Clues by Sam Apr 30, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👩🌾
farmer
C1
👨🍳
cook
D1
👷♀️
builder
A2
👮♀️
cop
B2
👨🍳
cook
C2
👨🍳
cook
D2
👨🌾
farmer
A3
👮♀️
cop
B3
💂♂️
guard
C3
💂♂️
guard
D3
👨🌾
farmer
A4
🕵️♀️
sleuth
B4
👨🎨
painter
C4
👩🎨
painter
D4
👨🎨
painter
A5
🕵️♀️
sleuth
B5
💂♀️
guard
C5
🕵️♀️
sleuth
D5
👷♂️
builder
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 Apr 30, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Debra (D1), Ike (D2), Jane (A3) and Wanda (C5); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.C1 · Chuck → INNOCENT, B2 · Frank → INNOCENT
Olof's clue says Barb and Gabe have exactly 2 innocent neighbors in common. The shared neighbors are only Chuck and Frank. Since those 2 common-neighbor innocent spots are still both needed, and Chuck and Frank are the only people in that shared group, both of them have to fill them. So Chuck and Frank must be innocent.
02.A4 · Nala → INNOCENT
Chuck’s clue directly says that Nala is one of the four innocents in column A. That makes Nala at A4 innocent.
03.B1 · Barb → INNOCENT, B5 · Vera → INNOCENT
Nala’s clue says exactly 2 corner people have an innocent directly to the right of them. Those 2 cases still have to come from the remaining direct-right spots that matter for this clue. The only such people are Barb and Vera, since they are the only unresolved direct-right neighbors still affecting whether those corner cases count. So Barb and Vera must be innocent.
04.C5 · Wanda → CRIMINAL
Frank’s clue says Olof has exactly 2 criminal neighbors, and exactly 1 of those criminals is to the right of Uma. In the part of Olof’s neighbors that is to the right of Uma, the only people are Vera and Wanda. Vera is already innocent, so that group still needs 1 criminal and Wanda is the only unknown left there. So Wanda must be criminal.
05.A5 · Uma → INNOCENT, D5 · Xavi → INNOCENT
Vera’s clue says every row has at least 3 innocents, so row 5 can have at most 1 criminal. Row 5 already has 1 known criminal, Wanda. The only people in that row whose status was not yet fixed are Uma and Xavi, so neither of them can be the row’s second criminal. That makes Uma and Xavi innocent.
06.A2 · Emily → INNOCENT
Vera’s clue says every row has at least 3 innocents, and Wanda’s clue says that among the people to the right of Frank there is exactly 1 innocent. Right of Frank are Gabe and Ike, and at the moment neither of them is already known innocent, so row 2 still has to fit that exact one-innocent requirement there while also reaching at least 3 innocents in the row. If Emily were criminal, the remaining unidentified people would have to satisfy both of those clue requirements at the same time, but they cannot. So Emily cannot be criminal. That makes Emily innocent.
07.B3 · Kyle → INNOCENT
Chuck’s clue says column A has exactly 4 innocents. In column A, Emily, Nala, and Uma are already innocent, so among the two unknown people there, Amy and Jane, exactly one can be criminal. Emily’s clue says column A has more criminals than column B. Right now both columns have 0 known criminals, so if Kyle were criminal in column B, column A would need more than 1 criminal. But Amy and Jane are the only unknown people in column A, and Chuck’s clue allows only one criminal there, not two. That conflict means Kyle cannot be criminal. So Kyle must be innocent.
08.C3 · Logan → INNOCENT
Row 3 contains Jane, Kyle, Logan, and Mark, and Kyle is the only person there already known to be innocent. Vera says every row has at least 3 innocents, while Kyle says all innocents in row 3 must be connected in one orthogonal block. If Logan were criminal, the remaining unknown people would have to satisfy both of those clues at once, and that cannot be done. So Logan at C3 must be innocent.
09.D3 · Mark → INNOCENT, D4 · Thor → INNOCENT
Column A has exactly four innocents, and Emily, Nala, and Uma are already three of them, so that clue tightly restricts Amy and Jane. At the same time, every row must have at least three innocents, and the innocents in column D must form one connected block that includes Xavi. If Mark and Thor were both criminals, then Amy, Debra, Gabe, Ike, Jane, and Paula would have to satisfy all of those requirements at once, and they cannot. So Mark and Thor must be innocent.
10.C4 · Paula → INNOCENT
Frank has exactly one innocent to his right, and the only people there are Gabe and Ike, so among Gabe and Ike, exactly one is innocent. Logan has an odd number of criminal neighbors, and the only unknown neighbors there are Gabe, Ike, and Paula. If Paula were criminal, then Gabe and Ike would still have to include exactly one innocent, which means the other one would be criminal. That would give Logan’s neighbors exactly two criminals among Gabe, Ike, and Paula, which is even, not odd. So Paula cannot be criminal. That makes Paula innocent.
11.A3 · Jane → CRIMINAL
Frank’s clue says Olof has exactly 2 criminal neighbors, and exactly 1 of those criminals is to the right of Uma. Among Olof’s neighbors who are to the right of Uma, Vera is innocent and Wanda is criminal, so that one criminal on the right is already Wanda. That means Olof’s neighbors who are not to the right of Uma must contain exactly 1 criminal. In that group, Kyle, Logan, Nala, Paula, and Uma are all innocent, and the only unknown person left there is Jane. So Jane must be criminal.
12.A1 · Amy → INNOCENT
Chuck’s clue says Nala is one of exactly 4 innocents in column A. In column A, Emily, Nala, and Uma are already known innocents, and Jane is a known criminal, so Amy is the only person there not yet identified. If Amy were criminal, column A would not be able to have the 4 innocents Chuck’s clue requires. So Amy must be innocent.
13.C2 · Gabe → INNOCENT
In column C, Wanda is already a known criminal, and Amy's clue says only one criminal in that column has an innocent directly to the right. If Gabe were also a criminal, then the remaining people involved here, Debra and Ike, would have to make Amy's clue and Vera's rule that every row has at least 3 innocents hold at the same time, but they cannot do that. So Gabe cannot be criminal. That makes Gabe innocent.
14.D2 · Ike → CRIMINAL
Wanda’s clue says the people to the right of Frank contain exactly one innocent. That group already has one known innocent, Gabe. The only person there whose identity was still unknown is Ike, so Ike cannot also be innocent. So Ike must be criminal.
15.D1 · Debra → CRIMINAL
Logan’s clue says that all innocents in column D have to be connected as one orthogonally connected block. In column D, Ike is criminal, while Mark, Thor, and Xavi are innocent, and Debra is the only person there not yet identified. If Debra were innocent, then the innocents in column D would include Debra along with Mark, Thor, and Xavi, but that cannot satisfy the clue’s requirement that all innocents in that column form one connected block. So Debra cannot be innocent. So Debra must be criminal.