Clues by Sam Feb 14, 2026 Answer – Full Solution Explained
Hard·Solved
A1
🥷
Amy
ninja
B1
🕵️♀️
Bonnie
sleuth
C1
👩⚖️
Claire
judge
D1
🕵️♀️
Dana
sleuth
A2
👨🌾
Ethan
farmer
B2
👩🎨
Flora
painter
C2
🥷
Gus
ninja
D2
👨⚖️
Hank
judge
A3
💀
Ike
skeleton
B3
👩🏫
Joy
teacher
C3
👨🚀
Kumar
astronaut
D3
👩💻
Mary
coder
A4
👨✈️
Nick
pilot
B4
👨🏫
Ollie
teacher
C4
👨🌾
Paul
farmer
D4
👨🔧
Rohan
mech
A5
👮♀️
Uma
cop
B5
👮♀️
Wanda
cop
C5
👩💻
Xia
coder
D5
👨✈️
Ziad
pilot
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 11Criminal 9Unknown 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 · 11
[ A1 ] [ D1 ] [ A2 ] [ B2 ] [ B3 ] [ C3 ] [ D3 ] [ B4 ] [ D4 ] [ B5 ] [ D5 ]
Criminal · 9
[ B1 ] [ C1 ] [ C2 ] [ D2 ] [ A3 ] [ A4 ] [ C4 ] [ A5 ] [ C5 ]
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 · Amy
"Guess I was hiding in plain sight. Luckily I'm innocent!"
B1 · Bonnie
"I like tag more. Don't catch me even if you can!"
C1 · Claire
"Only 1 of the 4 innocents neighboring Joy is Uma's neighbor"
D1 · Dana
"There were ninjas? Guess I need glasses."
A2 · Ethan
"Uma is one of Wanda's 4 criminal neighbors"
B2 · Flora
"Hey, why is there a ninja hiding behind my painting?"
C2 · Gus
"How did you find me!!?"
D2 · Hank
"Each column has at least one criminal"
A3 · Ike
"Someone hid me in the closet..."
B3 · Joy
"Are we playing hide 'n seek? Fun!"
C3 · Kumar
"I found someone hiding on the moon"
D3 · Mary
"Gus is one of my 3 criminal neighbors"
A4 · Nick
"I'd rathe play tag. Catch me if you can!"
B4 · Ollie
"Only one row has exactly 3 innocents"
C4 · Paul
"There are as many innocent coders as there are innocent sleuths"
D4 · Rohan
"There's an odd number of innocents neighboring Ollie"
A5 · Uma
"Each row has at least 2 innocents"
B5 · Wanda
"There are as many criminal cops as there are criminal sleuths"
C5 · Xia
"There's an odd number of criminals neighboring Gus"
D5 · Ziad
"Exactly 1 innocent in row 1 is neighboring Gus"
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)
A5 · Uma → CRIMINAL
Because: Ethan’s clue says that Uma is one of Wanda’s 4 criminal neighbors, which directly states that Uma is a criminal neighbor of Wanda. Therefore, we can determine that A5 Uma is CRIMINAL.
Clue:
"Uma is one of Wanda's 4 criminal neighbors" — Ethan (A2)
C5 · Xia → CRIMINAL, D4 · Rohan → INNOCENT
Because: Wanda is at B5, and her neighbors are A4 Nick, B4 Ollie, C4 Paul, A5 Uma, and C5 Xia. Ethan’s clue says Uma is one of Wanda’s 4 criminal neighbors, so among those five neighbors exactly four are criminals and exactly one is innocent; since Uma is already a criminal, the one innocent must be among Nick, Ollie, Paul, and Xia. Uma’s clue says row 4 must contain at least two innocents, but row 4’s only people who could match that “one innocent among Nick, Ollie, Paul, Xia” are Nick, Ollie, and Paul, because Xia is not in row 4 and Rohan is not Wanda’s neighbor. That means the single innocent among Nick, Ollie, Paul, and Xia must be in row 4 (so Xia cannot be the innocent there), which forces Xia to be a criminal, and then row 4 still needs a second innocent beyond the one among Nick/Ollie/Paul, so Rohan must be innocent. Therefore, we can determine that C5 Xia is CRIMINAL and D4 Rohan is INNOCENT.
Clue:
"Uma is one of Wanda's 4 criminal neighbors" — Ethan (A2)
"Each row has at least 2 innocents" — Uma (A5)
D5 · Ziad → INNOCENT, B5 · Wanda → INNOCENT
Because: In row 5, the people are Uma at A5, Wanda at B5, Xia at C5, and Ziad at D5. Uma’s clue says that each row must contain at least 2 innocents. Since Uma is already known to be a criminal and Xia is already known to be a criminal, row 5 already has two criminals, so the only way for row 5 to still have at least two innocents is for both remaining people in that row, Wanda and Ziad, to be innocents. Therefore, we can determine that B5 Wanda is INNOCENT and D5 Ziad is INNOCENT.
Clue:
"Each row has at least 2 innocents" — Uma (A5)
A1 · Amy → INNOCENT
Because: In row 1, the people who are neighboring Gus at C2 are B1, C1, and D1, because they sit directly above or diagonally above him. Ziad’s clue says that exactly 1 innocent in row 1 is neighboring Gus, so among B1, C1, and D1 there is exactly one innocent. Uma’s clue says each row has at least 2 innocents, so row 1 must contain at least two innocents total; since only one of B1, C1, and D1 can be an innocent, the other innocent in row 1 must be A1. Therefore, we can determine that A1 Amy is INNOCENT.
Clue:
"Each row has at least 2 innocents" — Uma (A5)
"Exactly 1 innocent in row 1 is neighboring Gus" — Ziad (D5)
C1 · Claire → CRIMINAL
Because: The people in row 1 who neighbor Gus at C2 are Bonnie at B1, Claire at C1, and Dana at D1. Ziad’s clue says that among those three row-1 neighbors of Gus, exactly one is innocent. Wanda’s clue says the number of criminal cops equals the number of criminal sleuths; since the only cops are Uma (criminal) and Wanda (innocent), there is exactly 1 criminal cop, so there must be exactly 1 criminal sleuth, meaning exactly one of the two sleuths Bonnie (B1) and Dana (D1) is criminal and the other is innocent. So within the three people B1, C1, and D1, we already know exactly one of B1 or D1 is the single innocent required, which forces Claire at C1 to be criminal to keep the total number of innocents among the three equal to exactly one. Therefore, we can determine that C1 Claire is CRIMINAL.
Clue:
"Exactly 1 innocent in row 1 is neighboring Gus" — Ziad (D5)
"There are as many criminal cops as there are criminal sleuths" — Wanda (B5)
C4 · Paul → CRIMINAL
Because: Wanda is at B5, and her neighbors are A4 Nick, B4 Ollie, C4 Paul, A5 Uma, and C5 Xia. Ethan’s clue says Wanda has 4 criminal neighbors and that Uma is one of them, so among Nick, Ollie, and Paul exactly one is innocent and the other two are criminals (since Uma and Xia are already criminals). Joy is at B3, and the only neighbors Joy has who are also neighbors of Uma (A5) are Nick (A4) and Ollie (B4); Claire’s clue says that among Joy’s four innocent neighbors, exactly one is a neighbor of Uma, so exactly one of Nick and Ollie is innocent. That means the single innocent neighbor of Wanda among {Nick, Ollie, Paul} must be Nick or Ollie, so Paul cannot be that innocent neighbor and must be criminal. Therefore, we can determine that C4 Paul is CRIMINAL.
Clue:
"Uma is one of Wanda's 4 criminal neighbors" — Ethan (A2)
"Only 1 of the 4 innocents neighboring Joy is Uma's neighbor" — Claire (C1)
D3 · Mary → INNOCENT
Because: Bonnie at B1 and Dana at D1 are the only sleuths, and Mary at D3 is the only coder who could possibly be innocent because Xia at C5 is already a criminal. Uma’s clue says each row has at least 2 innocents, and in row 1 we already have Amy at A1 as innocent and Claire at C1 as criminal, so at least one of Bonnie or Dana must be innocent. Paul’s clue says the number of innocent coders equals the number of innocent sleuths, so since there is at least one innocent sleuth, there must be at least one innocent coder. That forces Mary to be innocent. Therefore, we can determine that D3 Mary is INNOCENT.
Clue:
"Each row has at least 2 innocents" — Uma (A5)
"There are as many innocent coders as there are innocent sleuths" — Paul (C4)
C2 · Gus → CRIMINAL
Because: Mary at D3 says, “Gus is one of my 3 criminal neighbors,” which directly states that Gus is a criminal neighbor of hers. Since all clues are truthful, Gus’s identity is fixed by this statement. Therefore, we can determine that C2 Gus is CRIMINAL.
Clue:
"Gus is one of my 3 criminal neighbors" — Mary (D3)
B2 · Flora → INNOCENT
Because: Mary at D3 says she has exactly three criminal neighbors and that Gus at C2 is one of them; since Paul at C4 is also already a criminal and Rohan at D4 is innocent, the third criminal neighbor of Mary must be exactly one of Hank at D2 or Kumar at C3. Claire’s clue is about “the 4 innocents neighboring Joy” at B3, and the only neighbors Joy shares with Uma at A5 are Nick at A4 and Ollie at B4, so exactly one of Nick or Ollie is an innocent neighbor of Joy; with Ethan at A2 already an innocent neighbor of Joy and Gus and Paul already criminal neighbors of Joy, that forces exactly two of the three people Flora (B2), Ike (A3), and Kumar (C3) to be innocents. If Kumar is the criminal among those three then Flora is immediately one of those forced innocents; if Kumar is innocent instead, then Hank must be Mary’s third criminal neighbor, and row 2 would then have Ethan as its only known innocent unless Flora is innocent, which Uma’s “each row has at least 2 innocents” does not allow. Therefore, we can determine that B2 Flora is INNOCENT.
Clue:
"Each row has at least 2 innocents" — Uma (A5)
"Only 1 of the 4 innocents neighboring Joy is Uma's neighbor" — Claire (C1)
"Gus is one of my 3 criminal neighbors" — Mary (D3)
B3 · Joy → INNOCENT
Because: Gus is at C2, so his neighbors include B1 and D1 in row 1, and also Joy at B3 plus Hank at D2 and Kumar at C3. Ziad’s clue says that among the row 1 neighbors of Gus (B1, C1, and D1), exactly one is innocent; since Claire at C1 is already a criminal, that means exactly one of Bonnie at B1 and Dana at D1 is a criminal. Mary’s clue says she has exactly three criminal neighbors and Gus is one of them; with Paul also a criminal neighbor and Rohan an innocent neighbor, that forces exactly one of Hank and Kumar to be criminal. Xia’s clue says Gus has an odd number of criminal neighbors, and since Claire already makes that count odd by herself, the five unknown neighbors (B1, D1, Hank, Joy, and Kumar) must contain an even number of criminals; but Hank and Kumar contribute exactly one criminal, so Bonnie, Dana, and Joy together must contribute an odd number of criminals. Because Bonnie and Dana together contribute exactly one criminal, Joy must contribute zero criminals, so she is innocent. Therefore, we can determine that B3 Joy is INNOCENT.
Clue:
"Exactly 1 innocent in row 1 is neighboring Gus" — Ziad (D5)
"Gus is one of my 3 criminal neighbors" — Mary (D3)
"There's an odd number of criminals neighboring Gus" — Xia (C5)
B4 · Ollie → INNOCENT
Because: Ollie is at B4, and his neighbors include Joy and Wanda (both innocent) plus the three undecided people Ike, Kumar, and Nick. Rohan’s clue says the number of innocents neighboring Ollie is odd, so besides Joy and Wanda, an odd number of Ike, Kumar, and Nick must be innocent. Claire’s clue says Joy has exactly four innocent neighbors, and exactly one of those four is also a neighbor of Uma. The only neighbors Joy has who are also neighbors of Uma are Nick and Ollie, so exactly one of Nick and Ollie is innocent, and since Joy needs two more innocent neighbors besides Ethan and Flora, exactly one of Ike and Kumar is innocent. That means among Ike, Kumar, and Nick, there is exactly one innocent from {Ike, Kumar} plus whatever Nick is; to make the total number of Ollie’s innocent neighbors odd, Nick must be a criminal, which forces Ollie to be the innocent one among {Nick, Ollie}. Therefore, we can determine that B4 Ollie is INNOCENT.
Clue:
"There's an odd number of innocents neighboring Ollie" — Rohan (D4)
"Only 1 of the 4 innocents neighboring Joy is Uma's neighbor" — Claire (C1)
A4 · Nick → CRIMINAL
Because: Wanda is at B5, so her neighbors are A4 Nick, B4 Ollie, C4 Paul, A5 Uma, and C5 Xia. Ethan’s clue says that Uma is one of Wanda’s 4 criminal neighbors, meaning Wanda has exactly four neighbors who are criminals. Among Wanda’s neighbors, Ollie is already known innocent, and Paul, Uma, and Xia are already known criminals, so Wanda currently has three criminal neighbors with only Nick undecided. To reach the required total of four criminal neighbors, Nick must be the remaining criminal neighbor. Therefore, we can determine that A4 Nick is CRIMINAL.
Clue:
"Uma is one of Wanda's 4 criminal neighbors" — Ethan (A2)
D2 · Hank → CRIMINAL
Because: Ollie is at B4, and his neighbors are A3 Ike, B3 Joy, C3 Kumar, A4 Nick, C4 Paul, A5 Uma, B5 Wanda, and C5 Xia. Among these, Joy and Wanda are already known innocents, while Nick, Paul, Uma, and Xia are already known criminals, so the only neighbors whose innocence could change the count are Ike and Kumar. Rohan’s clue says there is an odd number of innocents neighboring Ollie, so exactly one of Ike and Kumar must be innocent, which makes row 3 (A3–D3) have exactly three innocents total because Joy and Mary are innocent and only one of Ike/Kumar adds one more innocent. Ollie’s clue says only one row has exactly 3 innocents, so since row 3 already must be that row, row 2 cannot also have 3 innocents. Row 2 would have 3 innocents only if Hank at D2 were innocent, so Hank must be criminal. Therefore, we can determine that D2 Hank is CRIMINAL.
Clue:
"There's an odd number of innocents neighboring Ollie" — Rohan (D4)
"Only one row has exactly 3 innocents" — Ollie (B4)
C3 · Kumar → INNOCENT
Because: Mary is at D3, so her neighbors are C2 (Gus), D2 (Hank), C3 (Kumar), C4 (Paul), and D4 (Rohan). Her clue says that she has exactly 3 criminal neighbors, and that Gus is one of those 3. We already know Gus is a criminal, and Mary’s other neighbors Hank and Paul are also criminals, making the total of 3 criminals among her neighbors. Since Rohan is already known to be innocent, the only remaining neighbor is Kumar, and he cannot be an additional criminal without exceeding the “3 criminal neighbors” count. Therefore, we can determine that C3 Kumar is INNOCENT.
Clue:
"Gus is one of my 3 criminal neighbors" — Mary (D3)
A3 · Ike → CRIMINAL
Because: Joy is at B3, so her neighbors are Ethan (A2), Flora (B2), Gus (C2), Ike (A3), Kumar (C3), Nick (A4), Ollie (B4), and Paul (C4). Claire’s clue talks about “the 4 innocents neighboring Joy,” which can only be true if Joy has exactly four innocent neighbors in total. Right now, we already know four of Joy’s neighbors are innocent: Ethan, Flora, Kumar, and Ollie, so the remaining unknown neighbor, Ike, cannot also be innocent. Therefore, we can determine that A3 Ike is CRIMINAL.
Clue:
"Only 1 of the 4 innocents neighboring Joy is Uma's neighbor" — Claire (C1)
B1 · Bonnie → CRIMINAL
Because: Look down column B: Flora at B2, Joy at B3, Ollie at B4, and Wanda at B5 are all already known to be innocent, leaving only Bonnie at B1 in that column still undecided. Hank’s clue says that each column has at least one criminal, so column B must contain a criminal somewhere. Since Bonnie is the only person in column B who could still be a criminal, she must be the one. Therefore, we can determine that B1 Bonnie is CRIMINAL.
Clue:
"Each column has at least one criminal" — Hank (D2)
D1 · Dana → INNOCENT
Because: In row 1, the people are Amy at A1, Bonnie at B1, Claire at C1, and Dana at D1. Uma’s clue says each row has at least 2 innocents. Since Amy is already innocent and Bonnie and Claire are both criminals, the only way for row 1 to reach at least 2 innocents is for Dana to be innocent. Therefore, we can determine that D1 Dana is INNOCENT.
Clue:
"Each row has at least 2 innocents" — Uma (A5)