Clues by Sam Sep 02, 2026 Answer – Full Solution Explained
A1
🕵️♂️
sleuth
B1
👨🔧
mech
C1
👩🔧
mech
D1
🕵️♀️
sleuth
A2
👨⚖️
judge
B2
👨⚖️
judge
C2
👷♂️
builder
D2
👷♀️
builder
A3
👩⚖️
judge
B3
👨🔧
mech
C3
👷♂️
builder
D3
🕵️♂️
sleuth
A4
👩🌾
farmer
B4
👩🌾
farmer
C4
👨💻
coder
D4
👨💻
coder
A5
👩🌾
farmer
B5
👩🍳
cook
C5
👩🍳
cook
D5
👩🍳
cook
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 02, 2026 — a Medium solved in 12 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Isaac (C2), Janet (D2), Olsi (C3) and Salil (C4); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.C3 · Olsi → CRIMINAL
Yara says that exactly one of Peter's neighboring criminals is in row 3. Among Peter's neighbors, the only person who is also in row 3 is Olsi. That means the one row-3 neighboring criminal has to be Olsi. So Olsi must be criminal.
02.C2 · Isaac → CRIMINAL
Olsi’s clue says row 2 contains exactly 2 criminals, and exactly 1 of those row 2 criminals is above Yara. In row 2, the only person who is above Yara is Isaac. So the one row 2 criminal who must be above Yara has to be Isaac. That makes Isaac criminal.
03.D3 · Peter → INNOCENT, C4 · Salil → CRIMINAL
Peter and Salil are tied together by these two clues. Peter’s neighbors must contain exactly 4 criminals, and exactly 1 of those criminals is in row 3, which is already Olsi. Also, the people between Eve and Zara must contain exactly 2 innocents, and right now those three people are Janet, Peter, and Tyler. If Peter were criminal and Salil were innocent, then Janet and Tyler would have to satisfy both clues at the same time, but that cannot be done. So that opposite assignment is impossible. Therefore Peter must be innocent and Salil must be criminal.
04.A2 · Frank → INNOCENT
To the right of Frank, there is exactly one innocent among Gus, Isaac, and Janet. In row 2, there are exactly two innocents among Frank, Gus, Isaac, and Janet. The only person who is in row 2 but not to the right of Frank is Frank himself. So if the right-of-Frank group accounts for one innocent, the remaining difference needed to reach two innocents in the whole row must be Frank. So Frank must be innocent.
05.D1 · Eve → INNOCENT
Peter has exactly 4 criminal neighbors, and exactly 1 of those criminal neighbors is in row 3. That one row-3 neighbor is already Olsi, so Janet and Tyler cannot both be criminals, because they are also Peter’s neighbors. Above Zara there must be an odd number of criminals, and the only people there not yet identified are Eve, Janet, and Tyler. If Eve were a criminal, Janet and Tyler would still have to fit both clues at once, but that cannot be done. So Eve must be innocent.
06.B1 · Bobby → INNOCENT
Row 2 has exactly 2 criminals, and exactly 1 of those row 2 criminals is above Yara; that one is Isaac. So the second row 2 criminal has to be either Gus or Janet, not another person. Dana’s neighbors must contain an odd number of innocents. Among Dana’s neighbors, Eve is already 1 known innocent, and the unknown neighbors are Bobby, Gus, and Janet. If Bobby were criminal, then Gus and Janet would have to fit both clues at once along with the fixed row 2 requirement, and that combination is impossible. So Bobby cannot be criminal. So Bobby must be innocent.
07.D5 · Zara → INNOCENT
If Zara were criminal, then Peter’s clue would fix both of Peter’s unknown neighboring spots. Peter has exactly 4 criminal neighbors in total, and exactly 1 of those criminals is in row 3; that one is already Olsi, while Isaac and Salil are the other two known criminal neighbors not in row 3. So Janet and Tyler cannot be criminals, which makes Janet innocent and Tyler innocent. Then column D would contain Eve, Peter, Janet, and Tyler as innocents, giving column D exactly 4 innocents even without Zara. But Bobby’s clue says column D is the only column with exactly 4 innocents, so Zara cannot be criminal here. That makes Zara innocent.
08.C1 · Dana → INNOCENT, B3 · Nick → INNOCENT
Isaac’s clue says his neighbors contain exactly 6 innocents, so among his neighbors there must be exactly 2 criminals. One of those is already Olsi, which means only one more of Isaac’s neighbors can be a criminal. From the row 2 clue, row 2 has exactly 2 criminals in total, and Isaac is already one of them, so among Gus and Janet there is exactly 1 criminal. That means the row 2 neighbors of Isaac, Gus and Janet, already account for the only remaining criminal spot among Isaac’s neighbors. So the other neighbor candidates outside that pair, Dana and Nick, cannot be criminals. That makes Dana and Nick innocent.
09.A3 · Kiran → INNOCENT
Olsi’s clue says row 2 has exactly 2 criminals, and exactly 1 of those row 2 criminals is above Yara. Isaac is already that one criminal in row 2 who is above Yara, so the second criminal in row 2 has to be someone else in that row who is not above Yara. Dana’s clue also says only one row has exactly 2 criminals, so once row 2 is the row with exactly 2 criminals, no other row can also have exactly 2. If Kiran were a criminal, the remaining unknown people named here could not satisfy both of those clue requirements at the same time. So Kiran at A3 must be innocent.
10.A1 · Austin → INNOCENT, A4 · Quita → INNOCENT, A5 · Wendy → INNOCENT
Kiran says column A has at least 4 innocents, and column A already has 2 known innocents: Frank and Kiran. The only other people in column A are Austin, Quita, and Wendy. If Austin, Quita, and Wendy were all criminals, then column A would stay at only 2 innocents, which cannot satisfy Kiran's clue. So Austin, Quita, and Wendy must be innocent.
11.B5 · Xena → INNOCENT
The edge cells must contain an odd number of innocents, and there are already 11 known innocents on the edge. So the unknown edge people together must add an even number of additional innocents. Among those edge unknowns, Janet and Tyler are in column D. Column D must contain exactly 4 innocents, and it already has 3 known innocents, so Janet and Tyler contribute exactly 1 more innocent. That leaves Xena as the only other unknown edge person, and she must supply one more innocent so the edge total stays odd. So Xena must be innocent.
12.B4 · Rose → INNOCENT, D4 · Tyler → INNOCENT, B2 · Gus → INNOCENT, D2 · Janet → CRIMINAL
Yara’s clue says her neighbors include at least 3 innocents, and right now those neighbors already have exactly 2 known innocents. If Rose and Tyler were both criminals, then Yara’s neighbors would not reach the required 3 innocents. At the same time, row 2 must contain exactly 2 criminals in total, with exactly 1 of them above Yara. Isaac is already a criminal in row 2 and is above Yara, so if Gus were also a criminal then Janet would have to be innocent to keep that row-2 clue satisfied. Those opposite assignments together are impossible because they break Yara’s neighbor clue. So Rose cannot be criminal, Tyler cannot be criminal, Gus cannot be criminal, and Janet cannot be innocent. That makes Rose innocent, Tyler innocent, Gus innocent, and Janet criminal.