Clues by Sam Aug 26, 2026 Answer – Full Solution Explained
A1
👷♀️
builder
B1
👷♂️
builder
C1
👨🌾
farmer
D1
👩🌾
farmer
A2
👷♀️
builder
B2
🕵️♂️
sleuth
C2
👮♂️
cop
D2
👨🌾
farmer
A3
👨💼
clerk
B3
🕵️♀️
sleuth
C3
👮♀️
cop
D3
👩🎤
singer
A4
👩💼
clerk
B4
👩💻
coder
C4
👮♂️
cop
D4
👨🎤
singer
A5
👨💻
coder
B5
👨💻
coder
C5
🕵️♀️
sleuth
D5
👩🎤
singer
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 Aug 26, 2026 — a Medium solved in 16 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Bobby (B1), Carl (C1), Debra (D1), Ellie (A2), Gary (B2), Henry (C2), Maren (D3) and Quita (B4); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.A4 · Olga → INNOCENT
Xola’s clue directly says that Olga is one of the four innocents in column A. Since Olga is named as one of those innocents, Olga at A4 must be innocent.
02.A1 · Anna → INNOCENT
Olga’s clue says column A has exactly 4 innocents. Olga’s own clue says the people below Anna have exactly 3 innocents, and those people are Ellie, John, Olga, and Umar. The only person in column A who is not among the people below Anna is Anna herself. So if the people below Anna account for 3 of the 4 innocents in column A, the remaining innocent in that column has to be Anna. So Anna must be innocent.
03.A5 · Umar → INNOCENT
Anna says there are exactly 3 innocents above Umar, and those people are Anna, Ellie, John, and Olga. Xola says column A has exactly 4 innocents, and column A is Anna, Ellie, John, Olga, and Umar. The only person in column A who is not already in the group above Umar is Umar, so he has to provide the one extra innocent needed to raise that count from 3 to 4. So Umar must be innocent.
04.C1 · Carl → CRIMINAL, A2 · Ellie → CRIMINAL, B2 · Gary → CRIMINAL, C2 · Henry → CRIMINAL
Umar’s clue says Bobby has exactly 4 criminal neighbors. Among Bobby’s neighbors, there are currently 0 known criminals, and the only people there whose identities are still unknown are Carl, Ellie, Gary, and Henry. Since Bobby’s neighbors still need all 4 of those criminals, those four unknown neighbors must all be criminals. So Carl, Ellie, Gary, and Henry must be criminals.
05.A3 · John → INNOCENT
Xola’s clue says Olga is one of exactly 4 innocents in column A. In column A, Anna, Olga, and Umar are already known innocents, Ellie is a criminal, and John is the only person there not yet identified. If John were a criminal, column A would have only those 3 innocents, which conflicts with the clue saying there are exactly 4. So John must be innocent.
06.D2 · Ike → INNOCENT
Henry’s clue says that Ike is one of the exactly 3 innocents below Debra. That directly identifies Ike as an innocent. So Ike must be innocent.
07.C3 · Lucy → INNOCENT
Ike’s clue says that Lucy is one of Rohan’s 6 innocent neighbors. That directly identifies Lucy as innocent. So Lucy must be innocent.
08.B1 · Bobby → CRIMINAL
Henry’s clue says that among the people below Debra, exactly 3 are innocent, and Ike is one of them. So that group is D2 Ike, D3 Maren, D4 Tom, and D5 Zara, with exactly three innocents among those four. Gary’s clue says exactly 2 edge criminals have a criminal directly below them. The edge people involved here are Carl and Ellie as known criminals, plus the unknowns Bobby, Debra, Maren, Tom, Vince, and Zara. If Bobby were innocent, then Debra, Maren, Tom, Vince, and Zara would have to make both clues true at the same time, and they cannot. So Bobby cannot be innocent. That makes Bobby criminal.
09.D4 · Tom → INNOCENT
Below Debra, there is exactly 1 criminal among Ike, Maren, Tom, and Zara. Tom’s edge neighbors contribute exactly 1 criminal in the overlap group Maren and Zara. That overlap is the smaller group, and it already accounts for all the criminals allowed in the larger group below Debra. The only person below Debra who is not in that overlap is Tom, so Tom cannot be criminal. So Tom must be innocent.
10.B3 · Katie → INNOCENT
Rohan’s clue says he has 6 innocent neighbors, and among his neighbors there are already 3 known innocents: Lucy, Tom, and Xola. That means the other 5 neighbors contain exactly 2 criminals. Those 5 are Katie, Maren, Quita, Vince, and Zara. But the 2 criminals among them have to come from Maren, Quita, Vince, and Zara, so Katie is not one of those criminals. So Katie must be innocent.
11.C4 · Rohan → INNOCENT
Below Debra there must be exactly 3 innocents, and among Debra’s below-people we already have Ike and Tom innocent, with only Maren and Zara left unknown there. Rohan’s neighbors must contain exactly 6 innocents, and among those neighbors Katie, Lucy, Tom, and Xola are already innocent, leaving Maren, Quita, Vince, and Zara to supply the rest. Now test Rohan as criminal. Then Maren, Quita, Vince, and Zara would have to satisfy those innocence totals while also fitting Tom’s clue that exactly one person in row 4 has exactly 2 criminal neighbors, but those requirements clash. So Rohan cannot be criminal. That makes Rohan innocent.
12.D1 · Debra → CRIMINAL
Column D must contain an odd number of innocents. It already has 2 known innocents, Ike and Tom, so Debra, Maren, and Zara together must contribute an odd number of additional innocents. Henry’s clue says there are exactly 3 innocents below Debra. Since Ike and Tom are already 2 innocents below Debra, Maren and Zara must contain exactly 1 more innocent between them. That means Maren and Zara already provide the needed odd extra innocent count for column D, so Debra cannot also be innocent. So Debra must be criminal.
13.B5 · Vince → INNOCENT
Lucy’s clue says there are exactly 5 criminals on the edge, and exactly 1 of those edge criminals is Tom’s neighbor. So among the edge people who are not Tom’s neighbors, there must be exactly 4 criminals. That non-neighbor edge group is Anna, Bobby, Carl, Debra, Ellie, Ike, John, Olga, Tom, Umar, and Vince. In that group, Bobby, Carl, Debra, and Ellie are already the 4 known criminals, so there is no room for any other criminal there. That makes Vince innocent.
14.B4 · Quita → CRIMINAL
Bobby's clue says exactly 1 innocent is in the overlap between Xola's neighbors and the coders. That shared group is just Quita and Vince, and Vince is already known to be innocent. Since the one innocent in that group is already accounted for, Quita cannot also be innocent. So Quita must be criminal.
15.D5 · Zara → INNOCENT
Vince’s clue says exactly 2 sleuths have an innocent directly to the right. Gary does not qualify, because the person directly to his right is Henry, who is criminal, while Katie does qualify because Lucy is innocent. That leaves one more qualifying sleuth, and the only remaining sleuth case is Xola, whose person directly to the right is Zara. So Zara must be innocent.
16.D3 · Maren → CRIMINAL
Henry’s clue says Ike is one of exactly 3 innocents below Debra. Below Debra are Ike, Maren, Tom, and Zara, and Ike, Tom, and Zara are already the 3 known innocents there. If Maren were also innocent, that group would have 4 innocents below Debra, which clashes with the clue. So Maren must be criminal.