Clues by Sam Sep 25, 2026 Answer – Full Solution Explained
A1
🕵️♂️
sleuth
B1
🕵️♀️
sleuth
C1
👨🎨
painter
D1
👩🎨
painter
A2
👨🏫
teacher
B2
😬
worm
C2
👩💻
coder
D2
👨🍳
cook
A3
👷♀️
builder
B3
😬
worm
C3
👩💻
coder
D3
👨🍳
cook
A4
💂♂️
guard
B4
👩🌾
farmer
C4
😬
worm
D4
👨🍳
cook
A5
💂♀️
guard
B5
💂♂️
guard
C5
😬
worm
D5
😬
worm
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 Sep 25, 2026 — a Hard solved in 16 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 Barb (B1), Dana (D1), Franco (A2), Gary (B2), Julie (A3), Nick (D3), Penny (B4), Sam (D4) and Uma (A5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.A2 · Franco → CRIMINAL
Olof’s clue says that the two criminals above Uma are connected. Above Uma, Olof is innocent, and if Franco were innocent too, then the only people left there who could be those two criminals would be Alex and Julie. But with Alex at A1 and Julie at A3, those two criminals would not be connected because innocent Franco would separate them. So Franco at A2 must be criminal.
02.C2 · Hazel → INNOCENT
Franco says the two innocents in row 2 are connected. If Hazel were criminal, then row 2 would be Franco criminal, Gary unknown, Hazel criminal, and Ivan unknown, so the only way to have exactly two innocents in that row is for Gary and Ivan to be the two innocents. But then the two innocents in row 2 would be Gary and Ivan, with Hazel between them, so they would not be connected. That contradiction means Hazel cannot be criminal. So Hazel must be innocent.
03.A3 · Julie → CRIMINAL
Hazel’s clue says there are no innocents among the people who are both between Alex and Uma and neighbors of Gary. That shared group is just Franco and Julie. Franco is already criminal, and there are already 0 known innocents in that group, so Julie cannot be innocent. So Julie must be criminal.
04.A1 · Alex → INNOCENT
Above Uma are Alex, Franco, Julie, and Olof. Olof’s clue says the two criminals in that group are connected, and Franco and Julie are already the two known criminals there. If Alex were also a criminal, there would be more than those two criminals above Uma, which conflicts with the clue. So Alex must be innocent.
05.B1 · Barb → CRIMINAL, D1 · Dana → CRIMINAL
Clyde’s clue says his neighbors contain exactly 3 criminals, and among those neighbors there are currently no known criminals, only Hazel as an innocent plus the four unknowns Barb, Dana, Gary, and Ivan. If Gary and Ivan were both criminals, then row 2 would be Franco criminal, Gary criminal, Hazel innocent, and Ivan criminal. But Franco’s clue says the two innocents in row 2 are connected, and that row would not allow that, so Gary and Ivan cannot both be criminals. That means Clyde’s three criminal neighbors cannot be made by using both Gary and Ivan. Since Hazel is already innocent and Clyde still needs exactly 3 criminal neighbors, Barb and Dana have to be criminals. So Barb and Dana must be criminal.
06.C1 · Clyde → INNOCENT
Hazel has exactly 4 innocent neighbors, and among her neighbors Barb and Dana are already known criminals. That means Hazel’s neighbor group has exactly 2 criminals total, so those 2 criminals must be chosen from Gary, Ivan, Karen, Mary, and Nick. Clyde is not one of the possible places for those 2 criminals, so Clyde cannot be a criminal. So Clyde must be innocent.
07.C3 · Mary → INNOCENT
Among Hazel’s neighbors, there must be exactly 4 criminals. In that same group, the four people who are also Mary’s neighbors are Gary, Ivan, Karen, and Nick, and exactly 2 of Mary’s 4 criminal neighbors are in that shared set. That means the criminals Hazel can have from Mary’s side are already accounted for within Gary, Ivan, Karen, and Nick. The only other person singled out in Hazel’s neighbor group beyond that shared set is Mary, so Mary cannot be a criminal without exceeding what that larger group allows. So Mary must be innocent.
08.B5 · Will → INNOCENT, D5 · Yara → INNOCENT
Dana’s clue says Ruth has exactly 3 criminal neighbors, and exactly 1 of those is in row 3. That means Ruth’s neighbors not in row 3 contain exactly 2 criminals, in the group B4 Penny, D4 Sam, B5 Will, C5 Xavi, and D5 Yara. Those 2 criminals must be Penny and Sam and Xavi’s group, so Will and Yara cannot be part of the 2 criminals outside row 3. Since Will and Yara are both in that outside-row-3 group, they cannot be criminals. So Will and Yara must be innocent.
09.B4 · Penny → CRIMINAL, D2 · Ivan → INNOCENT
Mary’s clue says her neighbors contain exactly 4 criminals, with exactly 2 of those criminals in the Hazel-sharing group B2 Gary, D2 Ivan, B3 Karen, and D3 Nick. Will’s clue says Mary and Nick have only 1 criminal neighbor in common, and that common-neighbor group is C2 Hazel, D2 Ivan, C4 Ruth, and D4 Sam, where Hazel is already innocent. Now test the opposite identities for the targets: Penny innocent and Ivan criminal. That would leave Gary, Karen, Nick, Ruth, and Sam having to satisfy all of those exact counts at the same time, but they cannot do it. So that opposite assignment is impossible. That makes Penny criminal and Ivan innocent.
10.B2 · Gary → CRIMINAL
Franco’s clue says the two innocents in row 2 are connected. In row 2, Franco is criminal, Hazel is innocent, Ivan is innocent, and Gary is the only person not yet identified. Since Hazel and Ivan are already the two innocents in that row, Gary cannot also be innocent. So Gary must be criminal.
11.C5 · Xavi → INNOCENT
Among Mary’s neighbors who are not neighbors of Hazel, there is exactly 1 criminal, and in this step that exact-count group is Ruth and Sam. Yara’s neighbors are Ruth, Sam, and Xavi, and Ivan’s clue says Yara must have an odd number of criminal neighbors. So the Ruth-and-Sam pair already supplies the odd criminal count for Yara’s neighborhood. Xavi is the only one in Yara’s neighborhood outside that exact-count pair, so Xavi cannot also be a criminal. That makes Xavi innocent.
12.D4 · Sam → CRIMINAL
Dana’s clue says Ruth’s neighbors contain exactly 3 criminals, and exactly 1 of those criminals is in row 3. So among Ruth’s neighbors, exactly 2 criminals must be not in row 3. The neighbors of Ruth who are not in row 3 are Penny, Sam, Will, Xavi, and Yara. Penny is already a known criminal, while Will, Xavi, and Yara are innocent, so that group still needs one more criminal. Sam is the only unknown person left in that not-in-row-3 group, so Sam at D4 must be criminal.
13.C4 · Ruth → INNOCENT
Barb's clue says there is an odd number of innocents among the people who are both in column C and neighboring Penny. That group is exactly Mary, Ruth, and Xavi. Mary and Xavi are already innocent, so the group currently has two innocents, which is even. To make the total odd, Ruth must also be innocent. So Ruth is innocent.
14.B3 · Karen → INNOCENT
Sam says there are exactly 2 innocents below Barb. Among the people below Barb, 1 is already known to be innocent, and the only person there whose identity is still unknown is Karen. That means the one remaining innocent below Barb has to be Karen. So Karen must be innocent.
15.D3 · Nick → CRIMINAL
Mary has exactly four criminal neighbors, and exactly two of those criminals also neighbor Hazel. Among the people who are both neighbors of Mary and neighbors of Hazel, Gary is already a criminal, while Ivan and Karen are innocent, leaving Nick as the only unknown person in that group. Since that group still needs one more criminal to make the required two, Nick must be criminal.
16.A5 · Uma → CRIMINAL
Nick’s clue says there are exactly 11 innocents in total. The board already has 11 people identified as innocent: Alex, Clyde, Hazel, Ivan, Karen, Mary, Olof, Will, Xavi, and Yara, plus one more already counted among the known innocents. That uses up the full total of 11 innocents, so the only person not yet identified cannot also be innocent. So Uma must be criminal.