Clues by Sam Sep 04, 2026 Answer – Full Solution Explained
A1
🕵️♀️
sleuth
B1
🕵️♀️
sleuth
C1
👩🔧
mech
D1
👨🌾
farmer
A2
👨🎨
painter
B2
👩🎨
painter
C2
👨🔧
mech
D2
👨🌾
farmer
A3
👨🎨
painter
B3
👮♂️
cop
C3
👨⚕️
doctor
D3
👩⚕️
doctor
A4
👮♂️
cop
B4
👮♂️
cop
C4
😬
goose
D4
👩💻
coder
A5
👨🍳
cook
B5
👩🍳
cook
C5
👩💻
coder
D5
👩🍳
cook
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 — 14 criminals.
Clues by Sam answer for Sep 04, 2026 — a Tricky solved in 16 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 14 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Betsy (A1), Celia (B1), Eve (C1), Gus (D1), Igor (A2), Kumar (C2), Ollie (B3), Pip (C3), Quita (D3), Rob (A4), Shaun (B4), Vince (A5), Wanda (B5) and Zara (D5); the remaining 6 suspects are innocent.
The deduction chain, in plain English
01.B1 · Celia → CRIMINAL
Uma’s clue says Betsy has exactly 2 criminal neighbors, and exactly 1 of those criminals is in row 1. Among Betsy’s neighbors, the only person who is both a neighbor of Betsy and in row 1 is Celia. So the row 1 part of the clue still needs 1 criminal, and Celia is the only person who can fill it. That makes Celia criminal.
02.C1 · Eve → CRIMINAL
Celia's clue explicitly says that Eve is one of the three criminals in column C. Since Eve is directly named as part of that criminal group, Eve must be criminal.
03.C2 · Kumar → CRIMINAL
Below Kumar, Pip, Tina, and Xia contain exactly 2 innocents. Column C also contains exactly 2 innocents in total, since Eve is one of 3 criminals in that column. That means the two innocents for all of column C are already accounted for among Pip, Tina, and Xia. The only person in column C who is not in that lower group, besides Eve who is already criminal, is Kumar, so Kumar cannot be innocent. So Kumar must be criminal.
04.A3 · Martin → INNOCENT
If Martin were criminal, then the painters would have 1 criminal while the mechs have 2 criminal. Betsy’s clue says her neighbors contain exactly 2 criminals in total, and exactly 1 of those criminals is in row 1, namely Celia. That leaves exactly 1 more criminal among Betsy’s other neighbors, Igor and Janet. But with Martin as a criminal, neither Igor nor Janet can be criminal, so Igor and Janet would both have to be innocent, which conflicts with Betsy needing one more criminal neighbor outside row 1. So Martin must be innocent.
05.A4 · Rob → CRIMINAL, B4 · Shaun → CRIMINAL, C5 · Xia → INNOCENT
Column C must contain exactly 3 criminals, and it already has Eve and Kumar as known criminals. Ollie’s neighbors must contain exactly 5 criminals, and among those neighbors only Kumar is currently known to be one, leaving A2 Igor, B2 Janet, C3 Pip, A4 Rob, B4 Shaun, and C4 Tina to supply the rest. If Rob were innocent, Shaun were innocent, and Xia were criminal, then Igor, Janet, Pip, and Tina would have to satisfy both of those clue totals at the same time, and they cannot. So Rob and Shaun cannot be innocent, and Xia cannot be criminal. That makes Rob and Shaun criminal, and Xia innocent.
06.C4 · Tina → INNOCENT
Column D already has one known innocent, Uma, and Shaun’s clue says there is exactly one innocent in column D who has a criminal directly to the left. If Tina were criminal, then the remaining people involved here, Gus, Logan, Pip, Quita, and Zara, would have to make both that column D condition and Rob’s “only one innocent farmer” condition come out right at the same time, but they cannot. So Tina cannot be criminal. That makes Tina innocent.
07.C3 · Pip → CRIMINAL
Uma's clue fixes Betsy's neighbors at exactly 2 criminals in total, and since Celia is already one of them, the other criminal among Betsy's neighbors must be either Igor or Janet. That means Igor and Janet together contribute exactly 1 criminal. Martin's clue says Ollie has exactly 5 criminal neighbors. Around Ollie, Kumar, Rob, and Shaun are already criminals, so 1 more criminal is needed from Igor, Janet, and Pip. Since Igor and Janet already account for exactly 1 criminal between them, Pip has to be the remaining criminal needed for Ollie's total. So Pip must be criminal.
08.B5 · Wanda → CRIMINAL
Uma’s clue fixes Betsy’s neighborhood at exactly 2 criminals, and the only row 1 criminal among Betsy’s neighbors is Celia. Pip’s clue also says exactly one person in column A has exactly 2 criminal neighbors, while Betsy, Igor, Martin, Rob, and Vince are the column A people affected here. If Wanda were innocent, then Betsy, Igor, Janet, Ollie, and Vince would have to satisfy all of those counts at the same time, but they cannot. That rules out Wanda being innocent. So Wanda must be criminal.
09.D3 · Quita → CRIMINAL
Column D already has one known innocent, Uma. Shaun’s clue says there is exactly one innocent in column D who has a criminal directly to the left, and Quita is the person being tested against that limit. If Quita were innocent, then the remaining people involved in these clues, Gus, Logan, and Zara, would also have to fit Rob’s farmer clue and Shaun’s column D clue at the same time, but they cannot. That makes Quita impossible as an innocent. So Quita must be criminal.
10.A1 · Betsy → CRIMINAL
Uma’s clue says Betsy has exactly 2 criminal neighbors, and exactly 1 of those criminals is in row 1. Since Celia is already that one criminal neighbor in row 1, Betsy’s other criminal neighbor has to come from Igor or Janet. At the same time, Pip’s clue allows exactly one person in column A to have exactly 2 criminal neighbors, while Xia’s clue says column B has exactly one innocent, so Janet and Ollie must account for that restriction in column B. If Betsy were innocent, then Betsy, Igor, Janet, Ollie, and Vince would have to satisfy all of those requirements together, but they cannot. So Betsy must be criminal.
11.A5 · Vince → CRIMINAL
If Vince were innocent, the remaining people involved here would be Igor, Janet, Ollie, and Zara. But these clues all have to hold at once: Betsy’s neighbors must contain exactly 2 criminals, with exactly 1 of them in row 1; exactly one person in column A must have exactly 2 criminal neighbors; and the cooks and cops must contain the same number of innocents. Vince already has 3 known criminal neighbors with no unknown neighbors affecting that count, so making Vince innocent does not help satisfy the column A clue, while the other listed people would still have to make all three clue conditions fit together and they cannot. So Vince must be criminal.
12.D1 · Gus → CRIMINAL, D2 · Logan → INNOCENT
Betsy’s clue says her neighbors include exactly two criminals, and exactly one of those criminals is in row 1. Celia is already that one row 1 criminal neighbor, so among Igor and Janet there must be exactly one more criminal. Row 1 already has three known criminals, while row 2 has only one. If Gus were innocent and Logan were criminal, then with exactly one of Igor and Janet criminal, row 2 would rise to three criminals while row 1 would stay at three, which breaks the clue that row 1 has more criminals than row 2. So Gus must be criminal and Logan must be innocent.
13.D5 · Zara → CRIMINAL
If Zara were innocent, then row 5 would already have 2 innocents: Xia and Zara. Logan says row 2 has more innocents than row 5, so row 2 would then need more than 2 innocents. But row 2 has Logan already innocent, and Uma’s clue about Betsy’s neighbors says Betsy has exactly 2 criminal neighbors with exactly 1 of them in row 1; since Celia is that one row-1 criminal neighbor, Igor and Janet cannot be criminal, so both Igor and Janet must be innocent. That would make row 2 exactly 3 innocents, creating the conflict reached from assuming Zara innocent, so Zara at D5 must be criminal.
14.B3 · Ollie → CRIMINAL
Betsy’s clue says the number of innocent cooks must equal the number of innocent cops. The cooks currently have 0 innocents, and there are no unknown cooks left to change that. The cops also currently have 0 innocents, so if Ollie were innocent then the cops would have 1 innocent while the cooks would still have 0. So Ollie at B3 must be criminal.
15.B2 · Janet → INNOCENT
Xia’s clue says column B contains exactly one innocent. In column B, there are currently no known innocents, and the only person there whose identity is still unknown is Janet. Since that column still needs one innocent and Janet is the only available person left to fill that spot, Janet must be innocent.
16.A2 · Igor → CRIMINAL
Uma’s clue says Betsy has exactly 2 criminal neighbors, and exactly 1 of those criminal neighbors is in row 1. The only neighbor of Betsy who is in row 1 is Celia, and Celia is criminal, so that uses up the 1 criminal neighbor in row 1. That means Betsy must have exactly 1 criminal neighbor who is not in row 1. Betsy’s neighbors not in row 1 are Igor and Janet, and Janet is innocent, so the remaining criminal spot has to be Igor. So Igor must be criminal.