Clues by Sam Mar 14, 2026 Answer – Full Solution Explained
Hard·Solved
A1
👮♂️
Andre
cop
B1
👨🍳
Chase
cook
C1
👩🌾
Emma
farmer
D1
👩🌾
Freya
farmer
A2
👮♂️
Gary
cop
B2
🕵️♀️
Hazel
sleuth
C2
👨⚕️
Isaac
doctor
D2
👷♂️
Jose
builder
A3
👮♂️
Larry
cop
B3
🕵️♂️
Mark
sleuth
C3
👷♀️
Nicole
builder
D3
👷♀️
Olivia
builder
A4
👨⚕️
Phil
doctor
B4
👨🏫
Rohan
teacher
C4
👩🏫
Susan
teacher
D4
👩🏫
Tina
teacher
A5
🕵️♀️
Uma
sleuth
B5
💂♂️
Vince
guard
C5
💂♀️
Wanda
guard
D5
👩🍳
Zoe
cook
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 8Criminal 12Unknown 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 · 8
[ B1 ] [ D1 ] [ A2 ] [ B2 ] [ C2 ] [ A3 ] [ B3 ] [ B5 ]
Criminal · 12
[ A1 ] [ C1 ] [ D2 ] [ C3 ] [ D3 ] [ A4 ] [ B4 ] [ C4 ] [ D4 ] [ A5 ] [ C5 ] [ D5 ]
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 · Andre
"I might know something... What is it worth to you?"
B1 · Chase
"There's an odd number of innocents neighboring me"
C1 · Emma
"There's an odd number of criminals neighboring Jose"
D1 · Freya
"Chase is one of Hazel's 5 innocent neighbors"
A2 · Gary
"There are no innocents above Zoe who neighbor Olivia"
B2 · Hazel
"Column B has more innocents than any other column"
C2 · Isaac
"Luckily most criminals give clues for free."
D2 · Jose
"Only 1 of the 2 criminals neighboring Uma is on the edges"
A3 · Larry
"There is only one innocent in column C"
B3 · Mark
"You are a fine example of high morale!"
C3 · Nicole
"If you had a jar opener, I could trade it for a clue..."
D3 · Olivia
"Exactly 2 of the 4 criminals neighboring Mark are below Emma"
A4 · Phil
"How about I give you a clue and you let me go?"
B4 · Rohan
"Exactly 1 innocent in row 5 is neighboring Wanda"
C4 · Susan
"There's an odd number of innocents to the right of Uma"
D4 · Tina
"There's an odd number of innocents neighboring Wanda"
A5 · Uma
"Susan has exactly 6 criminal neighbors"
B5 · Vince
"You seem to be fine without bribes."
C5 · Wanda
"Only 1 of the 5 criminals neighboring Rohan is in row 3"
D5 · Zoe
"A coconut for freedom? Just a bit scorched."
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 (17 steps)
D4 · Tina → CRIMINAL, D2 · Jose → CRIMINAL
Because: Above Zoe in column D, the only people who also neighbor Olivia at D3 are Jose at D2 and Tina at D4. Gary’s clue says there are no innocents among those people, so Jose and Tina cannot be innocent. Therefore, we can determine that D4 Tina is CRIMINAL and D2 Jose is CRIMINAL.
Clue:
"There are no innocents above Zoe who neighbor Olivia" — Gary (A2)
B4 · Rohan → CRIMINAL
Because: Uma is at A5, so her only neighbors are Phil at A4, Rohan at B4, and Vince at B5. Jose’s clue says that among Uma’s neighbors there are exactly two criminals, and exactly one of those two criminals is on the edge. Of Uma’s three neighbors, A4 and B5 are on the edge but B4 is the only one that is not on the edge, so the only way to have two criminal neighbors with exactly one of them on the edge is for B4 to be one of the criminals. Therefore, we can determine that B4 Rohan is CRIMINAL.
Clue:
"Only 1 of the 2 criminals neighboring Uma is on the edges" — Jose (D2)
C4 · Susan → CRIMINAL
Because: Wanda is at C5, so her neighbors are B4 Rohan, C4 Susan, D4 Tina, B5 Vince, and D5 Zoe. Rohan and Tina are already known criminals, and Rohan’s clue says that exactly one of the two row 5 neighbors of Wanda (Vince and Zoe) is innocent. That means Wanda already has exactly 1 innocent among Vince and Zoe, so to keep the total number of innocent neighbors of Wanda odd, Susan cannot be innocent, because 1 innocent plus Susan would make 2 innocents, which is even. Therefore, we can determine that C4 Susan is CRIMINAL.
Clue:
"There's an odd number of innocents neighboring Wanda" — Tina (D4)
"Exactly 1 innocent in row 5 is neighboring Wanda" — Rohan (B4)
C5 · Wanda → CRIMINAL
Because: Wanda is at C5, and her neighbors are Rohan (B4), Susan (C4), Tina (D4), Vince (B5), and Zoe (D5). Tina’s clue says there is an odd number of innocents neighboring Wanda; since Rohan, Susan, and Tina are already criminals, the only way for that neighbor count to be odd is for exactly one of Vince and Zoe to be innocent. Susan’s clue says there is an odd number of innocents to the right of Uma, meaning among Vince, Wanda, and Zoe the number of innocents is odd; with Vince and Zoe contributing exactly one innocent in total, Wanda cannot also be innocent or the total would become even. Therefore, we can determine that C5 Wanda is CRIMINAL.
Clue:
"There's an odd number of innocents neighboring Wanda" — Tina (D4)
"There's an odd number of innocents to the right of Uma" — Susan (C4)
A5 · Uma → CRIMINAL
Because: Uma is at A5, so her only neighbors are Phil at A4, Rohan at B4, and Vince at B5. Jose’s clue says Uma has exactly two criminal neighbors, and exactly one of those two is on the edge; since Rohan is already a known criminal and is not on the edge, the second criminal neighbor must be on the edge, so exactly one of Phil and Vince is criminal. Wanda’s clue says Rohan has exactly five criminal neighbors, with exactly one of them in row 3; among Rohan’s neighbors we already know Susan and Wanda are criminals, so the remaining three criminal neighbors must include exactly one of Larry/Mark/Nicole in row 3 and therefore exactly two of Phil, Uma, and Vince. Since Phil and Vince contribute exactly one criminal between them, Uma must be the second criminal in that trio. Therefore, we can determine that A5 Uma is CRIMINAL.
Clue:
"Only 1 of the 2 criminals neighboring Uma is on the edges" — Jose (D2)
"Only 1 of the 5 criminals neighboring Rohan is in row 3" — Wanda (C5)
D3 · Olivia → CRIMINAL
Because: Wanda at C5 neighbors Rohan, Susan, and Tina, who are already all criminals, plus Vince at B5 and Zoe at D5. Tina’s clue says Wanda has an odd number of innocent neighbors, so among Vince and Zoe exactly one must be innocent and the other must be criminal. Uma’s clue says Susan at C4 has exactly 6 criminal neighbors; since Rohan, Tina, and Wanda are already three of Susan’s neighbors and are criminals, that forces exactly 3 of Susan’s other neighboring people (Mark at B3, Nicole at C3, Olivia at D3, Vince at B5, and Zoe at D5) to be criminals. Because Vince and Zoe contribute exactly 1 criminal between them, that means Mark, Nicole, and Olivia together must contribute the other 2 criminals. Wanda’s clue about Rohan says that among the five criminal neighbors of Rohan, only one is in row 3, so Mark and Nicole (both in row 3 next to Rohan) cannot both be criminals; therefore Olivia must be criminal to make two criminals among Mark, Nicole, and Olivia. Therefore, we can determine that D3 · Olivia is CRIMINAL.
Clue:
"There's an odd number of innocents neighboring Wanda" — Tina (D4)
"Only 1 of the 5 criminals neighboring Rohan is in row 3" — Wanda (C5)
"Susan has exactly 6 criminal neighbors" — Uma (A5)
A3 · Larry → INNOCENT
Because: The key people here are Rohan at B4, Susan at C4, and Wanda at C5, since their clues all involve the overlapping group of nearby squares in rows 3–5. Wanda says Rohan has exactly five criminal neighbors and only one of those five is in row 3, so among A3, B3, and C3 there is exactly one criminal. Tina says Wanda has an odd number of innocent neighbors, and since Rohan, Susan, and Tina are already criminals next to Wanda, that forces exactly one of B5 and D5 to be innocent and the other to be criminal; then Uma’s clue that Susan has exactly six criminal neighbors means that among B3, C3, B5, and D5 there must be exactly two criminals, so B3 and C3 together must contain exactly one criminal. Since B3 and C3 already account for the one criminal allowed in row 3 next to Rohan, A3 cannot be criminal. Therefore, we can determine that A3 Larry is INNOCENT.
Clue:
"There's an odd number of innocents neighboring Wanda" — Tina (D4)
"Only 1 of the 5 criminals neighboring Rohan is in row 3" — Wanda (C5)
"Susan has exactly 6 criminal neighbors" — Uma (A5)
C1 · Emma → CRIMINAL
Because: Mark is at B3, so his neighbors include C2 (Isaac), C3 (Nicole), and C4 (Susan), and “below Emma” means the people in column C under C1. Olivia’s clue says Mark has exactly four criminal neighbors, and exactly two of those four are below Emma; since Susan at C4 is already a criminal and is below Emma, exactly one of Isaac (C2) or Nicole (C3) must be a criminal and the other must be innocent. Larry’s clue says there is only one innocent in column C, and since Susan (C4) and Wanda (C5) are already criminals, that one innocent must be among Emma (C1), Isaac (C2), and Nicole (C3). Because Isaac and Nicole already account for exactly one innocent between them, Emma cannot also be innocent. Therefore, we can determine that C1 (Emma) is CRIMINAL.
Clue:
"Exactly 2 of the 4 criminals neighboring Mark are below Emma" — Olivia (D3)
"There is only one innocent in column C" — Larry (A3)
D1 · Freya → INNOCENT
Because: Mark is at B3, and the only neighbors of Mark who are below Emma (in column C below C1) are C2 Isaac, C3 Nicole, and C4 Susan. Olivia’s clue says Mark has exactly four neighboring criminals, and exactly two of those four are below Emma; since Susan is already a criminal and is below Emma, exactly one of Isaac and Nicole must be a criminal to make the “below Emma” count equal to two. Jose at D2 has neighbors C1 Emma, D3 Olivia, C2 Isaac, C3 Nicole, and D1 Freya, and Emma’s clue says the number of criminals among Jose’s neighbors is odd; Emma and Olivia already contribute two criminals, and Isaac and Nicole contribute exactly one criminal, so Freya must contribute zero criminals to keep the total odd. Therefore, we can determine that D1 Freya is INNOCENT.
Clue:
"Exactly 2 of the 4 criminals neighboring Mark are below Emma" — Olivia (D3)
"There's an odd number of criminals neighboring Jose" — Emma (C1)
B1 · Chase → INNOCENT
Because: Freya’s clue says that Chase is one of Hazel’s 5 innocent neighbors, which directly states that Chase is an innocent person. Therefore, we can determine that B1 Chase is INNOCENT.
Clue:
"Chase is one of Hazel's 5 innocent neighbors" — Freya (D1)
B2 · Hazel → INNOCENT
Because: Hazel is at B2, so her neighbors are A1, B1 (Chase), C1, A2, C2, A3, B3, and C3. Freya’s clue says Chase is one of Hazel’s 5 innocent neighbors, which means Hazel has exactly five innocent neighbors; since Chase, Gary (A2), and Larry (A3) are already innocent and Emma (C1) is criminal, exactly two of A1, C2, B3, and C3 must be innocent. Wanda’s clue about Rohan forces exactly one of B3 and C3 to be criminal, so exactly one of B3 and C3 is innocent, which means exactly one of A1 and C2 must be innocent to reach those two innocents. Chase’s clue says there is an odd number of innocents neighboring him, and his neighbors include Gary (innocent) and Emma (criminal), so among A1, Hazel (B2), and Isaac (C2) there must be an even number of innocents; since exactly one of A1 and C2 is innocent, Hazel must also be innocent to make that count even. Therefore, we can determine that B2 Hazel is INNOCENT.
Clue:
"Only 1 of the 5 criminals neighboring Rohan is in row 3" — Wanda (C5)
"Chase is one of Hazel's 5 innocent neighbors" — Freya (D1)
"There's an odd number of innocents neighboring me" — Chase (B1)
A4 · Phil → CRIMINAL
Because: Mark is at B3, and his neighbors are Gary (A2), Hazel (B2), Isaac (C2), Larry (A3), Nicole (C3), Phil (A4), Rohan (B4), and Susan (C4). Olivia’s clue says that Mark has exactly four neighboring criminals, and exactly two of those four are below Emma; below Emma (C1) means in column C, so among Mark’s neighbors only Isaac (C2), Nicole (C3), and Susan (C4) qualify. Susan is already a criminal, so to make exactly two neighboring criminals below Emma, exactly one of Isaac or Nicole must also be a criminal; that accounts for only one additional criminal beyond the already-criminal Rohan and Susan, so the fourth neighboring criminal must be Phil. Therefore, we can determine that A4 Phil is CRIMINAL.
Clue:
"Exactly 2 of the 4 criminals neighboring Mark are below Emma" — Olivia (D3)
B5 · Vince → INNOCENT
Because: Uma is at A5, so her neighbors are Phil at A4, Rohan at B4, and Vince at B5. Jose’s clue says that Uma has exactly two neighboring criminals, and among those two criminals exactly one is on an edge. Phil is a criminal on an edge (A4), and Rohan is a criminal not on an edge (B4), which already matches the clue’s “two criminals with only one on the edges.” That leaves no room for Vince to also be a criminal neighbor of Uma, so Vince must be innocent. Therefore, we can determine that B5 Vince is INNOCENT.
Clue:
"Only 1 of the 2 criminals neighboring Uma is on the edges" — Jose (D2)
D5 · Zoe → CRIMINAL
Because: Wanda is at C5, so her neighbors are B4 Rohan, C4 Susan, D4 Tina, B5 Vince, and D5 Zoe. Tina’s clue says the number of innocents among Wanda’s neighbors is odd. Right now, Vince is the only known innocent among those neighbors, while Rohan, Susan, and Tina are all criminals, so the innocence count is 1 plus whatever Zoe is. If Zoe were innocent, that would make 2 innocents total, which is even, so Zoe cannot be innocent and must be criminal to keep the total odd. Therefore, we can determine that D5 Zoe is CRIMINAL.
Clue:
"There's an odd number of innocents neighboring Wanda" — Tina (D4)
A1 · Andre → CRIMINAL, B3 · Mark → INNOCENT
Because: Hazel at B2 says Chase is one of her 5 innocent neighbors, so among Hazel’s eight neighboring spaces there are exactly five innocents. Around Hazel, we already see three definite innocents (Chase at B1, Gary at A2, and Larry at A3) and one definite criminal (Emma at C1), so among the four unknown neighbors A1, C2, B3, and C3, exactly two must be innocent. Olivia’s clue about Mark at B3 says Mark has exactly four criminal neighbors; since A4, B4, and C4 are already criminals, exactly one of C2 and C3 must also be criminal, meaning exactly one of C2 and C3 is innocent. That means Hazel’s required “two innocents among A1, C2, B3, C3” can only happen if exactly one of A1 and Mark (B3) is innocent. Finally, Hazel also says column B has more innocents than any other column; if Mark (B3) were criminal, column B would have only three innocents, and then A1 would have to be criminal to keep column A from tying column B, but we just established exactly one of A1 and Mark is innocent. So Mark must be innocent, which forces A1 to be criminal. Therefore, we can determine that A1 (Andre) is CRIMINAL and B3 (Mark) is INNOCENT.
Clue:
"Exactly 2 of the 4 criminals neighboring Mark are below Emma" — Olivia (D3)
"Chase is one of Hazel's 5 innocent neighbors" — Freya (D1)
"Column B has more innocents than any other column" — Hazel (B2)
C2 · Isaac → INNOCENT
Because: Chase is at B1, so his neighbors are A1 Andre, C1 Emma, A2 Gary, B2 Hazel, and C2 Isaac. His clue says the number of innocents among those neighbors is odd. Andre and Emma are criminals, while Gary and Hazel are innocents, so we currently have exactly 2 innocent neighbors, which is even. The only remaining neighbor who can change that count is Isaac, so Isaac must be innocent to make the total 3, which is odd. Therefore, we can determine that C2 Isaac is INNOCENT.
Clue:
"There's an odd number of innocents neighboring me" — Chase (B1)
C3 · Nicole → CRIMINAL
Because: Mark is at B3, so his eight neighbors are Gary, Hazel, Isaac, Larry, Nicole, Phil, Rohan, and Susan. Olivia’s clue talks about “the 4 criminals neighboring Mark,” which means Mark must have exactly four criminal neighbors. Right now, Phil, Rohan, and Susan are already criminals, and the only remaining neighbor who could make the total reach four is Nicole at C3. Therefore, we can determine that C3 Nicole is CRIMINAL.
Clue:
"Exactly 2 of the 4 criminals neighboring Mark are below Emma" — Olivia (D3)