocents. Therefore, we can determine that C4 Sofia is INNOCENT.
11.C3 · Nancy → CRIMINAL
Maria is at B3; her neighbors are A2, B2, C2, A3 (Karen), C3 (Nancy), A4, B4, and C4. The clue says she has exactly four innocent neighbors, and Karen is one of them. Among these neighbors, A2 (Gabe), C2 (Isaac), A3 (Karen), and C4 (Sofia) are already known innocents—four in total. That fills the quota, so every other neighbor of Maria must be criminal, including Nancy at C3. Therefore, we can determine that C3 · Nancy is CRIMINAL.
12.D4 · Thor → INNOCENT
Look at the four corners: A1, D1, A5, and D5. A1 has two criminal neighbors (Barb and Hazel), and A5 has two criminal neighbors (Phil and Ruby), so neither corner has exactly one criminal neighbor. D1 has exactly one criminal neighbor (Chloe), because Isaac and Joy are innocent. Since Joy says only one corner has exactly one criminal neighbor, D5 must not have exactly one; D5’s other neighbors (Sofia and Xavi) are innocent, so the only way for D5 not to have exactly one is for Thor not to be a criminal. Therefore, we can determine that D4 Thor is INNOCENT.
13.D3 · Ollie → INNOCENT
We focus on rows 3 and 4. Row 4 has exactly two criminals (Phil and Ruby), while in row 3 Maria and Nancy are already criminals, Karen is innocent, and only Ollie is unknown. The clue says rows 3 and 4 must have the same number of criminals, so row 3 must also total two criminals; if Ollie were a criminal, row 3 would have three. Therefore, we can determine that D3 Ollie is INNOCENT.
14.D1 · Eric → CRIMINAL
We compare rows 1 and 3. Row 3 already has exactly two criminals (Maria and Nancy). The clue says row 1 must have more criminals than row 3, so row 1 needs at least three; row 1 already has two criminals (Barb and Chloe) and one known innocent (Alice), leaving Eric as the only way to reach three. Therefore, we can determine that Eric is CRIMINAL.
15.D5 · Ziad → CRIMINAL
We focus on column D: Joy (D2), Ollie (D3), and Thor (D4) are already confirmed innocents. The clue says there are exactly 3 innocents in column D. Since those three already fill that quota, D5 cannot also be innocent, or there would be four. Therefore, we can determine that D5 Ziad is CRIMINAL.
16.B5 · Wally → CRIMINAL
The corners are A1 (Alice), D1 (Eric), A5 (Vince), and D5 (Ziad). Ziad’s clue says exactly one corner has exactly two criminal neighbors. Counting neighbors: A1 has exactly two criminal neighbors (Barb and Hazel), D1 has one (Chloe), and D5 has none (all three of its neighbors are innocent). A5’s neighbors are Phil (criminal), Ruby (criminal), and Wally, so A5 has two criminals if Wally is innocent, but three if Wally is criminal. Since A1 already satisfies the “exactly two” condition and it must be unique, A5 cannot also have exactly two, so Wally cannot be innocent and must be criminal. Therefore, we can determine that B5 Wally is CRIMINAL.
17.A5 · Vince → INNOCENT
Row 1 already has exactly three criminals (Barb, Chloe, and Eric). Wally’s clue says only one row can have exactly three criminals, so no other row may also total three. In Row 5, Wally and Ziad are criminals while Xavi is innocent; to prevent Row 5 from reaching three criminals, Vince must be innocent. Therefore, we can determine that A5 Vince is INNOCENT.