Clues by Sam Mar 01, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
🕵️♂️
sleuth
C1
👷♂️
builder
D1
👨🎤
singer
A2
👮♀️
cop
B2
👨🌾
farmer
C2
👩🎤
singer
D2
💂♂️
guard
A3
🕵️♀️
sleuth
B3
👷♀️
builder
C3
👩🏫
teacher
D3
💂♂️
guard
A4
👮♀️
cop
B4
👨🏫
teacher
C4
👨🏫
teacher
D4
💂♂️
guard
A5
🕵️♀️
sleuth
B5
👩⚕️
doctor
C5
👩🌾
farmer
D5
👨⚕️
doctor
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 — 4 criminals.
Clues by Sam answer for Mar 01, 2026 — a Evil solved in 15 steps
Today's Clues by Sam puzzle is rated Evil and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Brian (B1), Gabe (B2), Hazel (C2) and Xavi (D5); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.C4 · Raul → INNOCENT, C5 · Wanda → INNOCENT
Xavi is at D5, and the only people in column C who neighbor him are Wanda at C5 (directly left) and Raul at C4 (diagonally up-left). Uma’s clue says that exactly 2 innocents in column C are neighboring Xavi, and there are exactly two such neighbors in column C. So both of them must be innocents. Therefore, we can determine that C4 Raul is INNOCENT and C5 Wanda is INNOCENT, and so on.
02.D3 · Nick → INNOCENT, D4 · Steve → INNOCENT, D2 · Ike → INNOCENT
Laura is at C3, so the only people in column D who are her neighbors are D2 (Ike), D3 (Nick), and D4 (Steve). Raul’s clue says that, among those column D neighbors of Laura, the number who are innocent is odd, so it must be either 1 or 3. Wanda’s clue says there are exactly 4 innocents in column D in total, meaning there is only 1 criminal anywhere in column D, so D2–D4 cannot contain two criminals. That rules out the “1 innocent” possibility, so all three of D2, D3, and D4 must be innocent. Therefore, we can determine that D3 Nick is INNOCENT, D4 Steve is INNOCENT, and D2 Ike is INNOCENT.
03.C3 · Laura → INNOCENT
Hazel is at C2, and her neighbors include Brian (B1), Derek (C1), Erwin (D1) in row 1, plus Gabe (B2), Katie (B3), and Laura (C3) outside row 1 (since Ike and Nick are already known innocents). Nick’s clue says there are exactly two criminals neighboring Hazel, and exactly one of those two is in row 1, so the other criminal neighbor of Hazel must be one of Gabe, Katie, or Laura. Steve’s clue looks at people below Brian (B2, B3, B4, B5) who also neighbor Flora (A2), and the only ones that fit are Gabe (B2) and Katie (B3); it says exactly one of them is innocent, so the other one is a criminal. That means the “not in row 1” criminal neighbor of Hazel is already forced to be Gabe or Katie, so Laura cannot be that criminal neighbor of Hazel. Therefore, we can determine that C3 · Laura is INNOCENT.
04.A1 · Anna → INNOCENT
Hazel is at C2, and her row 1 neighbors are Brian at B1, Derek at C1, and Erwin at D1. Nick’s clue says Hazel has exactly two neighboring criminals, and only one of those two is in row 1, so among Brian, Derek, and Erwin there is exactly one criminal. If Anna at A1 were a criminal, then row 1 would contain exactly two criminals: Anna plus that one criminal among Brian, Derek, and Erwin. Laura’s clue says row 2 is the only row with exactly two criminals, so row 1 cannot have exactly two criminals. Therefore, we can determine that A1 Anna is INNOCENT.
05.C2 · Hazel → CRIMINAL
In column C, we already have three confirmed innocents at C3 (Laura), C4 (Raul), and C5 (Wanda). Ike’s clue says there are more innocents in column A than in column C, so column A must have at least one more innocent than that, meaning column A must have at least 4 innocents. Laura’s clue says row 2 has exactly 2 criminals, and since D2 (Ike) is innocent, that means exactly two of A2 (Flora), B2 (Gabe), and C2 (Hazel) are criminals. If Hazel were innocent, then column C would have at least 4 innocents, which would force column A to have 5 innocents to still be “more,” but row 2 would then require Flora to be a criminal, preventing column A from having 5 innocents. Therefore, we can determine that C2 (Hazel) is CRIMINAL.
06.A5 · Tina → INNOCENT
Wanda says there are exactly 4 innocents in column D, and we already know D2, D3, and D4 are innocent, so D1 and D5 must be one criminal and one innocent. Laura says row 2 is the only row with exactly 2 criminals, so row 5 cannot have exactly 2 criminals; since B5 and C5 are already innocent, the only way row 5 could have 2 criminals would be if both A5 and D5 were criminals, which is not allowed. That means if D5 is the criminal in column D, then A5 cannot be criminal, so A5 is innocent; and if D5 is not the criminal, then D1 is the criminal, and Nick’s clue forces the other two row-1 neighbors of Hazel (B1 and C1) to be innocent, making column C have four innocents (C1, C3, C4, C5), so Ike’s clue then forces every person in column A, including A5, to be innocent. Therefore, we can determine that A5 Tina is INNOCENT.
07.A4 · Olive → INNOCENT
In column D, Wanda says there are exactly 4 innocents, and we already see D2, D3, and D4 are innocent, so exactly one of D1 and D5 is criminal and column D has exactly 1 criminal total. Anna says column B has more criminals than column D, so column B must have at least 2 criminals. Nick’s clue about Hazel means Hazel has exactly two criminal neighbors, with exactly one of them in row 1, so among B2 and B3 there is exactly one criminal, which gives column B one criminal already; therefore column B must contain another criminal somewhere else, and since B5 is innocent, that extra criminal has to be either B1 or B4. If the extra criminal is at B4, then row 4 cannot also have A4 as criminal because Laura says row 2 is the only row with exactly 2 criminals, and row 4 would become exactly two criminals (A4 and B4) since C4 and D4 are innocent; if the extra criminal is at B1 instead, then Nick’s clue forces C1 to be innocent, making column C have 4 innocents (C1, C3, C4, C5), and Ike’s clue then forces column A to have 5 innocents, including A4. Therefore, we can determine that A4 (Olive) is INNOCENT.
08.D1 · Erwin → INNOCENT
Hazel is at C2, so her row 1 neighbors are B1, C1, and D1. Ike says there are more innocents in column A than in column C, but column A already has exactly three confirmed innocents (A1, A4, A5), while column C already has three confirmed innocents (C3, C4, C5), so C1 cannot also be innocent or column A could not end up with strictly more; this forces C1 to be a criminal. Nick says that of the two criminals neighboring Hazel, only one is in row 1, so among B1, C1, and D1 there can be only one criminal; since C1 is that criminal, D1 cannot be. Therefore, we can determine that D1 is INNOCENT.
09.D5 · Xavi → CRIMINAL
Column D contains Erwin at D1, Ike at D2, Nick at D3, Steve at D4, and Xavi at D5. Wanda’s clue says there are exactly 4 innocents in column D, and we can already see that Erwin, Ike, Nick, and Steve are innocents, which fills all four innocent spots in that column. That means Xavi cannot also be innocent. Therefore, we can determine that D5 Xavi is CRIMINAL.
10.B4 · Peter → INNOCENT
In row 2, Hazel at C2 is already a criminal and Ike at D2 is already innocent, so Laura’s clue that “Row 2 is the only row with exactly 2 criminals” forces exactly one of Flora (A2) and Gabe (B2) to be a criminal. That means Anna in the A1 corner would have exactly one criminal neighbor whenever Brian (B1) is innocent, because her only other neighbors are A2 and B2 and we now know exactly one of those two is criminal. Similarly, Erwin in the D1 corner already neighbors Hazel (C2) as a criminal, so he would have exactly one criminal neighbor whenever Derek (C1) is innocent, since Ike (D2) is innocent. Brian and Derek also cannot both be criminals, because then row 1 would have exactly two criminals (B1 and C1), which is forbidden by Laura’s clue since row 2 is the only row allowed to have exactly two criminals. And Brian and Derek cannot both be innocents either, because then both corners A1 and D1 would each have exactly one criminal neighbor, but Xavi’s clue says only one corner person can have exactly one criminal neighbor. So exactly one of the corners A1 or D1 is already the unique corner with exactly one criminal neighbor. Tina in the A5 corner has only three neighbors (Olive at A4, Peter at B4, and Uma at B5), and Olive and Uma are already innocent, so Tina would have exactly one criminal neighbor if and only if Peter were a criminal. Since the unique “exactly one criminal neighbor” corner is already A1 or D1, Tina cannot also have that property, so Peter cannot be a criminal. Therefore, we can determine that B4 · Peter is INNOCENT.
11.C1 · Derek → INNOCENT
Hazel is at C2, and her row 1 neighbors are B1 (Brian), C1 (Derek), and D1 (Erwin). Nick’s clue says Hazel has exactly two criminal neighbors, and only one of those two is in row 1; since Erwin at D1 is already INNOCENT, that means exactly one of Brian at B1 or Derek at C1 is a CRIMINAL. Anna’s clue says column B has more criminals than column D, and column D has only one criminal (Xavi at D5), so column B must have at least two criminals; since only one of B2 (Gabe) and B3 (Katie) can be criminal (the second criminal neighbor of Hazel that is not in row 1), Brian at B1 must be the other criminal in column B. That forces Derek at C1 to be the non-criminal among the two row 1 candidates next to Hazel. Therefore, we can determine that C1 Derek is INNOCENT.
12.B1 · Brian → CRIMINAL
Hazel is at C2, so her row 1 neighbors are B1 (Brian), C1 (Derek), and D1 (Erwin). Nick’s clue says that among the two criminals neighboring Hazel, exactly one of them is in row 1. Since Derek at C1 and Erwin at D1 are already known to be innocent, the only way for row 1 to contain exactly one criminal neighbor of Hazel is for Brian at B1 to be that criminal. Therefore, we can determine that B1 Brian is CRIMINAL.
13.A2 · Flora → INNOCENT, A3 · Joy → INNOCENT
Look at column C first: C1 is innocent, C2 is criminal, and C3, C4, and C5 are all innocent, so column C has exactly 4 innocents. Ike’s clue says column A has more innocents than column C, so column A must have at least 5 innocents. Column A already has A1, A4, and A5 as innocents, and the only way for column A to reach 5 innocents is for both unknowns A2 and A3 to be innocent as well. Therefore, we can determine that A2 Flora is INNOCENT, A3 Joy is INNOCENT, and so on.
14.B2 · Gabe → CRIMINAL
Row 2 contains Flora at A2, Gabe at B2, Hazel at C2, and Ike at D2. Laura’s clue says that row 2 is the only row that has exactly 2 criminals, so row 2 must have exactly 2 criminals in total. In row 2 we already know Hazel is a criminal, while Flora and Ike are both known innocents, so the only person who can be the second criminal in that row is Gabe at B2. Therefore, we can determine that B2 Gabe is CRIMINAL.
15.B3 · Katie → INNOCENT
Hazel is at C2, and Nick’s clue talks about “the 2 criminals neighboring Hazel,” which means Hazel has exactly two neighboring criminals in total. We can already see that Brian at B1 and Gabe at B2 are both criminals and both neighbor Hazel, so they fill those two criminal-neighbor slots. The clue’s extra detail also matches this: Brian is the one in row 1, and Gabe is the one not in row 1. Since Hazel’s two neighboring criminals are already accounted for, Katie at B3, who is also a neighbor of Hazel, cannot be a criminal. Therefore, we can determine that B3 Katie is INNOCENT.