Clues by Sam Sep 06, 2026 Answer – Full Solution Explained
A1
💂♀️
guard
B1
👮♂️
cop
C1
👩🏫
teacher
D1
👨🏫
teacher
A2
🕵️♀️
sleuth
B2
💂♂️
guard
C2
👨🎨
painter
D2
👨🏫
teacher
A3
👮♂️
cop
B3
👨🎤
singer
C3
👩💻
coder
D3
👩🌾
farmer
A4
👩🌾
farmer
B4
👩💻
coder
C4
👩💻
coder
D4
👨🎤
singer
A5
🕵️♂️
sleuth
B5
🕵️♂️
sleuth
C5
👩🌾
farmer
D5
👩🌾
farmer
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 — 4 criminals.
Clues by Sam answer for Sep 06, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Freya (A2), Keith (A3), Martin (B3) and Vera (C5); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.C3 · Nancy → INNOCENT
Igor’s clue says Nancy is one of the exactly 2 innocents in row 3. That directly identifies Nancy as innocent. So Nancy must be innocent.
02.B3 · Martin → CRIMINAL
Nancy's clue says Rose has exactly 2 criminal neighbors, and exactly 1 of those criminals is above Quita. Among Rose's neighbors, the only person who is above Quita is Martin. So the one criminal neighbor of Rose who is above Quita has to be Martin. That makes Martin criminal.
03.D3 · Olga → INNOCENT, D4 · Salil → INNOCENT
Nancy says Rose has exactly two criminal neighbors, and exactly one of those criminals is above Quita. Among Rose's neighbors, Martin is already a criminal, and he is the only neighbor who is above Quita, so he fills that “above Quita” criminal spot by himself. That leaves exactly one more criminal neighbor of Rose, and it must come from the neighbors who are not above Quita: Quita, Umar, Vera, or Zara. Olga and Salil are not in that remaining group, so they cannot be that criminal neighbor. So Olga and Salil must be innocent.
04.A3 · Keith → CRIMINAL
Igor’s clue says Nancy is one of exactly 2 innocents in row 3. In row 3, the people are Keith, Martin, Nancy, and Olga, and Nancy and Olga are already the 2 innocents there. That leaves no innocent place for Keith in that row. So Keith must be criminal.
05.A1 · Barb → INNOCENT
Salil’s clue says Barb is one of Chola’s 4 innocent neighbors. That directly identifies Barb as innocent. So Barb must be innocent.
06.C1 · Dana → INNOCENT
Chola’s neighbors must contain exactly 4 innocents, and Barb is already one of them. If Dana were criminal, then among Chola’s other three row 2 neighbors, Freya, Gary, and Henry, all three would have to be innocent to bring Chola’s neighbor total up to 4 innocents. But then row 2 would be Freya, Gary, Henry, and Igor all innocent, for a total of 4 innocents in row 2. Olga’s clue says row 2 has an odd number of innocents, and 4 is even, so that cannot happen. So Dana must be innocent.
07.A4 · Penny → INNOCENT
Dana’s clue says Penny is one of Umar’s 4 innocent neighbors. That directly includes Penny among the innocent people in Umar’s neighborhood, so Penny cannot be anything else. So Penny must be innocent.
08.B4 · Quita → INNOCENT
If Quita were criminal, then Rose's clue would make Rose have exactly 2 criminal neighbors, with exactly 1 of those criminals above Quita, and that one above Quita is Martin. Since Martin is already that one criminal above Quita, the other criminal neighbor of Rose cannot be above Quita. Umar must have exactly 4 innocent neighbors, and Penny is one of them. With Quita assumed criminal, that leaves Umar, Rose, Terry, and Vera to supply the other innocents there, so Umar is innocent, Vera is innocent, and Rose is innocent. Keith's clue says there are exactly 3 criminals on the edge, and exactly 2 of those edge criminals are Quita's neighbors. Keith is already one such edge criminal neighbor of Quita. With Umar innocent, Vera innocent, and Rose's clue leaving Zara innocent, the only edge neighbors of Quita left are Penny, Terry, Umar, and Vera, so Terry also has to be innocent. But then there is no second edge criminal neighbor of Quita to go with Keith, which breaks Keith's clue. So Quita at B4 must be innocent.
09.B1 · Chola → INNOCENT, D1 · Erwin → INNOCENT
Keith’s clue says there are exactly 3 criminals on the edge, and exactly 2 of them are Quita’s neighbors. Since Keith is already an edge criminal and already a neighbor of Quita, that leaves exactly 1 edge criminal who is not a neighbor of Quita. Among the edge people who are not Quita’s neighbors, the only possible place for that 1 criminal is from A2 Freya or D5 Zara, so B1 Chola and D1 Erwin cannot be that criminal. So Chola and Erwin must be innocent.
10.B5 · Umar → INNOCENT
Keith says there are exactly 3 criminals on the edge, and exactly 2 of those edge criminals are neighbors of Quita. Among edge cells that are neighbors of Quita, Keith is already one known criminal, so there is room for exactly 1 more criminal there, and that extra criminal has to come from A5 Terry, B5 Umar, or C5 Vera. From the remaining step here, that 1 criminal is accounted for by A5 Terry or C5 Vera, so Umar cannot be that criminal neighbor on the edge. That makes Umar innocent.
11.C4 · Rose → INNOCENT
Umar’s neighbors must contain exactly 1 criminal. Among those neighbors, the smaller set A4 Penny, A5 Terry, and C5 Vera already accounts for that 1 criminal. The only person in Umar’s neighbor group who is not in that smaller set is Rose, so Rose cannot be criminal. So Rose must be innocent.
12.B2 · Gary → INNOCENT
If Gary were criminal, then column B would already have 2 criminals: Gary and Martin. Erwin’s clue says column A has more criminals than column B, so column A would then need more than 2 criminals. Salil’s clue says Chola has exactly 4 innocent neighbors. Among Chola’s neighbors, Barb and Dana are already innocent, and with Gary criminal, the other two innocent neighbors would have to come from A2 Freya and C2 Henry, which makes Freya not one of those innocents. Then column A would need both of its unknown people, Freya and Terry, to be criminals to get above column B. That would leave column A with 2 known criminals against 2 in column B, which clashes with Erwin’s clue that column A has more criminals than column B. So Gary at B2 must be innocent.
13.C2 · Henry → INNOCENT
The edge clue fixes exactly 3 criminals on the edge, with exactly 2 of those edge criminals neighboring Quita. Rose’s clue also fixes exactly 2 criminals among Rose’s neighbors, and exactly 1 of those is above Quita, which is already Martin. If Henry were criminal, then Freya, Terry, Vera, and Zara would still have to satisfy all of those counts at the same time, while Gary’s clue also says exactly 1 edge innocent has a criminal directly to the left. Those remaining edge positions cannot meet all of those requirements together if Henry is criminal. So Henry must be innocent.
14.A2 · Freya → CRIMINAL
Salil's clue says Barb is one of Chola's exactly 4 innocent neighbors. Chola's neighbors are Barb, Dana, Freya, Gary, and Henry, and among them Barb, Dana, Gary, and Henry are already innocent. That already fills all 4 innocent-neighbor spots, so Freya cannot be innocent. So Freya must be criminal.
15.D5 · Zara → INNOCENT
Keith's clue says there are exactly 3 criminals on the edge, and exactly 2 of those edge criminals are neighbors of Quita. Among the edge people who are not neighbors of Quita, Freya is already a known criminal. That means the one edge criminal who is not a neighbor of Quita is already accounted for, so no other unknown person in that non-neighbor edge group can be a criminal. Zara is in that group, so Zara must be innocent.
16.C5 · Vera → CRIMINAL
Nancy’s clue says Rose has exactly 2 criminal neighbors, and exactly 1 of those criminal neighbors is above Quita. The only criminal neighbor of Rose who is above Quita is Martin, so the other criminal neighbor must be someone neighboring Rose who is not above Quita. Among Rose’s neighbors who are not above Quita, Nancy, Olga, Quita, Salil, Umar, and Zara are all innocent. Vera is the only person left there who could fill that remaining criminal spot. So Vera must be criminal.
17.A5 · Terry → INNOCENT
Barb's clue says column A has exactly 3 innocents. In column A, 2 people are already known to be innocent, and the only person there whose identity is still unknown is Terry. That means column A still needs exactly 1 more innocent, so Terry has to fill that last innocent spot. So Terry must be innocent.