Clues by Sam Aug 22, 2026 Answer – Full Solution Explained
A1
👩🎤
singer
B1
👩🎤
singer
C1
🕵️♂️
sleuth
D1
🕵️♀️
sleuth
A2
👨🎤
singer
B2
🕵️♂️
sleuth
C2
👨💼
clerk
D2
👩💼
clerk
A3
👨✈️
pilot
B3
👨🍳
cook
C3
👩🍳
cook
D3
👩💼
clerk
A4
👨✈️
pilot
B4
👩🍳
cook
C4
👮♂️
cop
D4
👩⚖️
judge
A5
👨⚖️
judge
B5
👨⚖️
judge
C5
👮♀️
cop
D5
👮♀️
cop
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 22, 2026 — a Hard solved in 19 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Ethan (A2), Ivan (C2), Janet (D2), Quita (B4), Steve (C4), Vince (B5), Wanda (C5) and Xola (D5); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.C1 · Chase → INNOCENT
Paul’s clue says Chase is one of the exactly 2 innocents above Wanda. That directly identifies Chase as an innocent. So Chase must be innocent.
02.C4 · Steve → CRIMINAL
Chase’s clue says that the two criminals in row 4 must be connected. In row 4, Paul is innocent, and the other people there are Quita, Steve, and Tina. If Steve were innocent, then the only possible criminals in that row would be Quita and Tina. But with Paul innocent between that side of the row and Tina’s side, those two would not satisfy the clue that both criminals in row 4 are connected. So Steve must be criminal.
03.C5 · Wanda → CRIMINAL
Column C has exactly 2 innocents. Chase is already one innocent there, so column C needs exactly 1 more innocent. Among the people above Wanda, Chase is one of exactly 2 innocents, and Steve is criminal, so the only remaining innocent in that group must be either Ivan or Nicole. That means the second innocent required for column C is already accounted for among the people above Wanda. Wanda is in column C but not in that above-Wanda group, so she cannot also be innocent. So Wanda must be criminal.
04.B5 · Vince → CRIMINAL
Wanda's clue says there are exactly 5 criminals on the edge, and exactly 1 of those edge criminals is a neighbor of Umar. Among the edge people who are neighbors of Umar, Paul is already innocent, so he cannot be that 1 criminal. The only unknown person left in that edge-and-neighbor group is Vince. So Vince must be criminal.
05.B1 · Betsy → INNOCENT
Steve’s clue says column C is the only column with exactly 2 innocents, and Vince’s clue says exactly 2 innocents are in the overlap of the people below Betsy and Nicole’s neighbors. That overlap is specifically Ghani, Martin, and Quita. If Betsy were criminal, the remaining people involved in these two clues would have to make both of those exact statements true at the same time, while column C still stayed the only column with exactly 2 innocents. But with Betsy as criminal, those same people cannot satisfy both requirements together. So Betsy must be innocent.
06.D5 · Xola → CRIMINAL
Wanda’s neighbors must contain exactly one innocent. Among those neighbors, the possible innocent slots named here are Quita, Tina, and Xola. But the one innocent neighbor has to come from Quita and Tina. That means Xola cannot be Wanda’s innocent neighbor. So Xola must be criminal.
07.B2 · Ghani → INNOCENT
Paul’s clue fixes the people above Wanda at exactly two innocents, and Chase is already one of them. So among Ivan and Nicole, exactly one is innocent. Xola’s clue means the clerks who have a criminal directly to the left account for exactly two innocents among Ghani, Ivan, and Nicole. Since Ivan and Nicole together contribute exactly one innocent, the remaining innocent in that group has to be Ghani. So Ghani must be innocent.
08.A5 · Umar → INNOCENT
Ghani’s clue says every row has at least one innocent. In row 5, Vince, Wanda, and Xola are already known criminals, so that row already has 3 criminals. That means the only person left in row 5, Umar, cannot also be a criminal, or row 5 would have no innocent at all. So Umar must be innocent.
09.A2 · Ethan → CRIMINAL
Betsy’s clue says exactly two innocents are in the shared group of people below Betsy who also neighbor Nicole, and that group is Ghani, Martin, and Quita. Since Ghani is already innocent, Martin and Quita have to supply the rest of that requirement. Umar’s clue says Klay has an odd number of innocent neighbors. Klay’s neighbors already include two known innocents, Ghani and Paul, and the only unknown neighbors are Ethan, Martin, and Quita. If Ethan were innocent, then Ethan, Martin, and Quita would have to satisfy both clues at once, but that is impossible. So Ethan must be criminal.
10.D1 · Donna → INNOCENT
Vince's clue fixes the shared group below Betsy and neighboring Nicole: B2 Ghani, B3 Martin, and B4 Quita must contain exactly 2 innocents, and Ghani is already one of them. Ethan's clue also requires exactly 4 edge people to have a criminal directly to their right, and the edge people involved here include Donna along with Amy, Janet, Klay, Olivia, and Tina. If Donna were a criminal, then these same remaining people would have to satisfy both of those exact requirements at once, but they cannot. So Donna cannot be a criminal. That makes Donna innocent.
11.B4 · Quita → CRIMINAL
Donna’s clue says exactly one corner person has exactly one criminal neighbor. Amy already has exactly one criminal neighbor, and no unknown neighbors, so Amy is one such corner for certain. If Quita were innocent, Umar would also have exactly one criminal neighbor. But Xola already has two known criminal neighbors, so Xola cannot be the one, and Donna starts with no known criminal neighbors and would depend on Ivan and Janet instead. That would make the corner requirement impossible to satisfy with Amy already fixed as the one corner that has exactly one criminal neighbor. So Quita must be criminal.
12.B3 · Martin → INNOCENT
Vince’s clue says exactly 2 innocents are both below Betsy and neighboring Nicole. That shared group is Ghani, Martin, and Quita. Ghani is already innocent and Quita is already criminal, so the group still needs one more innocent, and Martin is the only person left there. So Martin must be innocent.
13.D4 · Tina → INNOCENT
Chase’s clue says both criminals in row 4 are connected, so row 4 must contain exactly two criminals. In row 4, Quita and Steve are already the two known criminals, while Paul is innocent and Tina is the only unknown there. If Tina were a criminal, row 4 would no longer have just those two criminals required by the clue. So Tina cannot be a criminal. That makes Tina innocent.
14.A1 · Amy → INNOCENT
Quita’s clue says the corner cells contain exactly 3 innocents. The corners already include 2 known innocents, and Amy is the only corner person whose identity is still unknown. Since the corners still need exactly 1 more innocent, that remaining corner spot has to be innocent. So Amy must be innocent.
15.A3 · Klay → INNOCENT
Amy’s clue says the number of innocent judges must equal the number of innocent pilots. The judges already have 2 innocents, while the pilots currently have only 1 innocent, with Klay the only pilot not yet identified. If Klay were criminal, the pilots would stay at 1 innocent, which would not match the judges’ 2 innocents. So Klay must be innocent.
16.D3 · Olivia → INNOCENT
Row 4 already has 2 innocents, and row 3 also currently has 2 known innocents. Klay’s clue says row 3 must have more innocents than row 4, so row 3 has to gain at least one more innocent. If Olivia were criminal, then the only person left in row 3 who could raise that innocent count would be Nicole. At the same time, Steve’s and Donna’s clues would still have to be satisfied by Ivan, Janet, and Nicole, but that combination cannot be made to work. So Olivia cannot be criminal. That makes Olivia innocent.
17.D2 · Janet → CRIMINAL
Wanda’s clue says there are exactly 5 criminals on the edge, and exactly 1 of those edge criminals is a neighbor of Umar. Since the only edge neighbor of Umar who is criminal is Vince, that means the other 4 edge criminals must all be edge people who are not neighbors of Umar. Among those edge non-neighbors, Ethan, Wanda, and Xola are already known criminals, so that group still needs 1 more criminal. The only unknown person left in that group is Janet, so Janet must be criminal.
18.C2 · Ivan → CRIMINAL
Donna’s clue says exactly one corner person has exactly one criminal neighbor. Amy already fits that exactly, because she has 1 known criminal neighbor and no unknown neighbors. Umar and Xola each already have 2 known criminal neighbors, so neither of them can be that one. That leaves Donna. She currently has 1 known criminal neighbor, and the only unknown affecting her count is Ivan. If Ivan were innocent, Donna would also have exactly one criminal neighbor, which would make both Amy and Donna fit the clue. So Ivan must be criminal.
19.C3 · Nicole → INNOCENT
Steve’s clue says column C is the only column with exactly 2 innocents. In column C, Chase is already the one known innocent, and Nicole is the only person there whose identity is not yet fixed. If Nicole were criminal, then column C would have only 1 innocent, not 2, which clashes with Steve’s clue. So Nicole cannot be criminal. That makes Nicole innocent.