Clues by Sam Aug 06, 2026 Answer – Full Solution Explained
A1
👷♀️
builder
B1
👷♀️
builder
C1
👩⚕️
doctor
D1
👨⚕️
doctor
A2
👮♀️
cop
B2
👮♂️
cop
C2
👮♂️
cop
D2
👩🌾
farmer
A3
👨🔬
scientist
B3
👨🏫
teacher
C3
👨🌾
farmer
D3
👩🌾
farmer
A4
👨🔬
scientist
B4
👩🔬
scientist
C4
💂♀️
guard
D4
💂♀️
guard
A5
👨✈️
pilot
B5
👨✈️
pilot
C5
👨✈️
pilot
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 — 7 criminals.
Clues by Sam answer for Aug 06, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 7 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Gabe (B2), Julie (D2), Nick (C3), Umar (A5), Vince (B5), Wally (C5) and Zoe (D5); the remaining 13 suspects are innocent.
The deduction chain, in plain English
01.A1 · Alice → INNOCENT
Martin’s clue says there is exactly 1 innocent in the overlap of column A and row 1. That overlap contains only one person: Alice at A1. Since that shared group still needs its 1 innocent, and Alice is the only person there, Alice at A1 must be innocent.
02.B5 · Vince → CRIMINAL
Alice’s clue says the two innocents below Barb must be orthogonally connected. Martin is already a known innocent among the people below Barb, so the other innocent in that group has to be next to Martin. That limits the other innocent to Gabe or Quita, because Vince is not next to Martin in that line. So Vince must be criminal.
03.B1 · Barb → INNOCENT
Alice’s clue says there are exactly two innocents below Barb, and those two must be connected. Below Barb are Gabe, Martin, Quita, and Vince; among them, Martin is already innocent and Vince is already criminal, so the two innocents below Barb have to come from Martin together with Gabe or Quita. Vince’s clue also requires every column to have at least 3 innocents. If Barb were criminal, the remaining people would have to satisfy both that column requirement and Alice’s exact connected pair below Barb at the same time, but they cannot. So Barb cannot be criminal. So Barb must be innocent.
04.C2 · Hal → INNOCENT
Barb’s clue says Hal is one of Martin’s six innocent neighbors. That directly identifies Hal as innocent. So Hal must be innocent.
05.D1 · Daniel → INNOCENT
Hal’s clue says the people below Daniel contain exactly 2 innocents, and those four people are Julie, Olga, Sofia, and Zoe. Vince’s clue also requires every column to have at least 3 innocents. So in column D, if Daniel were criminal, then the only possible innocents in that column would have to come from Julie, Olga, Sofia, and Zoe. But Hal’s clue allows exactly 2 innocents among those four, which is not enough to reach the required 3 innocents in column D. That makes Daniel innocent.
06.C1 · Celia → INNOCENT
Daniel’s clue says row 3 is the only row with exactly 3 innocents, so no other row can end with exactly 3 innocents. Row 1 already has 3 known innocents: Alice, Barb, and Daniel. Celia is the only person in row 1 not yet identified, so if Celia were criminal, row 1 would remain at exactly 3 innocents, which the clue does not allow. So Celia must be innocent.
07.A2 · Esha → INNOCENT, A3 · Luigi → INNOCENT
Vince says every column has at least 3 innocents, and Celia says all innocents in column A must be connected. In column A, Alice at A1 is already known to be innocent, while Esha at A2 and Luigi at A3 are the two people being tested. If Esha and Luigi were both criminals, then the remaining people would have to make column A reach at least 3 innocents while also keeping the innocents in column A as one connected group, and that cannot be done. So Esha and Luigi cannot both have that opposite identity. So Esha and Luigi must be innocent.
08.A4 · Paul → INNOCENT, C5 · Wally → CRIMINAL
Martin must have exactly 6 innocent neighbors, and three of them are already fixed: Esha, Hal, and Luigi. Sofia must have exactly 3 criminal neighbors, with exactly 2 of those criminals in column C among Nick, Ruth, and Wally. If Paul were criminal and Wally were innocent, then Gabe, Nick, Olga, Quita, Ruth, and Zoe would have to cover everything those two clues still require. But those remaining people cannot satisfy both Martin's neighbor count and Sofia's criminal-neighbor pattern at the same time. So Paul must be innocent, and Wally must be criminal.
09.A5 · Umar → CRIMINAL
Vince’s clue says every column has at least 3 innocents, and Wally’s clue says everyone in row 4 can have at most 4 innocent neighbors. In row 4, Paul already has 2 known innocent neighbors, and Quita already has 3 known innocent neighbors; Umar is one of the unknown neighbors counted for both of them. If Umar were innocent, then the remaining unknown people named here would have to meet both clues at the same time, but they cannot do that. So Umar cannot be innocent. That makes Umar criminal.
10.D5 · Zoe → CRIMINAL
Hal says there are exactly 2 innocents below Daniel. In that group, the two innocents have to be Julie, Olga, and Sofia, because those are the people below Daniel who are also above Zoe. That means Zoe cannot be one of the 2 innocents below Daniel. So Zoe must be criminal.
11.D3 · Olga → INNOCENT
Luigi's clue says Sofia has exactly 3 criminal neighbors, and exactly 2 of those criminals are in column C. Among Sofia's neighbors not in column C, the only people are Olga and Zoe, and Zoe is already known to be a criminal. That means the one criminal neighbor outside column C is already accounted for by Zoe, so Olga cannot also be a criminal. So Olga must be innocent.
12.C3 · Nick → CRIMINAL
Daniel’s clue says row 3 is the only row with exactly 3 innocents. Row 3 already has Luigi, Martin, and Olga as innocents, so it already has those 3 innocents before deciding Nick. If Nick were also innocent, then the remaining people involved here, Gabe, Julie, Quita, Ruth, and Sofia, would have to satisfy Daniel’s clue along with the counts in the other rows, but that cannot be done. So Nick cannot be innocent. That makes Nick criminal.
13.C4 · Ruth → INNOCENT
Luigi's clue says Sofia has exactly 3 criminal neighbors, and exactly 2 of those criminals are in column C. Among Sofia's neighbors who are in column C, Nick and Wally are already known criminals. That already fills the clue's total of 2 criminals in column C, so the only other such neighbor, Ruth, cannot be a criminal. So Ruth must be innocent.
14.B2 · Gabe → CRIMINAL, D2 · Julie → CRIMINAL
Zoe’s clue says exactly 2 people in row 3 have a criminal directly above them. To make that count reach 2, there must be 2 more such cases beyond the ones already known. The only direct-above positions still not identified for this clue are B2 Gabe and D2 Julie, so those are the two people who have to provide the needed criminal-above matches for row 3. So Gabe and Julie must be criminal.
15.B4 · Quita → INNOCENT, D4 · Sofia → INNOCENT
Vince’s clue says each column has at least 3 innocents, so that requirement has to hold for columns A, B, C, and D. The only people left to test here are Quita in column B and Sofia in column D. If Quita were criminal and Sofia were criminal, those remaining people would have to make all four columns satisfy the clue at the same time, and that is impossible. So they cannot both have that opposite identity. Therefore Quita and Sofia must be innocent.