Clues by Sam Feb 05, 2026 Answer – Full Solution Explained
Tricky·Solved
A1
👩🎨
Alice
painter
B1
👨🍳
Bobby
cook
C1
👨⚖️
Chuck
judge
D1
👩⚖️
Diane
judge
A2
👨🏫
Ethan
teacher
B2
👩🏫
Flora
teacher
C2
👨🏫
Gary
teacher
D2
💂♂️
Hank
guard
A3
👨🔧
Ike
mech
B3
👩🎨
Joyce
painter
C3
👨🌾
Kyle
farmer
D3
👩⚕️
Laura
doctor
A4
👩🍳
Petra
cook
B4
👨🔧
Rob
mech
C4
👩🌾
Tina
farmer
D4
💂♀️
Uma
guard
A5
👷♀️
Vera
builder
B5
👷♂️
Wally
builder
C5
👨🔧
Xavi
mech
D5
👩⚕️
Zara
doctor
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 12Criminal 8Unknown 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 · 12
[ A1 ] [ B1 ] [ C1 ] [ C2 ] [ D2 ] [ A3 ] [ B3 ] [ D4 ] [ A5 ] [ B5 ] [ C5 ] [ D5 ]
Criminal · 8
[ D1 ] [ A2 ] [ B2 ] [ C3 ] [ D3 ] [ A4 ] [ B4 ] [ C4 ]
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 · Alice
"Tina is one of 3 criminals in row 4"
B1 · Bobby
"Everyone has a new beginning tomorrow!"
C1 · Chuck
"Exactly 2 of the 4 innocents neighboring Gary are in row 1"
D1 · Diane
"How about we stick to the beginning tomorrow? Don't catch anyone?"
A2 · Ethan
"An odd number of innocents on the edges neighbor Gary"
B2 · Flora
"I'm beginning to understand this puzzle. I'm a criminal, right?"
C2 · Gary
"Exactly 1 innocent in row 4 is neighboring Xavi"
D2 · Hank
"I'm liking this ending more. Doesn't hurt my head."
A3 · Ike
"Xavi has exactly 3 innocent neighbors"
B3 · Joyce
"There's an odd number of innocents neighboring Tina"
C3 · Kyle
"An odd number of innocents above Zara neighbor Gary"
D3 · Laura
"Only one row has exactly 3 criminals"
A4 · Petra
"Only 1 of the 3 innocents neighboring Chuck is above Uma"
B4 · Rob
"I liked the beginning. I was still a free man back then."
C4 · Tina
"Both criminals above Xavi are connected"
D4 · Uma
"There's an equal number of criminals in columns A and D"
A5 · Vera
"Each column has at least 3 innocents"
B5 · Wally
"Is it just me, or was that beginning super tough?"
C5 · Xavi
"There are more criminals in row 1 than row 5"
D5 · Zara
"Today's beginning reminds me of my mornings. Slow."
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)
C4 · Tina → CRIMINAL
Because: Alice’s clue says that Tina is one of the three criminals in row 4, which directly includes Tina among the criminals. Therefore, we can determine that C4 Tina is CRIMINAL.
Clue:
"Tina is one of 3 criminals in row 4" — Alice (A1)
C1 · Chuck → INNOCENT
Because: “Above Xavi” means the four people in column C above C5: Chuck at C1, Gary at C2, Kyle at C3, and Tina at C4. The clue says there are exactly two criminals among those four, and those two criminals are connected by orthogonal adjacency. Since Tina at C4 is already a criminal, the only way for there to be exactly one more criminal above Xavi and still have the two criminals connected is for Kyle at C3 to be the other criminal, because any criminal at C1 or C2 would be separated from Tina by at least one person and would need an additional criminal in between to make a connected chain. That means Chuck at C1 cannot be one of the criminals above Xavi. Therefore, we can determine that C1 Chuck is INNOCENT.
Clue:
"Both criminals above Xavi are connected" — Tina (C4)
C3 · Kyle → CRIMINAL
Because: The clue talks about Xavi at C5, so “above Xavi” means the four people in column C at C1, C2, C3, and C4. It says there are exactly two criminals among those four, and that those two criminals are connected by up/down/left/right adjacency. Since Tina at C4 is already a criminal, the second criminal above Xavi must be either Gary at C2 or Kyle at C3 because Chuck at C1 is innocent. If the second criminal were Gary at C2, then Tina at C4 and Gary at C2 would not be connected unless Kyle at C3 were also a criminal, which would create three criminals above Xavi and violate “both.” Therefore, we can determine that C3 Kyle is CRIMINAL.
Clue:
"Both criminals above Xavi are connected" — Tina (C4)
C2 · Gary → INNOCENT
Because: The clue talks about everyone above Xavi, who is in column C at C5, so it refers to C1 Chuck, C2 Gary, C3 Kyle, and C4 Tina. It says “both criminals above Xavi,” which means there are exactly two criminals among those four people. Since Kyle at C3 and Tina at C4 are already known criminals (and they are also directly connected to each other, satisfying the “connected” part), there cannot be any other criminal above Xavi. Chuck at C1 is already innocent, so the remaining person, Gary at C2, must also be innocent. Therefore, we can determine that C2 Gary is INNOCENT.
Clue:
"Both criminals above Xavi are connected" — Tina (C4)
A4 · Petra → CRIMINAL
Because: In row 4, the people are Petra at A4, Rob at B4, Tina at C4, and Uma at D4. Alice’s clue says Tina is one of exactly 3 criminals in row 4, so row 4 has exactly 1 innocent total. Gary’s clue says that exactly 1 innocent in row 4 is neighboring Xavi at C5, and the only row 4 positions that neighbor C5 are B4, C4, and D4, not A4. Since the only innocent in row 4 must be one of B4, C4, or D4, Petra at A4 cannot be the innocent and must be a criminal. Therefore, we can determine that A4 Petra is CRIMINAL.
Clue:
"Tina is one of 3 criminals in row 4" — Alice (A1)
"Exactly 1 innocent in row 4 is neighboring Xavi" — Gary (C2)
D3 · Laura → CRIMINAL
Because: Chuck is at C1, and his neighbors are B1, B2, C2 (Gary), D1, and D2. Petra’s clue says there are exactly three innocents among Chuck’s neighbors, and only one of those three is above Uma; since the only Chuck-neighbors above Uma are Diane at D1 and Hank at D2, exactly one of D1 and D2 is innocent. Kyle’s clue looks at the people above Zara in column D and says an odd number of the innocents among them neighbor Gary; among D1, D2, and D3 (Laura), all three neighbor Gary, so the number of innocents in D1–D3 must be odd. With exactly one innocent already forced in D1–D2, Laura cannot be innocent (that would make two innocents in D1–D3), so Laura must be criminal. Therefore, we can determine that D3 Laura is CRIMINAL.
Clue:
"Only 1 of the 3 innocents neighboring Chuck is above Uma" — Petra (A4)
"An odd number of innocents above Zara neighbor Gary" — Kyle (C3)
B3 · Joyce → INNOCENT
Because: Gary is at C2, and the neighbors of Gary in row 1 are B1, C1, and D1. Chuck at C1 is already an innocent, and Chuck’s clue says exactly 2 of the 4 innocents neighboring Gary are in row 1, so exactly one of B1 or D1 is also an innocent (not both). Also, because Kyle at C3 and Laura at D3 are criminals, the remaining two innocents neighboring Gary (to make 4 total) must come from B2, D2, and B3, so exactly 2 of B2, D2, and Joyce at B3 are innocents. Chuck is at C1, so Chuck’s neighbors are B1, D1, B2, Gary (C2), and D2, and Petra’s clue says exactly 3 of those neighbors are innocents. Since Gary is already an innocent, exactly 2 of B1, D1, B2, and D2 are innocents, and Petra also says only 1 of those 3 innocent neighbors is above Uma; the only neighbors of Chuck that are above Uma are D1 and D2, so exactly one of D1 and D2 is innocent. With exactly one of B1 and D1 innocent, that forces exactly one of B2 and D2 to be innocent, so among B2 and D2 there is only 1 innocent total. But we already needed 2 innocents among B2, D2, and B3, and B2 plus D2 can only supply 1 of them, so Joyce at B3 must be the other innocent. Therefore, we can determine that B3 Joyce is INNOCENT.
Clue:
"Exactly 2 of the 4 innocents neighboring Gary are in row 1" — Chuck (C1)
"Only 1 of the 3 innocents neighboring Chuck is above Uma" — Petra (A4)
A3 · Ike → INNOCENT
Because: Row 4 contains Petra and Tina as known criminals, and Alice’s clue says Tina is one of exactly 3 criminals in row 4, so row 4 is a row with exactly 3 criminals. Laura’s clue says only one row has exactly 3 criminals, so no other row is allowed to have exactly 3 criminals. In row 3, Kyle and Laura are criminals and Joyce is innocent, so the only way row 3 could reach exactly 3 criminals would be if Ike were a criminal; since row 3 is not allowed to have exactly 3 criminals, Ike cannot be a criminal. Therefore, we can determine that A3 Ike is INNOCENT.
Clue:
"Tina is one of 3 criminals in row 4" — Alice (A1)
"Only one row has exactly 3 criminals" — Laura (D3)
B5 · Wally → INNOCENT, D5 · Zara → INNOCENT
Because: Xavi is at C5, so his neighbors are exactly the five people at B4 (Rob), C4 (Tina), D4 (Uma), B5 (Wally), and D5 (Zara). Ike’s clue says Xavi has exactly 3 innocent neighbors, which means among those five neighbors there are exactly 3 innocents and 2 criminals; since Tina is already a criminal, exactly one of Rob, Uma, Wally, and Zara must be a criminal. Alice’s clue says Tina is one of 3 criminals in row 4, and row 4 already contains Petra and Tina as criminals, so exactly one of Rob or Uma is the third criminal in that row. That uses up the single criminal that is allowed among Rob, Uma, Wally, and Zara, so Wally and Zara cannot be criminals and must both be innocents. Therefore, we can determine that B5 Wally is INNOCENT and D5 Zara is INNOCENT.
Clue:
"Tina is one of 3 criminals in row 4" — Alice (A1)
"Xavi has exactly 3 innocent neighbors" — Ike (A3)
C5 · Xavi → INNOCENT
Because: Tina is at C4, and Alice’s clue says Tina is one of exactly three criminals in row 4. Since Petra at A4 is already a criminal and Tina at C4 is a criminal, that forces exactly one of Rob at B4 and Uma at D4 to be a criminal and the other to be an innocent. Joyce’s clue says Tina has an odd number of innocent neighbors, and Tina’s neighbors are Joyce, Kyle, Laura, Rob, Uma, Wally, Xavi, and Zara. Joyce, Wally, and Zara are already innocent (3 innocents), and Kyle and Laura are already criminals, so the only neighbors whose innocence could change the count are Rob, Uma, and Xavi. Because exactly one of Rob and Uma is innocent, the number of innocents among Rob, Uma, and Xavi is 1 plus whether Xavi is innocent, and this total must be even to keep 3 plus that number odd; that only happens if Xavi is innocent. Therefore, we can determine that C5 (Xavi) is INNOCENT.
Clue:
"Tina is one of 3 criminals in row 4" — Alice (A1)
"There's an odd number of innocents neighboring Tina" — Joyce (B3)
A5 · Vera → INNOCENT
Because: Gary is at C2, and Chuck’s clue says that among Gary’s three neighbors in row 1 (B1, C1, and D1), exactly two are innocents. Since C1 (Chuck) is already an innocent, that means exactly one of B1 and D1 is an innocent, so row 1 cannot contain two criminals. Xavi’s clue says there are more criminals in row 1 than in row 5. Since row 1 can have at most one criminal, row 5 must have zero criminals; otherwise row 1 would need at least two criminals to be “more.” In row 5, Wally, Xavi, and Zara are already innocents, so the only way for row 5 to have zero criminals is for Vera at A5 to be an innocent. Therefore, we can determine that A5 Vera is INNOCENT.
Clue:
"Exactly 2 of the 4 innocents neighboring Gary are in row 1" — Chuck (C1)
"There are more criminals in row 1 than row 5" — Xavi (C5)
D4 · Uma → INNOCENT
Because: Chuck is at C1, so his neighbors are B1, B2, C2, D1, and D2. Petra’s clue says that among the three innocent neighbors Chuck has, exactly one is above Uma; the only neighbors of Chuck that are above Uma (in column D and above row 4) are D1 and D2, so exactly one of D1 and D2 is innocent. Vera’s clue says each column has at least three innocents, and in column D we already have Zara at D5 as innocent and Laura at D3 as criminal, so we still need two more innocents among D1, D2, and D4; since D1 and D2 provide exactly one innocent, D4 must be the other one. Therefore, we can determine that D4 Uma is INNOCENT.
Clue:
"Only 1 of the 3 innocents neighboring Chuck is above Uma" — Petra (A4)
"Each column has at least 3 innocents" — Vera (A5)
B4 · Rob → CRIMINAL
Because: Row 4 contains Petra at A4, Rob at B4, Tina at C4, and Uma at D4. Alice’s clue says that Tina is one of 3 criminals in row 4, so there must be exactly three criminals somewhere in that row including Tina. We already know Petra is a criminal and Tina is a criminal, and Uma is confirmed innocent, so the only remaining person who can be the third criminal in row 4 is Rob at B4. Therefore, we can determine that B4 Rob is CRIMINAL.
Clue:
"Tina is one of 3 criminals in row 4" — Alice (A1)
A2 · Ethan → CRIMINAL
Because: Chuck is at C1, and his only neighbors in column D are Diane at D1 and Hank at D2, which are both above Uma at D4. Petra’s clue says Chuck has exactly three innocent neighbors in total, and only one of those three is above Uma, so exactly one of Diane and Hank is an innocent neighbor and the other must be a criminal neighbor. That means column D has Laura as a criminal plus exactly one of Diane or Hank as a criminal, so there are exactly two criminals in column D. Uma’s clue says columns A and D have the same number of criminals, and since column A already has Petra as a criminal and everyone else there is known innocent except Ethan at A2, Ethan must be the second criminal in column A. Therefore, we can determine that A2 Ethan is CRIMINAL.
Clue:
"Only 1 of the 3 innocents neighboring Chuck is above Uma" — Petra (A4)
"There's an equal number of criminals in columns A and D" — Uma (D4)
D2 · Hank → INNOCENT, B2 · Flora → CRIMINAL
Because: The relevant person is Gary at C2, whose eight neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. Chuck’s clue says there are exactly four innocents among these neighbors, and exactly two of those four are in row 1; since C1 is already innocent, that forces exactly one of B1 or D1 to be innocent, and it also forces exactly one of B2 or D2 to be innocent because B3 is innocent while C3 and D3 are criminals. Ethan’s clue only looks at Gary’s edge-neighbors, which are B1, C1, D1, D2, and D3; among these, C1 is innocent and D3 is criminal, so the parity depends on B1, D1, and D2. Because exactly one of B1 or D1 is innocent, B1 and D1 contribute an odd count of innocents, so to make the total number of innocent edge-neighbors odd as the clue requires, D2 must be innocent. With D2 fixed as innocent and exactly one of B2 or D2 allowed to be innocent, B2 must be criminal. Therefore, we can determine that D2 Hank is INNOCENT and B2 Flora is CRIMINAL.
Clue:
"Exactly 2 of the 4 innocents neighboring Gary are in row 1" — Chuck (C1)
"An odd number of innocents on the edges neighbor Gary" — Ethan (A2)
D1 · Diane → CRIMINAL, B1 · Bobby → INNOCENT
Because: Chuck is at C1, and his neighbors are Bobby (B1), Diane (D1), Flora (B2), Gary (C2), and Hank (D2). Petra’s clue says that Chuck has exactly three innocent neighbors, and exactly one of those three is above Uma; since Flora is already known to be a criminal, the third innocent neighbor (besides Gary and Hank) must be either Bobby or Diane. Hank is an innocent neighbor of Chuck and he is above Uma because he is in column D above D4, so the clue’s “only 1” means none of the other innocent neighbors of Chuck can be above Uma. Diane is also in column D above Uma, so Diane cannot be the remaining innocent neighbor, which forces Bobby to be that third innocent neighbor instead. Therefore, we can determine that Bobby is INNOCENT and Diane is CRIMINAL.
Clue:
"Only 1 of the 3 innocents neighboring Chuck is above Uma" — Petra (A4)