Clues by Sam May 16, 2026 Answer – Full Solution Explained
A1
😬
robot
B1
👩⚖️
judge
C1
👨🍳
cook
D1
👮♂️
cop
A2
👩🍳
cook
B2
👩🍳
cook
C2
👨💻
coder
D2
🕵️♀️
sleuth
A3
👷♂️
builder
B3
👷♂️
builder
C3
👨🌾
farmer
D3
👮♀️
cop
A4
👩🌾
farmer
B4
👷♀️
builder
C4
💂♀️
guard
D4
👩🔧
mech
A5
👮♂️
cop
B5
👨🔧
mech
C5
💂♀️
guard
D5
👩🔧
mech
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 — 6 criminals.
Clues by Sam answer for May 16, 2026 — a Hard solved in 16 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alex (A1), Betsy (B1), David (D1), Luigi (C3), Olive (A4) and Ruth (D4); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.C5 · Vicky → INNOCENT, C1 · Carl → INNOCENT
Freya’s clue says exactly 2 corner people have an innocent directly to the left of them. There must be 2 more such corner cases, and the only direct-left neighbors still relevant to this clue are Carl at C1 and Vicky at C5. For those two corner cases to count, those left neighbors have to be innocent. So Carl and Vicky must be innocent.
02.A2 · Eve → INNOCENT
Vicky’s clue says that Eve is one of the exactly 3 innocents in column A. Since the clue directly includes Eve among the innocents in that column, Eve must be innocent. So Eve is innocent.
03.B1 · Betsy → CRIMINAL
Among Hendrik's neighbors who also neighbor Carl, the innocent total is exactly 2. In that group, D1 David and D2 Isi are the people who are above Ruth and neighbor Carl, and that smaller group contains exactly 1 innocent; Freya is already the other known innocent in the larger group. That means the smaller group plus Freya already accounts for all 2 innocents allowed in the larger group, so the only remaining person there, Betsy, cannot be innocent. So Betsy must be criminal.
04.A1 · Alex → CRIMINAL
Betsy’s clue says that Alex is one of the exactly 2 criminals among the people above Salil. Since Alex is explicitly included in that criminal pair, his identity is fixed by the clue itself. So Alex must be criminal.
05.A5 · Salil → INNOCENT
Betsy's clue says there are exactly 2 criminals above Salil, and Alex is one of them. Vicky's clue says there are exactly 3 innocents in column A, so column A contains exactly 2 criminals in total. The people above Salil are all in column A, and the only extra person in column A beyond that smaller group is Salil. Since the smaller group already accounts for all 2 criminals allowed in the whole column, Salil cannot be a criminal. So Salil must be innocent.
06.D3 · Nala → INNOCENT
Salil’s clue says there are exactly 5 criminals on the edge, and only 1 of those edge criminals is a neighbor of Vicky. That means the edge people who are not neighbors of Vicky contain exactly 4 edge criminals in total. Among those non-neighbors on the edge, Alex and Betsy are already known criminals. So the remaining 2 criminals in that non-neighbor group must come from David, Isi, Jerry, and Olive. That leaves Nala outside the two criminal spots in that group, so Nala must be innocent.
07.D4 · Ruth → CRIMINAL
Carl’s clue makes the shared pair D1 David and D2 Isi contain exactly 1 innocent. Nala’s clue says the people above Wanda, namely D1 David, D2 Isi, D3 Nala, and D4 Ruth, contain exactly 2 innocents in total. Nala is already one innocent in that larger group, and David plus Isi contribute exactly one more, so the full total of 2 innocents is already used up by D1, D2, and D3. The only remaining person in that group is Ruth, so Ruth cannot be innocent. So Ruth must be criminal.
08.B5 · Terry → INNOCENT, D5 · Wanda → INNOCENT
Salil's clue says that among the 5 criminals on the edge, exactly 1 is a neighbor of Vicky. The edge people who are neighbors of Vicky are D4 Ruth, B5 Terry, and D5 Wanda. Ruth is already known to be a criminal, so that one allowed criminal neighbor of Vicky is already accounted for. That means the other two edge neighbors of Vicky cannot be criminals. So Terry and Wanda must be innocent.
09.B4 · Penny → INNOCENT
Ruth says Kumar has exactly 2 criminal neighbors, and only 1 of those 2 is in row 4. Since Ruth herself is a criminal but is not a neighbor of Kumar, the single criminal in Kumar's row 4 neighbors has to be among A4 Olive, B4 Penny, and C4 Quita. From the step's stated restriction on that row 4 group, that one criminal comes from Olive or Quita, not Penny. So Penny cannot be the criminal in that group. That makes Penny innocent.
10.D2 · Isi → INNOCENT
Penny’s clue says Nala has exactly 3 innocent neighbors. Among Nala’s neighbors, Ruth is already a criminal, so the remaining four people there contain exactly 1 criminal: Hendrik, Isi, Luigi, and Quita. Wanda’s edge clue accounts for the edge criminal needed in this overlap without using Isi, so that 1 criminal among those four must come from Hendrik, Luigi, or Quita. That means Isi cannot be the criminal in Nala’s neighbor group. So Isi must be innocent.
11.D1 · David → CRIMINAL
Carl’s clue says there is exactly 1 innocent in the overlap between the people above Ruth and Carl’s neighbors. That shared group is just David and Isi. Isi is already known to be innocent, so the one innocent allowed in that shared group is already accounted for. David therefore cannot be innocent, so David must be criminal.
12.C2 · Hendrik → INNOCENT
Betsy's clue says there are exactly 2 criminals above Salil, and Alex is one of them. Above Salil, the only people not yet identified are Jerry and Olive, so those two spots have to account for the rest of that clue. David's clue also fixes the edge-cell count: exactly 2 edge people have a criminal directly below them. If Hendrik were criminal, then Jerry and Olive would have to fit these clue requirements at the same time, but they cannot do that without conflicting with the counts those two clues require. So Hendrik must be innocent.
13.A4 · Olive → CRIMINAL, C4 · Quita → INNOCENT
Kumar’s clue says his neighbors contain exactly 2 criminals, with exactly 1 of those in row 4. Hendrik’s clue fixes another exact count elsewhere, and the column A clue says exactly 1 criminal in column A has an innocent directly to the right; in column A, Alex is already a known criminal. Now test the opposite pairing for the targets: Olive innocent and Quita criminal. Then the remaining people involved here, Jerry, Kumar, and Luigi, would have to make Kumar’s count, Hendrik’s count, and the column A condition all true at the same time, but they cannot. So that opposite pairing is impossible. That makes Olive criminal and Quita innocent.
14.A3 · Jerry → INNOCENT
Betsy's clue says Alex is one of exactly 2 criminals above Salil. Above Salil, the people are Alex, Eve, Jerry, and Olive, and Alex and Olive are already the 2 criminals there. If Jerry were criminal too, that group would have 3 criminals, which clashes with the clue's exact total of 2. So Jerry must be innocent.
15.C3 · Luigi → CRIMINAL
Ruth's clue says Kumar's neighbors contain exactly 2 criminals, and exactly 1 of those criminals is in row 4. Among Kumar's neighbors in row 4, the only criminal already there is Olive, so the second criminal neighbor has to be outside row 4. Kumar's neighbors outside row 4 are Eve, Freya, Hendrik, Jerry, and Luigi, and Eve, Freya, Hendrik, and Jerry are all innocent. That leaves Luigi as the only person who can fill that remaining criminal spot, so Luigi must be criminal.
16.B3 · Kumar → INNOCENT
Hendrik has exactly 3 criminal neighbors in total, and exactly 2 of those criminals are also neighbors of Carl. In the group of Hendrik's neighbors who are not neighbors of Carl, the people are Carl, Kumar, Luigi, and Nala, and that group already contains 1 known criminal: Luigi. That fills the one criminal allowed in that non-neighboring group, so Kumar cannot be criminal. So Kumar must be innocent.