Clues by Sam Mar 13, 2026 Answer – Full Solution Explained
Hard·Solved
A1
👮♂️
Alex
cop
B1
👮♀️
Barb
cop
C1
👨⚕️
Ethan
doctor
D1
💂♀️
Freya
guard
A2
🕵️♂️
Gary
sleuth
B2
🕵️♀️
Helen
sleuth
C2
👷♂️
Isaac
builder
D2
👷♀️
Janet
builder
A3
🕵️♀️
Lisa
sleuth
B3
💂♂️
Martin
guard
C3
👷♂️
Nick
builder
D3
👨⚕️
Ollie
doctor
A4
👮♀️
Pam
cop
B4
👩🍳
Ruby
cook
C4
👨🌾
Steve
farmer
D4
👩🍳
Tina
cook
A5
👩✈️
Uma
pilot
B5
👩✈️
Wanda
pilot
C5
👨✈️
Xavi
pilot
D5
👨🌾
Zed
farmer
Final Board State
This puzzle is fully solved.
All characters have been identified as innocent or criminal based on today's clues.
Final Result
Innocent 15Criminal 5Unknown 0
See how each clue leads to the final result
Answer (spoilers)
A quick reference of the final identities. For explanations, see the reasoning above.
▶ Answer list (spoilers)
Innocent · 15
[ A1 ] [ B1 ] [ A2 ] [ B2 ] [ C2 ] [ D2 ] [ A3 ] [ C3 ] [ D3 ] [ A4 ] [ C4 ] [ A5 ] [ B5 ] [ C5 ] [ D5 ]
Criminal · 5
[ C1 ] [ D1 ] [ B3 ] [ B4 ] [ D4 ]
Clues
Raw text reference from the original puzzle
Original clue texts as provided in today's puzzle. No deductions or interpretations are applied here.
▶ Raw clues (original text)
A1 · Alex
"Only one person in row 2 has exactly 3 innocent neighbors"
B1 · Barb
"Maybe the weekend will be chill for a change?"
C1 · Ethan
"Exactly 2 of the 3 criminals neighboring Isaac are in row 1"
D1 · Freya
"I think we all learned something this week"
A2 · Gary
"There are more innocents in column A than column D"
B2 · Helen
"There are exactly 2 innocents to the right of Wanda"
C2 · Isaac
"There's an odd number of innocents above Xavi"
D2 · Janet
"Only one column has exactly 4 innocents"
A3 · Lisa
"Only one person in row 2 has exactly 5 innocent neighbors"
B3 · Martin
"Nothing is tougher than opening those coconuts or jars..."
C3 · Nick
"Only 1 of the 2 criminals neighboring Janet is in column D"
D3 · Ollie
"I'm unsure who's had it tougher this week..."
A4 · Pam
"The puzzles this week have been very tough!"
B4 · Ruby
"Only one person in row 1 has exactly 3 innocent neighbors"
C4 · Steve
"There are exactly 2 innocents above Ruby"
D4 · Tina
"Each row has at least 2 innocents"
A5 · Uma
"Only 1 of the 2 criminals in column B is Martin's neighbor"
B5 · Wanda
"I'm afraid of the weekend puzzles..."
C5 · Xavi
"There are at least 2 innocents above Steve"
D5 · Zed
"All innocents above Xavi are connected"
Answer Explanation
Full reasoning transcript (reference)
This is the full reasoning transcript for today's puzzle. For an interactive walkthrough, use Replay above.
▶ View full transcript (16 steps)
C5 · Xavi → INNOCENT, D5 · Zed → INNOCENT
Because: Wanda is at B5, so the only people to the right of Wanda in the same row are C5 (Xavi) and D5 (Zed). Helen’s clue says there are exactly 2 innocents to the right of Wanda, which must mean both of those right-side people are innocents, since there are only two positions there. Therefore, we can determine that C5 Xavi is INNOCENT and D5 Zed is INNOCENT.
Clue:
"There are exactly 2 innocents to the right of Wanda" — Helen (B2)
C2 · Isaac → INNOCENT
Because: The people above Steve are the three in column C at C1 (Ethan), C2 (Isaac), and C3 (Nick). Xavi’s clue says that at least two of those three are innocents. Zed’s clue says that any innocents in column C above Xavi (so among C1–C4) must all be connected, which in a single column means you cannot have two innocents with a gap between them. If Isaac at C2 were not innocent, then the only way to still have at least two innocents above Steve would be for Ethan at C1 and Nick at C3 to be innocents, but they would be separated by C2 and therefore not connected, breaking Zed’s clue. Therefore, we can determine that C2 Isaac is INNOCENT.
Clue:
"There are at least 2 innocents above Steve" — Xavi (C5)
"All innocents above Xavi are connected" — Zed (D5)
C3 · Nick → INNOCENT
Because: The people above Xavi in column C are Ethan at C1, Isaac at C2, Nick at C3, and Steve at C4. Xavi’s clue about “at least 2 innocents above Steve” refers to C1–C3, so among Ethan, Isaac, and Nick there are at least two innocents; since Isaac is already INNOCENT, that guarantees there is at least one more innocent in C1 or C3. Isaac also says there is an odd number of innocents above Xavi, so among C1–C4 the total cannot be 1 and must be 3. Zed says all innocents above Xavi are connected, and with exactly three innocents including Isaac at C2, the only possible connected groups in that column are C1–C3 or C2–C4, and both include Nick at C3. Therefore, we can determine that C3 Nick is INNOCENT.
Clue:
"There are at least 2 innocents above Steve" — Xavi (C5)
"All innocents above Xavi are connected" — Zed (D5)
"There's an odd number of innocents above Xavi" — Isaac (C2)
C1 · Ethan → CRIMINAL
Because: Janet is at D2, so her neighbors are C1 Ethan, C2 Isaac, D1 Freya, C3 Nick, and D3 Ollie. Nick’s clue says Janet has exactly two criminal neighbors in total, and exactly one of those criminals is in column D, meaning the other criminal neighbor must be in column C. Isaac at C2 and Nick at C3 are already confirmed INNOCENT, so the only remaining neighbor of Janet in column C who could be that required criminal is Ethan at C1. Therefore, we can determine that C1 Ethan is CRIMINAL.
Clue:
"Only 1 of the 2 criminals neighboring Janet is in column D" — Nick (C3)
C4 · Steve → INNOCENT
Because: Xavi is at C5, so the people above Xavi are the four cells in column C: C1 Ethan, C2 Isaac, C3 Nick, and C4 Steve. Isaac’s clue says the number of innocents among those four people is odd. We already know Isaac and Nick are innocents, and Ethan is a criminal, so there are currently exactly 2 innocents above Xavi; the only way to make that total odd is for Steve to also be an innocent, making 3. Therefore, we can determine that C4 Steve is INNOCENT.
Clue:
"There's an odd number of innocents above Xavi" — Isaac (C2)
D2 · Janet → INNOCENT
Because: Janet is at D2, and her neighbors include Ethan at C1, Freya at D1, and Ollie at D3. Nick’s clue says Janet has exactly two neighboring criminals, and only one of those two is in column D; since Ethan at C1 is already a criminal and is not in column D, exactly one of Freya (D1) or Ollie (D3) must be the other criminal neighbor of Janet. Ethan’s clue says Isaac at C2 has exactly two criminal neighbors in row 1, so with Ethan (C1) already a row 1 criminal, exactly one of Barb (B1) or Freya (D1) is criminal. Steve’s clue says there are exactly two innocents above Ruby (B4), and since Helen (B2) is already innocent, exactly one of Barb (B1) or Martin (B3) is innocent, so Barb and Martin must have opposite statuses; combining that with “Barb and Freya have opposite statuses” means Freya and Martin must have the same status. Since Freya and Ollie have opposite statuses, Martin and Ollie must also have opposite statuses, so exactly one of Martin or Ollie is criminal; Ethan’s clue also says that among Janet (D2), Martin (B3), and Ollie (D3), exactly one is criminal, so Janet cannot be criminal. Therefore, we can determine that D2 Janet is INNOCENT.
Clue:
"Only 1 of the 2 criminals neighboring Janet is in column D" — Nick (C3)
"There are exactly 2 innocents above Ruby" — Steve (C4)
"Exactly 2 of the 3 criminals neighboring Isaac are in row 1" — Ethan (C1)
D4 · Tina → CRIMINAL
Because: Janet is at D2, and her neighbors are C1 Ethan, C2 Isaac, C3 Nick, D1 Freya, and D3 Ollie. Nick’s clue says there are exactly two criminals among those neighbors, and exactly one of those two is in column D; since Ethan at C1 is already a criminal and is not in column D, the other criminal neighbor must be either Freya (D1) or Ollie (D3), making the other one of them innocent. That means column D already contains three innocents for sure: Janet, Zed, and whichever of Freya or Ollie is the innocent one. Janet’s own clue says only one column has exactly four innocents, and column C already has exactly four innocents (Isaac, Nick, Steve, and Xavi), so column D cannot also end up with exactly four innocents; the only way to prevent column D from reaching four innocents is for Tina at D4 to be a criminal. Therefore, we can determine that D4 Tina is CRIMINAL.
Clue:
"Only 1 of the 2 criminals neighboring Janet is in column D" — Nick (C3)
"Only one column has exactly 4 innocents" — Janet (D2)
A1 · Alex → INNOCENT
Because: Isaac is at C2, so his row 1 neighbors are B1, C1, and D1, and Ethan’s clue says Isaac has exactly three criminal neighbors in total, with exactly two of those criminals in row 1. Since C1 (Ethan) is already a criminal, exactly one of B1 or D1 must also be a criminal, and the third criminal neighbor must be elsewhere around Isaac (at B3 or D3, because B2, C3, and D2 are already innocents). That means row 1 already contains at least two criminals: C1 and whichever of B1 or D1 is the second row 1 criminal. Tina’s clue says each row has at least two innocents, so row 1 cannot have three criminals, which forces A1 to be innocent. Therefore, we can determine that A1 Alex is INNOCENT.
Clue:
"Exactly 2 of the 3 criminals neighboring Isaac are in row 1" — Ethan (C1)
"Each row has at least 2 innocents" — Tina (D4)
A3 · Lisa → INNOCENT
Because: Janet at D2 neighbors Ethan at C1, plus D1 and D3, and Nick at C3 and Isaac at C2. Nick’s clue says Janet has exactly two criminal neighbors, and only one of those two is in column D; since Ethan at C1 is already a criminal and not in column D, exactly one of D1 or D3 must be the second criminal and the other one must be innocent, so Janet has exactly three innocent neighbors (C2, C3, and the innocent one of D1/D3). Alex’s clue then makes Janet the only person in row 2 who has exactly three innocent neighbors, so Gary at A2 cannot have exactly three innocent neighbors. Gary already has two innocent neighbors (Alex at A1 and Helen at B2), and Steve’s clue makes exactly one of Barb at B1 or Martin at B3 innocent, so the only way for Gary to avoid landing on exactly three innocent neighbors is for Lisa at A3 to be innocent as well. Therefore, we can determine that A3 Lisa is INNOCENT.
Clue:
"Only 1 of the 2 criminals neighboring Janet is in column D" — Nick (C3)
"There are exactly 2 innocents above Ruby" — Steve (C4)
"Only one person in row 2 has exactly 3 innocent neighbors" — Alex (A1)
A2 · Gary → INNOCENT
Because: In row 2, Lisa says that only one person has exactly 5 innocent neighbors. Ethan’s clue about Isaac means Isaac has exactly three criminal neighbors in total, with exactly two of those criminals in row 1, so among B1 and D1 exactly one is criminal, and among B3 and D3 exactly one is criminal. Since Isaac’s neighbors also include the already-known innocents B2, D2, and C3, this guarantees that Isaac has exactly 5 innocent neighbors (those three, plus the one innocent from B1/D1, plus the one innocent from B3/D3). That means nobody else in row 2 is allowed to have exactly 5 innocent neighbors. Helen at B2 would have exactly 5 innocent neighbors precisely when exactly one of B1, A2, and B3 is innocent; Steve’s clue makes exactly one of B1 and B3 innocent (because B2 is already one of the two innocents above Ruby), so B2 would end up with exactly 5 innocent neighbors if A2 were criminal. Therefore A2 cannot be criminal, and Gary must be innocent. Therefore, we can determine that A2 Gary is INNOCENT.
Clue:
"There are exactly 2 innocents above Ruby" — Steve (C4)
"Exactly 2 of the 3 criminals neighboring Isaac are in row 1" — Ethan (C1)
"Only one person in row 2 has exactly 5 innocent neighbors" — Lisa (A3)
A4 · Pam → INNOCENT, A5 · Uma → INNOCENT
Because: Janet is at D2, and her neighbors are C1, C2, C3, D1, and D3. Nick’s clue says that Janet has exactly two criminal neighbors, and only one of those two is in column D; since C1 (Ethan) is already a criminal neighbor in column C, exactly one of D1 and D3 must be a criminal and the other must be innocent, so column D has exactly three innocents (D2, D5, and the innocent one of D1/D3). Gary’s clue says column A has more innocents than column D, so column A must have at least four innocents; but column C already has exactly four innocents (C2, C3, C4, C5), and Janet’s clue says only one column can have exactly four innocents, so column A cannot stop at four and must have five. That forces both A4 (Pam) and A5 (Uma) to be innocent. Therefore, we can determine that A4 Pam is INNOCENT and A5 Uma is INNOCENT.
Clue:
"Only 1 of the 2 criminals neighboring Janet is in column D" — Nick (C3)
"Only one column has exactly 4 innocents" — Janet (D2)
"There are more innocents in column A than column D" — Gary (A2)
B4 · Ruby → CRIMINAL
Because: The clue says there are exactly two criminals in column B, and among those two, exactly one is a neighbor of Martin at B3. The only positions in column B that are neighbors of Martin are B2 and B4, but Helen at B2 is already known to be innocent. That means the one column-B criminal who is Martin’s neighbor must be the person at B4, Ruby. Therefore, we can determine that B4 Ruby is CRIMINAL.
Clue:
"Only 1 of the 2 criminals in column B is Martin's neighbor" — Uma (A5)
B5 · Wanda → INNOCENT
Because: Ruby is at B4, so the people above her are Barb at B1, Helen at B2, and Martin at B3. Steve’s clue says exactly two of those three are innocents, and since Helen is already known to be innocent, that forces exactly one of Barb and Martin to be a criminal. Uma’s clue says there are exactly two criminals in column B in total, and Ruby is already one of them, so there can be only one more criminal anywhere else in column B. Since that one more criminal must be either Barb or Martin, Wanda at B5 cannot be the second criminal and must be innocent. Therefore, we can determine that B5 Wanda is INNOCENT.
Clue:
"There are exactly 2 innocents above Ruby" — Steve (C4)
"Only 1 of the 2 criminals in column B is Martin's neighbor" — Uma (A5)
B1 · Barb → INNOCENT, D1 · Freya → CRIMINAL
Because: Isaac is at C2, and his row 1 neighbors are B1 (Barb), C1 (Ethan), and D1 (Freya). Ethan’s clue says Isaac has exactly three criminal neighbors in total, and exactly two of those three are in row 1; since Ethan at C1 is already a criminal, that forces exactly one of Barb or Freya to be the second row 1 criminal, and the other must be innocent. Now use Ruby’s clue about row 1: Barb at B1 already has four innocent neighbors (Alex at A1, Gary at A2, Helen at B2, and Isaac at C2), so she cannot be the person in row 1 with exactly three innocent neighbors; Ethan at C1 has at least three innocent neighbors from B2, C2, and D2 and, with exactly one of Barb or Freya innocent, he would have four; Freya at D1 has only two innocent neighbors (Isaac at C2 and Janet at D2). That means the only way for there to be exactly one person in row 1 with exactly three innocent neighbors is for Alex at A1 to have three innocent neighbors, which happens only if Barb at B1 is innocent. With Barb fixed as innocent, Ethan’s clue then forces Freya at D1 to be the second row 1 criminal neighbor of Isaac. Therefore, we can determine that B1 Barb is INNOCENT and D1 Freya is CRIMINAL.
Clue:
"Exactly 2 of the 3 criminals neighboring Isaac are in row 1" — Ethan (C1)
"Only one person in row 1 has exactly 3 innocent neighbors" — Ruby (B4)
B3 · Martin → CRIMINAL
Because: Ruby is at B4, so the people above Ruby are exactly the three squares in column B above that: B1 Barb, B2 Helen, and B3 Martin. Steve’s clue says there are exactly 2 innocents above Ruby. Since Barb and Helen are already known to be innocent, those two already account for the full total of 2 innocents above Ruby. That means Martin cannot also be innocent, so he must be criminal. Therefore, we can determine that B3 Martin is CRIMINAL.
Clue:
"There are exactly 2 innocents above Ruby" — Steve (C4)
D3 · Ollie → INNOCENT
Because: Isaac is at C2, so his neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. The clue says “the 3 criminals neighboring Isaac,” which fixes that Isaac has exactly three criminal neighbors in total, and exactly two of those three are in row 1. We already see three criminals among Isaac’s neighbors: Ethan at C1 and Freya at D1 (both in row 1), and Martin at B3 (not in row 1), so those must be the full set of Isaac’s criminal neighbors. That leaves no room for Ollie at D3 to also be a criminal. Therefore, we can determine that D3 Ollie is INNOCENT.
Clue:
"Exactly 2 of the 3 criminals neighboring Isaac are in row 1" — Ethan (C1)