Clues by Sam Jun 12, 2026 Answer – Full Solution Explained
A1
😬
dog
B1
👨💼
clerk
C1
👩💼
clerk
D1
😬
dog
A2
🕵️♂️
sleuth
B2
👩🌾
farmer
C2
👨💻
coder
D2
👩🎤
singer
A3
🕵️♀️
sleuth
B3
👨🌾
farmer
C3
👩💻
coder
D3
👩🎤
singer
A4
🕵️♂️
sleuth
B4
👩🌾
farmer
C4
👨🏫
teacher
D4
💂♀️
guard
A5
👨🏫
teacher
B5
👩💻
coder
C5
💂♂️
guard
D5
💂♂️
guard
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 — 16 criminals.
Clues by Sam answer for Jun 12, 2026 — a Hard solved in 15 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 16 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alex (A1), Clyde (B1), Diane (C1), Gus (A2), Hilda (B2), Isaac (C2), Joyce (D2), Kay (A3), Mary (C3), Nancy (D3), Pip (A4), Ruby (B4), Tina (D4), Umar (A5), Xavi (C5) and Zulu (D5); the remaining 4 suspects are innocent.
The deduction chain, in plain English
01.D4 · Tina → CRIMINAL
Vera's clue directly says that Tina is one of the three criminals in row 4. Since Tina is explicitly included among those criminals, Tina at D4 must be criminal.
02.A5 · Umar → CRIMINAL, C5 · Xavi → CRIMINAL
Vera’s clue says row 4 has exactly 3 criminals, and Tina is already one of them, so among Pip, Ruby, and Sam there must be exactly 2 criminals. Tina’s own clue says Vera has exactly 4 criminal neighbors, and those neighbors are Pip, Ruby, Sam, Umar, and Xavi. Since Pip, Ruby, and Sam can contribute only 2 of those 4 criminals, the other 2 must come from the only remaining neighbors, Umar and Xavi. So Umar and Xavi must be criminal.
03.A4 · Pip → CRIMINAL, B4 · Ruby → CRIMINAL
Umar’s clue says the people to the left of Sam contain exactly 0 innocents. That group already has 0 known innocents, and the only people there who are not yet identified are Pip and Ruby. So neither of those remaining people can be innocent. That makes Pip and Ruby criminals.
04.C4 · Sam → INNOCENT
Vera’s clue says Tina is one of exactly 3 criminals in row 4. In row 4, Pip, Ruby, and Tina are already the 3 known criminals, and Sam is the only person there not yet identified. If Sam were also a criminal, row 4 would have 4 criminals, which conflicts with the clue saying there are exactly 3. So Sam must be innocent.
05.C2 · Isaac → CRIMINAL
Diane’s neighbors contain exactly 4 criminals, and exactly 3 of those criminals are in the group that also neighbors Isaac: Clyde, Eve, Hilda, and Joyce. That means exactly 1 of Diane’s criminal neighbors must be someone who does not neighbor Isaac. In Diane’s neighbor list, the only person who does not neighbor Isaac is Isaac himself, and that group still needs that 1 criminal. So Isaac must be criminal.
06.C1 · Diane → CRIMINAL
Among Isaac's neighbors, Clyde, Eve, Hilda, and Joyce are the ones who also neighbor Diane, and exactly 3 of them are criminals. Among Isaac's neighbors that are not in row 3, the group is Clyde, Diane, Eve, Hilda, and Joyce, and exactly 4 of that group are criminals. So this larger group needs one more criminal than the smaller group has. The only extra person in the larger group is Diane, so Diane must be criminal.
07.C3 · Mary → CRIMINAL
Pip’s clue says column B is the only column with exactly 3 criminals. Column C already has 3 known criminals, and Mary is the only person in that column not yet identified. If Mary were innocent, then column C would remain at exactly 3 criminals, which the clue does not allow for any column other than B. So Mary must be criminal.
08.D5 · Zulu → CRIMINAL
If Zulu were innocent, the remaining unknown people named in these two clues would have to satisfy both of them at the same time. But the clues are very tight: column B has to be the only column with exactly 3 criminals, and Isaac’s neighbors have to contain exactly 6 criminals with exactly 2 of those in row 3 among Logan, Mary, and Nancy. With Alex, Clyde, Eve, Gus, Hilda, Joyce, Kay, Logan, and Nancy carrying all of those requirements, making Zulu innocent makes those facts clash. So Zulu must be criminal.
09.D2 · Joyce → CRIMINAL
Ruby’s clue makes Diane’s neighbors contain exactly 4 criminals, so among Diane’s 5 neighbors there is exactly 1 innocent. The only candidates for that one innocent are B1 Clyde, D1 Eve, B2 Hilda, and D2 Joyce. But the one innocent in that group has to come from Clyde, Eve, or Hilda, not Joyce. So Joyce cannot be the innocent person among Diane’s neighbors. That makes Joyce criminal.
10.A2 · Gus → CRIMINAL, A3 · Kay → CRIMINAL
Isaac’s clue says his neighbors contain exactly 6 criminals, and exactly 2 of those criminals are in row 3. In that row-3 part of Isaac’s neighborhood, the only people are Logan, Mary, and Nancy, with Mary already known to be a criminal, so that clue tightly fixes how many criminals can be in that group. Joyce’s clue also says exactly 3 people in row 3 are criminals who have a criminal directly above them, and row 3 consists of Kay, Logan, Mary, and Nancy, with only Mary already known as a criminal there. If Gus were innocent and Kay were innocent, then Clyde, Eve, Hilda, Logan, and Nancy would have to satisfy both exact-count clues at once, and they cannot. So Gus and Kay cannot be innocent. That makes Gus and Kay criminal.
11.B2 · Hilda → CRIMINAL
Isaac’s neighbors contain exactly 6 criminals, so among his neighbors not in row 3 there is room for exactly 1 innocent in the group B1 Clyde, D1 Eve, and B2 Hilda. That one innocent has to come from Clyde or Eve. So Hilda cannot be the innocent person in that group. That makes Hilda criminal.
12.A1 · Alex → CRIMINAL
Column B has to be the only column with exactly 3 criminals, and column A must have more criminals than column D. Right now column A already has 4 known criminals, while column D has 3, and the other people involved in these clues are Clyde, Logan, Eve, and Nancy. If Alex were innocent, those same remaining people would have to make both clues true at once, but they cannot. That rules out Alex being innocent. So Alex must be criminal.
13.D3 · Nancy → CRIMINAL, B3 · Logan → INNOCENT
Isaac’s neighbors contain exactly 6 criminals, and exactly 2 of those criminals are in row 3. In the row-3 part of Isaac’s neighborhood, the three people are Logan, Mary, and Nancy, and Mary is already one of those criminals. Now test the opposite assignment: Nancy innocent and Logan criminal. Diane’s clue also says that an odd number of Isaac’s edge-neighbors are innocent, and those edge-neighbors are Clyde, Diane, Eve, Joyce, and Nancy. With Diane and Joyce criminal, that would make Nancy one of the innocents in that edge-neighbor group, and Clyde and Eve would then have to meet that odd-innocent requirement at the same time as the row-3 criminal count, but they cannot do both. So Nancy must be criminal and Logan must be innocent.
14.B1 · Clyde → CRIMINAL
Pip says column B is the only column with exactly 3 criminals. In column B, Hilda and Ruby are already criminals, while Logan and Vera are innocents, so Clyde is the only person in that column who can supply the third criminal there. If Clyde were innocent, column B would stay at only 2 criminals, which conflicts with Pip’s clue. So Clyde must be criminal.
15.D1 · Eve → INNOCENT
Diane’s neighbors contain exactly 4 criminals in total. Among Diane’s neighbors who also neighbor Isaac, there are already 3 known criminals: Clyde, Hilda, and Joyce, and the only other person in that same group is Eve. Since Ruby’s clue says exactly 3 of Diane’s criminal neighbors also neighbor Isaac, Eve cannot be a criminal. So Eve must be innocent.