Clues by Sam Aug 08, 2026 Answer – Full Solution Explained
A1
👩🏫
teacher
B1
🕵️♂️
sleuth
C1
🕵️♂️
sleuth
D1
👩✈️
pilot
A2
👨⚕️
doctor
B2
👨⚕️
doctor
C2
👨✈️
pilot
D2
👩✈️
pilot
A3
👨🍳
cook
B3
👩⚕️
doctor
C3
👨🎨
painter
D3
👩🎨
painter
A4
👨🍳
cook
B4
😬
goat
C4
👩🏫
teacher
D4
👩🔧
mech
A5
👨🏫
teacher
B5
💂♀️
guard
C5
💂♀️
guard
D5
👩🔧
mech
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 — 6 criminals.
Clues by Sam answer for Aug 08, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Erwin (A2), Ivan (A3), Janet (B3), Logan (C3), Paul (B4) and Quita (C4); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.B1 · Brian → INNOCENT
Zoe’s clue directly says that Brian is one of the exactly 3 innocents in column B. Since Brian is explicitly included in that innocent group, Brian must be innocent.
02.B5 · Vicky → INNOCENT
Zoe says column B has exactly 3 innocents, so with 5 people in that column, it has exactly 2 criminals. Brian is already an innocent there, and Brian’s clue only counts column B innocents who neighbor Logan. The only people in column B who neighbor Logan are Frank, Janet, and Paul, while Brian and Vicky do not. So the 2 criminals in column B must come from Frank, Janet, and Paul, which means Vicky cannot be one of those 2 criminals. That makes Vicky innocent.
03.B4 · Paul → CRIMINAL, B3 · Janet → CRIMINAL
Vicky’s clue says the people below Frank contain exactly one innocent. That group already has one known innocent, Vicky, and the only other people there are Janet and Paul. Since the full innocent count for that group is already used up, Janet and Paul cannot be innocent. So Janet and Paul must be criminal.
04.B2 · Frank → INNOCENT
Zoe’s clue says Brian is one of exactly 3 innocents in column B. In that column, Brian and Vicky are already known innocents, while Janet and Paul are already criminals, so Frank is the only person there whose identity is not yet fixed. If Frank were criminal, column B would have only 2 innocents instead of the required 3. So Frank must be innocent.
05.A4 · Oscar → INNOCENT
Row 4 has exactly 2 innocents, and exactly 1 innocent in row 4 is a neighbor of Logan. The row 4 people who are neighbors of Logan are Paul, Quita, and Tina, and Paul is already a criminal, so that one innocent neighbor in row 4 has to be Quita or Tina. If Oscar were a criminal, then the only possible innocents in row 4 would have to come from Quita and Tina. But that would make both of row 4's innocents neighbors of Logan, which clashes with the requirement that exactly 1 innocent in row 4 neighbors Logan. So Oscar must be innocent.
06.A3 · Ivan → CRIMINAL
Row 3 cannot end with exactly 2 innocents, because row 4 is the only row allowed to have exactly 2 innocents. Also, Tina’s clue says her neighbors contain exactly 2 criminals, and exactly 1 of those criminals is in row 3, so among Logan and Nicole exactly one must be a criminal. If Ivan were innocent, then row 3 would have Janet as a criminal, Ivan as an innocent, and Logan and Nicole split so that exactly one of them is a criminal. That would leave row 3 with exactly 2 innocents, which clashes with row 4 being the only row with exactly 2 innocents. So Ivan must be criminal.
07.A1 · Alice → INNOCENT
Ivan’s clue says Alice is one of the exactly 12 innocents on the edges. That directly identifies Alice as an innocent, because she is explicitly included in that group. So Alice must be innocent.
08.C4 · Quita → CRIMINAL, D4 · Tina → INNOCENT, C5 · Wanda → INNOCENT
Frank’s row clue, Ivan’s edge clue, and Oscar’s clue about Tina’s neighbors all apply to the same small group here: Quita in row 4, Tina on the edge, and Wanda as one of Tina’s neighbors on the edge. If you try the opposite statuses for these three in this step, namely Quita innocent, Tina criminal, and Wanda criminal, then the remaining people involved in those clues would have to make all three statements true at the same time, and they cannot do that. So that opposite combination is impossible. Quita at C4 must be criminal, Tina at D4 must be innocent, and Wanda at C5 must be innocent.
09.D2 · Habi → INNOCENT
Row 4 already has exactly 2 innocents, and it is the only row allowed to have exactly 2 innocents. Frank’s clue also fixes his neighboring group at exactly 4 criminals, with exactly 1 of those criminals in row 2, so within row 2 only Erwin or Gary can be that criminal neighbor. If Habi were a criminal, then the remaining unknown people named here would have to satisfy both of those clue restrictions at the same time, and that cannot be done. So Habi cannot be a criminal. That makes Habi innocent.
10.C3 · Logan → CRIMINAL, C1 · Chuck → INNOCENT
Frank’s clue says he has exactly 4 criminal neighbors, and exactly 1 of those criminals is in row 2. Right now Frank already has 2 known criminal neighbors, Ivan and Janet, and neither of them is in row 2, so the remaining criminal neighbors have to come from Chuck, Erwin, Gary, and Logan in a way that leaves only 1 criminal in row 2. Now test the opposite statuses here: Chuck criminal and Logan innocent. Along with the row clue that row 4 is the only row with exactly 2 innocents, and the edge clue that there are exactly 12 innocents on the edges, the remaining people involved here, Donna, Erwin, Gary, Nicole, and Umar, cannot be assigned in any way that satisfies all three clues at once. So the opposite pairing fails, which means Chuck cannot be criminal and Logan cannot be innocent. That makes Logan criminal and Chuck innocent.
11.D3 · Nicole → INNOCENT
Oscar’s clue says Tina has exactly 2 criminal neighbors, and exactly 1 of those criminals is in row 3. Among Tina’s neighbors who are in row 3, Logan is already a known criminal and Nicole is the only unknown. Since the one row 3 criminal neighbor is already accounted for by Logan, Nicole cannot also be a criminal. So Nicole must be innocent.
12.D1 · Donna → INNOCENT
Ivan’s clue says there are exactly 12 innocents on the edge. The edge already has 10 known innocents and 1 known criminal, so among the three unknown edge people, exactly one must be a criminal. Nicole’s clue says columns A and B have the same number of criminals. Column B already has 2 criminals, while column A has 1 known criminal and its only unknowns are Erwin and Umar, so the one extra criminal needed for column A must be Erwin or Umar. That means the single criminal among the unknown edge people has to come from A2 Erwin and A5 Umar, not D1 Donna. So Donna must be innocent.
13.A5 · Umar → INNOCENT
Frank’s clue fixes his neighborhood very tightly: he has exactly 4 criminal neighbors, and exactly 1 of those criminals is in row 2. The only row-2 neighbors of Frank are Erwin and Gary, so one of Erwin and Gary must be criminal, and the other must be innocent. Donna’s clue says exactly one row has exactly 3 innocents. If Umar were criminal, then Erwin and Gary would still have to fit Frank’s clue, but there is no way to assign Erwin and Gary so that Frank’s neighborhood condition and the “exactly one row has exactly 3 innocents” condition both hold. So Umar cannot be criminal. That makes Umar innocent.
14.A2 · Erwin → CRIMINAL
Nicole's clue says columns A and B must have the same number of criminals. Column B already has 2 criminals, while column A currently has only 1 known criminal, and the only person in column A not yet identified is Erwin. If Erwin were innocent, column A would stay at 1 criminal and could not match column B's 2. So Erwin must be criminal.
15.C2 · Gary → INNOCENT
Tina’s clue says that among Frank’s neighbors, exactly one criminal is in row 2. The row-2 neighbors of Frank are only Erwin and Gary, and Erwin is already a known criminal there. That means the one row-2 criminal among Frank’s neighbors is already accounted for by Erwin, so Gary cannot also be a criminal. So Gary must be innocent.