Clues by Sam Jul 10, 2026 Answer – Full Solution Explained
A1
👩🎨
painter
B1
👩⚖️
judge
C1
👨🎨
painter
D1
👩🎨
painter
A2
👮♂️
cop
B2
👨⚕️
doctor
C2
👮♀️
cop
D2
👩🍳
cook
A3
👨✈️
pilot
B3
👷♂️
builder
C3
👩🍳
cook
D3
👩🍳
cook
A4
👨🏫
teacher
B4
👩⚖️
judge
C4
👩⚕️
doctor
D4
👷♂️
builder
A5
👨🏫
teacher
B5
👨✈️
pilot
C5
👨🏫
teacher
D5
👷♀️
builder
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 — 11 criminals.
Clues by Sam answer for Jul 10, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 11 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Derek (C1), Evie (D1), Gabe (B2), Hilda (C2), Joyce (D2), Kumar (A3), Larry (B3), Nicole (C3), Tina (C4), Wally (B5) and Xavi (C5); the remaining 9 suspects are innocent.
The deduction chain, in plain English
01.B3 · Larry → CRIMINAL
Floyd’s clue says Kumar has exactly two criminal neighbors, and exactly one of those two is also a neighbor of Gabe. Among the people who are both Kumar’s neighbors and Gabe’s neighbors, Floyd is innocent, so that shared group still needs one criminal. The only unknown person left in that shared group is Larry. So Larry must be criminal.
02.A4 · Ronald → INNOCENT
Between Betsy and Wally, there is exactly 1 innocent among Gabe, Larry, and Samira. Kumar’s neighbors contain exactly 3 innocents among Floyd, Gabe, Larry, Ronald, and Samira. Those two groups overlap in Gabe, Larry, and Samira, so the only person counted among Kumar’s neighbors but not among the between-Betsy-and-Wally group is Ronald. Since Kumar’s neighbor group needs 2 more innocents than the between group has, that remaining person has to be one of those innocents. So Ronald must be innocent.
03.B2 · Gabe → CRIMINAL
Ronald's clue says Hilda has exactly 6 criminal neighbors, and exactly 1 of those criminal neighbors is between Floyd and Joyce. Among the people between Floyd and Joyce, the only one who is also a neighbor of Hilda is Gabe. So the one criminal neighbor of Hilda who is between Floyd and Joyce has to be Gabe. That makes Gabe criminal.
04.B4 · Samira → INNOCENT
Kumar’s neighbors contain exactly 2 criminals in total. The clue also says exactly 1 of Kumar’s criminal neighbors is also a neighbor of Gabe, and among the people who are both Kumar’s neighbors and Gabe’s neighbors, Larry is already that 1 criminal. That leaves the rest of Kumar’s neighbors who are not Gabe’s neighbors as Gabe, Ronald, and Samira, and that group already contains its 1 criminal in Gabe. So Samira must be innocent.
05.D1 · Evie → CRIMINAL
To Amy’s right, Betsy, Derek, and Evie must contain an odd number of innocents. Between Amy and Evie, Betsy and Derek must contain exactly one innocent. That means Betsy and Derek already contribute one innocent to the people on Amy’s right, and one is odd. If Evie were also innocent, the total to Amy’s right would become two innocents, which is even and breaks Gabe’s clue. So Evie must be criminal.
06.D2 · Joyce → CRIMINAL
Hilda’s neighbors have exactly 6 criminals, so among her 8 neighbors there are exactly 2 innocents. Those two innocents must be found among Betsy, Derek, Joyce, Nicole, and Olivia. Amy has an odd number of innocents to her right, and the only unknown people there are Betsy and Derek, since Evie is already a criminal. Kumar has exactly one innocent to his right, and the only unknown people there are Nicole and Olivia, since Larry is already a criminal. So the two innocents among Hilda’s neighbors must be Betsy and Derek, or Nicole and Olivia, not Joyce. That makes Joyce criminal.
07.A1 · Amy → INNOCENT, A3 · Kumar → CRIMINAL
Gabe’s clue says the people to Amy’s right contain an odd number of innocents, and among those three people Evie is already a criminal, so Betsy and Derek have to supply that odd innocent count. Joyce’s clue says row 1 has more innocents than row 3, and right now both rows have 0 known innocents. If Amy were a criminal and Kumar were an innocent, then Betsy, Derek, Nicole, and Olivia would have to satisfy both of those clue demands at the same time, but they cannot. That rules out Amy as criminal and Kumar as innocent. So Amy must be innocent and Kumar must be criminal.
08.C5 · Xavi → CRIMINAL, C1 · Derek → CRIMINAL
Amy’s clue says there are exactly 6 criminals on the edge, and exactly 2 of those edge criminals are in column C. Among edge cells in column C, the only people are Derek at C1 and Xavi at C5. Since that column-C edge group still needs 2 criminals and those are the only two people there, both of them have to fill those 2 spots. So Derek and Xavi must be criminal.
09.B1 · Betsy → INNOCENT
Gabe’s clue says the people to the right of Amy contain an odd number of innocents. In that group, there are currently 0 known innocents, and the only person there whose identity is still unknown is Betsy. If Betsy were criminal, that group would still have 0 innocents, which is not odd. So Betsy must be innocent.
10.D5 · Zara → INNOCENT
Amy’s clue says there are exactly 6 criminals on the edge. The edge already has 5 known criminals, so among D3 Olivia, D4 Umar, A5 Vince, B5 Wally, and D5 Zara, exactly 1 is criminal. The same clue also says exactly 2 edge criminals are in column C, and those 2 are already Derek at C1 and Xavi at C5. So the one remaining edge criminal cannot be in column C. Zara is at D5, not in column C, so the one remaining edge criminal must come from D3 Olivia, D4 Umar, A5 Vince, or B5 Wally instead. That makes Zara innocent.
11.C4 · Tina → CRIMINAL
Ronald’s clue fixes Hilda’s neighbors at exactly 6 criminals. Five of Hilda’s neighbors are already known criminals, so among the two shared neighbors Nicole and Olivia, exactly one is criminal. Derek’s clue says Umar has exactly 2 innocent neighbors, so among Umar’s five neighbors there must be exactly 3 criminals. Umar already has one known criminal neighbor, Xavi, and one known innocent neighbor, Zara. Since Nicole and Olivia contribute exactly one criminal between them, the only remaining neighbor who can supply the last needed criminal is Tina. So Tina must be criminal.
12.D4 · Umar → INNOCENT
Ronald’s clue fixes Hilda’s neighborhood at exactly 6 criminals, and says exactly one of those criminals is between Floyd and Joyce; that one is Gabe. So the remaining people tied up in these clues, including Hilda, Nicole, and Olivia, have to fit that exact count without creating any extra criminal in that between-Floyd-and-Joyce slot. Now look at the builders clue: exactly one builder has exactly 1 innocent neighbor. Larry already has 3 known innocent neighbors, so he cannot be that builder. Zara has 0 known innocent neighbors, with only Umar unknown next to her, so she is not the one fixed at exactly 1 here; that leaves Umar, who already has 1 known innocent neighbor. If Umar were criminal, Hilda, Nicole, and Olivia could not satisfy both Ronald’s count and Kumar’s builder condition at the same time. So Umar must be innocent.
13.C2 · Hilda → CRIMINAL
Ronald’s clue says Hilda has exactly 6 criminal neighbors, and exactly 1 of those criminals is between Floyd and Joyce. That one is already Gabe. Umar’s clue says Joyce has an odd number of innocent neighbors. Joyce currently has 0 known innocent neighbors, and the people there whose identities are still not fixed are Hilda, Nicole, and Olivia. If Hilda were innocent, then Hilda, Nicole, and Olivia would have to make both of those clues true at the same time, but they cannot. So Hilda cannot be innocent. Hilda must be criminal.
14.A5 · Vince → INNOCENT
Hilda’s clue says there are more criminal cooks than criminal teachers. Right now each profession has 1 known criminal: Joyce among the cooks, and Xavi among the teachers. If Vince were criminal, then the teachers would rise to 2 criminal teachers. The only cooks not yet identified are Nicole and Olivia, so those same remaining people would have to make the cooks exceed 2 criminal cooks, but that cannot be done while also fitting the other clue facts tied to these same unknown people. So Vince must be innocent.
15.B5 · Wally → CRIMINAL
Hilda’s neighbors must contain exactly 6 criminals, and the only criminal among them who is strictly between Floyd and Joyce is Gabe. Right now Hilda’s neighbors already have 5 known criminals, so among Nicole and Olivia exactly one of them must also be a criminal to make that total 6. The whole board must contain exactly 11 criminals. There are already 9 known criminals, and the only people not yet identified are Nicole, Olivia, and Wally, so exactly 2 of those 3 must be criminals. If Wally were innocent, then Nicole and Olivia would both have to be criminals to reach 11 criminals on the board, but that would make Hilda’s neighbors contain 7 criminals instead of 6. So Wally must be criminal.
16.D3 · Olivia → INNOCENT
Amy’s clue says there are exactly 6 criminals on the edge, and exactly 2 of those edge criminals are in column C. Those 2 are already Derek and Xavi, so the edge cells outside column C must contain the other 4 edge criminals. In that outside-column-C edge group, the 4 criminals are already Evie, Joyce, Kumar, and Wally. That leaves Olivia unable to be one of the edge criminals, so Olivia must be innocent.
17.C3 · Nicole → CRIMINAL
Ronald’s clue says Hilda’s neighbors contain exactly 6 criminals, and exactly 1 of those criminals is strictly between Floyd and Joyce. That one criminal is Gabe, so the other 5 criminal neighbors of Hilda must be people who are not strictly between Floyd and Joyce. Among Hilda’s neighbors who are not strictly between Floyd and Joyce, Derek, Evie, Joyce, and Larry are already known criminals, while Betsy and Olivia are innocent. That group still needs 1 more criminal, and the only unknown person left there is Nicole. So Nicole must be criminal.