Clues by Sam Aug 15, 2026 Answer – Full Solution Explained
A1
👩🔧
mech
B1
👩🌾
farmer
C1
👩🌾
farmer
D1
👩🌾
farmer
A2
👩💼
clerk
B2
👨💼
clerk
C2
👩🔧
mech
D2
👮♂️
cop
A3
👨🎤
singer
B3
👩🍳
cook
C3
👨🍳
cook
D3
👮♀️
cop
A4
👨✈️
pilot
B4
👨✈️
pilot
C4
👨🎨
painter
D4
👨🎨
painter
A5
👨🎤
singer
B5
🕵️♂️
sleuth
C5
🕵️♀️
sleuth
D5
🕵️♀️
sleuth
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 — 9 criminals.
Clues by Sam answer for Aug 15, 2026 — a Hard solved in 15 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 9 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alice (A1), Chloe (C1), Dana (D1), Eve (A2), Ivan (D2), Kirsty (B3), Paul (A4), Rob (B4) and Xena (D5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.B2 · Franco → INNOCENT
Betty’s clue says Franco is one of the exactly 2 innocents to the left of Ivan. Since Franco is explicitly included in that innocent group, Franco must be innocent. So Franco must be innocent.
02.D2 · Ivan → CRIMINAL
Franco’s clue says Hazel has exactly 4 criminal neighbors, and exactly 1 of those criminals is to the right of Eve. Among Hazel’s neighbors who are to the right of Eve, the only people are Franco and Ivan. Franco is innocent, so that group still needs 1 criminal, and Ivan is the only person left who can fill it. So Ivan must be criminal.
03.C1 · Chloe → CRIMINAL, D1 · Dana → CRIMINAL, A3 · Jose → INNOCENT
Hazel’s clue says her neighbors contain exactly 4 criminals, and exactly 1 of those criminal neighbors is to the right of Eve. Since Franco is innocent and Ivan is the one known criminal to the right of Eve, that leaves the rest of Hazel’s required criminal neighbors to be found among the other people involved in these clues. At the same time, Ivan’s clue says row 2 has more criminals than row 3, with row 2 currently at 1 known criminal and row 3 at 0. If Chloe were innocent, Dana were innocent, and Jose were criminal, then Eve, Hazel, Kirsty, Martin, and Olga would have to make all of those counts work at once, but they cannot. So Chloe and Dana cannot be innocent, and Jose cannot be criminal. Chloe, Dana, and Jose must be criminal, criminal, and innocent respectively.
04.A1 · Alice → CRIMINAL, D3 · Olga → INNOCENT
Hazel’s neighbors must contain exactly 4 criminals, and exactly 1 of those criminals is to the right of Eve. In that right-of-Eve group, Franco is innocent and Ivan is criminal, so Ivan is already the one criminal there. Now test the opposite statuses: Alice innocent and Olga criminal. Then the same remaining people involved in these clues, Eve, Hazel, Kirsty, and Martin, would have to satisfy Hazel’s total of 4 criminal neighbors and Franco’s total of 4 criminal neighbors at the same time, and that cannot be done. So the opposite pairing is impossible. That makes Alice criminal and Olga innocent.
05.A4 · Paul → CRIMINAL
Olga’s clue says Paul is one of Kirsty’s 3 criminal neighbors. That directly places Paul among the criminal neighbors counted around Kirsty. So Paul must be criminal.
06.C3 · Martin → INNOCENT
Olga says Paul is one of Kirsty’s 3 criminal neighbors. Among Kirsty’s neighbors, Paul is already one known criminal, so the other two criminals must be chosen from Eve, Hazel, Martin, Rob, and Steve. But Ivan’s row clue and Dana’s clue about the people below Betty leave those two criminal spots among Eve, Hazel, Rob, and Steve instead. That means Martin cannot be one of Kirsty’s extra criminal neighbors. So Martin must be innocent.
07.B3 · Kirsty → CRIMINAL
Franco’s clue says Hazel’s neighbors contain exactly 4 criminals, and exactly 1 of those criminals is to the right of Eve. Among Hazel’s neighbors to the right of Eve, Franco is innocent and Ivan is criminal, so that 1 criminal on the right is already Ivan. That means the other 3 criminals among Hazel’s neighbors must be people who are not to the right of Eve. In that group, Chloe and Dana are already criminals, so one more is needed, and Kirsty is the only unknown person there. So Kirsty must be criminal.
08.C4 · Steve → INNOCENT
Below Betty there must be exactly two innocents, and Franco is already one of them, so among Rob and Vince there is exactly one innocent. Wanda’s neighbors contain exactly two criminals, with exactly one of those below Ivan, so among Wanda’s neighbors who are not below Ivan there are exactly two innocents: Rob, Steve, and Vince. Since Rob and Vince account for exactly one innocent between them, the remaining innocent in that three-person group has to be Steve. So Steve must be innocent.
09.B4 · Rob → CRIMINAL
Betty’s clue says the three people to the left of Ivan contain exactly 2 innocents: Eve, Franco, and Hazel. Olga’s clue means Kirsty’s neighbors contain exactly 3 criminals, so among those 8 neighbors there must be exactly 5 innocents. Jose, Martin, and Steve are already innocent there, and Franco is also innocent, so the only places left to supply the remaining innocent needed by Kirsty’s neighbors are Eve or Hazel. That uses up all 2 innocents allowed among Eve, Franco, and Hazel, leaving no room for Rob to be innocent. So Rob must be criminal.
10.B5 · Vince → INNOCENT
Dana’s clue says there are exactly 2 innocents below Betty. Below Betty, there is currently 1 known innocent, and the only person there whose status is still unknown is Vince. So the group still needs exactly 1 more innocent, and Vince is the only person who can fill it. That makes Vince innocent.
11.A5 · Umar → INNOCENT
Betty’s clue fixes that Franco is one of exactly two innocents to Ivan’s left, and Kirsty’s clue fixes Wanda’s neighbors as exactly two criminals with exactly one of those below Ivan. Steve’s clue also requires exactly four edge people to have a criminal directly below them. If Umar were criminal, then the remaining unknown people involved here, Eve, Hazel, Terry, Wanda, and Xena, would have to satisfy all of those exact requirements at the same time, and they cannot. That rules out Umar being criminal. So Umar must be innocent.
12.C5 · Wanda → INNOCENT
Wanda’s neighbors must contain exactly 2 criminals, and exactly 1 of those criminals is below Ivan. Steve’s neighbors must contain an odd number of criminals, and right now Steve’s neighborhood already has 2 known criminals, with only Terry, Wanda, and Xena still undecided there. If Wanda were criminal, then Terry and Xena would also have to fit both clues at the same time, because they are the other undecided people involved in Wanda’s neighborhood and Steve’s neighborhood. But making Wanda criminal makes those remaining requirements clash, so Wanda cannot be criminal. So Wanda must be innocent.
13.C2 · Hazel → INNOCENT
Row 3 is the key here. Kirsty already has exactly 4 known innocent neighbors, so she is the one person in row 3 who fits Wanda’s clue unless someone else can also reach exactly 4. If Hazel were criminal, Martin would stay at only 3 innocent neighbors, and Olga would stay at only 2, so neither of them could match Kirsty. At the same time, Eve, Terry, and Xena would still have to satisfy Chloe’s edge clue about exactly 2 edge innocents with an innocent directly below them, and those requirements cannot all be met with Hazel criminal. So Hazel cannot be criminal. That makes Hazel innocent.
14.A2 · Eve → CRIMINAL
Betty’s clue says Franco is one of exactly 2 innocents to the left of Ivan. The people there are Eve, Franco, and Hazel, and Franco and Hazel are already innocent. If Eve were also innocent, that group would contain 3 innocents, which clashes with the clue’s exact total of 2. So Eve must be criminal.
15.D4 · Terry → INNOCENT, D5 · Xena → CRIMINAL
Chloe’s clue says exactly 2 edge innocents have an innocent directly below them. Among the edge cells, the only undecided people are Terry and Xena. If Terry were criminal and Xena were innocent, then those remaining edge people would have to make Chloe’s exact count come out right, but they cannot do that. So that opposite pairing is impossible. That makes Terry innocent and Xena criminal.