Clues by Sam Apr 05, 2026 Answer – Full Solution Explained
Evil·Solved
A1
💂♀️
Anna
guard
B1
👩⚖️
Betty
judge
C1
👮♂️
Chase
cop
D1
👮♂️
Ethan
cop
A2
🕵️♀️
Flora
sleuth
B2
👨💻
Ghani
coder
C2
👨💻
Isaac
coder
D2
👮♀️
Jane
cop
A3
💂♀️
Kay
guard
B3
🕵️♂️
Logan
sleuth
C3
👨💻
Noah
coder
D3
👨⚖️
Olof
judge
A4
💂♀️
Paula
guard
B4
👨⚖️
Rob
judge
C4
👷♂️
Sam
builder
D4
👷♀️
Tina
builder
A5
🕵️♀️
Uma
sleuth
B5
👨🎨
Vince
painter
C5
👩🎨
Xena
painter
D5
👷♀️
Zara
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 9Criminal 11Unknown 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 · 9
[ A1 ] [ B2 ] [ C2 ] [ C3 ] [ A4 ] [ B4 ] [ A5 ] [ B5 ] [ C5 ]
Criminal · 11
[ B1 ] [ C1 ] [ D1 ] [ A2 ] [ D2 ] [ A3 ] [ B3 ] [ D3 ] [ C4 ] [ D4 ] [ 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 · Anna
"Betty has exactly 3 innocent neighbors"
B1 · Betty
"I just realized what this puzzle is missing..."
C1 · Chase
"There's an odd number of innocents in column C"
D1 · Ethan
"Only 1 of the 9 criminals on the edges is in column C"
A2 · Flora
"2 of the 3 guards have 3 innocent neighbors"
B2 · Ghani
"Sam is one of Tina's 3 criminal neighbors"
C2 · Isaac
"I think I know what they're thinking..."
D2 · Jane
"An odd number of innocents above Uma neighbor Logan"
A3 · Kay
"Paula... That doesn't count! It needs to be an obnoxious popup!"
B3 · Logan
"There are exactly 3 innocents in column B"
C3 · Noah
"Oh, no. I bet it's something evil."
D3 · Olof
"There are more criminals in row 2 than row 5"
A4 · Paula
"Oh, I have an ad. Go install Bee Sort by Sam!"
B4 · Rob
"Xena is one of 3 innocents in row 5"
C4 · Sam
"Oh yeah! I know what you mean, Betty!"
D4 · Tina
"There's an equal number of innocents in columns A and C"
A5 · Uma
"Ugh, I knew it..."
B5 · Vince
"Ethan is one of 2 criminals in the corners"
C5 · Xena
"Only 1 of the 2 criminals above Vince is Chase's neighbor"
D5 · Zara
"Ads! Why isn't this riddle riddled with ads!?"
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)
C5 · Xena → INNOCENT
Because: Rob’s clue explicitly says “Xena is one of 3 innocents in row 5,” which directly includes Xena among the innocents. Therefore, we can determine that C5 Xena is INNOCENT.
Clue:
"Xena is one of 3 innocents in row 5" — Rob (B4)
B3 · Logan → CRIMINAL
Because: The people above Vince are the four people in column B: Betty at B1, Ghani at B2, Logan at B3, and Rob at B4. Xena’s clue says that among these four there are exactly two criminals, and exactly one of those two criminals is a neighbor of Chase at C1. The only people above Vince who are neighbors of Chase are Betty (B1) and Ghani (B2), while Logan (B3) and Rob (B4) are not; since Rob is already known to be innocent, the one “not Chase’s neighbor” criminal above Vince must be Logan. Therefore, we can determine that B3 Logan is CRIMINAL.
Clue:
"Only 1 of the 2 criminals above Vince is Chase's neighbor" — Xena (C5)
B5 · Vince → INNOCENT
Because: Vince is in column B, and the clue from Logan says there are exactly 3 innocents in column B, so column B contains exactly 2 criminals in total. Xena’s clue talks about “the 2 criminals above Vince,” which means that among the four people above Vince in column B, there are exactly two criminals. Since column B only has two criminals altogether, those two criminals must already be above Vince, leaving no room for Vince to also be a criminal. Therefore, we can determine that B5 Vince is INNOCENT.
Clue:
"Only 1 of the 2 criminals above Vince is Chase's neighbor" — Xena (C5)
"There are exactly 3 innocents in column B" — Logan (B3)
D1 · Ethan → CRIMINAL
Because: Vince’s clue says that among the four corner people there are exactly two criminals, and that Ethan is one of those two. Since clues are always true, that directly fixes Ethan’s identity as a criminal. Therefore, we can determine that D1 Ethan is CRIMINAL.
Clue:
"Ethan is one of 2 criminals in the corners" — Vince (B5)
C1 · Chase → CRIMINAL
Because: The edge positions in column C are only C1 (Chase) and C5 (Xena), since the other C-column spots are not on the outer border. Ethan’s clue says that among the 9 edge criminals, exactly 1 is in column C, so exactly one of Chase or Xena must be a criminal. Xena at C5 is already known to be INNOCENT, so the only way for there to be exactly one edge criminal in column C is for Chase at C1 to be that criminal. Therefore, we can determine that C1 Chase is CRIMINAL.
Clue:
"Only 1 of the 9 criminals on the edges is in column C" — Ethan (D1)
A1 · Anna → INNOCENT
Because: The corner positions are A1, D1, A5, and D5, and Vince’s clue says Ethan at D1 is one of exactly two criminals in those corners. That means there is exactly one more criminal among A1, A5, and D5, and the other two of those three corners are innocent. Rob’s clue says Xena is one of three innocents in row 5; since Vince (B5) and Xena (C5) are already innocent, exactly one of the remaining row-5 people in corners A5 and D5 must also be innocent, so the other one (A5 or D5) must be the second corner criminal. That uses up the only remaining corner criminal besides Ethan, so A1 cannot be a corner criminal and must be innocent. Therefore, we can determine that A1 Anna is INNOCENT.
Clue:
"Xena is one of 3 innocents in row 5" — Rob (B4)
"Ethan is one of 2 criminals in the corners" — Vince (B5)
B1 · Betty → CRIMINAL, B2 · Ghani → INNOCENT
Because: Clue 4 says there are 9 criminals on the edges, so there are exactly 5 innocents on the edges. We already have three edge innocents for sure (Anna at A1, Vince at B5, and Xena at C5), and clue 1 says row 5 has exactly three innocents, so exactly one of A5 and D5 is also an edge innocent, bringing the edge-innocent total to 4. That leaves room for only one more edge innocent among the remaining edge positions B1, A2, A3, A4, D2, D3, and D4. Clue 2 says there are exactly two criminals above Vince in column B, and since Logan (B3) is one of them and Rob (B4) is not, exactly one of Betty (B1) and Ghani (B2) is the other criminal. Clue 3 says Betty has exactly 3 innocent neighbors; with Anna (A1) already an innocent neighbor and Chase (C1) already a criminal neighbor, this forces two of {A2, B2, C2} to be innocent, so if Ghani (B2) were the criminal then A2 would have to be innocent. But Ghani being the criminal would also make Betty (B1) innocent (because only one of B1 and B2 can be criminal), and that would create two edge innocents (B1 and A2) inside the set where only one more edge innocent is allowed. So Ghani cannot be the criminal above Vince, which forces Betty to be that criminal instead. Therefore, we can determine that B1 (Betty) is CRIMINAL and B2 (Ghani) is INNOCENT.
Clue:
"Xena is one of 3 innocents in row 5" — Rob (B4)
"Only 1 of the 2 criminals above Vince is Chase's neighbor" — Xena (C5)
"Betty has exactly 3 innocent neighbors" — Anna (A1)
"Only 1 of the 9 criminals on the edges is in column C" — Ethan (D1)
C4 · Sam → CRIMINAL
Because: Ghani’s clue says that Sam is one of Tina’s three criminal neighbors, which directly identifies Sam as a criminal. Therefore, we can determine that C4 Sam is CRIMINAL.
Clue:
"Sam is one of Tina's 3 criminal neighbors" — Ghani (B2)
D3 · Olof → CRIMINAL
Because: Tina is at D4, and her neighbors are C3, Sam at C4, Xena at C5, Olof at D3, and Zara at D5. Ghani’s clue says Tina has exactly 3 criminal neighbors and that Sam is one of them; since Xena is already an innocent neighbor, that forces exactly two of the three spots C3, D3, and D5 to be criminals, so exactly one of them is innocent. Ethan’s clue says there are exactly 9 criminals on the edge, so there are exactly 5 edge innocents; with Anna at A1 and Vince and Xena at B5 and C5 already innocent, that leaves exactly two more edge innocents to be found. Rob’s clue says row 5 has exactly 3 innocents, so exactly one of A5 and D5 is innocent, meaning at least one of those last two edge-innocent slots is taken by A5 or D5. If D3 were the one innocent among C3, D3, and D5, then D5 would have to be criminal, which would make A5 the innocent in row 5, and that would already use up both remaining edge-innocent slots (A5 and D3). That would force A2 to be a criminal, so by Anna’s clue C2 must be innocent, and then by Chase’s clue C3 must also be innocent, but D3 being the one innocent among C3, D3, and D5 requires C3 to be criminal. So D3 cannot be the innocent among C3, D3, and D5, which means D3 must be one of the criminals there. Therefore, we can determine that D3 (Olof) is CRIMINAL.
Clue:
"Xena is one of 3 innocents in row 5" — Rob (B4)
"Betty has exactly 3 innocent neighbors" — Anna (A1)
"Only 1 of the 9 criminals on the edges is in column C" — Ethan (D1)
"There's an odd number of innocents in column C" — Chase (C1)
"Sam is one of Tina's 3 criminal neighbors" — Ghani (B2)
D2 · Jane → CRIMINAL
Because: Betty is at B1, and her neighbors are A1, A2, B2, C1, and C2. Anna’s clue says Betty has exactly 3 innocent neighbors; we already know A1 and B2 are innocent and C1 is criminal, so exactly one of A2 and C2 must be innocent. Rob’s clue says row 5 has exactly 3 innocents, so row 5 has exactly 1 criminal, and Olof’s clue then forces row 2 to have more than 1 criminal, meaning at most one of A2, C2, and D2 can be innocent since B2 is already innocent. Because A2 and C2 already use up that single possible innocent between them, D2 cannot be innocent. Therefore, we can determine that D2 (Jane) is CRIMINAL.
Clue:
"Xena is one of 3 innocents in row 5" — Rob (B4)
"Betty has exactly 3 innocent neighbors" — Anna (A1)
"There are more criminals in row 2 than row 5" — Olof (D3)
D4 · Tina → CRIMINAL
Because: The edge of the board has 14 people, and Ethan’s clue says exactly 9 of those edge people are criminals. Right now we already have 5 edge criminals (B1, C1, D1, D2, D3), so the 6 remaining unknown edge people (A2, A3, A4, A5, D4, D5) must contain exactly 4 criminals. Rob’s clue says row 5 has exactly 3 innocents, and since B5 and C5 are already innocent, exactly one of A5 (Uma) and D5 (Zara) is a criminal. That means the other four unknown edge people (A2, A3, A4, and D4) must contain the remaining 3 criminals, so there is exactly 1 innocent among A2, A3, A4, and D4. Jane’s clue talks about the people above Uma who are neighbors of Logan; those are A2, A3, and A4, and it says an odd number of them are innocents. If D4 were the single innocent among A2, A3, A4, and D4, then A2, A3, and A4 would have 0 innocents, which is not odd, so D4 cannot be innocent and must be one of the criminals. Therefore, we can determine that D4 Tina is CRIMINAL.
Clue:
"Xena is one of 3 innocents in row 5" — Rob (B4)
"Only 1 of the 9 criminals on the edges is in column C" — Ethan (D1)
"An odd number of innocents above Uma neighbor Logan" — Jane (D2)
C3 · Noah → INNOCENT, D5 · Zara → CRIMINAL, C2 · Isaac → INNOCENT
Because: In column C, we currently have C1 as a criminal, C4 as a criminal, and C5 as an innocent, with only C2 and C3 unknown. Chase’s clue says column C has an odd number of innocents, so C2 and C3 must match each other: if exactly one of them were innocent, column C would have 2 innocents, which is even. Tina’s clue says columns A and C have the same number of innocents; since column C already has C5 plus C2 and C3, that forces column A (beyond A1) to have either 0 innocents (if C2 and C3 were both criminals) or 2 innocents (if C2 and C3 are both innocents). Ethan’s clue says there are exactly 9 criminals on the edges, and we already have 6 known edge criminals; if C2 and C3 were both criminals, then column A would contribute 0 more innocents beyond A1, meaning A2 through A5 would all be edge criminals, already pushing the edge-criminal total too high, so C2 and C3 must instead both be innocents. With C2 and C3 innocent, column A must have exactly 2 innocents among A2 through A5, so it has exactly 2 edge criminals there; that makes the current edge-criminal total 8, so the remaining edge unknown D5 must be a criminal to reach exactly 9 edge criminals. Therefore, we can determine that C3 Noah is INNOCENT, D5 Zara is CRIMINAL, and C2 Isaac is INNOCENT.
Clue:
"Only 1 of the 9 criminals on the edges is in column C" — Ethan (D1)
"There's an odd number of innocents in column C" — Chase (C1)
"There's an equal number of innocents in columns A and C" — Tina (D4)
A5 · Uma → INNOCENT
Because: In row 5, the people are Uma at A5, Vince at B5, Xena at C5, and Zara at D5. Rob’s clue says that Xena is one of 3 innocents in row 5, so row 5 must contain exactly three innocents in total. We already know Vince and Xena are innocents, and Zara is a criminal, so the only way for row 5 to have three innocents is for Uma to be the third innocent. Therefore, we can determine that A5 (Uma) is INNOCENT.
Clue:
"Xena is one of 3 innocents in row 5" — Rob (B4)
A2 · Flora → CRIMINAL
Because: Betty is at B1, so her neighbors (including diagonals) are A1 Anna, A2 Flora, B2 Ghani, C1 Chase, and C2 Isaac. Anna, Ghani, and Isaac are already known to be innocent, giving Betty exactly three innocent neighbors right away, and Chase is already known to be criminal. Since Anna’s clue says Betty has exactly 3 innocent neighbors, Flora cannot also be innocent or Betty would have four innocent neighbors. Therefore, we can determine that A2 Flora is CRIMINAL.
Clue:
"Betty has exactly 3 innocent neighbors" — Anna (A1)
A4 · Paula → INNOCENT
Because: The three guards are Anna at A1, Kay at A3, and Paula at A4. Flora’s clue says exactly two of these three guards have exactly three innocent neighbors. Anna’s neighbors are only Betty (B1), Flora (A2), and Ghani (B2), and only Ghani is innocent, so Anna cannot be one of the two guards with three innocent neighbors; that forces both Kay and Paula to be the two guards who do. Kay’s neighbors include Ghani (B2) and Rob (B4), which are two innocents, and the only remaining neighbor that could make Kay’s innocent-neighbor count reach three is Paula at A4, so Paula must be innocent. Therefore, we can determine that A4 Paula is INNOCENT.
Clue:
"2 of the 3 guards have 3 innocent neighbors" — Flora (A2)
A3 · Kay → CRIMINAL
Because: Look at all the edge positions: the whole top row (A1–D1), the whole bottom row (A5–D5), plus the left edge (A2–A4) and right edge (D2–D4). Among these, we already have 8 confirmed edge criminals: B1, C1, D1, A2, D2, D3, D4, and D5, and the only one of them in column C is C1 (since C5 is innocent). Ethan’s clue says there are 9 criminals on the edges in total, and only 1 of those edge criminals is in column C, so the extra edge criminal cannot be in column C and must be the only undetermined edge person, Kay at A3. Therefore, we can determine that A3 Kay is CRIMINAL.
Clue:
"Only 1 of the 9 criminals on the edges is in column C" — Ethan (D1)