Clues by Sam Feb 07, 2026 Answer – Full Solution Explained
A1
👮♂️
cop
B1
👨🔧
mech
C1
👨🏫
teacher
D1
💂♀️
guard
A2
👨🍳
cook
B2
👩🌾
farmer
C2
👨✈️
pilot
D2
💂♂️
guard
A3
👩🍳
cook
B3
👩🍳
cook
C3
👨✈️
pilot
D3
👮♀️
cop
A4
👩🔧
mech
B4
👨🏫
teacher
C4
🐨
koala
D4
👩🔧
mech
A5
👮♂️
cop
B5
🐨
koala
C5
👩🌾
farmer
D5
💂♀️
guard
Final Board State
This puzzle is fully solved.
All characters have been identified as innocent or criminal based on today's clues.
See how each clue leads to the final result
Skip the reasoning — 12 criminals.
Clues by Sam answer for Feb 07, 2026 — a Tricky solved in 18 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Andre (A1), Bruce (B1), Chuck (C1), Gary (A2), Hilda (B2), Isaac (C2), Kay (A3), Lisa (B3), Olivia (D3), Vince (A5), Wanda (B5) and Xena (C5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.A5 · Vince → CRIMINAL, A3 · Kay → CRIMINAL
Gary is at A2, so the people below him in the same column are Kay at A3, Pam at A4, and Vince at A5. Pam’s clue says there is only one innocent below Gary, meaning exactly one of those three people is innocent. Since Pam at A4 is already known to be INNOCENT, she must be the only innocent among them, so Kay at A3 and Vince at A5 cannot be innocent. Therefore, we can determine that A5 Vince is CRIMINAL and A3 Kay is CRIMINAL.
02.C2 · Isaac → CRIMINAL
Row 2 is Gary at A2, then Hilda at B2, Isaac at C2, and Jerry at D2. Vince’s clue says there is only one innocent to the right of Gary, so among Hilda, Isaac, and Jerry exactly one is innocent and the other two are criminals. Kay’s clue says all criminals in row 2 are connected, which means you cannot have two criminals in that row with an innocent sitting between them. If Isaac were the one innocent to Gary’s right, then Hilda and Jerry would both be criminals with Isaac in between them, breaking that connectedness, so Isaac cannot be innocent. Therefore, we can determine that C2 Isaac is CRIMINAL.
03.D1 · Flora → INNOCENT
Isaac’s clue says that Flora is one of Chuck’s 2 innocent neighbors, which directly states that Flora is innocent. Therefore, we can determine that D1 Flora is INNOCENT.
04.B1 · Bruce → CRIMINAL
Chuck is at C1, and his neighbors are Bruce at B1, Hilda at B2, Isaac at C2, Jerry at D2, and Flora at D1. Isaac says that Flora is one of Chuck’s 2 innocent neighbors, so besides Flora there is exactly one more innocent among Bruce, Hilda, and Jerry (since Isaac himself is a criminal). Vince says there is only one innocent to the right of Gary, and the people to the right of Gary at A2 are Hilda, Isaac, and Jerry; with Isaac already criminal, that means exactly one of Hilda and Jerry is innocent. Since Hilda and Jerry cannot both be criminal, Bruce cannot be the “one more” innocent neighbor of Chuck, so Bruce must be a criminal. Therefore, we can determine that B1 Bruce is CRIMINAL.
05.A2 · Gary → CRIMINAL
Gary is at A2, so the people to his right in row 2 are Hilda at B2, Isaac at C2, and Jerry at D2. Vince’s clue says there is only one innocent to the right of Gary, and since Isaac is already known to be a criminal, that means exactly one of Hilda or Jerry is innocent and the other is a criminal. So row 2 has Isaac plus exactly one of {Hilda, Jerry} as criminals, which makes 2 criminals in row 2 before counting Gary. Bruce’s clue says row 2 has an odd number of criminals, so Gary must be a criminal to make that total odd. Therefore, we can determine that A2 Gary is CRIMINAL.
06.B2 · Hilda → CRIMINAL
In row 2, we already know A2 (Gary) is a criminal and C2 (Isaac) is a criminal, with B2 (Hilda) sitting directly between them. Kay’s clue says that all criminals in row 2 form one orthogonally connected chain, so the criminals in that row cannot be split into separate groups. The only way for A2 and C2 to be connected within the row is if B2 is also a criminal, because B2 is the only orthogonal link between A2 and C2. Therefore, we can determine that B2 (Hilda) is CRIMINAL.
07.D2 · Jerry → INNOCENT
Gary is at A2, so the people to his right are the rest of row 2: B2 Hilda, C2 Isaac, and D2 Jerry. Vince’s clue says there is only one innocent to the right of Gary, meaning exactly one of those three is innocent. Hilda and Isaac are already known to be criminals, so neither of them can be that single innocent. That forces Jerry at D2 to be the one innocent to the right of Gary. Therefore, we can determine that D2 Jerry is INNOCENT.
08.A1 · Andre → CRIMINAL
Gary’s clue says that column C contains exactly 2 innocents. Flora’s clue says that only one column on the whole board has exactly 2 innocents, so column C must be that unique column. In column A, Pam at A4 is already innocent and the other known people in that column (Gary at A2, Kay at A3, and Vince at A5) are criminals, so Andre at A1 is the only one who could change the innocent count there; if Andre were innocent, column A would have exactly 2 innocents, creating a second column with exactly 2 innocents. Therefore, we can determine that A1 Andre is CRIMINAL.
09.B5 · Wanda → CRIMINAL
Nick is at C3, and his only neighbors in column B are B2, B3, and B4. Hilda’s clue says Nick has exactly four innocent neighbors and only one of those is in column B; since B2 (Hilda) is already a criminal, this forces exactly one of B3 (Lisa) and B4 (Raul) to be innocent. Gary’s clue tells us column C has exactly two innocents, and Flora’s clue says only one column has exactly two innocents, so no other column, including column B, can have exactly two innocents. But if B5 (Wanda) were innocent, then column B would have exactly two innocents in total (Wanda plus the one innocent among B3 and B4), which is not allowed, so Wanda cannot be innocent. Therefore, we can determine that B5 Wanda is CRIMINAL.
10.D4 · Uma → INNOCENT
Zara is at D5, so her neighbors are exactly C4, D4, and C5. Gary’s clue says there are exactly two innocents in column C, and only one of those two is Zara’s neighbor, so among C4 and C5 exactly one is innocent and the other is a criminal. That means Zara already has exactly one criminal among her column-C neighbors, and Wanda’s clue says the total number of criminals neighboring Zara is odd, so D4 cannot also be a criminal or it would make two criminals (an even number). Therefore, we can determine that D4 (Uma) is INNOCENT.
11.B3 · Lisa → CRIMINAL
Isaac is at C2, and the only people in row 3 who neighbor him are Lisa at B3, Nick at C3, and Olivia at D3. Uma at D4 says she has exactly three innocent neighbors, and exactly one of those innocent neighbors is to the right of Kay; among Uma’s neighbors, the only ones that are to the right of Kay (same row as Kay, row 3) are Nick at C3 and Olivia at D3, so exactly one of Nick and Olivia is innocent. Jerry at D2 says an odd number of the row 3 neighbors of Isaac are innocent, so among Lisa, Nick, and Olivia the number of innocents must be 1 or 3, and it cannot be 3 because Nick and Olivia cannot both be innocent. That forces the total to be exactly 1 innocent among the three, so Lisa cannot be innocent and must be criminal. Therefore, we can determine that B3 Lisa is CRIMINAL.
12.B4 · Raul → INNOCENT
Nick is at C3, so his column B neighbors are exactly B2 Hilda, B3 Lisa, and B4 Raul. Hilda’s clue says that among the four innocents neighboring Nick, exactly one of those innocents is in column B, so at least one of Nick’s column B neighbors must be innocent. Since B2 Hilda and B3 Lisa are already known to be criminals, the only remaining way for there to be an innocent neighbor of Nick in column B is for B4 Raul to be that one. Therefore, we can determine that B4 · Raul is INNOCENT.
13.D5 · Zara → INNOCENT
Zara at D5 has two neighbors in column C, Tom at C4 and Xena at C5. Gary’s clue says there are exactly two innocents in column C, and exactly one of those two column-C innocents is a neighbor of Zara, so exactly one of Tom and Xena is innocent. Uma’s clue says she has exactly three innocent neighbors, and exactly one of those is to the right of Kay; among Uma’s neighbors, the only people to the right of Kay are Nick at C3 and Olivia at D3, so exactly one of Nick and Olivia is innocent, leaving exactly two innocents among Uma’s other neighbors Tom, Xena, and Zara. Since Tom and Xena contain exactly one innocent, the other required innocent among Tom, Xena, and Zara must be Zara. Therefore, we can determine that D5 Zara is INNOCENT.
14.C5 · Xena → CRIMINAL
Jerry’s clue looks at the people in row 3 who neighbor Isaac at C2, which are Lisa at B3, Nick at C3, and Olivia at D3, and it says an odd number of those are innocents. Since Lisa is already a criminal, the only way for the innocent count among those three to be odd is for exactly one of Nick and Olivia to be innocent, meaning Nick and Olivia are one innocent and one criminal. Lisa’s clue says only one koala has exactly 4 criminal neighbors, and Wanda at B5 cannot be that koala because she has only five neighbors and two of them (Pam at A4 and Raul at B4) are already innocents, so she can have at most 3 criminal neighbors. That forces Tom at C4 to be the koala with exactly 4 criminal neighbors; Tom already has Lisa and Wanda as criminal neighbors, so among Tom’s remaining unknown neighbors Nick, Olivia, and Xena, exactly two must be criminals. But Nick and Olivia together contribute exactly one criminal, so Xena must be the other criminal to reach two. Therefore, we can determine that C5 Xena is CRIMINAL.
15.C4 · Tom → INNOCENT
Zara is at D5, so the only people in column C who can be her neighbors are C4 (Tom), which is diagonally adjacent, and C5 (Xena), which is directly left. Gary’s clue says there are exactly two innocents in column C, and exactly one of those two is Zara’s neighbor. Since Xena at C5 is already known to be a criminal, C5 cannot be that neighboring innocent, so the only way for Zara to have exactly one neighboring innocent in column C is for Tom at C4 to be innocent. Therefore, we can determine that C4 Tom is INNOCENT.
16.D3 · Olivia → CRIMINAL
Nick is at C3, so his neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Among these, we already know four innocents: Jerry at D2, Raul at B4, Tom at C4, and Uma at D4, and only one of them is in column B (Raul). Hilda’s clue specifically says “the 4 innocents neighboring Nick,” which means Nick has exactly four innocent neighbors, so D3 cannot be an additional innocent neighbor. Therefore, we can determine that D3 Olivia is CRIMINAL.
17.C3 · Nick → INNOCENT
Isaac is at C2, and the row 3 people who neighbor him are B3 Lisa, C3 Nick, and D3 Olivia, since neighbors include diagonals. The clue says that among the row 3 neighbors of Isaac, the number of innocents is odd. Lisa and Olivia are both already known to be criminals, so the only way for the count of innocents among those three to be odd is for Nick to be innocent. Therefore, we can determine that C3 Nick is INNOCENT.
18.C1 · Chuck → CRIMINAL
Column C currently contains Nick at C3 and Tom at C4 as INNOCENT, while Chuck at C1 is the only unknown in that column. Gary’s clue talks about “the 2 innocents in column C,” which means there are exactly two innocents in column C total. Since Nick and Tom already account for those two innocents, Chuck cannot be an additional innocent in column C. Therefore, we can determine that C1 Chuck is CRIMINAL.