BrutalAug 23, 2026Solved

Clues by Sam Aug 23, 2026 Answer – Full Solution Explained

A1

👨‍🏫

Alex

teacher

B1

👩‍🎤

Betsy

singer

C1

👨‍🎨

Chase

painter

D1

👨‍🎨

Daniel

painter

A2

🕵️‍♂️

Eric

sleuth

B2

👩‍🎤

Frida

singer

C2

💂‍♂️

Hal

guard

D2

👨‍🎨

Isaac

painter

A3

🕵️‍♂️

John

sleuth

B3

👩‍🎤

Kay

singer

C3

💂‍♂️

Larry

guard

D3

💂‍♂️

Martin

guard

A4

👩‍💻

Olivia

coder

B4

👩‍💻

Quita

coder

C4

👩‍🏫

Saga

teacher

D4

👩‍🌾

Uma

farmer

A5

👨‍🏫

Vince

teacher

B5

👩‍💻

Wanda

coder

C5

👩‍🌾

Xola

farmer

D5

👩‍🌾

Zoe

farmer

Final Board State

This puzzle is fully solved.

All characters have been identified as innocent or criminal based on today's clues.

Final Result
Innocent 6Criminal 14Unknown 0

See how each clue leads to the final result

Just the answer

Skip the reasoning — 14 criminals.

Full walkthrough · Sunday Aug 23, 2026

Clues by Sam answer for Aug 23, 2026 — a Brutal solved in 17 steps

Today's Clues by Sam puzzle is rated Brutal and resolves with 14 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alex (A1), Betsy (B1), Chase (C1), Daniel (D1), Eric (A2), Kay (B3), Larry (C3), Martin (D3), Quita (B4), Saga (C4), Uma (D4), Wanda (B5), Xola (C5) and Zoe (D5); the remaining 6 suspects are innocent.

The deduction chain, in plain English

01.C5 · Xola CRIMINAL

Hal’s clue says that Xola is one of the exactly 10 criminals on the edge. That directly places Xola among the criminals counted on the edge. So Xola must be criminal.

02.D4 · Uma CRIMINAL

Xola's clue says row 4 has exactly 3 criminals, and exactly 2 of those criminals are Kay's neighbors. That means exactly 1 criminal in row 4 is not a neighbor of Kay. In row 4, the people who are Kay's neighbors are Olivia, Quita, and Saga, so the only person in row 4 who is not a neighbor of Kay is Uma. Since that non-neighbor part of row 4 still needs its 1 criminal, Uma must fill it. So Uma must be criminal.

03.C4 · Saga CRIMINAL

John and Vince have exactly one innocent neighbor in common, and that common-neighbor group is just Olivia and Quita. So among Olivia and Quita, exactly one is innocent. Row 4 has exactly one innocent in total, and the only people in row 4 besides Olivia and Quita are Saga and Uma. Since Uma is already criminal, the single innocent allowed in row 4 is already fully accounted for by Olivia and Quita, which leaves no room for Saga to be innocent. So Saga must be criminal.

04.B3 · Kay CRIMINAL

Row 4 has exactly three criminals in total, and Saga and Uma are already two of them, so among Olivia and Quita there is exactly one criminal. John's neighbors contain exactly three criminals, and exactly one of those is in row 2, so among John's neighbors who are not in row 2 there must be exactly two criminals: Kay, Olivia, and Quita. Since Olivia and Quita contribute exactly one criminal, the remaining criminal in that group has to be Kay. So Kay must be criminal.

05.D2 · Isaac INNOCENT

John’s neighbors must contain exactly 3 criminals, and exactly 1 of those criminals is in row 2. The only row-2 neighbors of John are Eric and Frida, so the row-2 criminal among John’s neighbors has to come from those two. Row 2 is also the only row with exactly 3 innocents. If Isaac were criminal, then the remaining people named in these clues could not satisfy both of those row-2 requirements at the same time. That contradiction rules out Isaac being criminal. So Isaac must be innocent.

06.A3 · John INNOCENT

Isaac’s clue says John is one of Quita’s 3 innocent neighbors. That directly identifies John as an innocent member of that neighbor group. So John must be innocent.

07.D3 · Martin CRIMINAL

Hal’s clue says there are exactly 10 criminals on the edge, and the edge has 14 people total, so exactly 4 edge people are innocents. On the edge, Isaac and John are already known innocents, which leaves only 2 edge-innocent spots still available. Those 2 remaining edge innocents must come from Alex, Betsy, Chase, Daniel, Eric, Olivia, Vince, Wanda, and Zoe. That leaves Martin outside the set of people who can fill those 2 edge-innocent spots. So Martin must be criminal.

08.B1 · Betsy CRIMINAL, B5 · Wanda CRIMINAL

Column B has exactly 4 criminals, and Kay is already one of them, so there is exactly 1 innocent person in column B. Among the unknown people in column B, that one innocent must come from Frida or Quita, not from Betsy or Wanda. That means Betsy and Wanda cannot be innocent. So Betsy and Wanda must be criminal.

09.C1 · Chase CRIMINAL

Wanda’s clue says Daniel is the only person whose neighbors contain exactly 1 criminal. Daniel’s neighbors currently have two known innocents, no known criminals, and only one unknown neighbor: Chase, so that one criminal in Daniel’s neighbor set has to be Chase. If Chase were innocent, Daniel’s neighbors would contain 0 criminals instead of exactly 1, which breaks the clue immediately. So Chase at C1 must be criminal.

10.A1 · Alex CRIMINAL

Column A has exactly 2 criminals, and exactly 1 of those criminals is Eric's neighbor. Among the people in column A who are Eric's neighbors, John is already innocent, so that group still needs 1 criminal. The only unknown person left in that group is Alex. So Alex must be criminal.

11.D5 · Zoe CRIMINAL, D1 · Daniel CRIMINAL

Hal’s clue says there are exactly 10 criminals on the edge. On the edge, 7 people are already known to be criminals, and Chase’s clue fixes column A at exactly 2 criminals total, so among the edge people in column A there can be only one more criminal besides Alex. That accounts for 8 of the 10 edge criminals, leaving the two edge people outside column A who are still unknown, Daniel and Zoe, to provide the remaining two criminals. So Daniel and Zoe must be criminal.

12.A5 · Vince INNOCENT

Hal says there are exactly 10 criminals on the edges, and 9 edge criminals are already known. So among the three unknown edge people, A2 Eric, A4 Olivia, and A5 Vince, exactly one is a criminal. Saga's clue about John's neighbors says John's neighbors contain exactly 3 criminals, with exactly 1 of those in row 2. Since Kay is already the 1 known criminal neighbor, the other two criminal neighbors cannot be in row 2, so they must be A2 Eric and A4 Olivia. That uses up the one remaining edge criminal among Eric, Olivia, and Vince on Eric or Olivia, so Vince cannot be that edge criminal. So Vince must be innocent.

13.C3 · Larry CRIMINAL

Column B has exactly 4 criminals, and at the moment it has 3 known criminals with only Frida and Quita left unknown there. Martin’s clue therefore means exactly one of Frida and Quita must be a criminal, so column B must finish with 4 criminals. Vince says columns B and C have the same number of criminals. Column C currently also has 3 known criminals, and Larry is the only unknown person in that column. If Larry were innocent, column C would stay at 3 criminals while column B has to reach 4, so the two columns could not match. So Larry must be criminal.

14.A4 · Olivia INNOCENT

Isaac says John is one of Quita's 3 innocent neighbors. Among Quita's neighbors, John and Vince are already known to be innocent, while Kay, Larry, Saga, Wanda, and Xola are criminals. If Olivia were also criminal, then Quita's neighbors would contain only those 2 innocents, John and Vince, instead of the 3 the clue requires. So Olivia must be innocent.

15.A2 · Eric CRIMINAL

Hal's clue says there are exactly 10 criminals on the edge, and Xola is one of them. In the edge cells, 9 people are already known to be criminals, 4 are known to be innocent, and the only edge person not yet identified is Eric. If Eric were innocent, the edge would stay at only 9 criminals, which conflicts with Hal's clue that there are exactly 10. So Eric must be criminal.

16.B2 · Frida INNOCENT

John’s neighbors contain exactly 3 criminals, and exactly 1 of those criminals is in row 2. Among John’s neighbors in row 2, the people are Eric and Frida, and Eric is already a known criminal. That means the one row-2 criminal among John’s neighbors is already accounted for, so Frida cannot be a criminal. So Frida must be innocent.

17.B4 · Quita CRIMINAL

John’s neighbors contain exactly 3 criminals, and exactly 1 of those criminals is in row 2. Among John’s neighbors in row 2, Eric is the only criminal there, so the other 2 criminals among John’s neighbors must be outside row 2. The neighbors of John that are not in row 2 are Kay, Olivia, and Quita, and among them Kay is already a criminal while Olivia is innocent, so one more criminal is still needed there. The only person left to fill that spot is Quita, so Quita must be criminal.

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