TrickyMay 01, 2026Solved

Clues by Sam May 01, 2026 Answer – Full Solution Explained

A1

🤡

Adam

clown

B1

👩‍🔧

Cheryl

mech

C1

👩‍✈️

Diane

pilot

D1

👨‍🎨

Ethan

painter

A2

👩‍⚖️

Freya

judge

B2

👨‍🔧

Gary

mech

C2

👨‍🔧

Henry

mech

D2

👨‍🎨

Ike

painter

A3

😬

Cake

cake

B3

💂‍♂️

Kumar

guard

C3

👷‍♀️

Laura

builder

D3

👷‍♀️

Nancy

builder

A4

👩‍✈️

Olga

pilot

B4

💂‍♀️

Penny

guard

C4

👨‍💻

Scott

coder

D4

👮‍♀️

Tina

cop

A5

😬

Cake

cake

B5

👮‍♂️

Will

cop

C5

👨‍💻

Xavi

coder

D5

🤡

Zane

clown

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 11Criminal 9Unknown 0

See how each clue leads to the final result

Just the answer

Skip the reasoning — 8 criminals.

Full walkthrough · Friday May 01, 2026

Clues by Sam answer for May 01, 2026 — a Tricky solved in 13 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 Diane (C1), Gary (B2), Henry (C2), Ike (D2), Cake (A5), Kumar (B3), Olga (A4) and Xavi (C5); the remaining 12 suspects are innocent.

The deduction chain, in plain English

01.D4 · Tina INNOCENT, C5 · Xavi CRIMINAL, D5 · Zane INNOCENT

Scott’s clue says the people below Nancy must include more innocents than the people below Laura. Right now, below Nancy there are no known innocents, while below Laura there is already 1 known innocent, Scott. If Tina were criminal, Xavi were innocent, and Zane were criminal, then below Nancy would have 0 innocents and below Laura would have 2 innocents. That directly clashes with Scott’s clue, which requires below Nancy to have more innocents than below Laura. So Tina must be innocent, Xavi must be criminal, and Zane must be innocent.

02.B4 · Penny INNOCENT

Xavi’s clue says Penny is one of the exactly 3 innocents in column B. Since Penny is explicitly included in that innocent group, her identity is fixed by the clue itself. So Penny must be innocent.

03.B2 · Gary CRIMINAL

Penny’s clue says the two criminals above her must be connected, and the people above Penny are Cheryl at B1, Gary at B2, and Kumar at B3. If Gary were innocent, then the two criminals above Penny would have to be Cheryl and Kumar. But Cheryl at B1 and Kumar at B3 are not connected to each other unless Gary sits between them as a criminal. So Gary at B2 must be criminal.

04.B5 · Will INNOCENT

Xavi says Penny is one of exactly 3 innocents in column B, so column B has exactly 2 criminals. In that column, Penny is already innocent and Gary is already criminal, leaving Cheryl, Kumar, and Will to supply exactly 1 more criminal. Penny also says both criminals above her are connected, so that second criminal has to be above Penny, among Cheryl and Kumar, not at Will’s spot below her. That makes Will innocent.

05.D2 · Ike CRIMINAL

Will’s clue says that Ike is one of the exactly 5 criminals on the edges. That directly identifies Ike as a criminal. So Ike at D2 must be criminal.

06.A4 · Olga CRIMINAL

Ike’s clue says each row has at least one criminal. In row 4, Penny, Scott, and Tina are already known innocents, and Olga is the only person in that row whose identity was not yet fixed. That means row 4 cannot have Olga as innocent too, because then the whole row would be innocent and the clue would fail. So Olga must be criminal.

07.D3 · Nancy INNOCENT, A2 · Freya INNOCENT

Will’s clue says there are exactly 5 criminals on the edge. On the edge, 3 are already known criminals, so the remaining 8 edge people contain exactly 2 more criminals. Those 8 are Adam, Cheryl, Diane, Ethan, Freya, Cake at A3, Nancy, and Cake at A5. The two remaining edge criminals must come from Adam, Cheryl, Diane, Ethan, Cake at A3, and Cake at A5, so Freya and Nancy cannot be two of them. So Nancy and Freya must be innocent.

08.B1 · Cheryl INNOCENT, B3 · Kumar CRIMINAL

Olga’s clue says her neighbors include at least 2 criminals, and right now none of Olga’s neighbors is a known criminal. At the same time, the edge clue fixes the total edge criminals at 5, with Ike included, while the edge cells already have 3 known criminals and the remaining edge places involved here are A1 Adam, B1 Cheryl, C1 Diane, D1 Ethan, A3 Cake, and A5 Cake. If Cheryl were criminal and Kumar were innocent, those same remaining people would have to satisfy both clues at once, but they cannot. So Cheryl and Kumar cannot have those opposite identities. That makes Cheryl innocent and Kumar criminal.

09.C2 · Henry CRIMINAL

Cheryl’s clue says that in column C, exactly one criminal has a criminal directly above them. In that column, the only known criminal so far is Xavi, and Diane and Laura are the other people there whose identities are not yet fixed. If Henry were innocent, then Diane and Laura would have to account for that clue by themselves, together with Xavi, and that cannot be made to fit the requirement that there be exactly one such criminal in column C. So Henry cannot be innocent. That makes Henry criminal.

10.C3 · Laura INNOCENT

Will’s clue says there are exactly 5 criminals on the edge, and among the edge cells only three are already known criminals: Ike, Olga, and Xavi. So the five edge unknowns named here, Adam, Diane, Ethan, Cake at A3, and Cake at A5, have to account for exactly two more edge criminals. At the same time, every row must have at least one criminal, and only one row can have exactly 2 criminals. If Laura were a criminal, then Laura together with those five edge unknowns would have to satisfy all of those row requirements as well as the edge total, but they cannot do that without breaking one of those clues. So Laura must be innocent.

11.C1 · Diane CRIMINAL

Cheryl’s clue says exactly one criminal in column C has a criminal directly above them. In column C, the known criminals are Henry at C2 and Xavi at C5, and Diane at C1 is the only person there whose identity is still unknown. If Diane were innocent, column C would have to satisfy Cheryl’s clue without Diane being one of the criminals there, but that cannot be done. So Diane cannot be innocent. So Diane must be criminal.

12.A1 · Adam INNOCENT, D1 · Ethan INNOCENT

Will’s clue says there are exactly 5 criminals on the edge. The edge already has 4 known criminals, so among A1 Adam, D1 Ethan, A3 Cake, and A5 Cake, exactly 1 of them is a criminal. Tina’s clue says Olga has at least 2 criminal neighbors. Around Olga, Kumar is already a known criminal, so the needed extra criminal must be one of Olga’s unknown neighbors, A3 Cake or A5 Cake. That means the single remaining edge criminal has to come from A3 or A5, not from A1 or D1. So Adam and Ethan cannot be that edge criminal. That makes Adam and Ethan innocent.

13.A5 · Cake CRIMINAL

Ethan’s clue says there is exactly 1 innocent person who is both a neighbor of Will and a cake. In this case, that shared group contains only A5 Cake, and there are no known innocents in it yet. Since the group still needs 1 innocent and A5 Cake is the only person there who could fill it, A5 Cake must be innocent.

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