Clues by Sam Jan 20, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👩⚖️
judge
C1
👩💻
coder
D1
👩💼
clerk
A2
👩✈️
pilot
B2
👨⚖️
judge
C2
👩💼
clerk
D2
👩💼
clerk
A3
👩✈️
pilot
B3
👩🎨
painter
C3
👩🎨
painter
D3
👩🎨
painter
A4
💂♀️
guard
B4
👮♀️
cop
C4
👩⚖️
judge
D4
👨💻
coder
A5
👩🎤
singer
B5
👨🎤
singer
C5
👮♀️
cop
D5
💂♀️
guard
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 — 4 criminals.
Clues by Sam answer for Jan 20, 2026 — a Medium solved in 15 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Grams (C2), Jen (A3), Karen (B3) and Penny (C4); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.A5 · Steph → INNOCENT
Wendy’s clue explicitly says “Steph is one of 2 innocents below Elle,” which directly states Steph’s status as an innocent. Therefore, we can determine that A5 · Steph is INNOCENT.
02.B4 · Olivia → INNOCENT, A4 · Nicola → INNOCENT
Penny is at C4, so the only people to her left in row 4 are Nicola at A4 and Olivia at B4. The clue says there are exactly 2 innocents to the left of Penny. Since there are exactly two people there, both of them must be the two innocents. Therefore, we can determine that Olivia (B4) is INNOCENT and Nicola (A4) is INNOCENT.
03.A3 · Jen → CRIMINAL
In column A, Elle is at A2, and the people below her are Jen (A3), Nicola (A4, already Innocent), and Steph (A5, already Innocent). The clue says Steph is one of exactly two innocents below Elle. So among Jen, Nicola, and Steph, exactly two are Innocent, with Steph being one of them; since Nicola is also already Innocent, those two are Steph and Nicola. That leaves Jen as the only one below Elle who is not innocent. Therefore, we can determine that A3 Jen is CRIMINAL.
04.C1 · Clair → INNOCENT
The clue fixes two facts about column C: there are exactly three innocents there, and only one of those three is a neighbor of Martha at D3. In column C, Martha’s neighbors are C2, C3, and C4; C1 and C5 are not her neighbors. So exactly one of C2–C4 is innocent, meaning the other two innocents in column C must be the non-neighbors C1 and C5. Since C5 Wendy is already known innocent, the remaining required non-neighbor innocent is C1 Clair. Therefore, we can determine that C1 · Clair is INNOCENT.
05.C2 · Grams → CRIMINAL
Clair (C1) and Wendy (C5) are already the two known innocents in column C, leaving C2–C4 as the three people between them. Olivia’s clue says column C has exactly three innocents in total and only one of them is Martha’s neighbor; since C1 and C5 are not neighbors of Martha (D3), exactly one of C2–C4 is the third innocent. Clair’s clue adds that the innocent “in between” must be neighboring Olivia (B4). Of C2–C4, only C3 and C4 neighbor Olivia; C2 does not. Therefore, we can determine that C2 (Grams) is CRIMINAL.
06.D2 · Heike → INNOCENT
Focus on Dayna at D1, a corner who has only three neighbors: Clair at C1 (already Innocent), Grams at C2 (Criminal), and Heike at D2 (unknown). The clue says everyone must have at least two innocent neighbors. Dayna already has one (Clair), and since Grams is a Criminal, the only way for Dayna to reach two innocent neighbors is for Heike to be Innocent. Therefore, we can determine that D2 · Heike is INNOCENT.
07.A2 · Elle → INNOCENT, B2 · Ficus → INNOCENT
Look at the bottom-right corner D5 (Yamini), whose only neighbors are C4 (Penny), D4 (Ron), and C5 (Wendy). Since everyone must have at least 2 innocent neighbors and C5 is already innocent, at least one of C4 or D4 must also be innocent, making row 4 contain at least three innocents (A4, B4, plus one of C4/D4). Heike’s clue says rows 2 and 4 have the same number of innocents, so row 2 must also have at least three. In row 2, C2 is criminal and D2 is innocent, so the only way to reach at least three innocents is for both A2 and B2 to be innocent. Therefore, we can determine that A2 · Elle is INNOCENT, B2 · Ficus is INNOCENT, and so on.
08.A1 · Addie → INNOCENT
Relevant spots: column C, where Clair, Grams, Lynne, Penny, and Wendy sit; and column A, where Jen is a criminal and Addie is unknown. Olivia’s clue states there are exactly 3 innocents in column C, so column C must have exactly 2 criminals. Elle’s clue says only one column can have exactly 2 criminals. Since column C already does, no other column may also have 2 criminals. Column A already has one criminal (Jen). If Addie were also a criminal, column A would become a second column with exactly 2 criminals, which is not allowed. Therefore, Addie must be innocent. Therefore, we can determine that A1 Addie is INNOCENT.
09.B5 · Tom → INNOCENT, D5 · Yamini → INNOCENT
Look at Yamini in D5: she has only three neighbors—C4, D4, and C5—and C5 is already innocent. The rule that everyone has at least two innocent neighbors means at least one of C4 or D4 must also be innocent, so C4 and D4 cannot both be criminals, i.e., row 4 can have at most one criminal. Addie’s statement says row 4 has more criminals than row 5, so row 4 must have exactly one criminal and row 5 must have zero. In row 5 the only people who could possibly be criminals are Tom (B5) and Yamini (D5), so both must be innocent. Therefore, we can determine that B5 · Tom is INNOCENT, D5 · Yamini is INNOCENT, and so on.
10.D3 · Martha → INNOCENT
Column C already has Clair (C1) and Wendy (C5) confirmed innocent. Olivia says there are exactly 3 innocents in column C, and only one of those three is Martha’s neighbor, so among the two column‑C neighbors of Martha—Lynne (C3) and Penny (C4)—exactly one is innocent and the other is a criminal. Ron (D4) is neighbored by Martha (D3), Lynne (C3), Penny (C4), and two known innocents, Yamini (D5) and Wendy (C5). Yamini says the number of criminals around Ron is odd, so among {Martha, Lynne, Penny} the count of criminals must be odd. Since Lynne and Penny contribute exactly one criminal between them, the total around Ron is odd only if Martha adds zero criminals—that is, Martha is innocent. Therefore, we can determine that D3 · Martha is INNOCENT.
11.B1 · Binder → INNOCENT, D1 · Dayna → INNOCENT
D5 (Yamini) is a corner with neighbors C4, D4, and C5. From “everyone has at least 2 innocent neighbors,” and since C5 is already innocent, at least one of C4 or D4 must also be innocent so that D5 has two innocent neighbors. Row 4 already has A4 and B4 innocent; with at least one of C4 or D4 also innocent, Row 4 has at least 3 innocents. From “there are more innocents in row 1 than row 4,” Row 1 must therefore have 4 innocents. Row 1 already has A1 and C1 innocent, so the only way to reach 4 is for both B1 and D1 to be innocent. Therefore, we can determine that B1 (Binder) is INNOCENT, D1 (Dayna) is INNOCENT, and so on.
12.B3 · Karen → CRIMINAL
Relevant people: Tom at B5; those above him in column B (Binder B1, Ficus B2, Karen B3, Olivia B4); and Penny at C4. The clue says exactly one innocent above Tom is neighboring Penny. Only B3 and B4 are neighbors of Penny; B4 Olivia is already a known innocent, so she is that one. Since B3 Karen also neighbors Penny, she cannot be innocent or the count would exceed one. Therefore, we can determine that B3 · Karen is CRIMINAL.
13.C3 · Lynne → INNOCENT
In row 3, A3 Jen and B3 Karen are criminals, D3 Martha is innocent, and only C3 Lynne is unknown. The clue says there are exactly two innocents in row 3. With Martha already one innocent and the other two known to be criminals, Lynne must be the second innocent to make the total exactly two. Therefore, we can determine that C3 Lynne is INNOCENT.
14.C4 · Penny → CRIMINAL
Relevant spots are column C (Clair at C1, Lynne at C3, Penny at C4, Wendy at C5) and Martha at D3. The clue says there are exactly 3 innocents in column C, and only 1 of those three is Martha’s neighbor. We already have three confirmed innocents in column C—Clair, Lynne, and Wendy—so Penny cannot also be innocent without making four. Therefore, we can determine that C4 · Penny is CRIMINAL.
15.D4 · Ron → INNOCENT
Look at Yamini at D5, a corner with only three neighbors: Penny (C4), Wendy (C5), and Ron (D4). The clue says everyone must have at least two innocent neighbors. Penny is a criminal and Wendy is innocent, so Yamini has only one innocent neighbor unless Ron is also innocent. Therefore, we can determine that D4 Ron is INNOCENT.