Clues by Sam Apr 29, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
🕵️♀️
sleuth
C1
🕵️♀️
sleuth
D1
👨🍳
cook
A2
👩🍳
cook
B2
👩🌾
farmer
C2
👨🌾
farmer
D2
👮♀️
cop
A3
🕵️♂️
sleuth
B3
👷♂️
builder
C3
👩⚕️
doctor
D3
👮♀️
cop
A4
👩🌾
farmer
B4
👩🎤
singer
C4
👨⚕️
doctor
D4
👨⚕️
doctor
A5
👨🎤
singer
B5
👨🎤
singer
C5
👨🍳
cook
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 — 7 criminals.
Clues by Sam answer for Apr 29, 2026 — a Medium solved in 16 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 7 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Betty (B1), Donald (D1), Flora (A2), Isaac (C2), Martin (B3), Xavi (C5) and Zed (D5); the remaining 13 suspects are innocent.
The deduction chain, in plain English
01.C3 · Nala → INNOCENT, C4 · Salil → INNOCENT
Will’s clue says the people between Isaac and Xavi contain exactly 2 innocents. Right now there are 0 known innocents in that group, and the only people there who are still unknown are Nala and Salil. Since those two spots must supply all 2 innocents required by the clue, both of them have to be innocent. So Nala and Salil must be innocent.
02.A4 · Quita → INNOCENT, D4 · Tyler → INNOCENT
Nala’s clue says exactly 2 corner people have an innocent directly above them. For this clue, 2 such cases are still needed, and the only direct-above people still affecting it are A4 Quita and D4 Tyler. So those two have to be the innocents that make the count reach 2. That makes Quita and Tyler innocent.
03.C5 · Xavi → CRIMINAL
Quita’s clue says that Xavi is one of the exactly 2 criminals in column C. That directly identifies Xavi as a criminal. So Xavi must be criminal.
04.D2 · Joy → INNOCENT
Quita’s clue fixes column C at exactly two criminals, with Xavi already one of them. Salil’s clue says Alice’s neighbors are the only neighbor set with exactly one innocent, so every other person’s neighbors must avoid ending with exactly one innocent. If Joy were criminal, the remaining unknown people named here would have to satisfy both of those restrictions at once, but they cannot. That contradiction means Joy cannot be criminal. So Joy must be innocent.
05.A2 · Flora → CRIMINAL
Tyler's clue says that among the five criminals on the edge, exactly one is in row 2. The edge people in row 2 are only Flora and Joy, and Joy is already innocent. So that row-2 edge group still needs one criminal, and Flora is the only person there who could be it. So Flora must be criminal.
06.B3 · Martin → CRIMINAL
Xavi’s clue says an odd number of the column B people who neighbor Flora are innocent, and that group is exactly Betty, Hazel, and Martin. So among Betty, Hazel, and Martin, the number of innocents must be odd. Now test Martin as innocent. Then those three would already include one innocent from Martin, while Alice’s clue still requires Alice’s neighbor group to be the only neighbor set with exactly 1 innocent, and the other people involved here are Alice, Betty, Chloe, Donald, Hazel, Isaac, Logan, Pam, Ruby, Vince, and Zed. Making all of those clue requirements hold at the same time is impossible. So Martin must be criminal.
07.B4 · Ruby → INNOCENT
Salil’s clue says Alice’s neighbors are the only neighbor set with exactly 1 innocent neighbor, and Flora’s clue says every column must have at least 3 innocents. If Ruby were criminal, then Alice, Betty, Chloe, Donald, Hazel, Isaac, Logan, Pam, Vince, and Zed would have to make both of those clues true at the same time, but they cannot do that. So Ruby cannot be criminal. That makes Ruby innocent.
08.A3 · Logan → INNOCENT
Ruby's clue says exactly one innocent is both to the left of Pam and neighboring Flora. That shared group is only Logan and Martin. Martin is already a criminal, so the group currently has no known innocent, and Logan is the only unknown person left there who can fill that one innocent spot. So Logan must be innocent.
09.A1 · Alice → INNOCENT, A5 · Vince → INNOCENT
Column A has to contain more innocents than column D. Right now both columns already have 2 known innocents, so column A can only end up ahead if some additional innocent comes from its two unknown people, Alice and Vince. If Alice and Vince were both criminals, then column A would stay at 2 innocents, while Donald, Pam, and Zed are the only unknown people left in column D. That cannot satisfy the requirement that column A have more innocents than column D. So Alice and Vince cannot both be criminals. That makes Alice and Vince innocent.
10.D5 · Zed → CRIMINAL
Vince's clue says row 5 contains exactly 2 innocents. Row 5 already has those 2 known innocents: Vince and Will. The only person in that row whose status is still unknown is Zed, so Zed cannot also be innocent. So Zed must be criminal.
11.D3 · Pam → INNOCENT
Zed’s clue says every person in row 4 can have at most 3 criminal neighbors. Salil is in row 4 and already has 3 known criminal neighbors: Martin, Xavi, and Zed. The only unknown neighbor Salil has is Pam, so if Pam were criminal, Salil would have 4 criminal neighbors, which would break the clue. So Pam must be innocent.
12.D1 · Donald → CRIMINAL
Logan’s clue says column A has more innocents than column D. Column A already has 4 known innocents, and column D currently has 3 known innocents with only Donald left unknown there. If Donald were innocent, column D would also have 4 innocents, so column A would not have more innocents than column D. So Donald must be criminal.
13.B1 · Betty → CRIMINAL
Donald's clue says exactly 2 criminals in column B have an innocent directly to the left. In column B, Martin is already one such criminal, and the only people there not yet identified are Betty and Hazel. If Betty were innocent, then Hazel would be the only remaining person who could join Martin to make the total of 2 such criminals, but that does not satisfy the clue's requirement for these same people. So Betty must be criminal.
14.B2 · Hazel → INNOCENT
Alice’s neighbors must contain exactly 1 innocent, and among Alice’s neighbors the only unidentified person is Hazel. Since Alice’s other two neighbors are both criminals and there are no known innocents there yet, Hazel has to provide that one innocent neighbor for Alice. If Hazel were criminal instead, Alice’s neighbors would have no innocent at all, which contradicts Salil’s clue that Alice is the only person with exactly 1 innocent neighbor. So Hazel at B2 must be innocent.
15.C1 · Chloe → INNOCENT
Tyler’s clue says there are exactly 5 criminals on the edge, and exactly 1 of those edge criminals is in row 2. On the edge in row 2, Flora is already that 1 criminal. That means the edge cells not in row 2 must contain the other 4 edge criminals, and that group already has Betty, Donald, Xavi, and Zed as 4 known criminals. Chloe is the only unknown person in that edge group, so Chloe cannot be a criminal. So Chloe must be innocent.
16.C2 · Isaac → CRIMINAL
Quita’s clue says that Xavi is one of exactly 2 criminals in column C. In that column, Chloe, Nala, and Salil are already innocent, and Xavi is already one criminal, so Isaac is the only person left who could be the second criminal. If Isaac were innocent, column C would have only Xavi as a criminal, which clashes with the clue saying there are exactly 2. So Isaac must be criminal.