Clues by Sam Sep 19, 2026 Answer – Full Solution Explained
A1
🕵️♂️
sleuth
B1
👮♂️
cop
C1
👷♂️
builder
D1
💂♂️
guard
A2
👮♀️
cop
B2
🕵️♂️
sleuth
C2
💂♀️
guard
D2
👨🌾
farmer
A3
👮♀️
cop
B3
🕵️♂️
sleuth
C3
👷♀️
builder
D3
👨🌾
farmer
A4
👩💻
coder
B4
👩💻
coder
C4
👷♂️
builder
D4
👩🌾
farmer
A5
👩💻
coder
B5
👨🏫
teacher
C5
👩🏫
teacher
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 — 15 criminals.
Clues by Sam answer for Sep 19, 2026 — a Hard solved in 16 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 15 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Aaron (A1), Bruce (B1), Daniel (C1), Gus (B2), Igor (D2), Joyce (A3), Nick (D3), Paula (A4), Ruby (B4), Salil (C4), Uma (D4), Vicky (A5), Wally (B5), Xena (C5) and Zara (D5); the remaining 5 suspects are innocent.
The deduction chain, in plain English
01.B1 · Bruce → CRIMINAL, B2 · Gus → CRIMINAL
Kevin’s clue says there are no innocents among the people who are both in column B and neighbors of Aaron. That shared group is exactly Bruce and Gus, and there are already 0 innocents in it. So Bruce and Gus cannot be innocent. Therefore Bruce and Gus must be criminal.
02.A2 · Freya → INNOCENT, C2 · Helen → INNOCENT
Bruce’s clue says the people to the left of Igor contain exactly 2 innocents. In that group, there are currently 0 known innocents, and the only people there whose status is still unknown are Freya and Helen. So the 2 innocents required by the clue have to be those two people. That makes Freya and Helen innocent.
03.D2 · Igor → CRIMINAL
Helen’s clue says every row has at least 2 criminals, so in a 4-person row there can be at most 2 innocents. In row 2, Freya and Helen are already known innocents. That uses up the two innocent places allowed in that row, so the only person there not yet fixed, Igor, cannot be innocent. So Igor must be criminal.
04.C1 · Daniel → CRIMINAL
Igor’s clue says that exactly one of his two criminal neighbors is also a neighbor of Bruce. Among the people who are neighbors of both Igor and Bruce, the only ones are Daniel and Helen. Helen is innocent, so that shared pair still needs one criminal, and Daniel is the only possible person left to fill it. So Daniel must be criminal.
05.A3 · Joyce → CRIMINAL
Assume Joyce were innocent. Then the clue that each row has at least 2 criminals makes Maria and Nick criminals, and Freya’s clue says exactly 1 innocent is in the shared neighbor group of Igor and Salil. But that shared group is Maria and Nick, so if both Maria and Nick are criminals, there are 0 innocents there, not exactly 1. That contradiction means Joyce cannot be innocent. So Joyce must be criminal.
06.D1 · Ethan → INNOCENT
Igor’s neighbors are Daniel, Ethan, Helen, Maria, and Nick, and among the overlap of Igor’s neighbors with Salil’s neighbors, exactly one of Maria and Nick is innocent. Igor’s neighbor group must contain exactly three innocents in total. Helen is already one innocent, and the shared pair Maria and Nick contribute exactly one more, so that accounts for only two innocents among Igor’s neighbors. The only remaining person in Igor’s neighbor group outside that shared pair is Ethan, so Ethan has to be the third innocent. So Ethan must be innocent.
07.D3 · Nick → CRIMINAL
Column C and column D currently each have 1 known innocent. Helen’s clue says column C must end with more innocents than column D. If Nick were innocent, column D would go up to 2 innocents immediately, so column C would have to reach at least 3 innocents to stay ahead. At the same time, row 5 must contain at least 2 criminals, and Daniel’s clue says all criminals in row 5 have to be one connected block. With Maria, Salil, Uma, Vicky, Wally, Xena, and Zara also still needing to satisfy those clues, that combination cannot be made to work. So Nick must be criminal.
08.C3 · Maria → INNOCENT
Freya’s clue says there is exactly one innocent among the people who are neighbors of both Igor and Salil. That shared group is just Maria and Nick. Nick is already criminal, so the shared group still needs one innocent, and Maria is the only person there who could fill it. So Maria must be innocent.
09.D4 · Uma → CRIMINAL
Row 4 currently has no known criminals, and every row must have at least 2 criminals. At the same time, row 1 already has 1 innocent while row 4 has 0, so row 4 cannot simply gain innocents freely because row 1 must still end with more innocents than row 4. Column C also already has 2 innocents while column D has 1, so column D cannot freely add innocents either because column C must still finish with more innocents than column D. If Uma at D4 were innocent, then Paula, Ruby, Salil, Aaron, and the whole of row 5 would have to satisfy all of those requirements at once, including row 5 needing at least 2 criminals with its criminals connected, and that combination does not work. So Uma at D4 must be criminal.
10.A4 · Paula → CRIMINAL
Row 1 already has 1 known innocent, while row 4 has 0, and Maria says row 1 has more innocents than row 4. Uma also says columns A and B have the same number of criminals, but column A currently has 1 known criminal and column B already has 2. If Paula were innocent, the remaining people named in these clues would have to make both of those statements true at the same time, and that cannot be done. So Paula cannot be innocent. That makes Paula criminal.
11.B5 · Wally → CRIMINAL
Column B and column D have to contain the same number of criminals, but column B currently has 2 known criminals while column D already has 3. So column B needs at least one more criminal from its unknown people, Ruby or Wally. Now test Wally as innocent. Row 5 still has to contain at least 2 criminals, and all criminals in row 5 must be one connected block, but with Wally innocent the remaining people touched by these clues cannot satisfy those requirements while also keeping columns B and D equal in criminal count. So Wally at B5 must be criminal.
12.C4 · Salil → CRIMINAL, A5 · Vicky → CRIMINAL
Row 1 already has 1 known innocent, while row 4 has 0, so Maria's clue says row 4 cannot end up with as many innocents as row 1. Also, column A currently has 2 known criminals while column B already has 3, so Uma's clue means column A must gain a criminal and column B cannot gain another one. Now test the opposite statuses named here: Salil innocent and Vicky innocent. Then the remaining people involved in these clues are Aaron and Ruby, but they cannot make both clues true at the same time: row 4 would already have an innocent in Salil, and column A would still be short of the extra criminal it needs while column B could not afford another criminal. So those opposite statuses do not work. That makes Salil and Vicky criminals.
13.B4 · Ruby → CRIMINAL
If Ruby were innocent, then row 5 would have Vicky and Wally as criminals, Xena as innocent, and Zara as criminal. But Daniel’s clue says all criminals in row 5 must be connected in one orthogonal block, and Vicky, Wally, and Zara are not one connected block if Xena is innocent between Wally and Zara. That makes the assumption impossible, so Ruby cannot be innocent. So Ruby must be criminal.
14.A1 · Aaron → CRIMINAL
If Aaron were innocent, then row 5 would have Vicky and Wally as criminals, Xena as innocent, and Zara as criminal. But Daniel’s clue says all criminals in row 5 must be connected in one orthogonal block, and having criminals at A5, B5, and D5 with Xena innocent at C5 breaks that block. So that assumption cannot be right, and Aaron at A1 must be criminal.
15.D5 · Zara → CRIMINAL
Paula’s clue says columns B and D have the same number of criminals. Column B already has 4 criminals, while column D currently has 3. The only person in column D whose status is not yet known is Zara, so she has to supply the fourth criminal in that column. So Zara must be criminal.
16.C5 · Xena → CRIMINAL
Daniel’s clue says that all criminals in row 5 have to be connected as one orthogonally linked block. In row 5, Vicky, Wally, and Zara are already criminals, and Xena is the only person there not yet identified. If Xena were innocent, the criminals in that row would not form one connected block. So Xena must be criminal.