Clues by Sam Jan 28, 2026 Answer – Full Solution Explained
Tricky·Solved
A1
🕵️♀️
Bonnie
sleuth
B1
👩⚖️
Debra
judge
C1
👩⚖️
Evie
judge
D1
👩⚖️
Freya
judge
A2
🕵️♂️
Gary
sleuth
B2
👩💼
Hazel
clerk
C2
👨🌾
Ivan
farmer
D2
👩🌾
Janet
farmer
A3
👮♂️
Kyle
cop
B3
👩⚕️
Laura
doctor
C3
💂♂️
Mark
guard
D3
👨⚕️
Nick
doctor
A4
👨💻
Oscar
coder
B4
👮♂️
Peter
cop
C4
👩🔧
Saga
mech
D4
👨💻
Terry
coder
A5
👩⚕️
Uma
doctor
B5
👨💼
Wally
clerk
C5
👩🔧
Xena
mech
D5
💂♂️
Ziad
guard
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 5Criminal 15Unknown 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 · 5
[ C1 ] [ B2 ] [ B3 ] [ A5 ] [ C5 ]
Criminal · 15
[ A1 ] [ B1 ] [ D1 ] [ A2 ] [ C2 ] [ D2 ] [ A3 ] [ C3 ] [ D3 ] [ A4 ] [ B4 ] [ C4 ] [ D4 ] [ B5 ] [ 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 · Bonnie
"All criminals in column D are connected"
B1 · Debra
"Only one more innocent in the edges?"
C1 · Evie
"Uma is one of 3 innocents on the edges"
D1 · Freya
"There are exactly 2 innocents to the left of Ziad"
A2 · Gary
"Exactly 2 of Peter's 3 innocent neighbors also neighbor Oscar"
B2 · Hazel
"Terry is one of Saga's 6 criminal neighbors"
C2 · Ivan
"Janet has only one innocent neighbor"
D2 · Janet
"There's plenty of us criminals here, that's for sure."
A3 · Kyle
"There's an odd number of innocents below Hazel"
B3 · Laura
"I know who is it! I know! Oh... you too?"
C3 · Mark
"I bet it's either Wally or Xena"
D3 · Nick
"Exactly 1 innocent in row 5 is neighboring Saga"
A4 · Oscar
"Who could it be, that mysterious innocent living on the edge?"
B4 · Peter
"Only 1 of the 2 innocents in column B is Laura's neighbor"
C4 · Saga
"Row 5 has more innocents than any other row"
D4 · Terry
"There are more criminals in column A than column B"
A5 · Uma
"All criminals in column A are connected"
B5 · Wally
"3 innocents on the edges and 2 of them are my neighbors? Just my luck."
C5 · Xena
"It's me! Surprise!"
D5 · Ziad
"Not me! I'm far from innocent! Yesterday I stole candy from a kid."
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)
A5 · Uma → INNOCENT
Because: Evie’s clue says that “Uma is one of 3 innocents on the edges,” which directly states that Uma is an innocent (and Uma is indeed on an edge at A5). Therefore, we can determine that A5 Uma is INNOCENT.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)
A3 · Kyle → CRIMINAL, A2 · Gary → CRIMINAL
Because: The edge squares include column A, so Gary at A2 and Kyle at A3 are both on the edge. Evie’s clue says there are exactly three innocents on the edges, and we already have two edge innocents identified: Evie at C1 and Uma at A5, so among all other edge people there can be only one more innocent. If Gary at A2 were that remaining edge innocent, then Bonnie at A1, Kyle at A3, and Oscar at A4 would all have to be criminals, which would put criminals at A1 and A3 with an innocent between them, so the criminals in column A would not be connected as Uma’s clue requires. The same issue happens if Kyle at A3 were the remaining edge innocent, because then Bonnie at A1, Gary at A2, and Oscar at A4 would all have to be criminals, splitting the column-A criminals into two separated groups. Therefore, we can determine that A3 Kyle is CRIMINAL and A2 Gary is CRIMINAL.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)
"All criminals in column A are connected" — Uma (A5)
B4 · Peter → CRIMINAL
Because: Peter is at B4, and Hazel is at B2, so the people below Hazel are Laura at B3, Peter at B4, and Wally at B5. Gary’s clue says Peter has exactly three innocent neighbors, and exactly two of those three also neighbor Oscar at A4; since Uma at A5 is an innocent neighbor of Peter and also neighbors Oscar, exactly one of Laura (B3) and Wally (B5) must be innocent and the other must not be. That means among Laura and Wally there is exactly one innocent, so the number of innocents below Hazel is either 1 (if Peter is not innocent) or 2 (if Peter is innocent). Kyle’s clue says the number of innocents below Hazel is odd, so it cannot be 2, which forces Peter to not be innocent. Therefore, we can determine that B4 Peter is CRIMINAL.
Clue:
"Exactly 2 of Peter's 3 innocent neighbors also neighbor Oscar" — Gary (A2)
"There's an odd number of innocents below Hazel" — Kyle (A3)
B2 · Hazel → INNOCENT
Because: In column B the people are Debra at B1, Hazel at B2, Laura at B3, Peter at B4, and Wally at B5. Peter’s clue says there are exactly two innocents in column B, and among those two, exactly one is a neighbor of Laura. The only people in column B who are neighbors of Laura at B3 are Hazel at B2 and Peter at B4, but Peter is already known to be a criminal, so he cannot be that one innocent neighbor. That forces Hazel to be the one innocent in column B who is Laura’s neighbor. Therefore, we can determine that B2 Hazel is INNOCENT.
Clue:
"Only 1 of the 2 innocents in column B is Laura's neighbor" — Peter (B4)
D4 · Terry → CRIMINAL
Because: Hazel’s clue says that Terry is one of Saga’s criminal neighbors, meaning Terry is explicitly included in the set of criminals around Saga. Therefore, we can determine that D4 Terry is CRIMINAL.
Clue:
"Terry is one of Saga's 6 criminal neighbors" — Hazel (B2)
B1 · Debra → CRIMINAL
Because: Peter is at B4, and the neighbors of Peter who also neighbor Oscar at A4 are Laura (B3), Uma (A5), and Wally (B5). Gary’s clue says Peter has exactly three innocent neighbors, and exactly two of those innocents also neighbor Oscar; since Uma is already known to be innocent and is one of Oscar’s neighbors, exactly one of Laura or Wally must be the other innocent that neighbors Oscar. Peter’s clue says there are exactly two innocents in column B; Hazel at B2 is already one of them, so the only remaining innocent in column B must be whichever of Laura or Wally is innocent. That leaves no room for Debra at B1 to be innocent in column B. Therefore, we can determine that B1 Debra is CRIMINAL.
Clue:
"Exactly 2 of Peter's 3 innocent neighbors also neighbor Oscar" — Gary (A2)
"Only 1 of the 2 innocents in column B is Laura's neighbor" — Peter (B4)
A1 · Bonnie → CRIMINAL, A4 · Oscar → CRIMINAL
Because: Peter is at B4, and the only people who are both neighbors of Peter and neighbors of Oscar at A4 are Kyle (A3), Laura (B3), Uma (A5), and Wally (B5). Gary’s clue says Peter has exactly three innocent neighbors, and exactly two of those innocents are in that overlap set; since Kyle is already a criminal and Uma is already innocent, the second overlapping innocent must be either Laura or Wally, but not both. So exactly one of Laura (B3) and Wally (B5) is innocent, which means exactly one of them is a criminal, giving column B a total of three criminals (Debra, Peter, and exactly one of Laura/Wally). Terry’s clue says column A has more criminals than column B, and column A already has Gary and Kyle as criminals while Uma is innocent, so the only way for column A to exceed three criminals is for both remaining people in column A, Bonnie (A1) and Oscar (A4), to be criminals. Therefore, we can determine that A1 Bonnie is CRIMINAL and A4 Oscar is CRIMINAL.
Clue:
"Exactly 2 of Peter's 3 innocent neighbors also neighbor Oscar" — Gary (A2)
"There are more criminals in column A than column B" — Terry (D4)
D2 · Janet → CRIMINAL, D3 · Nick → CRIMINAL
Because: Evie’s clue says there are exactly 3 innocents on the edge, and Uma is one of them. Since Evie at C1 and Uma at A5 are already two edge innocents, only one of the six still-unknown edge people (D1, D2, D3, D5, B5, C5) can be innocent, so the other five must be criminals. That means among the four unknown edge people in column D (D1, D2, D3, D5), at least three are criminals, and with Terry at D4 already a criminal, column D contains at least four criminals in total. Bonnie’s clue says all criminals in column D are connected, which in a single column can only happen if they form one unbroken vertical run with no gaps. With at least four criminals in that column, the two middle positions D2 and D3 must be part of that unbroken run, so they must be criminals. Therefore, we can determine that D2 Janet is CRIMINAL and D3 Nick is CRIMINAL.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)
"All criminals in column D are connected" — Bonnie (A1)
D1 · Freya → CRIMINAL
Because: Evie’s clue says that there are exactly 3 innocents on the edges in total, and we already know two of them: Evie at C1 and Uma at A5. Nick’s clue says that in row 5, exactly 1 innocent is neighboring Saga at C4, so exactly one of B5, C5, and D5 must be innocent, and that person is also on the edge. That uses up the third and final “edge innocent” required by Evie’s clue, so no other edge position can be innocent. Since Freya at D1 is on the edge, Freya cannot be innocent. Therefore, we can determine that D1 Freya is CRIMINAL.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)
"Exactly 1 innocent in row 5 is neighboring Saga" — Nick (D3)
D5 · Ziad → CRIMINAL
Because: The edge squares are all of row 1, all of row 5, plus column A and column D in rows 2–4. Evie’s clue says there are exactly 3 innocents on the edges and Uma is one of them; we already know Evie at C1 is an edge innocent and Uma at A5 is an edge innocent, so among the remaining unknown edge people on row 5 (Wally at B5, Xena at C5, and Ziad at D5) exactly one can be innocent. Freya’s clue says there are exactly 2 innocents to the left of Ziad, and the three people to Ziad’s left in row 5 are Uma (A5), Wally (B5), and Xena (C5); since Uma is already innocent, exactly one of Wally and Xena must be innocent. That uses up the single remaining edge-innocent slot among B5, C5, and D5, so Ziad cannot be innocent. Therefore, we can determine that D5 Ziad is CRIMINAL.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)
"There are exactly 2 innocents to the left of Ziad" — Freya (D1)
C4 · Saga → CRIMINAL
Because: Peter is at B4, Oscar is at A4, and Saga is at C4. Hazel’s clue says Saga has exactly 6 criminal neighbors, and since four of Saga’s neighbors are already known criminals (Peter, Nick, Terry, and Ziad), the remaining four neighbors of Saga (Laura, Mark, Wally, and Xena) must contain exactly two innocents. Gary’s clue says Peter has exactly three innocent neighbors, and exactly two of those three also neighbor Oscar; since Uma is an innocent neighbor of both Peter and Oscar and Kyle is a criminal, exactly one of Laura or Wally is innocent, so to reach the “two innocents” required among Laura, Mark, Wally, and Xena, exactly one of Mark or Xena must also be innocent. But the only neighbors of Peter that do not neighbor Oscar are Mark, Saga, and Xena, and Gary’s clue allows only one innocent among those three, so Saga cannot be that innocent if Mark or Xena already is. Therefore, we can determine that C4 (Saga) is CRIMINAL.
Clue:
"Exactly 2 of Peter's 3 innocent neighbors also neighbor Oscar" — Gary (A2)
"Terry is one of Saga's 6 criminal neighbors" — Hazel (B2)
C2 · Ivan → CRIMINAL
Because: The only unknown edge people are Wally at B5 and Xena at C5, since every other edge position is already known. Evie’s clue says there are exactly 3 innocents on the edges and Uma is one of them; with Uma already innocent and Evie at C1 also an edge innocent, that leaves exactly one more edge innocent, so exactly one of Wally or Xena is innocent and the other is criminal. That means row 5 contains exactly 2 innocents in total: Uma plus exactly one of Wally or Xena. Since Saga’s clue says row 5 has more innocents than any other row, every other row must have at most 1 innocent, and row 2 already has Hazel as an innocent, so Ivan at C2 cannot also be innocent. Therefore, we can determine that Ivan is CRIMINAL.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)
"Row 5 has more innocents than any other row" — Saga (C4)
C3 · Mark → CRIMINAL
Because: Janet is at D2, so her neighbors are C1 Evie, D1 Freya, C2 Ivan, C3 Mark, and D3 Nick (including diagonals). Ivan’s clue says Janet has only one innocent neighbor, and we already know Evie at C1 is innocent. That uses up Janet’s single allowed innocent neighbor, so every other neighbor of Janet must be criminal, including Mark at C3. Therefore, we can determine that C3 Mark is CRIMINAL.
Clue:
"Janet has only one innocent neighbor" — Ivan (C2)
B3 · Laura → INNOCENT
Because: Peter is at B4, and his three unknown-status neighbors are Laura at B3, Wally at B5, and Xena at C5; Uma at A5 is already a known innocent neighbor of Peter. Gary’s clue says Peter has exactly three innocent neighbors, so besides Uma there must be exactly two innocents among Laura, Wally, and Xena. Evie’s clue says there are exactly three innocents on the edges, and since the only unknown edge people are Wally and Xena, exactly one of them is innocent. That means Wally and Xena can contribute only one innocent between them, so Laura must be the other innocent needed to reach Peter’s total of three. Therefore, we can determine that B3 Laura is INNOCENT.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)
"Exactly 2 of Peter's 3 innocent neighbors also neighbor Oscar" — Gary (A2)
C5 · Xena → INNOCENT
Because: Peter is at B4, and his eight neighbors are A3 Kyle, B3 Laura, C3 Mark, A4 Oscar, C4 Saga, A5 Uma, B5 Wally, and C5 Xena. Among these, Laura and Uma are already known to be innocent, and Gary’s clue says Peter has exactly three innocent neighbors in total, so exactly one of Wally or Xena must also be innocent. Oscar is at A4, and among Peter’s neighbors, Laura, Uma, and Wally all neighbor Oscar, but Xena does not. Since the clue says exactly two of Peter’s three innocent neighbors also neighbor Oscar, the third innocent neighbor cannot be Wally and must be Xena. Therefore, we can determine that C5 Xena is INNOCENT.
Clue:
"Exactly 2 of Peter's 3 innocent neighbors also neighbor Oscar" — Gary (A2)
B5 · Wally → CRIMINAL
Because: The clue from Evie says that there are exactly three innocents on the edge of the board, and Uma is one of them. We can already see three edge positions that are confirmed innocents: Evie at C1, Uma at A5, and Xena at C5, and each of these is on the edge. Since that already accounts for all three edge innocents allowed by the clue, every other edge position must be a criminal. Wally is at B5, which is also an edge position, so Wally cannot be innocent. Therefore, we can determine that B5 Wally is CRIMINAL.
Clue:
"Uma is one of 3 innocents on the edges" — Evie (C1)