Clues by Sam Feb 27, 2026 Answer – Full Solution Explained
Tricky·Solved
A1
🕵️♀️
Amy
sleuth
B1
👮♀️
Celia
cop
C1
👨🍳
Derek
cook
D1
👨🏫
Erwin
teacher
A2
👩🌾
Flora
farmer
B2
👮♂️
Gary
cop
C2
👩🍳
Hope
cook
D2
👨🏫
Isaac
teacher
A3
👩🌾
Megan
farmer
B3
👮♀️
Nicole
cop
C3
👩🍳
Olga
cook
D3
👨🏫
Pip
teacher
A4
👨🌾
Rob
farmer
B4
🕵️♂️
Scott
sleuth
C4
👨⚕️
Tyler
doctor
D4
👩💻
Uma
coder
A5
🕵️♂️
Vince
sleuth
B5
👩⚕️
Wanda
doctor
C5
👩⚕️
Xia
doctor
D5
👨💻
Zed
coder
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 ] [ B2 ] [ B3 ] [ D3 ] [ C4 ] [ A5 ] [ D5 ]
Criminal · 12
[ A1 ] [ C1 ] [ A2 ] [ C2 ] [ D2 ] [ A3 ] [ C3 ] [ A4 ] [ B4 ] [ D4 ] [ B5 ] [ 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
"Only 1 of the 3 criminals neighboring Rob is in column A"
B1 · Celia
"Zed has only one innocent neighbor"
C1 · Derek
"Are you hanging on to your hats so we don't steal them again?"
D1 · Erwin
"I hid my hat in a safe."
A2 · Flora
"Amy is one of 4 criminals in column A"
B2 · Gary
"I left my hat at home. No one will steal it again!"
C2 · Hope
"There are exactly 2 innocents in row 1"
D2 · Isaac
"Celia and Derek have only one innocent neighbor in common"
A3 · Megan
"Haha! We were always after your socks! The hats were just a diversion!"
B3 · Nicole
"I glued my hat onto my head"
C3 · Olga
"I can't believe we're able to fill two days with hat jokes. Hats off!"
D3 · Pip
"Only 1 of the 3 criminals in row 2 is Amy's neighbor"
A4 · Rob
"Only one person in a corner has exactly 2 criminal neighbors"
B4 · Scott
"Only 1 of the 2 criminals neighboring Uma is a doctor"
C4 · Tyler
"Hey! Where are my socks! They did it again!"
D4 · Uma
"Exactly 2 of the 5 criminals neighboring Scott are in row 5"
A5 · Vince
"All my socks were taken while I looked after my hat!"
B5 · Wanda
"I've always wanted to wear many hats"
C5 · Xia
"Off the top of my head, I don't know anything better than hats"
D5 · Zed
"There's an odd number of innocents on the edges"
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 (15 steps)
C2 · Hope → CRIMINAL, D2 · Isaac → CRIMINAL
Because: Row 2 contains Flora at A2, Gary at B2, Hope at C2, and Isaac at D2, and the clue tells us there are exactly three criminals in that row. Amy is at A1, so in row 2 the only people who are Amy’s neighbors are A2 and B2; C2 and D2 are not neighbors of Amy. The clue also says that among the three row-2 criminals, only one is Amy’s neighbor, so at most one of A2 and B2 can be a criminal. That means the other two criminals in row 2 must be the two people who are not Amy’s neighbors, which are C2 and D2. Therefore, we can determine that C2 Hope is CRIMINAL and D2 Isaac is CRIMINAL.
Clue:
"Only 1 of the 3 criminals in row 2 is Amy's neighbor" — Pip (D3)
B2 · Gary → INNOCENT
Because: Celia is at B1 and Derek is at C1, and the only people who are neighbors of both of them are Gary at B2 and Hope at C2. Isaac’s clue says that among these shared neighbors, there is only one innocent. Since Hope at C2 is already known to be a criminal, the one innocent shared neighbor must be Gary at B2. Therefore, we can determine that B2 Gary is INNOCENT.
Clue:
"Celia and Derek have only one innocent neighbor in common" — Isaac (D2)
A2 · Flora → CRIMINAL
Because: Row 2 is Flora at A2, Gary at B2, Hope at C2, and Isaac at D2, and Pip’s clue says there are exactly three criminals in that row. Since Gary at B2 is already known to be innocent, the other three people in row 2 must be the three criminals. That forces Flora at A2 to be one of the criminals (and it also matches the clue’s “only 1 is Amy’s neighbor” part, because Amy’s only possible criminal neighbor in row 2 would then be Flora). Therefore, we can determine that A2 Flora is CRIMINAL.
Clue:
"Only 1 of the 3 criminals in row 2 is Amy's neighbor" — Pip (D3)
A1 · Amy → CRIMINAL
Because: Flora’s clue explicitly says that Amy is one of the 4 criminals in column A, which directly states Amy’s identity. Therefore, we can determine that A1 Amy is CRIMINAL.
Clue:
"Amy is one of 4 criminals in column A" — Flora (A2)
A4 · Rob → CRIMINAL
Because: In column A there must be exactly 4 criminals, and we already know Amy at A1 and Flora at A2 are criminals, so two of the remaining three people in column A (Megan at A3, Rob at A4, and Vince at A5) must be criminals. Rob’s only neighbors in column A are Megan (A3) and Vince (A5), and Amy’s clue says that among Rob’s three criminal neighbors, only one is in column A, so at most one of Megan and Vince can be a criminal. Since Megan and Vince can contribute at most one of the two needed criminals among A3, A4, and A5, Rob must be the other one. Therefore, we can determine that A4 Rob is CRIMINAL.
Clue:
"Amy is one of 4 criminals in column A" — Flora (A2)
"Only 1 of the 3 criminals neighboring Rob is in column A" — Amy (A1)
B1 · Celia → INNOCENT
Because: In row 1, the clue from Hope says there are exactly 2 innocents, so with four people in that row and Amy at A1 already known to be a criminal, exactly one of B1, C1, and D1 is a criminal. If Celia at B1 were that one criminal, then C1 would have to be innocent, which would make the corner D1 have exactly 2 criminal neighbors (Hope at C2 and Isaac at D2). But B1 being criminal would also make the corner A1 have exactly 2 criminal neighbors (Flora at A2 and Celia at B1), giving two corners with exactly 2 criminal neighbors, which Rob’s clue says cannot happen because only one corner can have that property. Therefore, we can determine that B1 (Celia) is INNOCENT.
Clue:
"There are exactly 2 innocents in row 1" — Hope (C2)
"Only one person in a corner has exactly 2 criminal neighbors" — Rob (A4)
C1 · Derek → CRIMINAL
Because: Zed is in the D5 corner, and the only people who neighbor him are Tyler at C4, Uma at D4, and Xia at C5. Celia’s clue says Zed has only one innocent neighbor, so among those three neighbors exactly one is innocent and the other two are criminals, meaning Zed has exactly 2 criminal neighbors. Rob’s clue says only one corner person has exactly 2 criminal neighbors, so no other corner can have exactly 2 criminal neighbors. In the D1 corner, Erwin’s neighbors are Derek at C1, Hope at C2, and Isaac at D2, and since Hope and Isaac are already criminals, the only way for D1 to not have exactly 2 criminal neighbors is for Derek to be criminal as well, making 3 criminal neighbors there. Therefore, we can determine that C1 Derek is CRIMINAL.
Clue:
"Only one person in a corner has exactly 2 criminal neighbors" — Rob (A4)
"Zed has only one innocent neighbor" — Celia (B1)
D1 · Erwin → INNOCENT
Because: Row 1 contains Amy at A1, Celia at B1, Derek at C1, and Erwin at D1. Hope’s clue says there are exactly 2 innocents in row 1, and we already know Amy and Derek are criminals while Celia is innocent. That means row 1 currently has exactly 1 confirmed innocent, so Erwin must be the second innocent to make the total exactly 2. Therefore, we can determine that D1 Erwin is INNOCENT.
Clue:
"There are exactly 2 innocents in row 1" — Hope (C2)
B4 · Scott → CRIMINAL, B5 · Wanda → CRIMINAL, B3 · Nicole → INNOCENT
Because: Rob is at A4, and his neighbors are A3, B3, B4, A5, and B5. Amy’s clue says that exactly three of those neighbors are criminals, and that only one of those criminals is in column A, so among B3, B4, and B5 there must be exactly two criminals. Celia’s clue says Zed at D5 has only one innocent neighbor, so Zed has exactly two criminal neighbors, which makes Zed the one and only corner person with exactly two criminal neighbors as required by Rob’s corner clue. That means Vince at A5 cannot have exactly two criminal neighbors; since A5 is adjacent to A4 (already criminal) and also to B4 and B5, and we still need two criminals among B3, B4, and B5, the only possibility is that both B4 and B5 are criminals, giving A5 three criminal neighbors. With B4 and B5 now criminal, Amy’s “exactly two criminals among B3, B4, B5” forces B3 to be innocent. Therefore, we can determine that B4 Scott is CRIMINAL, B5 Wanda is CRIMINAL, and B3 Nicole is INNOCENT.
Clue:
"Only 1 of the 3 criminals neighboring Rob is in column A" — Amy (A1)
"Only one person in a corner has exactly 2 criminal neighbors" — Rob (A4)
"Zed has only one innocent neighbor" — Celia (B1)
D4 · Uma → CRIMINAL
Because: Uma at D4 has two doctor neighbors, Tyler at C4 and Xia at C5. Scott’s clue says that exactly two of Uma’s neighbors are criminals, and among those two criminals exactly one is a doctor, so between Tyler and Xia exactly one must be a criminal and the other must be an innocent. Zed at D5 is a neighbor of Uma, and Celia’s clue says Zed has only one innocent neighbor among his three neighbors (Tyler, Xia, and Uma). Since Tyler and Xia already include one innocent, Uma cannot also be innocent, so Uma must be a criminal. Therefore, we can determine that D4 Uma is CRIMINAL.
Clue:
"Zed has only one innocent neighbor" — Celia (B1)
"Only 1 of the 2 criminals neighboring Uma is a doctor" — Scott (B4)
C3 · Olga → CRIMINAL
Because: Scott is at B4, and his neighbors include Rob and Wanda (both already criminals) plus Megan (A3), Olga (C3), Tyler (C4), Vince (A5), and Xia (C5). Uma’s clue says there are exactly 5 criminals among Scott’s neighbors, so those five unknown neighbors must contain exactly 3 criminals in total. Uma’s clue also says exactly 2 of Scott’s neighboring criminals are in row 5; since Wanda at B5 is already a criminal, that forces exactly one of Vince (A5) and Xia (C5) to be a criminal. Celia’s clue says Zed has only one innocent neighbor, and since Uma (D4) is a criminal neighbor of Zed, exactly one of Tyler (C4) and Xia (C5) is innocent, which means Tyler and Vince must have the same status (they are both the opposite of Xia), and Flora’s clue then makes Megan the opposite of Vince because column A needs exactly one more criminal between Megan and Vince. That setup guarantees that among Megan, Tyler, Vince, and Xia there are always exactly two criminals, so to reach the required three criminals among the five unknown neighbors of Scott, Olga must be the remaining criminal. Therefore, we can determine that C3 Olga is CRIMINAL.
Clue:
"Amy is one of 4 criminals in column A" — Flora (A2)
"Zed has only one innocent neighbor" — Celia (B1)
"Exactly 2 of the 5 criminals neighboring Scott are in row 5" — Uma (D4)
D5 · Zed → INNOCENT
Because: Uma is at D4, and her neighboring people are Olga at C3, Pip at D3, Tyler at C4, Xia at C5, and Zed at D5. Scott’s clue says that Uma has exactly two neighboring criminals, and that exactly one of those two criminals is a doctor. Olga at C3 is already a known criminal, but she is a cook, so she cannot be the doctor among those two criminal neighbors; that means the other criminal neighbor must be a doctor, which can only be Tyler at C4 or Xia at C5. Since that uses up the only remaining “criminal neighbor” slot and it must be a doctor, Zed at D5 cannot be one of Uma’s neighboring criminals. Therefore, we can determine that D5 Zed is INNOCENT.
Clue:
"Only 1 of the 2 criminals neighboring Uma is a doctor" — Scott (B4)
C5 · Xia → CRIMINAL
Because: The only unknown people on the edge are Megan at A3, Vince at A5, and Xia at C5. Flora’s clue says there are exactly 4 criminals in column A; since Amy at A1, Flora at A2, and Rob at A4 are already criminals, exactly one of Megan (A3) and Vince (A5) must be a criminal and the other must be an innocent, so there is exactly 1 innocent among A3 and A5. Zed’s clue says the total number of innocents on the entire edge is odd, and we already have 4 known edge-innocents (Celia at B1, Erwin at D1, Pip at D3, and Zed at D5), so the number of innocents among A3, A5, and C5 must be odd. Since A3 and A5 contribute exactly 1 innocent, Xia at C5 cannot be an innocent (that would make 2), so Xia must be a criminal. Therefore, we can determine that C5 Xia is CRIMINAL.
Clue:
"Amy is one of 4 criminals in column A" — Flora (A2)
"There's an odd number of innocents on the edges" — Zed (D5)
C4 · Tyler → INNOCENT
Because: Zed is at D5, and the only neighbors of D5 are C5, D4, and the diagonal C4. Celia’s clue says Zed has only one innocent neighbor, meaning exactly one of those three neighbors is innocent. Since C5 (Xia) and D4 (Uma) are already known to be criminals, the only remaining neighbor who can be innocent is C4 (Tyler). Therefore, we can determine that C4 Tyler is INNOCENT.
Clue:
"Zed has only one innocent neighbor" — Celia (B1)
A5 · Vince → INNOCENT, A3 · Megan → CRIMINAL
Because: Scott is at B4, so his neighbors are Megan at A3, Nicole at B3, Olga at C3, Rob at A4, Tyler at C4, Vince at A5, Wanda at B5, and Xia at C5. Uma’s clue says that among Scott’s five neighboring criminals, exactly two of them are in row 5. In row 5, the neighbors of Scott are Vince, Wanda, and Xia, and Wanda and Xia are already criminals, so Vince cannot also be a criminal or there would be three row 5 criminals. That makes Vince innocent, and since Scott must have five criminal neighbors total while Rob, Olga, Wanda, and Xia already make four, the only remaining neighbor who can supply the fifth is Megan. Therefore, we can determine that A5 Vince is INNOCENT and A3 Megan is CRIMINAL.
Clue:
"Exactly 2 of the 5 criminals neighboring Scott are in row 5" — Uma (D4)