Clues by Sam Jul 23, 2026 Answer – Full Solution Explained
A1
👮♂️
cop
B1
👩🌾
farmer
C1
👨💻
coder
D1
👩💻
coder
A2
👨🌾
farmer
B2
👨✈️
pilot
C2
👩💻
coder
D2
👮♀️
cop
A3
👷♂️
builder
B3
👩✈️
pilot
C3
👨🎨
painter
D3
👩🎨
painter
A4
👩🎤
singer
B4
👨🎤
singer
C4
👩🎨
painter
D4
👩🎤
singer
A5
👷♂️
builder
B5
👨💼
clerk
C5
👩💼
clerk
D5
👨💼
clerk
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 — 12 criminals.
Clues by Sam answer for Jul 23, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Austin (A1), Chase (C1), Frida (D1), Hank (B2), Joy (C2), Kay (D2), Larry (A3), Ruby (A4), Tina (C4), Will (B5), Xia (C5) and Zach (D5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.C2 · Joy → CRIMINAL
Steve’s clue says Chase has exactly 4 criminal neighbors, and exactly 1 of those criminal neighbors is above Tina. Among Chase’s neighbors, the only person who is above Tina is Joy. So the one criminal neighbor of Chase who is above Tina has to be Joy. That makes Joy criminal.
02.B2 · Hank → CRIMINAL
Joy’s clue says the two innocents below Bonnie are connected. Steve is already known to be one of those innocents, so the other innocent below Bonnie has to be orthogonally next to Steve. In that group, that means the other innocent must be Olga or Will, because Hank is not next to Steve in that line. So Hank cannot be that second innocent, and Hank must be criminal.
03.A2 · Gabe → INNOCENT
Hank’s clue says Bonnie and Olga have exactly one innocent neighbor in common. Their shared neighbors are only Gabe, Hank, and Joy, and Hank and Joy are already criminals. That means the shared group still needs one innocent, and the only person left there who can fill it is Gabe. So Gabe must be innocent.
04.D2 · Kay → CRIMINAL
Gabe’s clue says row 4 is the only row with exactly 2 innocents. Row 2 already has 1 known innocent, Gabe, and Kay is the only person in that row not yet identified. If Kay were innocent, then row 2 would also have exactly 2 innocents, which would contradict the clue that only row 4 has that total. So Kay must be criminal.
05.D5 · Zach → CRIMINAL
Kay’s clue says that Zach is one of the exactly 3 criminals in the corner cells. Since Zach is explicitly included among those corner criminals, his identity is fixed by the clue itself. So Zach must be criminal.
06.A4 · Ruby → CRIMINAL
Row 4 must contain exactly 2 innocents, and Steve is already one of them, so row 4 needs exactly 1 more innocent. The only people to the right of Steve are Tina and Uma, and that group contains exactly 1 innocent. Ruby is in row 4 but not in that group, so Tina and Uma already account for the one remaining innocent that row 4 needs. That makes Ruby criminal.
07.B3 · Olga → INNOCENT
Ruby’s clue says the people strictly between Hank and Will contain exactly 2 innocents. In that group, Steve is already known to be innocent, so 1 innocent is still needed. The only person in that group whose identity is not yet known is Olga, so she has to be the second innocent. So Olga must be innocent.
08.B5 · Will → CRIMINAL
Joy’s clue says that the innocents below Bonnie are both connected, so there must be exactly two innocents in that group and they must form one orthogonally connected set. Below Bonnie are Hank, Olga, Steve, and Will, with Olga and Steve already known to be innocent and Hank known to be criminal. If Will were also innocent, there would be more than those two innocents below Bonnie, which conflicts with the clue’s “both innocents below Bonnie.” So Will at B5 must be criminal.
09.A1 · Austin → CRIMINAL
Will’s clue says exactly two people in row 1 are criminals with a criminal directly below them. In row 1, nobody is already known to be a criminal, so that clue has to be satisfied entirely by Austin, Bonnie, Chase, and Frida. If Austin were innocent, then Bonnie, Chase, and Frida would have to fit Will’s row 1 requirement while the other clue still keeps row 4 as the only row with exactly 2 innocents. But with the remaining people involved in these two clues, those facts cannot all be made true at the same time. So Austin must be criminal.
10.A3 · Larry → CRIMINAL
Austin’s clue says the people above Vince contain exactly one innocent. That group already has one known innocent, Gabe. The only person there whose identity was still unknown is Larry, so Larry cannot also be innocent. So Larry must be criminal.
11.C1 · Chase → CRIMINAL
Steve’s clue says Chase’s neighbors contain exactly 4 criminals, and exactly 1 of those criminals is above Tina. That one criminal above Tina is already Joy. If Chase were innocent, then Bonnie and Frida would have to account for the remaining criminal total around Chase without creating any additional criminal above Tina beyond Joy, while Gabe’s clue also still has to make row 4 the only row with exactly 2 innocents among row 1, row 2, row 3, row 4, and row 5. With Bonnie, Frida, Pip, Quita, Tina, Uma, Vince, and Xia left to satisfy all of that at once, those requirements clash. So Chase must be criminal.
12.C3 · Pip → INNOCENT, D3 · Quita → INNOCENT
Larry’s clue says every person in column D has at most 3 criminal neighbors. Frida is in column D and already has 3 known criminal neighbors, so none of her neighbors can be criminal; in particular, Pip cannot be criminal. Quita is also in column D and already has 2 known criminal neighbors. If both Pip and Quita were criminal, then the remaining nearby unknown people named here would have to fit all of column D’s neighbor limits at once, but they cannot. So that opposite assignment fails. That makes Pip and Quita innocent.
13.C5 · Xia → CRIMINAL
Zach’s neighbors must contain an odd number of innocents, and right now all three of those neighbors are unknown: Tina, Uma, and Xia. Row 4 has exactly 2 innocents, and since Ruby is criminal and Steve is innocent, Tina and Uma must contribute exactly 1 more innocent between them. That means Zach’s row 4 neighbors already contribute an odd number of innocents. If Xia were innocent too, Zach’s neighbors would have 2 innocents, which is even, not odd. So Xia must be criminal.
14.A5 · Vince → INNOCENT
Xia’s clue says every row has at least one innocent. That means row 5 cannot be all criminals, so row 5 can have at most 3 criminals. Row 5 already has 3 known criminals, Will, Xia, and Zach, and the only person there whose status was not fixed yet is Vince. So Vince at A5 must be innocent.
15.D1 · Frida → CRIMINAL
Kay’s clue says Zach is one of exactly 3 criminals in the corner cells. The corners are Austin, Frida, Vince, and Zach, and among them Austin and Zach are already criminals while Vince is already innocent. That means the only way to reach exactly 3 criminal corners is for the remaining corner, Frida, to be criminal. So Frida must be criminal.
16.B1 · Bonnie → INNOCENT
Steve’s clue says Chase has exactly 4 criminal neighbors, and exactly 1 of those criminals is above Tina. That one criminal above Tina is already Joy. So among Chase’s neighbors who are not above Tina, there must be exactly 3 criminals. Frida, Hank, and Kay are those 3 known criminals, and the only other person there is Bonnie. Bonnie therefore cannot also be a criminal. So Bonnie must be innocent.
17.C4 · Tina → CRIMINAL, D4 · Uma → INNOCENT
Frida’s clue says column C has more criminals than column D. Right now the known counts are tied: column C has 3 known criminals, and column D also has 3, with only Tina in column C and Uma in column D still undecided. If Tina were innocent and Uma were criminal, column C would stay at 3 criminals while column D would rise to 4, which directly contradicts the clue. So Tina must be criminal and Uma must be innocent.