Clues by Sam Jun 19, 2026 Answer – Full Solution Explained
A1
🕵️♀️
sleuth
B1
💂♂️
guard
C1
👨✈️
pilot
D1
👩🏫
teacher
A2
👨🎤
singer
B2
💂♀️
guard
C2
🕵️♂️
sleuth
D2
👩💻
coder
A3
👨🏫
teacher
B3
👨✈️
pilot
C3
💂♂️
guard
D3
👩🎤
singer
A4
👨✈️
pilot
B4
👩🎨
painter
C4
👨🎨
painter
D4
👩🎨
painter
A5
👷♂️
builder
B5
👷♀️
builder
C5
👷♂️
builder
D5
👩💻
coder
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 — 12 criminals.
Clues by Sam answer for Jun 19, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Bonnie (A1), Chuck (B1), Eric (C1), John (C2), Lucy (D2), Oscar (B3), Stella (B4), Terry (C4), Uma (D4), Vince (A5), Wanda (B5) and Zoe (D5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.D2 · Lucy → CRIMINAL
Habiba’s clue says the two innocents in row 2 are connected, and Habiba is already one of those innocents. That means the other innocent in row 2 has to be next to Habiba, so it can only be Graham or John. Lucy is not next to Habiba in that row, so Lucy cannot be the second innocent. So Lucy must be criminal.
02.A2 · Graham → INNOCENT, C2 · John → CRIMINAL
Lucy’s clue says the two criminals in row 2 must be connected. In row 2, Habiba is innocent and Lucy is criminal, so the only candidates for the second criminal are Graham and John. If Graham were criminal and John were innocent, then the two criminals in that row would be Graham and Lucy, with Habiba and John between them, so they would not be connected. That cannot fit Lucy’s clue. So Graham must be innocent and John must be criminal.
03.A1 · Bonnie → CRIMINAL, C1 · Eric → CRIMINAL
Graham's clue says row 1 has exactly 3 criminals, and exactly 1 of those criminals is Eric's neighbor. That means the other 2 criminals in row 1 must be people in row 1 who are not neighbors of Eric. The only people in row 1 who are not neighbors of Eric are Bonnie and Eric, and there are no known criminals among them yet. So Bonnie and Eric must fill those 2 criminal spots. Bonnie and Eric must be criminal.
04.D3 · Quita → INNOCENT
Row 1 has exactly 3 criminals, and Bonnie and Eric are already two of them, so exactly one of Chuck and Flora is a criminal. That means among John's neighbors on the edge, the row 1 part contributes exactly one criminal from Chuck and Flora, and with Eric and Lucy already criminal, the edge neighbors of John already account for exactly 3 criminals using Chuck or Flora, Eric, and Lucy. But John's edge neighbors are Chuck, Eric, Flora, Lucy, and Quita, and the clue says exactly 3 of John's 4 criminal neighbors are on the edge. Since those 3 edge criminals are already fully accounted for without Quita, the only remaining edge neighbor in that group cannot be criminal. So Quita must be innocent.
05.B1 · Chuck → CRIMINAL, D1 · Flora → INNOCENT
John’s neighborhood must contain exactly 4 criminals, and exactly 3 of those criminals must be on the edge. Also, Habiba and John’s common neighbors must contain exactly 1 innocent, and that common group is Chuck, Eric, Oscar, and Peter. If Chuck were innocent and Flora were criminal, then Chuck would already be the one innocent in that common-neighbor group, so Oscar and Peter could not be innocent there. But with Flora also criminal, John’s neighboring edge cells would then already supply Chuck, Eric, Flora, and Lucy for the edge part of Bonnie’s clue, which clashes with the required counts for these same people. So Chuck must be criminal and Flora must be innocent.
06.A3 · Michael → INNOCENT
John’s clue leaves exactly 1 criminal among the relevant unknown neighbors, namely Oscar and Peter. Stella’s clue says there is exactly 1 criminal among her row 3 neighbors, which are Michael, Oscar, and Peter. Since Oscar and Peter already account for the only criminal allowed in that row 3 group, the only person left there, Michael, cannot be criminal. So Michael must be innocent.
07.B5 · Wanda → CRIMINAL
Michael’s clue says that Wanda is one of the exactly 8 criminals on the edges. Since Wanda is named directly as one of those edge criminals, her identity is fixed by the clue itself. So Wanda must be criminal.
08.C5 · Xavi → INNOCENT
Wanda’s clue says that Xavi is one of Uma’s 3 innocent neighbors. Since the clue explicitly places Xavi among the innocent neighbors, his identity is fixed by the clue itself. So Xavi must be innocent.
09.D5 · Zoe → CRIMINAL
Uma’s clue says Xavi is one of Uma’s 3 innocent neighbors. In Uma’s neighbor list, Quita and Xavi are already innocent, so there is room for exactly 1 more innocent among the unknown neighbors C3 Peter, C4 Terry, and D5 Zoe. From there, that remaining innocent spot is accounted for by Peter or Terry, so Zoe cannot be the innocent in that group. That makes Zoe criminal.
10.A5 · Vince → CRIMINAL
Among Stella’s neighbors who are not in row 3, Wanda is already criminal and Xavi is already innocent, so the only places left for the one remaining innocent in that part are Ronald, Terry, and Vince. But that one innocent must come from Ronald or Terry. That means Vince cannot be the innocent person there. So Vince must be criminal.
11.B4 · Stella → CRIMINAL, D4 · Uma → CRIMINAL
Flora’s clue fixes Stella’s neighboring group at exactly 4 criminals, with exactly 1 of those criminals in row 3. Vince’s clue says there are 12 criminals on the whole board, and there are already 8 known criminals, so the six unknown people have to supply the remaining 4 criminals. If Stella and Uma were both innocent, then those 4 needed criminals would all have to come from Oscar, Peter, Ronald, and Terry. But those are exactly the other unknown people involved here, and that cannot fit Flora’s restriction on Stella’s neighbors at the same time. So Stella and Uma cannot both be innocent. That makes Stella and Uma criminal.
12.A4 · Ronald → INNOCENT
Michael’s clue says Wanda is one of exactly 8 criminals on the edge. In the edge cells, there are already 8 known criminals: Bonnie, Chuck, Eric, Lucy, Uma, Vince, Wanda, and Zoe. Ronald is the only edge person there whose identity was not yet fixed, so if Ronald were criminal, the edge would contain 9 criminals instead of 8. That makes Ronald innocent.
13.C4 · Terry → CRIMINAL
Flora’s clue says Stella has exactly 4 criminal neighbors, and exactly 1 of those criminals is in row 3. That means the neighbors of Stella who are not in row 3 must contain exactly 3 criminals. Those non-row-3 neighbors are Ronald, Terry, Vince, Wanda, and Xavi. Among them, Vince and Wanda are already known criminals, while Ronald and Xavi are innocent, so that group still needs 1 more criminal. Terry is the only unknown person left in that non-row-3 group, so Terry at C4 must be criminal.
14.C3 · Peter → INNOCENT
Wanda’s clue says Xavi is one of Uma’s exactly 3 innocent neighbors. Among Uma’s neighbors, Quita and Xavi are already innocent, Terry and Zoe are criminal, and Peter is the only neighbor not yet identified. If Peter were criminal, Uma would have only 2 innocent neighbors instead of 3, which clashes with the clue. So Peter must be innocent.
15.B3 · Oscar → CRIMINAL
Bonnie’s clue means John’s neighbors contain exactly 4 criminals, and exactly 3 of those criminals are on the edge. Among John’s edge neighbors, the 3 criminals are already Chuck, Eric, and Lucy, so the fourth criminal neighbor has to be someone not on the edge. John’s non-edge neighbors are Habiba, Oscar, and Peter, and Habiba and Peter are already innocent. That leaves Oscar as the only possible non-edge criminal neighbor, so Oscar must be criminal.