Clues by Sam Sep 29, 2026 Answer – Full Solution Explained
A1
👩💼
clerk
B1
👨🌾
farmer
C1
👮♂️
cop
D1
💂♀️
guard
A2
👨💼
clerk
B2
👩💼
clerk
C2
👷♂️
builder
D2
👨🎤
singer
A3
🕵️♀️
sleuth
B3
💂♂️
guard
C3
👷♂️
builder
D3
👩🎤
singer
A4
👨🍳
cook
B4
👩🍳
cook
C4
👷♂️
builder
D4
👩🎤
singer
A5
👮♂️
cop
B5
🕵️♂️
sleuth
C5
🕵️♀️
sleuth
D5
👩🌾
farmer
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 29, 2026 — a Medium solved in 15 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 9 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Bobby (B1), Chad (C1), Hrishi (D2), Joyce (A3), Pam (D3), Raul (A4), Uma (D4), Wally (B5) and Xena (C5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.C5 · Xena → CRIMINAL
Vince’s clue says the two criminals in row 5 must be connected. In row 5, Vince is innocent, and if Xena were innocent, then the two criminals would have to come from Wally and Zara. That would make Wally and Zara the two criminals in row 5, but they are not connected because Xena would be between them. So Xena at C5 must be criminal.
02.C3 · Larry → INNOCENT, C4 · Tyler → INNOCENT, C2 · Gary → INNOCENT
Xena’s clue says there are exactly 3 innocents below Chad. Below Chad there are currently no known innocents, and the only people there whose status is still unknown are Gary, Larry, and Tyler. That means those three spots must supply all 3 innocents required by the clue. So Gary, Larry, and Tyler must be innocent.
03.D4 · Uma → CRIMINAL, D2 · Hrishi → CRIMINAL
Gary’s clue says there are exactly 2 innocent people in the group shared by Larry’s neighbors and Pam’s neighbors. That shared group is Gary, Hrishi, Tyler, and Uma. Gary and Tyler are already known to be innocent, so the clue’s total of 2 innocents is already filled. That means Hrishi and Uma cannot be innocent. So Uma and Hrishi must be criminal.
04.D5 · Zara → INNOCENT
Uma’s clue says there is exactly one innocent to the right of Wally. To the right of Wally, there are currently no known innocents, and the only person there whose identity is still unknown is Zara. Since that group still needs exactly one innocent, Zara has to be that innocent. So Zara must be innocent.
05.B5 · Wally → CRIMINAL
Vince’s clue says the two criminals in row 5 are connected. In row 5, Vince is innocent, Xena is criminal, Zara is innocent, and Wally is the only person not yet identified. Since row 5 must contain exactly two criminals, Wally has to be the second one, and that also puts the two criminals together in that row. So Wally must be criminal.
06.A3 · Joyce → CRIMINAL
Row 3 has exactly 2 criminals. Of those 2, exactly 1 is Larry's neighbor, and the row 3 neighbors of Larry are Kevin and Pam, so exactly 1 criminal in row 3 must be someone who is not Larry's neighbor. The people in row 3 who are not Larry's neighbors are Joyce and Larry. Larry is already innocent, so that non-neighbor criminal spot cannot be Larry. So Joyce at A3 must be criminal.
07.C1 · Chad → CRIMINAL
Joyce’s clue says every column has at least 2 criminals. In column C, that means there can be at most 3 innocents total. But column C already has 3 known innocents: Gary, Larry, and Tyler. So the only person left in column C, Chad, cannot be innocent. That makes Chad criminal.
08.A2 · Eli → INNOCENT, B2 · Flora → INNOCENT
Chad’s clue says row 2 has more innocents than row 5. Row 5 already has 2 innocents, while row 2 currently has only 1 known innocent, Gary, and the only unknown people left in row 2 are Eli and Flora. So row 2 can get above 2 innocents only if both of those unknowns are innocents. That makes Eli and Flora innocent.
09.B4 · Samin → INNOCENT
Row 3 has exactly 2 criminals, and exactly 1 of those row 3 criminals is Larry's neighbor. Since Joyce is already a criminal in row 3 and is not Larry's neighbor there, Kevin and Pam have to account for the row 3 criminal who is Larry's neighbor. Larry's neighbors must contain an odd number of innocents. Those neighbors already include 3 known innocents, so the unknown neighbors Kevin, Pam, and Samin have to fit that odd-total requirement while also meeting the row 3 requirement above. If Samin were a criminal, Kevin and Pam could not satisfy both clues at the same time. So Samin must be innocent.
10.A4 · Raul → CRIMINAL
Samin’s clue says there is only one row with exactly one criminal. Row 2 already fits that exactly, because Eli, Flora, and Gary are innocent while Hrishi is criminal. Row 4 currently has one known criminal, Uma, and Raul is the only person in that row not yet identified. If Raul were innocent, then row 4 would also have exactly one criminal, which would clash with the clue that only one row can have that pattern. So Raul must be criminal.
11.A1 · Anna → INNOCENT
Raul’s clue says column D is the only column that has exactly 3 criminals. Column A already has 2 known criminals, and Anna is the only person in that column whose identity is not yet fixed. If Anna were a criminal, then column A would also have exactly 3 criminals, which would contradict the clue that only column D does. So Anna must be innocent.
12.B1 · Bobby → CRIMINAL
Anna’s clue applies to the edge-cell neighbors of Gary, which are Bobby, Chad, Dana, Hrishi, and Pam, and it says an odd number of that group are innocent. Chad and Hrishi are already criminal, so if Bobby were innocent, then Dana and Pam would have to fit that odd-innocent requirement along with Raul’s column clue at the same time. But with Dana and Pam also being the unknown people in column D, that combination cannot satisfy both clues together. So Bobby must be criminal.
13.B3 · Kevin → INNOCENT
Raul's clue says column D is the only column with exactly 3 criminals. Column B already has 2 known criminals, and Kevin is the only person in that column whose identity is not yet fixed. If Kevin were a criminal, then column B would also have exactly 3 criminals, which would break the clue. So Kevin must be innocent.
14.D3 · Pam → CRIMINAL
Row 3 must contain exactly 2 criminals. In row 3, the people who are Larry's neighbors are Kevin and Pam, and exactly 1 of the row 3 criminals must be in that neighbor group. Kevin is innocent, so that neighbor group still needs 1 criminal, and Pam is the only unknown person left in it. So Pam must be criminal.
15.D1 · Dana → INNOCENT
Raul’s clue says column D is the only column with exactly 3 criminals, and the other columns to compare with it are columns A, B, and C. In column D, Hrishi, Pam, and Uma are already the 3 known criminals. If Dana were also a criminal, then column D would have 4 criminals instead of exactly 3. So Dana must be innocent.