Clues by Sam Aug 20, 2026 Answer – Full Solution Explained
A1
🕵️♀️
sleuth
B1
👨💼
clerk
C1
👨⚕️
doctor
D1
👷♂️
builder
A2
🕵️♂️
sleuth
B2
👨🎤
singer
C2
👩⚖️
judge
D2
👩💼
clerk
A3
🕵️♂️
sleuth
B3
👨🎤
singer
C3
👩⚖️
judge
D3
👨⚖️
judge
A4
👩🎤
singer
B4
💂♀️
guard
C4
💂♀️
guard
D4
👩🌾
farmer
A5
👷♀️
builder
B5
💂♂️
guard
C5
👩🌾
farmer
D5
👨⚕️
doctor
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 Aug 20, 2026 — a Tricky solved in 16 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 16 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alice (A1), Ben (B1), Ethan (D1), Gary (B2), Habiba (C2), Jane (D2), Logan (B3), Nala (C3), Oscar (D3), Penny (A4), Rose (B4), Susan (C4), Tina (D4), Uma (A5), Vince (B5) and Wanda (C5); the remaining 4 suspects are innocent.
The deduction chain, in plain English
01.A2 · Frank → INNOCENT
Xavi’s clue says Frank is one of Gary’s 3 innocent neighbors. That directly identifies Frank as innocent. So Frank must be innocent.
02.A4 · Penny → CRIMINAL, A5 · Uma → CRIMINAL
Frank’s clue says there are exactly two innocents in column A, and those two innocents must be connected. In column A, the people involved are Alice, Frank, Kumar, Penny, and Uma, with Frank already known to be innocent. If Penny were innocent and Uma were innocent, then Alice and Kumar would also have to fit that same clue at the same time, and that cannot be done. So Penny and Uma cannot be innocent. So Penny and Uma must be criminal.
03.D2 · Jane → CRIMINAL
Uma’s clue says exactly one corner person has an innocent directly below them. That one case is already accounted for by Alice. The only other direct-below status that could still change this clue is Jane at D2, for the corner person above her. If Jane were innocent, there would be another corner with an innocent directly below, making more than one. So Jane must be criminal.
04.D1 · Ethan → CRIMINAL
Jane’s clue says the people above Oscar contain exactly 0 innocents. Above Oscar, there are already 0 known innocents, and the only person there whose identity is still unknown is Ethan. So Ethan cannot be innocent. So Ethan must be criminal.
05.B2 · Gary → CRIMINAL, B3 · Logan → CRIMINAL
Ethan’s clue says Habiba’s neighbors contain exactly 7 criminals, and exactly 2 of those criminals are below Ben. The only people who are both Habiba’s neighbors and below Ben are Gary and Logan. Since that below-Ben part still needs exactly 2 criminals, Gary and Logan have to fill both of those spots. So Gary and Logan must be criminal.
06.C4 · Susan → CRIMINAL
Penny’s clue says the people above Wanda must contain an odd number of innocents. Right now those four people are Daniel, Habiba, Nala, and Susan, with no known innocents among them. Logan’s clue also makes Daniel, Habiba, and Nala alone contain an odd number of innocents, because they are exactly the people in column C who neighbor Gary. If Susan were innocent, then Daniel, Habiba, and Nala would also have to make Penny’s group odd after Susan is added, while already making Logan’s group odd on their own, and that cannot fit both clues at once. So Susan cannot be innocent. So Susan must be criminal.
07.B1 · Ben → CRIMINAL
Gary has exactly three innocent neighbors, and Frank is already one of them, so among Gary's other neighbors there are exactly two more innocents. Frank's clue about column A and Penny's clue about the people above Wanda account for those two innocent spots among Alice, Daniel, Habiba, Kumar, and Nala. That leaves Ben outside the two innocent places in Gary's neighborhood. So Ben must be criminal.
08.C5 · Wanda → CRIMINAL
Ben’s clue says column A is the only column with exactly 2 innocents, and Xavi’s clue fixes one of Gary’s three innocent neighbors as Frank. The people these clues still depend on are Alice, Daniel, Habiba, Kumar, Nala, Oscar, Rose, Tina, Vince, and Wanda. If Wanda were innocent, those same people would have to fit both of those clue requirements at once, but that cannot be done. So Wanda cannot be innocent. That makes Wanda criminal.
09.C3 · Nala → CRIMINAL
Above Wanda, the only people not yet identified are Daniel, Habiba, and Nala, and Penny's clue says that group must contain an odd number of innocents. Wanda's clue also says exactly one innocent among the people above her is a neighbor of Ben, and that shared group is only Daniel and Habiba. If Nala were innocent, then the people above Wanda would already include one innocent outside the Daniel-Habiba pair. But Daniel and Habiba still have to supply exactly one innocent between them, which would make the total number of innocents above Wanda 2, an even number, not odd. So Nala must be criminal.
10.A3 · Kumar → INNOCENT
Gary’s clue says the people to the left of Oscar contain exactly one innocent. In that group, there are currently no known innocents, and the only person there whose identity is still unknown is Kumar. Since that group still needs one innocent, Kumar has to be the one. So Kumar must be innocent.
11.A1 · Alice → CRIMINAL
Frank’s clue says the two innocents in column A are connected. In column A, Frank at A2 and Kumar at A3 are already known to be innocent, while Penny at A4 and Uma at A5 are criminals. If Alice at A1 were also innocent, then column A would not have just those two innocents required by the clue. That conflicts with Frank’s statement about both innocents in column A. So Alice must be criminal.
12.B4 · Rose → CRIMINAL, B5 · Vince → CRIMINAL
Kumar’s clue says column B has more criminals than column C. Right now both columns already have 3 known criminals, so column B can be ahead only if one of its unknown people adds to that criminal count. If Rose and Vince were both innocent, then column B would stay at 3 criminals. The only other people involved here are Daniel and Habiba in column C, and they cannot make column B have more criminals than column C when column B has gained none. So Rose and Vince cannot both be innocent. That makes Rose and Vince criminal.
13.C1 · Daniel → INNOCENT
Rose’s clue says there is exactly one innocent in row 1 who has a criminal directly below them. In row 1, Alice, Ben, and Ethan are already criminals, so the only person there who could fill that “one innocent” spot is Daniel. If Daniel were also a criminal, then row 1 would contain no innocent at all, which clashes with the clue’s requirement of exactly one such innocent in that row. So Daniel at C1 must be innocent.
14.C2 · Habiba → CRIMINAL
Xavi’s clue says Frank is one of Gary’s exactly 3 innocent neighbors. Gary’s neighbors already include three identified innocents: Daniel, Frank, and Kumar. If Habiba were also innocent, Gary would have four innocent neighbors, which conflicts with the clue. So Habiba at C2 must be criminal.
15.D3 · Oscar → CRIMINAL
Habiba’s neighbors must contain exactly 7 criminals, and exactly 2 of those criminals are below Ben. Those 2 are already Gary and Logan, so the neighbors of Habiba who are not below Ben must contain exactly 5 criminals. Among Habiba’s neighbors who are not below Ben, Ben, Ethan, Jane, and Nala are already known criminals, while Daniel is innocent. That group therefore still needs 1 more criminal, and the only unknown person left there is Oscar. So Oscar must be criminal.
16.D4 · Tina → CRIMINAL
Ben’s clue says column A is the only column with exactly 2 innocents. Column D already has 1 known innocent, and Tina is the only person in that column not yet identified. If Tina were innocent, then column D would also have exactly 2 innocents, which would contradict Ben’s clue that only column A has that count. So Tina must be criminal.