Clues by Sam Jan 25, 2026 Answer – Full Solution Explained
Evil·Solved
A1
👮♂️
Aaron
cop
B1
👨🔧
Chuck
mech
C1
👩🍳
Betsy
cook
D1
👩🍳
Eve
cook
A2
👩🎨
Freya
painter
B2
👨💻
Gabe
coder
C2
👩🍳
Helen
cook
D2
👨⚖️
Igor
judge
A3
👨🎨
Jerry
painter
B3
👩🔧
Katie
mech
C3
👩🏫
Linda
teacher
D3
👨🔧
Mark
mech
A4
👮♀️
Penny
cop
B4
👨💻
Scott
coder
C4
👨🏫
Thor
teacher
D4
👩⚖️
Uma
judge
A5
👩🎤
Vera
singer
B5
👩🎤
Wanda
singer
C5
👩🎤
Xena
singer
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 10Criminal 10Unknown 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 · 10
[ A1 ] [ D1 ] [ B2 ] [ B3 ] [ C3 ] [ B4 ] [ C4 ] [ A5 ] [ B5 ] [ C5 ]
Criminal · 10
[ B1 ] [ C1 ] [ A2 ] [ C2 ] [ D2 ] [ A3 ] [ D3 ] [ A4 ] [ 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 · Aaron
"There is only one innocent above Zed"
B1 · Chuck
"How about Samogram?"
C1 · Betsy
"An odd number of innocents in column B are Thor's neighbors"
D1 · Eve
"I'd play SamSam if it existed."
A2 · Freya
"Nonogram is my type of game. I like drawing."
B2 · Gabe
"I prefer this to Sudoku. Numbers make me dizzy."
C2 · Helen
"Exactly 2 of the 6 innocents neighboring Scott are in column B"
D2 · Igor
"I prefer Sudoku. Criminals are left alone."
A3 · Jerry
"I like KenKen. It's like Sudoku but with even more numbers!"
B3 · Katie
"Mark is one of Uma's 2 criminal neighbors"
C3 · Linda
"There are no innocents above Xena who neighbor Chuck"
D3 · Mark
"All criminals below Aaron are connected"
A4 · Penny
"There's an odd number of criminals above Vera"
B4 · Scott
"Only one column has exactly 2 criminals"
C4 · Thor
"All criminals in column D are connected"
D4 · Uma
"Helen has exactly 4 innocent neighbors"
A5 · Vera
"There are more innocents in row 1 than row 2"
B5 · Wanda
"Only 1 of the 5 innocents on the edges is Betsy's neighbor"
C5 · Xena
"Makes me think. What would Sudoku by Sam be like?"
D5 · Zed
"I like Slitherlink. It has an evil ring to it."
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)
C1 · Betsy → CRIMINAL, C2 · Helen → CRIMINAL
Because: Xena is at C5, so the people above Xena in the same column are Betsy at C1, Helen at C2, Linda at C3, and Thor at C4. Chuck is at B1, and the only people among those who are neighbors of Chuck are Betsy at C1 (directly right of him) and Helen at C2 (diagonally down-right of him). Linda’s clue says that there are no innocents among the people above Xena who neighbor Chuck, so neither Betsy nor Helen can be innocent. Therefore, we can determine that C1 Betsy is CRIMINAL and C2 Helen is CRIMINAL.
Clue:
"There are no innocents above Xena who neighbor Chuck" — Linda (C3)
B5 · Wanda → INNOCENT, B3 · Katie → INNOCENT
Because: Scott is at B4, so the only neighbors of Scott that lie in column B are the two directly above and below him: Katie at B3 and Wanda at B5. Helen’s clue says that among Scott’s neighboring innocents, exactly 2 of them are in column B. Since column B contributes only those two neighboring positions, both Katie and Wanda must be those two column-B innocents. Therefore, we can determine that B5 Wanda is INNOCENT and B3 Katie is INNOCENT.
Clue:
"Exactly 2 of the 6 innocents neighboring Scott are in column B" — Helen (C2)
B4 · Scott → INNOCENT
Because: Thor is at C4, and the only people in column B who are Thor’s neighbors are B3 Katie, B4 Scott, and B5 Wanda. Betsy’s clue says that an odd number of innocents in column B are Thor’s neighbors, so among Katie, Scott, and Wanda the number of innocents must be 1 or 3. Since Katie and Wanda are already known to be innocent, there are already 2 innocents among those neighbors, so the total can only be odd if Scott is also innocent. Therefore, we can determine that B4 Scott is INNOCENT.
Clue:
"An odd number of innocents in column B are Thor's neighbors" — Betsy (C1)
D3 · Mark → CRIMINAL
Because: Katie’s clue says that Mark is one of Uma’s 2 criminal neighbors, which directly includes Mark in the set of criminals next to Uma. Since clues are always true, Mark must be a criminal. Therefore, we can determine that D3 Mark is CRIMINAL.
Clue:
"Mark is one of Uma's 2 criminal neighbors" — Katie (B3)
C4 · Thor → INNOCENT
Because: Scott is at B4, and his neighbors include Jerry A3, Penny A4, Thor C4, Vera A5, Wanda B5, and Xena C5; Wanda is already an innocent, and Helen’s clue says Scott has exactly 6 innocent neighbors, so among Jerry, Penny, Thor, Vera, and Xena there must be exactly 3 innocents. Uma is at D4, and Katie’s clue says Uma has exactly 2 criminal neighbors and Mark is one of them, so among Thor, Xena, and Zed exactly 1 is criminal. Wanda’s clue says there are exactly 5 innocents on the edges, and exactly 1 of those edge-innocents is a neighbor of Betsy, so only 4 edge-innocents can be people who are not next to Betsy. If Thor were the one criminal among Thor, Xena, and Zed, then Xena and Zed would both be edge-innocents, and Scott’s “exactly 3” would also force Xena plus two of Jerry, Penny, and Vera to be edge-innocents along with Wanda, which already makes five edge-innocents that are not neighbors of Betsy, leaving no edge-innocent available to be Betsy’s required edge-neighbor. So Thor cannot be that criminal, and Thor must be one of the innocents. Therefore, we can determine that C4 Thor is INNOCENT.
Clue:
"Exactly 2 of the 6 innocents neighboring Scott are in column B" — Helen (C2)
"Mark is one of Uma's 2 criminal neighbors" — Katie (B3)
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)
D2 · Igor → CRIMINAL
Because: Betsy is at C1, so her edge neighbors are B1, D1, and D2. Wanda’s clue says there are five innocents on the edges in total, and exactly one of those edge-innocents is Betsy’s neighbor, so exactly one of B1, D1, and D2 is innocent and the other two are criminals. Mark at D3 is already a criminal, and Thor’s clue says all criminals in column D must be connected, so we cannot have criminals at D1 and D3 with an innocent at D2 separating them. That rules out D2 being innocent, so D2 must be a criminal. Therefore, we can determine that D2 Igor is CRIMINAL.
Clue:
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)
"All criminals in column D are connected" — Thor (C4)
D4 · Uma → CRIMINAL
Because: Look at the edge count from Wanda’s clue: there are exactly 5 innocents on the edges, and exactly 1 of those edge-innocents is a neighbor of Betsy at C1, which forces exactly one of B1 or D1 to be an edge-innocent. Helen’s clue about Scott at B4 says Scott has 6 innocent neighbors; since Katie (B3), Linda (C3), Thor (C4), and Wanda (B5) are already innocent, exactly two more of Scott’s other neighbors (A3, A4, A5, C5) must be innocent, and all of those positions are on the edge. That already accounts for 4 edge-innocents in total (Wanda, exactly one of B1/D1, and two of A3/A4/A5/C5), so there is room for only 1 more edge-innocent anywhere else. Thor’s clue says all criminals in column D are connected, so D5 can only be a criminal if D4 is also a criminal; therefore, if D4 were an edge-innocent, D5 would be forced to be an edge-innocent too, which would require two remaining edge-innocent spots, not one. Therefore, we can determine that D4 (Uma) is CRIMINAL.
Clue:
"Exactly 2 of the 6 innocents neighboring Scott are in column B" — Helen (C2)
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)
"All criminals in column D are connected" — Thor (C4)
B2 · Gabe → INNOCENT
Because: Helen is at C2, so her neighbors are B1, C1 (Betsy), D1, B2 (Gabe), D2 (Igor), B3 (Katie), C3 (Linda), and D3 (Mark). Uma says Helen has exactly 4 innocent neighbors; Katie and Linda already account for 2, and Betsy, Igor, and Mark are already not innocent, so exactly 2 of the remaining three people B1, D1, and B2 must be innocent. Wanda says that among the 5 edge innocents, only 1 is Betsy’s neighbor, and Betsy’s edge neighbors are B1 and D1 (D2 is also an edge neighbor but is already not innocent), so B1 and D1 cannot both be innocent at the same time. That forces B2 (Gabe) to be one of the two required innocent neighbors of Helen. Therefore, we can determine that B2 (Gabe) is INNOCENT.
Clue:
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)
"Helen has exactly 4 innocent neighbors" — Uma (D4)
C5 · Xena → INNOCENT, D5 · Zed → CRIMINAL
Because: Betsy is at C1, and her only neighbors that are on the edge are B1 and D1 (since D2 is also an edge neighbor but Igor at D2 is already a criminal). Wanda’s clue says there are exactly 5 innocents on the edges in total, and exactly 1 of those edge-innocents is Betsy’s neighbor, so one of B1 or D1 must be an edge-innocent. Katie’s clue about Uma at D4 says Uma has exactly 2 criminal neighbors and Mark is one of them; since Linda at C3 and Thor at C4 are innocent, exactly one of C5 (Xena) and D5 (Zed) is criminal, so at least one of them is an edge-innocent. Now use Scott’s clue: only one column has exactly 2 criminals; column B cannot (it has at most 1 criminal because B2–B5 are all innocent), and column D cannot (it already has 3 criminals at D2, D3, and D4), so the “exactly 2 criminals” column must be either A or C. But if column A had exactly 2 criminals, that would force 3 edge-innocents in column A, and together with Wanda at B5, the required edge-innocent at B1 or D1, and the guaranteed edge-innocent among C5/D5, we would already have at least 6 edge-innocents, which is impossible because there are exactly 5; therefore column A cannot be the one, so column C must have exactly 2 criminals, forcing Xena at C5 to be innocent. With C5 innocent, Uma’s “exactly 2 criminal neighbors” clue then forces Zed at D5 to be the other criminal neighbor besides Mark. Therefore, we can determine that C5 Xena is INNOCENT and D5 Zed is CRIMINAL.
Clue:
"Only one column has exactly 2 criminals" — Scott (B4)
"Mark is one of Uma's 2 criminal neighbors" — Katie (B3)
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)
A4 · Penny → CRIMINAL
Because: Scott is at B4, and his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Helen says there are exactly 6 innocents among those neighbors, and we already know five of them are innocents (B3 Katie, C3 Linda, C4 Thor, B5 Wanda, and C5 Xena), so exactly one of A3, A4, and A5 is innocent and the other two are criminals. Mark says all criminals below Aaron (A2–A5) are orthogonally connected, and since B5 is innocent, anyone criminal at A5 can only connect to the rest through A4; so the two criminals among A3/A4/A5 must be adjacent, which is only possible if A4 is one of them. Therefore, we can determine that A4 Penny is CRIMINAL.
Clue:
"Exactly 2 of the 6 innocents neighboring Scott are in column B" — Helen (C2)
"All criminals below Aaron are connected" — Mark (D3)
A5 · Vera → INNOCENT
Because: Betsy is at C1, and the only edge positions that are her neighbors are B1, D1, and D2 (since neighbors include diagonals). Wanda’s clue says there are exactly five innocents on the edge, and exactly one of those five is Betsy’s neighbor; because D2 (Igor) is a criminal, that forces exactly one of B1 or D1 to be an edge-innocent. We already have two edge-innocents for sure (Wanda at B5 and Xena at C5), so the remaining three edge-innocents must come from A1, A2, A3, A5, B1, and D1. Penny’s clue says the number of criminals above Vera (A1–A4) is odd; since A4 (Penny) is already a criminal, A1–A3 must contain an even number of criminals, so A1–A3 contribute either three edge-innocents or only one edge-innocent. They cannot contribute three edge-innocents, because then Wanda, Xena, and A1–A3 would already make all five edge-innocents, leaving no way for B1 or D1 to be the required edge-innocent neighbor of Betsy, so A1–A3 must contribute only one edge-innocent. That means A5, B1, and D1 together must contribute the remaining two edge-innocents, and since exactly one of B1 or D1 is an edge-innocent, the other one has to be A5. Therefore, we can determine that A5 Vera is INNOCENT.
Clue:
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)
"There's an odd number of criminals above Vera" — Penny (A4)
A1 · Aaron → INNOCENT, A2 · Freya → CRIMINAL
Because: Betsy is at C1, and her only edge-position neighbors are B1 and D1 (the other neighbors B2, C2, and D2 are not on the edge or are already known criminals). Wanda’s clue says that exactly one of the five edge innocents is a neighbor of Betsy, and none of the already-known edge innocents (A5 Vera, B5 Wanda, C5 Xena) are near C1, so the only way to have exactly one edge innocent neighboring Betsy is for exactly one of B1 or D1 to be innocent. That means row 1 has exactly one innocent among B1 and D1, plus whatever A1 is. Vera’s clue says row 1 has more innocents than row 2, but row 2 already has Gabe as an innocent and the only other possible innocent there is A2, so the only way for row 1 to beat row 2 is for A1 to be innocent while A2 is not innocent. Therefore, we can determine that A1 (Aaron) is INNOCENT and A2 (Freya) is CRIMINAL.
Clue:
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)
"There are more innocents in row 1 than row 2" — Vera (A5)
A3 · Jerry → CRIMINAL
Because: Scott is at B4, so his neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Helen’s clue says that among Scott’s neighbors there are exactly 6 innocents, and exactly 2 of those innocents are in column B; the only column B neighbors are B3 (Katie) and B5 (Wanda), and both are already known innocents, so they account for those 2. Among the other six neighbors, C3 (Linda), C4 (Thor), A5 (Vera), and C5 (Xena) are already known innocents, while A4 (Penny) is a known criminal, so the only remaining neighbor A3 (Jerry) cannot also be innocent or Scott would have 7 innocent neighbors instead of 6. Therefore, we can determine that A3 (Jerry) is CRIMINAL.
Clue:
"Exactly 2 of the 6 innocents neighboring Scott are in column B" — Helen (C2)
D1 · Eve → INNOCENT
Because: Zed is at D5, so “above Zed” means the four people in column D at D1, D2, D3, and D4. Aaron’s clue says there is only one innocent among those four. We already know D2 (Igor), D3 (Mark), and D4 (Uma) are criminals, so none of them can be that one innocent. That forces the remaining spot, D1 (Eve), to be the only innocent above Zed. Therefore, we can determine that D1 Eve is INNOCENT.
Clue:
"There is only one innocent above Zed" — Aaron (A1)
B1 · Chuck → CRIMINAL
Because: The five innocents who are on the edge right now are Aaron at A1, Eve at D1, Vera at A5, Wanda at B5, and Xena at C5. Wanda’s clue says that among these five edge-innocents, exactly one is a neighbor of Betsy at C1. Betsy’s neighbors are B1, D1, B2, C2, and D2, so Eve at D1 is already one of those neighbors, and none of the other listed edge-innocents are adjacent to Betsy. If Chuck at B1 were innocent, he would become an edge-innocent who is also Betsy’s neighbor, making the total two, which would break the clue. Therefore, we can determine that B1 Chuck is CRIMINAL.
Clue:
"Only 1 of the 5 innocents on the edges is Betsy's neighbor" — Wanda (B5)