TrickyMay 21, 2026Solved

Clues by Sam May 21, 2026 Answer – Full Solution Explained

A1

👮‍♀️

Amy

cop

B1

👨‍🏫

Bobby

teacher

C1

👨‍🍳

Erwin

cook

D1

👩‍🏫

Frida

teacher

A2

👨‍🎨

Gary

painter

B2

🕵️‍♀️

Hilda

sleuth

C2

👨‍⚕️

Ivan

doctor

D2

👩‍💻

Julie

coder

A3

👩‍🔧

Katie

mech

B3

👨‍🍳

Logan

cook

C3

👩‍🍳

Maria

cook

D3

👩‍🔧

Nicole

mech

A4

👨‍⚕️

Ollie

doctor

B4

👩‍💻

Quita

coder

C4

👨‍🌾

Ronald

farmer

D4

🕵️‍♀️

Sofia

sleuth

A5

👷‍♂️

Thor

builder

B5

👷‍♂️

Vince

builder

C5

👷‍♀️

Wanda

builder

D5

👮‍♂️

Ziad

cop

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 12Criminal 8Unknown 0

See how each clue leads to the final result

Just the answer

Skip the reasoning — 8 criminals.

Full walkthrough · Thursday May 21, 2026

Clues by Sam answer for May 21, 2026 — a Tricky solved in 16 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 Amy (A1), Erwin (C1), Frida (D1), Hilda (B2), Maria (C3), Sofia (D4), Thor (A5) and Ziad (D5); the remaining 12 suspects are innocent.

The deduction chain, in plain English

01.A3 · Katie INNOCENT, B3 · Logan INNOCENT

Ivan’s clue says exactly 2 innocents are in the overlap between the people to the left of Nicole and the people neighboring Gary. That shared group is only Katie and Logan. Since the clue needs 2 innocents there, and Katie and Logan are the only people in that group, both of them have to be the innocents required by the clue. So Katie and Logan must be innocent.

02.A5 · Thor CRIMINAL

Row 5 has exactly 2 innocents, and it currently has none identified yet, so it needs exactly 2 innocents in total. Those 2 innocents are already accounted for among Vince, Wanda, and Ziad. Thor is also in row 5, but not in that group, so if Thor were innocent then row 5 would have more than 2 innocents. So Thor must be criminal.

03.A1 · Amy CRIMINAL

To the left of Frida, there are exactly two criminals in total: Amy, Bobby, and Erwin. The clue also says exactly one of those criminals is Ivan's neighbor, and among those three the neighbors of Ivan are only Bobby and Erwin. So exactly one criminal must be to the left of Frida and not be Ivan's neighbor. The only person there who is not Ivan's neighbor is Amy, so Amy must be criminal.

04.B5 · Vince INNOCENT

Katie’s clue says there is exactly one builder who has a criminal directly to the right, and among the builders that fit that condition the relevant people are Vince, Wanda, and Ziad. Amy’s clue gives exactly one criminal among Sofia’s neighbors who are in row 5, which here are Wanda and Ziad. So Wanda and Ziad already account for the single criminal allowed in that builder group. That leaves Vince unable to be a criminal. So Vince must be innocent.

05.C4 · Ronald INNOCENT

Maria and Nicole are the only people shared by the two relevant groups. Ivan's row 3 neighbors contain exactly 1 criminal, so among Maria and Nicole there is exactly 1 criminal. Sofia's neighbors who are not in row 5 also contain exactly 1 criminal, and that larger group is Maria, Nicole, and Ronald. Since the single criminal allowed in that larger group is already accounted for by Maria and Nicole, Ronald cannot be a criminal. So Ronald is innocent.

06.D1 · Frida CRIMINAL

Thor’s clue says there are exactly 2 criminals to the left of Frida, and exactly 1 of those criminals is Ivan’s neighbor. Amy is already one of the criminals to Frida’s left, so Bobby and Erwin have to supply the rest of that clue in just the right way. If Frida were innocent, then Bobby, Erwin, and the other people involved in these two clues would have to fit both Logan’s row condition and Thor’s left-of-Frida condition at the same time, but they cannot. That makes the innocent option for Frida impossible. So Frida must be criminal.

07.A2 · Gary INNOCENT

Logan’s clue says row 5 is the only row with exactly 2 innocents, while Thor’s and Vince’s clues also fix exact criminal totals for the people left of Frida and for Ivan’s neighbors, including how many of those criminals can be in Ivan’s neighboring part of Frida’s-left group and how many can be in row 3. If Gary were a criminal, then Bobby, Erwin, Hilda, Julie, Maria, Nicole, Ollie, Quita, Sofia, Wanda, and Ziad would have to satisfy all of those exact row and neighbor counts at the same time. Those requirements conflict, so Gary cannot be criminal. So Gary must be innocent.

08.D3 · Nicole INNOCENT

Gary’s clue directly says that Nicole is one of Maria’s 6 innocent neighbors. Since the clue explicitly places Nicole among the innocent neighbors, her identity is fixed by the clue itself. So Nicole must be innocent.

09.C3 · Maria CRIMINAL

Vince's clue says Ivan has exactly 4 criminal neighbors, and exactly 1 of those criminals is in row 3. Among Ivan's neighbors in row 3, Logan and Nicole are already innocent, so that row-3 group still needs 1 criminal and the only unknown person left there is Maria. So Maria must be criminal.

10.C1 · Erwin CRIMINAL

Vince’s clue says Ivan’s neighbors contain exactly 4 criminals, and exactly 1 of those criminals is in row 3. In Ivan’s row 3 neighbors, Maria is already that 1 criminal, while Logan and Nicole are innocent, so the other 3 criminals around Ivan must all be outside row 3. That means among Ivan’s non-row-3 neighbors, only 2 can be innocent. Those non-row-3 unknowns are Bobby, Erwin, Hilda, and Julie, so the 2 innocents must come from Bobby, Hilda, and Julie, not from Erwin. So Erwin must be criminal.

11.B1 · Bobby INNOCENT

To the left of Frida, there are exactly two criminals in total. Among those left-of-Frida people, exactly one criminal is Ivan's neighbor, and in that neighbor group the people are Bobby and Erwin. Since Erwin is already a known criminal there, that one criminal is already accounted for, so Bobby cannot also be a criminal. So Bobby must be innocent.

12.A4 · Ollie INNOCENT, B4 · Quita INNOCENT, B2 · Hilda CRIMINAL

Bobby’s clue says Thor has more innocent neighbors than Amy. Right now Thor’s neighbors have only 1 known innocent, while Amy’s neighbors already have 2 known innocents. If Ollie and Quita were criminals and Hilda were innocent, then Thor’s side would stay at 1 innocent and Amy’s side would rise to 3 innocents. That directly clashes with the clue that Thor has more innocent neighbors than Amy. So Ollie and Quita cannot be criminals, and Hilda cannot be innocent. Ollie and Quita must be innocent, and Hilda must be criminal.

13.D2 · Julie INNOCENT

Logan’s clue says row 5 is the only row with exactly 2 innocents. Row 2 already has 2 known innocents, and Julie is the only person left in that row whose identity is not yet known. If Julie were criminal, then row 2 would remain on exactly 2 innocents, which the clue does not allow for any row except row 5. So Julie must be innocent.

14.D4 · Sofia CRIMINAL

Gary’s clue says Nicole is one of Maria’s exactly 6 innocent neighbors. Around Maria, the neighbors are Hilda, Ivan, Julie, Logan, Nicole, Quita, Ronald, and Sofia, and among them Hilda is criminal while Ivan, Julie, Logan, Nicole, Quita, and Ronald are already innocent. That already fills all 6 innocent-neighbor spots named by the clue, so Sofia cannot also be innocent. So Sofia must be criminal.

15.C5 · Wanda INNOCENT

Julie’s clue says columns A and C must have the same number of criminals. Column A already has 2 criminals, and column C also already has 2: Erwin and Maria. The only unknown left in column C is Wanda, so if Wanda were a criminal, column C would rise to 3 criminals while column A would stay at 2, breaking the clue. So Wanda must be innocent.

16.D5 · Ziad CRIMINAL

Logan’s clue says row 5 is the only row with exactly 2 innocents. In row 5, Vince and Wanda are already innocent, so row 5 already has those 2 innocents. If Ziad were also innocent, then row 5 would have 3 innocents instead of exactly 2, which clashes with the clue. So Ziad cannot be innocent. That makes Ziad criminal.

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