Clues by Sam Aug 09, 2026 Answer – Full Solution Explained
A1
👨🎨
painter
B1
👨🔧
mech
C1
👨🔧
mech
D1
💂♀️
guard
A2
👨🎨
painter
B2
👩🎨
painter
C2
👨🏫
teacher
D2
💂♂️
guard
A3
👩💼
clerk
B3
👨💼
clerk
C3
👩🏫
teacher
D3
💂♀️
guard
A4
👨🍳
cook
B4
👨💼
clerk
C4
👩🏫
teacher
D4
👨⚕️
doctor
A5
👩🍳
cook
B5
👩🍳
cook
C5
👨⚕️
doctor
D5
👩⚕️
doctor
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 Aug 09, 2026 — a Hard solved in 18 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 Denis (B1), Eric (C1), Ivan (C2), Luigi (B3), Peter (B4), Quita (C4), Scott (D4), Tina (A5) and Xola (D5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.B1 · Denis → CRIMINAL
Kiran’s clue says there are exactly 5 criminals on the edge, and exactly 1 of those edge criminals is above Luigi. Among edge cells above Luigi, the only person there is Denis. Since that group still needs its 1 criminal, Denis has to be that person. So Denis must be criminal.
02.C4 · Quita → CRIMINAL
Denis’s clue says row 4 has exactly 3 criminals, and exactly 1 of those row 4 criminals is below Eric. In row 4, the only person who is below Eric is Quita. So the one criminal in row 4 who is below Eric has to be Quita. That makes Quita criminal.
03.D4 · Scott → CRIMINAL
Quita’s clue says 2 of the 3 cooks have an innocent directly above them. The people directly above cooks are Ollie and Peter, so that gives exactly 1 innocent among Ollie and Peter. In row 4, the only possible innocent is also exactly 1, and the row 4 people are Ollie, Peter, Quita, and Scott. Since Ollie and Peter already account for that one innocent in row 4, the remaining row 4 person outside that pair cannot be innocent. So Scott must be criminal.
04.B5 · Uma → INNOCENT
Tina’s neighbors contain exactly 1 criminal among A4 Ollie, B4 Peter, and B5 Uma. The smaller group inside that set, Ollie and Peter, already accounts for that 1 criminal. That leaves the only person outside that smaller group, Uma, unable to be a criminal. So Uma must be innocent.
05.A2 · Ghani → INNOCENT, B2 · Helen → INNOCENT
Row 4 has exactly three criminals, and Quita and Scott are already two of them, so among A4 Ollie and B4 Peter there is exactly one criminal. Luigi's neighbors who are not below Eric must also contain exactly one criminal, and that group is A2 Ghani, B2 Helen, A3 Kiran, A4 Ollie, and B4 Peter. Since A4 Ollie and B4 Peter already account for the one criminal allowed in that larger group, the remaining people there cannot be criminals. So Ghani and Helen must be innocent.
06.D1 · Frida → INNOCENT
Helen’s clue says Frida is one of the exactly 2 innocents in the corner cells. Since Frida is explicitly included in that innocent corner group, her identity is fixed by the clue itself. So Frida must be innocent.
07.B3 · Luigi → CRIMINAL
Frida's clue says that Peter has exactly 3 criminal neighbors, and exactly 1 of those criminals is in column B. Among Peter's neighbors in column B, Uma is already innocent, so that column-B part of the clue still needs 1 criminal there. The only unknown neighbor of Peter who is also in column B is Luigi. So Luigi must be criminal.
08.D3 · Nicole → INNOCENT
Luigi’s clue says directly that Nicole is innocent. So Nicole must be innocent.
09.A4 · Ollie → INNOCENT
Peter’s neighbors must contain exactly 3 criminals. Among those neighbors, Luigi and Quita are already known criminals, so there is room for exactly 1 more criminal among Mary, Ollie, Tina, and Wally. That remaining criminal has to come from Mary, Tina, or Wally, not from Ollie. So Ollie cannot be a criminal. So Ollie must be innocent.
10.B4 · Peter → CRIMINAL
Denis's clue says row 4 has exactly 3 criminals, and exactly 1 of those row 4 criminals is below Eric. The only person in row 4 who is below Eric is Quita, and she is already a criminal, so the one criminal below Eric is already accounted for. That means the other 2 criminals in row 4 must be among the people in row 4 who are not below Eric: Ollie, Peter, and Scott. Ollie is innocent and Scott is already a criminal, so that group still needs 1 more criminal, and Peter is the only person there left who can fill it. So Peter must be criminal.
11.D2 · Jerry → INNOCENT
Kiran’s clue says there are exactly 5 criminals on the edges. Two edge criminals are already known there, Denis and Scott, so the other 3 edge criminals have to come from the remaining unknown edge people. Those remaining edge-criminal spots are accounted for by Ben, Eric, Tina, Wally, and Xola. That leaves Jerry outside that set, so Jerry cannot be one of the 3 needed edge criminals. So Jerry must be innocent.
12.A5 · Tina → CRIMINAL
The edge cells must contain exactly 5 criminals, and among the edge people there are already 7 known innocents. Since edge cells make up 14 people total, that leaves exactly 2 innocents among the five unknown edge people: Ben, Eric, Tina, Wally, and Xola. But the two edge innocents must come from Ben, Eric, Wally, and Xola, so Tina cannot be one of those two innocents. So Tina must be criminal.
13.C5 · Wally → INNOCENT
Peter’s neighbors must contain exactly 3 criminals, and exactly 1 of those criminal neighbors is in column B. Among Peter’s neighbors, the known criminals are Luigi, Quita, and Tina, and the only one of those in column B is Luigi, so that full requirement is already used up. If Wally were a criminal, Peter’s neighbors would have more than 3 criminals, which clashes with Frida’s clue. So Wally at C5 must be innocent.
14.C3 · Mary → INNOCENT
Frida’s clue says Peter’s neighbors contain exactly 3 criminals, and exactly 1 of those criminals is in column B. Among Peter’s neighbors in column B, Luigi is the one criminal there. Outside column B, Peter’s neighboring group already contains 2 known criminals, Quita and Tina, which fills the other two criminal spots among Peter’s neighbors. That leaves no room for Mary to be a criminal, so Mary must be innocent.
15.D5 · Xola → CRIMINAL
Kiran’s clue says there are exactly 5 criminals on the edge, and the edge already has 3 known criminals: Denis, Scott, and Tina. That means the remaining 2 edge criminals must come from the only unknown edge people, Ben, Eric, and Xola. The step here uses that the one edge innocent among those three must be Ben or Eric, not Xola. So Xola cannot be the innocent one in that group. Therefore Xola must be criminal.
16.C2 · Ivan → CRIMINAL
Uma’s clue says Luigi’s neighbors contain exactly 3 criminals, and exactly 2 of those criminals are below Eric. Among Luigi’s neighbors who are below Eric, we already have Quita as a criminal and Mary as innocent, so that group still needs 1 more criminal. The only unknown person left in that below-Eric group is Ivan. So Ivan must be criminal.
17.C1 · Eric → CRIMINAL
Kiran’s clue says there are exactly 5 criminals on the edge, and exactly 1 of those edge criminals is above Luigi. Denis is already that one edge criminal above Luigi, and there are already 3 known edge criminals not above Luigi: Scott, Tina, and Xola. Since there are 4 known edge criminals on the edge already, only one of the two unknown edge people, Ben or Eric, can be innocent. Nicole’s clue says there are more innocents in column A than in column C. Column A already has 3 innocents while column C has 2, with Ben the only unknown in column A and Eric the only unknown in column C. So the one innocent between Ben and Eric has to be Ben, not Eric. That makes Eric criminal.
18.A1 · Ben → INNOCENT
Helen’s clue says Frida is one of exactly 2 innocents in the corner cells. The corners are Ben, Frida, Tina, and Xola, and among them Tina and Xola are criminals while Frida is already the one known innocent. If Ben were criminal too, then there would be only 1 innocent in the corners, which conflicts with the clue requiring exactly 2. So Ben at A1 must be innocent.