Clues by Sam Feb 21, 2026 Answer – Full Solution Explained
Hard·Solved
A1
🕵️♂️
Austin
sleuth
B1
💂♀️
Chloe
guard
C1
👮♂️
Frank
cop
D1
👨⚕️
Hank
doctor
A2
👮♂️
Ivan
cop
B2
💂♀️
Julie
guard
C2
👨⚖️
Klay
judge
D2
👨⚖️
Luigi
judge
A3
👮♀️
Max
cop
B3
👩🍳
Nicole
cook
C3
🕵️♀️
Olga
sleuth
D3
👨⚕️
Peter
doctor
A4
💂♀️
Ruby
guard
B4
👨🍳
Steve
cook
C4
🕵️♀️
Tina
sleuth
D4
👩⚕️
Uma
doctor
A5
👩🍳
Vicky
cook
B5
👷♂️
Wally
builder
C5
👷♀️
Xia
builder
D5
👷♂️
Zach
builder
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
[ B1 ] [ C1 ] [ D1 ] [ B2 ] [ C3 ] [ D3 ] [ A4 ] [ C4 ] [ B5 ] [ C5 ] [ D5 ]
Criminal · 9
[ A1 ] [ A2 ] [ C2 ] [ D2 ] [ A3 ] [ B3 ] [ B4 ] [ D4 ] [ A5 ]
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 · Austin
"I want to be important too! I'm never in the center..."
B1 · Chloe
"Exactly 1 innocent in row 2 is neighboring Olga"
C1 · Frank
"There's an odd number of innocents neighboring Wally"
D1 · Hank
"Only one person in a corner has exactly one innocent neighbor"
A2 · Ivan
"Only 1 of the 4 criminals neighboring Nicole is in row 3"
B2 · Julie
"Finally! The excitement was killing me!"
C2 · Klay
"It was fun while it lasted! Good run!"
D2 · Luigi
"There are exactly 3 innocents in row 1"
A3 · Max
"This neighborhood is so confusing"
B3 · Nicole
"Row 2 is the only row with exactly one innocent"
C3 · Olga
"There's an equal number of criminals in rows 1 and 5"
D3 · Peter
"There are exactly 2 criminals in row 4"
A4 · Ruby
"Wally is one of 3 innocents in column B"
B4 · Steve
"Blast it! I thought I'd make it this time"
C4 · Tina
"Austin, maybe if you changed your name? How about Naustin?"
D4 · Uma
"There are exactly 3 innocents below Frank"
A5 · Vicky
"Everything seems to revolve around Nicole today. Not that I'm jealous."
B5 · Wally
"Only 1 of the 2 criminals below Chloe is Nicole's neighbor"
C5 · Xia
"There are more criminals in column B than column C"
D5 · Zach
"There's an equal number of innocents in columns B and D"
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)
B5 · Wally → INNOCENT
Because: Ruby’s clue directly says that Wally is one of the three innocents in column B, which means Wally himself must be an innocent. Therefore, we can determine that B5 Wally is INNOCENT.
Clue:
"Wally is one of 3 innocents in column B" — Ruby (A4)
B3 · Nicole → CRIMINAL
Because: Chloe is at B1, so the people below her are Julie at B2, Nicole at B3, Steve at B4, and Wally at B5, and Wally’s clue says there are exactly two criminals among those four. Since Wally is already known to be innocent, that means exactly two of Julie, Nicole, and Steve are criminals. Nicole’s neighbors include Julie and Steve, but a person is not their own neighbor and Wally is not a neighbor of Nicole. If Nicole were innocent, then the two criminals below Chloe would have to be Julie and Steve, and then both of those criminals would be Nicole’s neighbors, not “only 1,” so Nicole cannot be innocent. Therefore, we can determine that B3 Nicole is CRIMINAL.
Clue:
"Only 1 of the 2 criminals below Chloe is Nicole's neighbor" — Wally (B5)
B1 · Chloe → INNOCENT
Because: In column B we have Chloe at B1, Julie at B2, Nicole at B3, Steve at B4, and Wally at B5. Ruby says that Wally is one of 3 innocents in column B, so column B contains exactly 3 innocents and therefore exactly 2 criminals, and since Nicole is already known to be a criminal, there is only room for one more criminal anywhere else in column B. Wally says there are 2 criminals below Chloe, meaning among B2–B5 there must be exactly two criminals; one of them is Nicole, so the other must be either Julie or Steve. That uses up both criminals that column B is allowed to have, so Chloe cannot be a criminal. Therefore, we can determine that B1 Chloe is INNOCENT.
Clue:
"Wally is one of 3 innocents in column B" — Ruby (A4)
"Only 1 of the 2 criminals below Chloe is Nicole's neighbor" — Wally (B5)
A2 · Ivan → CRIMINAL
Because: Olga is at C3, so the only people in row 2 who neighbor her are B2, C2, and D2. Chloe’s clue says that among those three, exactly one is innocent. Nicole’s clue says row 2 has exactly one innocent in total across A2, B2, C2, and D2. Since the single row-2 innocent is already accounted for within B2/C2/D2, A2 cannot be innocent. Therefore, we can determine that A2 Ivan is CRIMINAL.
Clue:
"Exactly 1 innocent in row 2 is neighboring Olga" — Chloe (B1)
"Row 2 is the only row with exactly one innocent" — Nicole (B3)
D3 · Peter → INNOCENT
Because: Nicole is at B3, and the only people in row 3 who neighbor her are A3 (Max) and C3 (Olga). Ivan’s clue says that among the four criminal neighbors of Nicole, only one is in row 3, so exactly one of Max and Olga is a criminal and the other is an innocent. That means row 3 already has Nicole as a criminal, plus exactly one criminal among Max and Olga; if Peter at D3 were also a criminal, then row 3 would have three criminals and only one innocent (the innocent among Max and Olga). Nicole’s other clue says row 2 is the only row with exactly one innocent, so row 3 cannot have exactly one innocent. Therefore, we can determine that D3 Peter is INNOCENT.
Clue:
"Row 2 is the only row with exactly one innocent" — Nicole (B3)
"Only 1 of the 4 criminals neighboring Nicole is in row 3" — Ivan (A2)
D4 · Uma → CRIMINAL, D2 · Luigi → CRIMINAL
Because: Nicole at B3 has eight neighbors, and Ivan’s clue tells us that exactly four of those neighbors are criminals, with exactly one of those criminal neighbors being in row 3 (so exactly one of A3 and C3 is a criminal neighbor of Nicole). Since Ivan at A2 is already one criminal neighbor of Nicole, that leaves exactly two more criminal neighbors among B2, C2, B4, and C4. Chloe’s clue says that among the row 2 neighbors of Olga (B2, C2, and D2), exactly one is innocent, so exactly two of them are criminals. If both B2 and C2 were criminals, then B4 and C4 would both have to be innocent to keep the total at two criminals among B2, C2, B4, and C4, but then row 4 could not have exactly two criminals as Peter’s clue requires; so exactly one of B2 and C2 is a criminal, which forces D2 (Luigi) to be the other criminal in {B2, C2, D2}. With exactly one criminal among B2 and C2, we also get exactly one criminal among B4 and C4, and then Peter’s clue (exactly two criminals in row 4) forces D4 (Uma) to be a criminal as well to reach two criminals in {B4, C4, D4}. Therefore, we can determine that D4 (Uma) is CRIMINAL and D2 (Luigi) is CRIMINAL.
Clue:
"Exactly 1 innocent in row 2 is neighboring Olga" — Chloe (B1)
"Only 1 of the 4 criminals neighboring Nicole is in row 3" — Ivan (A2)
"There are exactly 2 criminals in row 4" — Peter (D3)
C3 · Olga → INNOCENT, C5 · Xia → INNOCENT, A3 · Max → CRIMINAL
Because: Nicole is at B3, so her neighbors include B2, C2, A3, C3, B4, and C4 (among others). Ivan’s clue says Nicole has exactly four criminal neighbors, and only one of those is in row 3, so exactly one of A3 (Max) and C3 (Olga) is a criminal, and exactly two of B2, C2, B4, and C4 are criminals. Ruby’s clue says column B has exactly three innocents, and since B1 (Chloe) and B5 (Wally) are innocents while B3 (Nicole) is a criminal, exactly one of B2 and B4 is a criminal; that forces exactly one of C2 and C4 to be a criminal as well to make two criminals total among B2, C2, B4, and C4. Uma’s clue says there are exactly three innocents below Frank in column C, meaning among C2, C3, C4, and C5 there is exactly one criminal; since we already know exactly one of C2 and C4 is a criminal, C3 (Olga) and C5 (Xia) must both be innocent. With Olga now fixed as innocent, Ivan’s clue’s “only one criminal in row 3” among Nicole’s row-3 neighbors forces A3 (Max) to be the criminal. Therefore, we can determine that C3 Olga is INNOCENT, C5 Xia is INNOCENT, and A3 Max is CRIMINAL.
Clue:
"Wally is one of 3 innocents in column B" — Ruby (A4)
"Only 1 of the 4 criminals neighboring Nicole is in row 3" — Ivan (A2)
"There are exactly 3 innocents below Frank" — Uma (D4)
C1 · Frank → INNOCENT
Because: Frank is at C1, so the people below him in column C are Klay (C2), Olga (C3), Tina (C4), and Xia (C5). Uma says there are exactly 3 innocents below Frank, and since Olga and Xia are already innocent, that forces exactly one of Klay and Tina to be a criminal. Ruby’s clue makes column B have exactly 3 innocents, so column B has exactly 2 criminals in total, and Xia says column B has more criminals than column C, so column C must have fewer than 2 criminals. But column C already has at least 1 criminal (either Klay or Tina), so column C must have exactly 1 criminal, which means Frank cannot be a criminal as well. Therefore, we can determine that C1 Frank is INNOCENT.
Clue:
"Wally is one of 3 innocents in column B" — Ruby (A4)
"There are exactly 3 innocents below Frank" — Uma (D4)
"There are more criminals in column B than column C" — Xia (C5)
A5 · Vicky → CRIMINAL
Because: Wally is at B5, so his neighbors are Ruby (A4), Steve (B4), Tina (C4), Vicky (A5), and Xia (C5). Frank’s clue says Wally has an odd number of innocent neighbors, and we already know Ruby and Xia are innocent, so the remaining three neighbors (Steve, Tina, Vicky) must contain an odd number of innocents. Peter’s clue says row 4 has exactly 2 criminals; since Ruby (A4) is innocent and Uma (D4) is criminal, exactly one of Steve (B4) and Tina (C4) is innocent. If Vicky were innocent, then Steve and Tina would contribute one innocent and Vicky would add another, making two innocents among those three neighbors, which is even and would break Frank’s clue. Therefore, we can determine that A5 Vicky is CRIMINAL.
Clue:
"There are exactly 2 criminals in row 4" — Peter (D3)
"There's an odd number of innocents neighboring Wally" — Frank (C1)
D5 · Zach → INNOCENT
Because: In row 1, Chloe at B1 and Frank at C1 are already known innocents, and Luigi’s clue says there are exactly 3 innocents in row 1. That means only one of the two unknowns in row 1 (Austin at A1 and Hank at D1) can be a criminal, so row 1 has exactly 1 criminal total. Olga’s clue says rows 1 and 5 have an equal number of criminals, so row 5 must also have exactly 1 criminal. Since Vicky at A5 is already a criminal, Zach at D5 cannot be a criminal and must be innocent. Therefore, we can determine that D5 Zach is INNOCENT.
Clue:
"There's an equal number of criminals in rows 1 and 5" — Olga (C3)
"There are exactly 3 innocents in row 1" — Luigi (D2)
D1 · Hank → INNOCENT
Because: In column B, we already see two confirmed innocents (Chloe at B1 and Wally at B5) and one confirmed criminal (Nicole at B3). Ruby’s clue says that Wally is one of three innocents in column B, so column B must contain exactly three innocents total, which means exactly one of Julie (B2) or Steve (B4) is also innocent. Zach’s clue says columns B and D have an equal number of innocents, so column D must also have three innocents. In column D, Luigi (D2) and Uma (D4) are already criminals, while Peter (D3) and Zach (D5) are already innocents, so the only way for column D to reach three innocents is for Hank at D1 to be innocent. Therefore, we can determine that D1 Hank is INNOCENT.
Clue:
"Wally is one of 3 innocents in column B" — Ruby (A4)
"There's an equal number of innocents in columns B and D" — Zach (D5)
A1 · Austin → CRIMINAL
Because: Row 5 has exactly one criminal, because Vicky at A5 is criminal while Wally at B5, Xia at C5, and Zach at D5 are all innocent. Olga’s clue says rows 1 and 5 have an equal number of criminals, so row 1 must also have exactly one criminal. In row 1, Chloe at B1, Frank at C1, and Hank at D1 are already known to be innocent, leaving only Austin at A1 to account for that one criminal in the row. Therefore, we can determine that A1 Austin is CRIMINAL.
Clue:
"There's an equal number of criminals in rows 1 and 5" — Olga (C3)
C4 · Tina → INNOCENT
Because: Olga is at C3, and in row 2 the neighbors of Olga are B2 (Julie), C2 (Klay), and D2 (Luigi). Chloe’s clue says that exactly one innocent in row 2 is neighboring Olga, and since Luigi at D2 is already a criminal, that forces exactly one of Julie and Klay to be innocent and the other to be criminal. Julie is a neighbor of the corner A1 (Austin), and Klay is a neighbor of the corner D1 (Hank), so because one of Julie or Klay must be criminal, at least one of those two corners (A1 or D1) has exactly one innocent neighbor. Hank’s clue says only one corner person has exactly one innocent neighbor, so D5 (Zach) cannot also have exactly one innocent neighbor. Zach’s only neighbors are C4 (Tina), D4 (Uma, a criminal), and C5 (Xia, an innocent), so the only way for Zach to have more than one innocent neighbor is for Tina to be innocent. Therefore, we can determine that C4 Tina is INNOCENT.
Clue:
"Exactly 1 innocent in row 2 is neighboring Olga" — Chloe (B1)
"Only one person in a corner has exactly one innocent neighbor" — Hank (D1)
B4 · Steve → CRIMINAL
Because: Row 4 contains Ruby at A4, Steve at B4, Tina at C4, and Uma at D4. Peter’s clue says there are exactly 2 criminals in row 4, and we already know Uma at D4 is a criminal while Ruby at A4 and Tina at C4 are innocents. Since row 4 must have one more criminal besides Uma, and the only remaining person in that row is Steve at B4, Steve must be the second criminal. Therefore, we can determine that B4 Steve is CRIMINAL.
Clue:
"There are exactly 2 criminals in row 4" — Peter (D3)
B2 · Julie → INNOCENT
Because: In column B, we have Chloe at B1 who is already known to be innocent, Nicole at B3 who is criminal, Steve at B4 who is criminal, and Wally at B5 who is already known to be innocent, leaving only Julie at B2 unknown. Ruby’s clue says that Wally is one of 3 innocents in column B, so there must be exactly three innocents somewhere in that column. Since B3 and B4 are already criminals, the only way to reach three innocents is for B2 to be the third innocent alongside B1 and B5. Therefore, we can determine that B2 Julie is INNOCENT.
Clue:
"Wally is one of 3 innocents in column B" — Ruby (A4)
C2 · Klay → CRIMINAL
Because: Olga is at C3, and the people in row 2 who neighbor her are B2 (diagonally up-left), C2 (directly above), and D2 (diagonally up-right). Chloe’s clue says that among the innocents in row 2, exactly one of them is neighboring Olga, meaning only one of B2, C2, and D2 can be innocent. We already know Julie at B2 is innocent and she does neighbor Olga, and Luigi at D2 is criminal, so C2 cannot also be innocent without making the total more than one. Therefore, we can determine that C2 Klay is CRIMINAL.
Clue:
"Exactly 1 innocent in row 2 is neighboring Olga" — Chloe (B1)