Clues by Sam Jan 21, 2026 Answer – Full Solution Explained
A1
👩🌾
farmer
B1
🕵️♂️
sleuth
C1
👷♀️
builder
D1
🕵️♂️
sleuth
A2
👨🏫
teacher
B2
👨🏫
teacher
C2
👷♀️
builder
D2
👨🏫
teacher
A3
👩🌾
farmer
B3
👨💼
clerk
C3
👷♂️
builder
D3
👩🔧
mech
A4
👩🌾
farmer
B4
👩✈️
pilot
C4
👨💼
clerk
D4
👩🔧
mech
A5
👩⚕️
doctor
B5
👩✈️
pilot
C5
👨🔧
mech
D5
👨⚕️
doctor
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 Jan 21, 2026 — a Medium solved in 16 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Ben (B1), Chloe (C1), Frank (A2), Henry (B2), Keith (D2), Nick (C3), Olive (D3), Penny (A4), Vera (A5), Xavi (C5) and Zach (D5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.C2 · Julie → INNOCENT
The clue explicitly says “Julie is one of 2 innocents below Chloe,” which directly identifies Julie as an innocent. Therefore, we can determine that C2 Julie is INNOCENT.
02.C1 · Chloe → CRIMINAL
Relevant people are in column C: Chloe at C1, and below her Julie (C2), Nick (C3), Salil (C4), and Xavi (C5). Wanda says Julie is one of exactly two innocents below Chloe, so among C2–C5 there are exactly two innocents and therefore exactly two criminals. Julie says column B is the only column with exactly two criminals, so column C cannot also total two criminals. Since C2–C5 already include exactly two criminals, the only way for column C’s total not to be two is if Chloe adds another criminal, making three. Therefore, we can determine that C1 Chloe is CRIMINAL.
03.A2 · Frank → CRIMINAL
Keith is at D2 and Henry is at B2; the people to Keith’s left in row 2 are A2 Frank, B2 Henry, and C2 Julie. Among these, the ones who neighbor Henry are A2 Frank and C2 Julie. The clue says exactly one innocent to the left of Keith is neighboring Henry; since Julie (C2) is already known to be innocent and neighbors Henry, Frank (A2) cannot also be innocent. Therefore, we can determine that A2 Frank is CRIMINAL.
04.A5 · Vera → CRIMINAL
Xavi is at C5, so the people to his left in row 5 are A5 (Vera) and B5 (Wanda). The clue says there is only one innocent to the left of Xavi. B5 Wanda is already known to be innocent, so she must be that one, which means A5 Vera cannot be innocent. Therefore, we can determine that A5 Vera is CRIMINAL.
05.B4 · Ruby → INNOCENT
Relevant people: Column B has Ben (B1), Henry (B2), Mark (B3), Ruby (B4), and Wanda (B5), with Wanda already known to be innocent. Julie says column B has exactly 2 criminals. Vera says there’s an odd number of innocents above Ruby—that is, among Ben, Henry, and Mark. Among three people, an odd count of innocents means 1 or 3; to end with exactly 2 criminals in column B while Wanda is innocent, the trio above Ruby must contribute exactly 2 criminals, which means exactly 1 of them is innocent. With those 2 criminals already in B1–B3, Ruby cannot add another criminal. Therefore, we can determine that B4 · Ruby is INNOCENT.
06.B2 · Henry → CRIMINAL
The people “above Wanda” are the four in column B at rows 1–4: Ben (B1), Henry (B2), Mark (B3), and Ruby (B4), and Ruby is already known to be innocent. Ruby’s clue says there are exactly two criminals among those above Wanda and they are connected, which in this straight column segment means they form a single continuous block with no innocent between them. Since B4 is innocent, the only way to place two connected criminals among B1–B3 is for one of them to be the middle spot B2, adjacent to either B1 or B3. Therefore, we can determine that B2 · Henry is CRIMINAL.
07.A4 · Penny → CRIMINAL
Consider Vera at A5: her only neighbors are Penny at A4, Ruby at B4 (Innocent), and Wanda at B5 (Innocent). The clue says everyone must have at least one criminal neighbor, so Vera must have at least one criminal neighbor. Since Ruby and Wanda are Innocent, the only possible criminal neighbor for Vera is Penny. Therefore, we can determine that A4 Penny is CRIMINAL.
08.C5 · Xavi → CRIMINAL
The two clerks are Chloe at C1 and Salil at C4; the people directly below them are Julie at C2 (already known INNOCENT) and Xavi at C5 (unknown). The clue says exactly one clerk has an innocent directly below. Chloe already does, so Salil must be the one who does not. The person directly below Salil is Xavi, so he cannot be innocent. Therefore, we can determine that C5 Xavi is CRIMINAL.
09.D1 · Ethan → INNOCENT
Keith is at D2, and the only people in row 1 who neighbor him are C1 (Chloe) and D1 (Ethan). The clue says exactly one innocent in row 1 is neighboring Keith. Chloe at C1 is already known to be a criminal, so she cannot be that innocent, leaving Ethan as the only possibility to satisfy the “exactly one” requirement. Therefore, we can determine that D1 · Ethan is INNOCENT.
10.D5 · Zach → CRIMINAL
Row 5 contains Vera (criminal), Wanda (innocent), Xavi (criminal), and Zach (unknown). Ethan’s clue says there is only one innocent in row 5. Wanda already fills that single innocent spot, so no one else in the row can be innocent; Zach must be a criminal. Therefore, we can determine that D5 Zach is CRIMINAL.
11.D2 · Keith → CRIMINAL
We’re focusing on Keith at D2 and the group “teachers.” The clue says that every teacher must be a criminal. Since Keith is known to be a teacher from earlier information, he must be a criminal under this rule. Therefore, we can determine that D2 Keith is CRIMINAL.
12.A3 · Lucy → INNOCENT
Below Chloe in column C are Julie (C2), Nick (C3), Salil (C4), and Xavi (C5). Wanda says Julie is one of exactly two innocents below Chloe; since Xavi is known criminal, exactly one of Nick and Salil is innocent. Keith says Mark (B3) has exactly four innocent neighbors; among his neighbors, Julie and Ruby are already innocent, leaving Lucy, Nick, and Salil to supply exactly two more innocents. Because only one of Nick or Salil can be innocent, Lucy must be the second. Therefore, we can determine that A3 · Lucy is INNOCENT.
13.A1 · Anna → CRIMINAL
The key people are in column C (Chloe at C1, Julie at C2, Nick at C3, Salil at C4, Xavi at C5) and column A (Anna at A1, with Frank, Lucy, Penny, and Vera below). Wanda says there are exactly two innocents below Chloe and Julie is one of them; since Julie is innocent and Xavi is criminal, exactly one of Nick or Salil is innocent and the other is criminal, making column C have exactly three criminals (Chloe, Xavi, and one of Nick/Salil). Lucy says column A has more criminals than any other column, so column A must have more than three criminals. Column A already has three criminals (Frank, Penny, Vera) and Lucy is innocent, so the only way to exceed three is for Anna to be a criminal. Therefore, we can determine that Anna is CRIMINAL.
14.B1 · Ben → CRIMINAL
Consider the four corners: A1 (Anna), D1 (Ethan), A5 (Vera), and D5 (Zach), and how many innocent neighbors each has. D1 Ethan has neighbors C1 (criminal), C2 (innocent), and D2 (criminal), so he already has exactly one innocent neighbor. A5 Vera’s neighbors A4 (criminal), B4 (innocent), and B5 (innocent) give her two innocent neighbors, so she is not the unique “one-innocent-neighbor” corner. The clue says only one corner has exactly one innocent neighbor, so no other corner besides D1 can have that count. At A1, both A2 and B2 are criminals, so Anna would have exactly one innocent neighbor only if Ben (B1) were innocent—which the clue forbids. Therefore, we can determine that B1 Ben is CRIMINAL.
15.B3 · Mark → INNOCENT
In column B, Ben (B1) and Henry (B2) are already criminals, Ruby (B4) and Wanda (B5) are innocents, and only Mark (B3) is unknown. The clue says column B is the only column with exactly 2 criminals, so column B must have exactly two criminals total. Since it already has two, Mark cannot also be a criminal and must be innocent. Therefore, we can determine that B3 · Mark is INNOCENT.
16.C3 · Nick → CRIMINAL, D3 · Olive → CRIMINAL, C4 · Salil → INNOCENT, D4 · Uma → INNOCENT
Row 3 has Lucy and Mark already innocent, so only Nick (C3) and Olive (D3) could be criminals there; row 4 has Penny already criminal and Ruby innocent, so only Salil (C4) and Uma (D4) could change that row’s count. Ben’s clue says row 3 must have more criminals than row 4. Since row 3 can have at most two criminals, row 4 must be held to at most one criminal; but row 4 already has one (Penny), so Salil and Uma must both be innocent. That means row 3 must have more than one criminal, so both Nick and Olive must be criminals. Therefore, we can determine that C3 Nick is CRIMINAL, D3 Olive is CRIMINAL, C4 Salil is INNOCENT, and D4 Uma is INNOCENT.