Clues by Sam Sep 17, 2026 Answer – Full Solution Explained
A1
👮♂️
cop
B1
👷♂️
builder
C1
👩🌾
farmer
D1
👩🍳
cook
A2
👷♂️
builder
B2
👨🔧
mech
C2
👨💻
coder
D2
👩🌾
farmer
A3
👷♂️
builder
B3
👨🔧
mech
C3
👩🔧
mech
D3
👩🌾
farmer
A4
👨💻
coder
B4
👮♀️
cop
C4
💂♀️
guard
D4
👮♀️
cop
A5
👩🍳
cook
B5
👨💻
coder
C5
💂♂️
guard
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 — 8 criminals.
Clues by Sam answer for Sep 17, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chris (B1), Ivan (C2), Joy (D2), Keith (A3), Petra (B4), Sofia (D4), Uma (A5) and Vince (B5); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.D4 · Sofia → CRIMINAL
Hank's clue says row 4 contains exactly 2 criminals, and exactly 1 of those criminals is Logan's neighbor. Since Oscar, Petra, and Rose are the row 4 people who are Logan's neighbors, that leaves exactly 1 criminal in the row 4 people who are not Logan's neighbors. The only row 4 person who is not Logan's neighbor is Sofia, and that group still needs that 1 criminal. So Sofia must be criminal.
02.B4 · Petra → CRIMINAL
Sofia’s clue says Petra is one of Mary’s 4 criminal neighbors. That directly places Petra among the neighbors who are criminal. So Petra must be criminal.
03.A4 · Oscar → INNOCENT, C4 · Rose → INNOCENT
Row 4 has exactly two criminals in total, and exactly one of those row 4 criminals is Logan's neighbor. Among the people who are both in row 4 and neighbors of Logan, Petra is already a known criminal. That already fills the clue's one allowed criminal in that shared group, so the other two people there cannot be criminals. So Oscar and Rose must be innocent.
04.C1 · Debra → INNOCENT
Aaron and Chris are the only people who are both to the left of Freya and neighbors of Gabe, and that group contains exactly 1 innocent. Aaron, Chris, and Debra are the people in row 1 who are neighbors of Hank, and that larger group contains exactly 2 innocents. So after the 1 innocent among Aaron and Chris is accounted for, the remaining person in the larger group, Debra, must provide the difference and be innocent. That makes Debra innocent.
05.C2 · Ivan → CRIMINAL
Ivan and Mary’s shared neighbor group has exactly 3 innocents, and those people are among Hank, Joy, Logan, and Nicole. Mary’s full neighbor group has exactly 4 innocents in total, and compared with that shared group the only extra person who could add to the innocent count is Ivan. Since the shared group already accounts for the innocents allowed there, Ivan cannot be innocent without pushing Mary’s total past 4. So Ivan must be criminal.
06.B3 · Logan → INNOCENT
Oscar’s clue says there are exactly 2 innocents in the row-1 group that also neighbors Hank: Aaron, Chris, and Debra. Since Debra is already innocent, Aaron and Chris account for the other innocent needed there. Now look at Hank’s neighbors who are not neighbors of Logan: Aaron, Chris, Debra, and Logan. Among these, Aaron, Chris, and Debra already make 3 innocents. The only extra person in this larger group, compared with the row-1 group, is Logan, so Logan must match the innocent difference here. So Logan must be innocent.
07.A2 · Gabe → INNOCENT
Petra’s clue fixes the row 3 part of her neighboring criminals at exactly one, and in that row 3 group the only possible people are Keith and Mary. Ivan’s clue says exactly two of Hank’s criminal neighbors also neighbor Logan, and that larger group is Gabe, Ivan, Keith, and Mary. Since Keith and Mary can contribute only one criminal from Petra’s row 3 restriction, the two criminals required in Ivan’s group are already accounted for by Ivan plus exactly one of Keith or Mary. That leaves Gabe as the only remaining person in that larger group who cannot be criminal. So Gabe must be innocent.
08.D1 · Freya → INNOCENT
Mary has exactly 4 criminal neighbors, and three of them are already known: Ivan, Petra, and Sofia. So among the two unknown neighbors of Mary, Joy and Nicole, exactly one is criminal. Ivan has exactly 2 criminal neighbors, and exactly 1 of those criminals is above Zoe. The people above Zoe who are also Ivan's neighbors are Freya, Joy, and Nicole, so that group contains exactly 1 criminal. Since Joy and Nicole already account for that one criminal in this above-Zoe group, the only remaining person in the larger group, Freya, cannot be criminal. So Freya is innocent.
09.D5 · Zoe → INNOCENT
Mary’s neighbors must contain exactly 4 criminals, and among those neighbors Petra, Ivan, and Sofia are already known criminals. That means the only remaining criminal neighbor of Mary has to be either Joy or Nicole. Column D must contain an odd number of innocents. Freya is already an innocent there, so with Zoe assumed criminal, the only undecided people left to change that innocent count would be Joy and Nicole. But from Mary’s clue, Joy and Nicole cannot both be innocents, because one of them has to be Mary’s fourth criminal neighbor. So Zoe cannot be criminal. That makes Zoe innocent.
10.A5 · Uma → CRIMINAL
To the right of Uma, there must be exactly 2 innocents, and Zoe is already one of them, so Vince and Wally contribute exactly 1 innocent between them. Among Petra's neighbors who are not in row 3, there must be exactly 3 innocents; Oscar and Rose are already innocent, and Vince and Wally together supply the only remaining innocent allowed in that group. That leaves Uma as the only other person in Petra's non-row-3 neighbor group, so Uma cannot be innocent. So Uma must be criminal.
11.D3 · Nicole → INNOCENT
Petra’s clue fixes the row 3 neighbors A3 Keith, B3 Logan, and C3 Mary as exactly 2 innocents. Row 3 is Keith, Logan, Mary, and Nicole, and Uma’s clue says row 3 contains an odd number of innocents. Those first three people already contribute an even number of innocents, since they contribute exactly 2. Nicole is the only person in row 3 outside that exact-count group, so she has to be innocent to make the total in row 3 odd. That makes Nicole innocent.
12.D2 · Joy → CRIMINAL
Sofia’s clue says Petra is one of Mary’s exactly 4 criminal neighbors. Among Mary’s neighbors, the criminals already identified are Ivan, Petra, and Sofia, while Hank, Logan, Nicole, and Rose are innocent. That leaves Joy as the only neighbor of Mary not yet identified, and Mary still needs a fourth criminal neighbor. So Joy must be criminal.
13.B5 · Vince → CRIMINAL
Joy's clue says there are exactly 2 criminals below Chris. Below Chris, there is already 1 known criminal, Petra, and the only person there whose identity is still unknown is Vince. That group still needs exactly 1 more criminal, so Vince has to supply it. So Vince must be criminal.
14.C5 · Wally → INNOCENT
Petra says exactly 3 of her neighbors are criminals, and exactly 1 of those criminals is in row 3. That means the neighbors not in row 3 must contain exactly 2 criminals. Among Petra's neighbors not in row 3, Uma and Vince are already known criminals, while Oscar and Rose are innocent and Wally is the only unknown there. Since those 2 criminal places are already filled by Uma and Vince, Wally cannot also be a criminal. So Wally must be innocent.
15.A1 · Aaron → INNOCENT, B1 · Chris → CRIMINAL, C3 · Mary → INNOCENT, A3 · Keith → CRIMINAL
Freya’s clue says only one column has exactly 4 innocents, so columns A, B, C, and D have to fit that count pattern in one unique way. If Aaron were criminal, Chris innocent, Mary criminal, and Keith innocent, then those four people would have to make the column counts work while also matching the other clues, and they cannot. That means this opposite assignment is impossible. So Aaron must be innocent, Chris must be criminal, Mary must be innocent, and Keith must be criminal.