HardAug 29, 2026Solved

Clues by Sam Aug 29, 2026 Answer – Full Solution Explained

A1

👩‍⚖️

Amy

judge

B1

👷‍♂️

Bruce

builder

C1

👮‍♀️

Cheryl

cop

D1

👮‍♀️

Debra

cop

A2

👨‍🎨

Erwin

painter

B2

👮‍♂️

Frank

cop

C2

👨‍🎨

Gary

painter

D2

👨‍🎨

Hank

painter

A3

👨‍🎤

Igor

singer

B3

👩‍✈️

Mary

pilot

C3

👨‍✈️

Nick

pilot

D3

👨‍✈️

Olof

pilot

A4

👩‍⚖️

Paula

judge

B4

👷‍♀️

Rose

builder

C4

👨‍🎤

Scott

singer

D4

👨‍💻

Terry

coder

A5

👩‍⚖️

Vicky

judge

B5

👩‍💻

Wanda

coder

C5

👩‍🎤

Xena

singer

D5

👩‍💻

Zoe

coder

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 4Criminal 16Unknown 0

See how each clue leads to the final result

Just the answer

Skip the reasoning — 16 criminals.

Full walkthrough · Saturday Aug 29, 2026

Clues by Sam answer for Aug 29, 2026 — a Hard solved in 15 steps

Today's Clues by Sam puzzle is rated Hard and resolves with 16 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Amy (A1), Bruce (B1), Cheryl (C1), Erwin (A2), Frank (B2), Gary (C2), Hank (D2), Mary (B3), Nick (C3), Paula (A4), Rose (B4), Scott (C4), Terry (D4), Vicky (A5), Xena (C5) and Zoe (D5); the remaining 4 suspects are innocent.

The deduction chain, in plain English

01.B3 · Mary CRIMINAL

Olof’s clue says that the two criminals in row 3 must be connected. In row 3, Olof is already innocent, so the only people who could fill those two criminal spots are Igor, Mary, and Nick. If Mary were innocent, then Igor and Nick would have to be the two criminals in that row. But then the two criminals in row 3 would not satisfy Olof’s requirement that both criminals there are connected. So Mary must be criminal.

02.B4 · Rose CRIMINAL

Mary’s clue says that the two innocents neighboring Paula are also neighbors of Rose. Among Paula’s neighbors, the only person who is not a neighbor of Rose is Rose herself. So every innocent neighbor of Paula has to be someone other than Rose, and Rose cannot be one of Paula’s two innocent neighbors. That makes Rose criminal.

03.A3 · Igor INNOCENT

Row 3 has to contain more innocents than any other row. Right now row 3 already has Olof as an innocent, and its only unknown people are Igor and Nick. If Igor were criminal, then row 3 would have to rely on the remaining people to keep row 3 ahead of every other row, but that cannot be done while also satisfying these clue facts. So Igor cannot be criminal. That makes Igor innocent.

04.C3 · Nick CRIMINAL

Olof’s clue says that the two criminals in row 3 must be connected. In row 3, Igor is innocent, Mary is criminal, Nick is unknown, and Olof is innocent. If Nick were innocent, then row 3 would not have the required pair of connected criminals. So Nick at C3 must be criminal.

05.C5 · Xena CRIMINAL, D5 · Zoe CRIMINAL

Rose says row 3 has more innocents than any other row, and row 3 already has exactly two innocents: Igor and Olof. So no other row can also reach two innocents. If Xena and Zoe were innocent, then row 5 would have two innocents in that same row. That would tie row 5 with row 3, which contradicts Rose's clue that row 3 has more innocents than any other row. So Xena and Zoe must be criminals.

06.A4 · Paula CRIMINAL

Mary says the two innocents neighboring Paula are also neighbors of Rose. Paula’s neighbors already include Igor as one known innocent, so the other innocent neighbor of Paula must be either Vicky or Wanda, and that person is also a neighbor of Rose. Among the edge people who neighbor Rose, Igor is already one innocent, so the group can have only one more innocent. That remaining innocent must therefore be Vicky or Wanda, not Paula. So Paula must be criminal.

07.C4 · Scott CRIMINAL

Paula has two innocent neighbors, and Mary says both of those innocents are also neighbors of Rose. Igor is already one innocent neighbor of Paula, so Paula’s other innocent neighbor must be either Vicky or Wanda, and that person is also a neighbor of Rose. Paula says Rose has exactly two innocent neighbors, and exactly one of them is in row 3. Among Rose’s neighbors in row 3, the only innocent is Igor, so Igor is that one. That means Rose’s other innocent neighbor must be outside row 3, and from Mary’s clue it has to be Vicky or Wanda, not Scott. So Scott must be criminal.

08.B1 · Bruce CRIMINAL, A2 · Erwin CRIMINAL, B2 · Frank CRIMINAL

Paula's neighbors and Amy's neighbors must contain the same number of criminals. Right now Paula's neighbors already contain 2 known criminals, while Amy's neighbors contain 0 known criminals. If Bruce, Erwin, and Frank were all innocent, then Amy's entire neighbor group would have no criminals at all. The only other people involved here are Vicky and Wanda, so those remaining people would have to make the clues fit, but they cannot make Amy's neighbors match Paula's neighbors' criminal count. So Bruce, Erwin, and Frank must be criminal.

09.D4 · Terry CRIMINAL

Bruce’s clue says Paula’s neighbors and Zoe’s neighbors must contain the same number of criminal people. Right now each of those groups already has 2 known criminals. If Terry were innocent, then Zoe’s neighbor group would stay at 2 criminals. But Paula still has Vicky and Wanda not yet identified, and Mary’s clue also has to be satisfied by those same remaining people. That combination cannot be met if Terry is innocent. So Terry must be criminal.

10.D2 · Hank CRIMINAL

Igor’s clue says there are exactly 10 criminals on the edge, and exactly 3 of those edge criminals are Rose’s neighbors, so exactly 7 edge criminals are not Rose’s neighbors. In the edge cells that are not neighboring Rose, there are already 4 known criminals, which means among A1 Amy, C1 Cheryl, D1 Debra, and D2 Hank there is exactly 1 innocent. That one innocent must come from Amy, Cheryl, and Debra, so Hank cannot be that innocent person. So Hank must be criminal.

11.C2 · Gary CRIMINAL

Olof’s neighbors must contain more criminals than Erwin’s neighbors. Right now that count is 4 around Olof and 3 around Erwin, and the only unknown in those two groups is Gary around Olof and Amy around Erwin. At the same time, Bruce and Wanda must have the same number of criminal neighbors. But Bruce currently has 2 known criminal neighbors while Wanda already has 4, with only Amy, Cheryl, and Gary affecting Bruce’s side and only Vicky affecting Wanda’s side. If Gary were innocent, Amy, Cheryl, and Vicky would have to make both clues true at once, and that cannot be done. So Gary cannot be innocent. That makes Gary at C2 criminal.

12.A1 · Amy CRIMINAL

Bruce’s neighbors must have the same number of criminal neighbors as Wanda’s neighbors. Right now Bruce’s neighbors have 3 known criminals, while Wanda’s neighbors already have 4 known criminals. Column A must contain more criminals than innocents, so with Erwin and Paula already criminal and Igor innocent, column A needs at least one of Amy or Vicky to be criminal. If Amy were innocent, then Cheryl and Vicky would be the only remaining people who could fix both clues at once, but that cannot be done. So Amy cannot be innocent. So Amy must be criminal.

13.C1 · Cheryl CRIMINAL, D1 · Debra INNOCENT

Amy’s clue says all criminals in row 1 must be connected. In row 1, Amy and Bruce are already criminals, while Cheryl and Debra are the two unknowns. If Cheryl were innocent and Debra were criminal, then the criminals in row 1 would be Amy, Bruce, and Debra. Cheryl would sit between Bruce and Debra, so those row 1 criminals would not form one connected block. That clashes with Amy’s clue. So Cheryl must be criminal and Debra must be innocent.

14.A5 · Vicky CRIMINAL

Zoe’s clue says Bruce and Wanda must have the same number of criminal neighbors. Bruce’s neighbors already give a total of 5 criminals, while Wanda’s neighbors currently give 4 criminals, with Vicky the only unknown among them. If Vicky were innocent, Wanda would stay at 4 criminal neighbors, which would not match Bruce’s 5. So Vicky must be criminal.

15.B5 · Wanda INNOCENT

Mary’s clue says that the two innocents among Paula’s neighbors are also neighbors of Rose. Among Paula’s neighbors, anyone outside Rose’s neighboring group is already identified as criminal, so the second innocent cannot be outside that shared group. In the shared group, Igor is already one innocent, and Paula’s neighbors still need one more innocent there; the only unknown person left in that shared group is Wanda. So Wanda must be innocent.

More answers

Full archive →