Clues by Sam Jun 14, 2026 Answer – Full Solution Explained
A1
👩🌾
farmer
B1
👩🎤
singer
C1
👨🔧
mech
D1
😬
dog
A2
👩🌾
farmer
B2
💂♂️
guard
C2
💂♀️
guard
D2
👨⚕️
doctor
A3
👩⚕️
doctor
B3
🕵️♂️
sleuth
C3
👩🎤
singer
D3
👮♂️
cop
A4
👩🎨
painter
B4
🕵️♂️
sleuth
C4
👨🏫
teacher
D4
👩🏫
teacher
A5
👩🔧
mech
B5
🕵️♂️
sleuth
C5
👨🎨
painter
D5
👩🏫
teacher
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 — 9 criminals.
Clues by Sam answer for Jun 14, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 9 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Amy (A1), Eric (D1), Frida (A2), Hope (C2), Ike (D2), Joy (A3), Vicky (A5), Xavi (C5) and Zoe (D5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.A1 · Amy → CRIMINAL, A3 · Joy → CRIMINAL
Uma’s clue says column A has exactly 4 criminals, and exactly 2 of those criminals are Frida’s neighbors. In column A, the only people who are neighbors of Frida are Amy and Joy. Since the two criminal neighbor spots in column A still have to be filled, and Amy and Joy are the only people who can fill them, both of them have to be criminals. So Amy and Joy must be criminal.
02.C4 · Salil → INNOCENT
Joy’s clue says that Salil is one of Martin’s 5 innocent neighbors. That directly places Salil among the innocent people in Martin’s neighborhood. So Salil must be innocent.
03.A2 · Frida → CRIMINAL
Uma’s clue says column A has exactly 4 criminals, so among the 5 people in column A there is exactly 1 innocent. Amy and Joy are already criminals, which leaves A2 Frida, A4 Quita, and A5 Vicky as the three candidates for that single innocent spot. But exactly 2 of the 4 criminals in column A are Frida’s neighbors, and those two are already Amy and Joy. So the other two criminals in column A must be the people in column A who are not Frida’s neighbors, namely Quita and Vicky. That means the one innocent in column A has to be among Quita and Vicky, not Frida. So Frida must be criminal.
04.C3 · Olivia → INNOCENT
Joy says Salil is one of Martin's 5 innocent neighbors, so among Martin's neighbors there must be exactly 5 innocents and therefore exactly 3 criminals. Frida, Joy, and Salil are already fixed there as two criminals and one innocent, leaving exactly 1 criminal among the five unknown neighbors B2 Ghani, C2 Hope, C3 Olivia, A4 Quita, and B4 Rohan. Amy says only one criminal in row 3 has an innocent directly below. In the overlap of Martin's neighbors with row 3, Joy is already the criminal, so the remaining one criminal among Martin's unknown neighbors cannot be C3 Olivia. That makes Olivia innocent.
05.B4 · Rohan → INNOCENT
Joy says Martin has exactly 5 innocent neighbors. Among Martin's neighbors, Frida and Joy are already criminal, and Olivia and Salil are already innocent, so among the four unknown neighbors there can be only 1 criminal. Those four unknown neighbors are Ghani, Hope, Quita, and Rohan, but that one criminal must come from Ghani, Hope, or Quita, not from Rohan. So Rohan must be innocent.
06.B3 · Martin → INNOCENT
Joy’s clue says Martin has exactly 5 innocent neighbors. Around Martin, Frida and Joy are already criminals, while Olivia, Rohan, and Salil are already innocent, so the only way to reach 5 innocent neighbors is through the remaining unknown neighbors Ghani, Hope, and Quita. Now test Martin as criminal. Amy’s clue says exactly one criminal in row 3 has an innocent directly below, and Frida’s clue says all innocents in column B must form one connected block with Rohan in that column. With Martin set to criminal, the remaining people involved here, Bunty, Ghani, Hope, Paul, Quita, and Wally, cannot satisfy those requirements at the same time. So Martin must be innocent.
07.B5 · Wally → INNOCENT
Martin and Rohan are already the two known innocent sleuths, and Salil's clue says only one innocent sleuth has an innocent directly to the left. Martin already fits that description, so Wally cannot also be an innocent sleuth with an innocent directly to the left. Uma's clue fixes column A at exactly four criminals, with exactly two of those criminals being Frida's neighbors, and those two are already Amy and Joy. Martin's clue also says there is at least one innocent to the left of Xavi, but among the people to Xavi's left there is not yet any known innocent. If Wally were criminal, the remaining people from these clues, Quita and Vicky, would have to satisfy all of that at once, and they cannot. So Wally cannot be criminal. That makes Wally innocent.
08.B2 · Ghani → INNOCENT
Martin has exactly 5 innocent neighbors. Among his neighbors, Frida and Joy are already criminals, and Olivia, Rohan, and Salil are already innocents, so the only neighbors still undecided are Ghani, Hope, and Quita. That means there is room for exactly one more criminal among those three undecided neighbors. But Wally says column B has more innocents than column C, and column B already has 3 known innocents while column C has 2, so the one criminal among Martin's undecided neighbors must be Hope or Quita, not Ghani. So Ghani must be innocent.
09.C1 · Daniel → INNOCENT
Ghani’s clue says that Daniel is one of the 6 innocents on the edge. Since Daniel is named directly as part of that innocent edge group, his identity is fixed by the clue itself. So Daniel must be innocent.
10.C5 · Xavi → CRIMINAL, D5 · Zoe → CRIMINAL
Ghani says there are exactly 6 innocents on the edge, and 3 edge innocents are already known: Daniel, Uma, and Wally. That means the other 3 edge innocents have to come from the remaining edge people other than C5 Xavi and D5 Zoe, namely Bunty, Eric, Ike, Paul, Quita, and Vicky. So Xavi and Zoe cannot be among those 3 edge innocents. That makes Xavi and Zoe criminal.
11.B1 · Bunty → INNOCENT
Ghani’s clue says there are exactly 6 innocents on the edge. On the edge, 5 people are already known criminals and 3 are already known innocents, so among the 6 unknown edge people, exactly 3 must be criminals. Daniel’s clue says exactly 1 person above Uma is innocent, so the three people above Uma on the edge, Eric, Ike, and Paul, account for 2 of those 3 edge criminals. That leaves just 1 more edge criminal to come from Quita and Vicky, so Bunty cannot be one of the 3 edge criminals. So Bunty must be innocent.
12.D1 · Eric → CRIMINAL
Martin’s clue fixes his neighborhood at exactly 5 innocents, and among Martin’s neighbors the only unknown people are Hope and Quita. Amy’s clue also has to be satisfied in row 3, where Paul is the only unknown person alongside Joy, Martin, and Olivia. Bunty’s clue says Ike’s neighbors must contain an odd number of innocents, and right now Ike already has 2 known innocent neighbors, with Eric, Hope, and Paul the unknown neighbors there. If Eric were innocent, then Eric, Hope, Paul, and Quita would have to satisfy all three of those facts at once, but they cannot. So Eric cannot be innocent. That makes Eric criminal.
13.A5 · Vicky → CRIMINAL
Rohan’s clue says row 1 has more innocents than row 5. Row 1 already has 2 innocents, and row 5 currently has 1 innocent. If Vicky were innocent, then row 5 would also have 2 innocents, which would not be fewer than row 1. That clashes with the clue that row 1 has more innocents than row 5. So Vicky must be criminal.
14.A4 · Quita → INNOCENT
Salil’s clue says exactly one sleuth has an innocent directly to the left. That clue still needs one more such case, and the only remaining direct-left neighbor that can affect it is A4, Quita. So Quita has to be innocent to supply that required case. That makes Quita innocent.
15.D3 · Paul → INNOCENT
Amy’s clue says exactly one criminal in row 3 has an innocent directly below them. In row 3, the only known criminal there already is Joy. If Paul were also a criminal, then row 3 would have another criminal to fit into that same clue, and that makes the clue impossible to satisfy. So Paul at D3 must be innocent.
16.D2 · Ike → CRIMINAL
Daniel’s clue says the people above Uma contain exactly one innocent. That group already has its one known innocent, Paul. The only person there whose identity was still unknown is Ike, so Ike cannot also be innocent. So Ike must be criminal.
17.C2 · Hope → CRIMINAL
Ike's neighbors must contain an odd number of innocents. Among those neighbors, Daniel, Olivia, and Paul are already innocent, giving 3 innocents so far, which is already odd. That means the remaining unsettled neighbor cannot add 1 more innocent, because that would change the total to 4, which is even. So Hope cannot be innocent. That makes Hope criminal.