Clues by Sam Jul 31, 2026 Answer – Full Solution Explained
A1
😬
chicken
B1
👮♀️
cop
C1
👩🌾
farmer
D1
👩🍳
cook
A2
🕵️♀️
sleuth
B2
💂♂️
guard
C2
👩🍳
cook
D2
😬
parrot
A3
😬
pig
B3
👩🌾
farmer
C3
👨💻
coder
D3
👨💻
coder
A4
👮♀️
cop
B4
🤡
clown
C4
👨💻
coder
D4
👷♂️
builder
A5
👮♀️
cop
B5
🕵️♀️
sleuth
C5
👷♂️
builder
D5
😬
chicken
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 — 5 criminals.
Clues by Sam answer for Jul 31, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 5 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Ivan (D2), Joy (A3), Kay (B3), Pip (B4) and Vera (B5); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.C4 · Steve → INNOCENT, C5 · Xavi → INNOCENT
Cheryl’s clue says there are exactly 2 innocents below Martin. Below Martin there are currently 0 known innocents, and the only people there whose identities are not yet known are Steve and Xavi. Since those two spots still have to supply all 2 innocents, both of them have to be innocent. So Steve and Xavi must be innocent.
02.B3 · Kay → CRIMINAL
Steve’s clue says an odd number of the people who are both above Vera and neighbors of Martin are innocent, and that group is Gus, Kay, and Pip. Xavi’s clue says exactly one of the people who are both between Betty and Vera and neighbors of Kay is innocent, and that group is Gus and Pip. If Kay were innocent, then Gus and Pip would have to make both clues true at the same time, but they cannot. So Kay cannot be innocent. That makes Kay criminal.
03.C2 · Helen → INNOCENT
Kay says Martin has exactly 3 criminal neighbors, and only 1 of those criminals is on an edge. Among Martin's neighbors, the non-edge people are Gus, Helen, and Pip, and there must be exactly 1 criminal in that non-edge group. That 1 criminal has to be Gus or Pip, so Helen cannot be the criminal there. So Helen must be innocent.
04.D3 · Nick → INNOCENT
Helen’s clue says the innocents below Eve must be one connected block. Among Martin’s edge neighbors, the two innocents have to come from D2 Ivan, D3 Nick, and D4 Thor, and the only connected innocent blocks that fit are D2 and D3, or D3 and D4, or D3, D4, and D5 Zed. In every one of those connected blocks, D3 Nick is included. So Nick must be innocent.
05.B5 · Vera → CRIMINAL
Steve’s clue makes the shared group B2 Gus, B3 Kay, and B4 Pip contain an odd number of innocents. Since Kay is already criminal, that odd innocent count has to come from Gus and Pip. Nick’s clue says the people below Betty contain an odd number of innocents. Right now that group has no known innocents, and its unknown members are Gus, Pip, and Vera. If Vera were innocent, then Gus and Pip would have to fit both odd-count requirements at the same time with Vera included in Betty’s below group, and that cannot be done. So Vera cannot be innocent. That makes Vera criminal.
06.B1 · Betty → INNOCENT
Steve’s clue says an odd number of the people who are both above Vera and neighbors of Martin are innocent. That shared group is Gus, Kay, and Pip, and Kay is already a criminal, so the innocence requirement has to be met by Gus and Pip. Vera’s clue says column B is the only column with exactly 2 innocents. If Betty were a criminal, the remaining people involved in these clues could not satisfy that column-B requirement at the same time as Steve’s odd-number requirement. So Betty must be innocent.
07.C3 · Martin → INNOCENT
Betty says Kay has exactly two criminal neighbors, and only one of those two is in column A. That means among Kay's neighbors who are not in column A, there is exactly one criminal. In that non-column-A group, Helen and Steve are already innocent, so the single criminal there has to be one of Gus or Pip. Martin cannot be that criminal, so Martin must be innocent.
08.A1 · Austin → INNOCENT, A5 · Uma → INNOCENT
Column B has exactly 2 innocents, Kay’s neighbors have exactly 2 criminals with exactly 1 of those in column A, and column A has an odd number of criminals. The column A people in Kay’s neighborhood are Freya, Joy, and Olga, so that clue already fixes the criminal count in that part of column A very tightly. If Austin and Uma were both criminals, then the remaining people involved in these same clues would have to satisfy all three requirements at once, but they cannot. That means Austin and Uma cannot both be criminals. So Austin and Uma must be innocent.
09.A3 · Joy → CRIMINAL
Kay says exactly one of her two criminal neighbors is in column A, so among Kay's three neighbors in column A, exactly two must be innocent. Those three people are Freya, Joy, and Olga. The two innocents in that group are Freya and Olga, so Joy cannot be one of them. So Joy must be criminal.
10.A4 · Olga → INNOCENT, A2 · Freya → INNOCENT
Betty’s clue says Kay has exactly 2 criminal neighbors, and exactly 1 of those criminal neighbors is in column A. Among Kay’s neighbors in column A, Joy is already a known criminal. That means the one column-A criminal allowed by the clue is already accounted for, so the other column-A neighbors of Kay cannot be criminals. So Olga and Freya must be innocent.
11.D5 · Zed → INNOCENT, D1 · Eve → INNOCENT
Kay’s clue fixes Martin’s neighborhood at exactly 3 criminals, with exactly 1 of those edge neighbors coming from D2, D3, and D4. Vera’s clue says column B is the only column with exactly 2 innocents, so column D cannot end up with exactly 2 innocents, and Freya’s clue says column D must contain an odd number of criminals. Now test the opposite for the two targets: suppose D1 Eve and D5 Zed were criminals. Then the remaining people involved here, B2 Gus, D2 Ivan, B4 Pip, and D4 Thor, would have to make Martin’s-neighbor count, the single edge-criminal condition there, column B’s innocent count, and column D’s odd criminal count all hold at the same time. That cannot be done. So Zed and Eve cannot be criminals. Therefore D5 Zed and D1 Eve must be innocent.
12.D4 · Thor → INNOCENT
Helen’s clue says that all innocents below Eve have to be connected as one orthogonally connected group. Below Eve are Ivan, Nick, Thor, and Zed, with Nick and Zed already known to be innocent. If Thor were criminal, then among the people below Eve the only other unknown left to satisfy that clue would be Ivan, and that cannot make the innocents below Eve fit the clue. So Thor cannot be criminal. That makes Thor innocent.
13.D2 · Ivan → CRIMINAL
Kay’s clue says Martin has exactly 3 criminal neighbors, and exactly 1 of those criminal neighbors is on the edge. Among Martin’s neighboring edge cells, D3 Nick and D4 Thor are already innocent, so the edge-neighbor group still needs 1 criminal. The only unknown person left in that edge-neighbor group is D2 Ivan, so Ivan at D2 must be criminal.
14.B2 · Gus → INNOCENT
Thor’s clue says only one row can have exactly 2 innocents. Row 3 already has exactly 2 innocents: Martin and Nick, with Joy and Kay criminal. Row 2 already has 2 known innocents, Freya and Helen, and Gus is the only person there not yet identified. If Gus were criminal, row 2 would also have exactly 2 innocents, which would clash with Thor’s clue. So Gus must be innocent.
15.B4 · Pip → CRIMINAL
Steve’s clue says an odd number of the people who are both above Vera and neighbors of Martin are innocents. That shared group is Gus, Kay, and Pip. Gus is already innocent and Kay is already criminal, so in that group there is currently exactly one innocent. If Pip were innocent too, the clue’s requirement would conflict for this group. So Pip cannot be innocent. That makes Pip criminal.