Clues by Sam Aug 28, 2026 Answer – Full Solution Explained
A1
💂♂️
guard
B1
👨🔧
mech
C1
👩💼
clerk
D1
🕵️♂️
sleuth
A2
👮♂️
cop
B2
👨⚖️
judge
C2
👷♂️
builder
D2
👷♀️
builder
A3
👩🔧
mech
B3
🕵️♀️
sleuth
C3
👨💼
clerk
D3
👩🎨
painter
A4
👮♀️
cop
B4
💂♀️
guard
C4
👩🎨
painter
D4
👨🍳
cook
A5
👩🍳
cook
B5
👨⚖️
judge
C5
👨🍳
cook
D5
👩⚖️
judge
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 — 15 criminals.
Clues by Sam answer for Aug 28, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 15 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Aaron (A1), Eric (D1), Floyd (A2), Gabe (B2), Igor (C2), Janet (D2), Katie (A3), Paula (A4), Ruth (B4), Sue (C4), Tom (D4), Vera (A5), Will (B5), Xavi (C5) and Zara (D5); the remaining 5 suspects are innocent.
The deduction chain, in plain English
01.C3 · Nick → INNOCENT
Lisa’s clue says that Nick is one of Sue’s three innocent neighbors. That directly identifies Nick as innocent. So Nick must be innocent.
02.A3 · Katie → CRIMINAL
Nick’s clue says there is exactly one innocent among the people who are neighbors of both Floyd and Paula. That shared group is only Katie and Lisa, and Lisa is already known to be innocent. So the one innocent in that shared group is already accounted for, which means Katie cannot also be innocent. So Katie must be criminal.
03.C2 · Igor → CRIMINAL, C4 · Sue → CRIMINAL
Katie’s clue says the people between Debra and Xavi contain exactly one innocent. That group already has its one known innocent, Nick. The only people there who are still unknown are Igor and Sue, so neither of them can also be innocent. So Igor and Sue must be criminal.
04.C5 · Xavi → CRIMINAL
Lisa says Nick is one of Sue's 3 innocent neighbors. Since Lisa and Nick are already the two known innocents among Sue's neighbors, there is room for exactly one more innocent there. That remaining innocent has to come from D3 Olive, B4 Ruth, D4 Tom, B5 Will, or D5 Zara, so it cannot be Xavi. That makes Xavi criminal.
05.C1 · Debra → INNOCENT
Xavi's clue says Janet has exactly 3 innocent neighbors, and exactly 2 of those innocents also neighbor Gabe. Among Janet's neighbors who also neighbor Gabe, Igor is already criminal and Nick is already innocent, so that group currently has only 1 known innocent. The only unknown person left in that same group is Debra, so she has to be the second innocent there. So Debra must be innocent.
06.D2 · Janet → CRIMINAL
Debra’s clue says every person has at least 2 criminal neighbors, so Eric’s neighboring group can include at most 1 innocent. Eric’s neighbors already include 1 known innocent, and the only neighbor there whose status is not yet fixed is Janet. That means Janet cannot be innocent, because then Eric’s neighbors would have too many innocents for the clue. So Janet must be criminal.
07.D4 · Tom → CRIMINAL
Zara’s neighbors already include two known criminals, Sue and Xavi, while Aaron’s neighbors currently have no known criminals. Janet’s clue says Zara has more criminal neighbors than Aaron. If Tom were innocent, Zara’s neighbor count would stay at 2, and that clashes with the required comparison once the same remaining unknown people also have to satisfy Debra’s rule that everyone has at least 2 criminal neighbors. So Tom must be criminal.
08.D5 · Zara → CRIMINAL
Janet’s clue says she has exactly 3 innocent neighbors, and exactly 2 of those innocents also neighbor Gabe. The two already known innocent neighbors of Janet are Debra and Nick, and both of them also neighbor Gabe, so the one remaining innocent neighbor Janet needs must be someone who does not neighbor Gabe. Tom’s clue says column D is the only column with exactly 4 criminals. If Zara were innocent, then the remaining people tied up in these clues would have to satisfy Janet’s count and also keep column D as that unique 4-criminal column, but those requirements clash. So Zara cannot be innocent. So Zara must be criminal.
09.A1 · Aaron → CRIMINAL
Zara's clue says the two innocents in row 1 are connected. Debra is already known to be one of those innocents, so the other innocent in row 1 has to be next to Debra, which limits that person to Brian or Eric. Aaron is not next to Debra in that row, so he cannot be the second innocent. So Aaron must be criminal.
10.A2 · Floyd → CRIMINAL
Aaron’s clue says Floyd and Zara must have the same number of criminal neighbors. Zara’s neighbors are already all known, and she has 3 criminal neighbors, while Floyd’s neighbor group currently has only 2 known criminals. So Floyd’s side has to make up that difference while also fitting Debra’s statement that everyone has at least 2 criminal neighbors. If Floyd were innocent, the remaining people involved here could not satisfy those requirements together. So Floyd must be criminal.
11.B5 · Will → CRIMINAL
In row 5, Xavi and Zara are already criminals, and Sue’s clue says all criminals in that row have to be one connected block. Tom’s clue also fixes column D at exactly 4 criminals, while Debra’s clue requires every person to have at least 2 criminal neighbors. If Will were innocent, the remaining unknown people involved here would be Brian, Eric, Gabe, Olive, Paula, Ruth, and Vera, and they would have to satisfy all of those facts at the same time. That cannot be done, so Will cannot be innocent. So Will must be criminal.
12.A4 · Paula → CRIMINAL, A5 · Vera → CRIMINAL
Will’s clue says column A has more criminals than every other column. Column A already has three known criminals, and the only undecided people there are Paula and Vera. If Paula and Vera were both innocent, column A would stay at three criminals, but that cannot fit with the clue once the other people tied to these clues are made to satisfy everything at once. So Paula at A4 and Vera at A5 must be criminal.
13.B1 · Brian → INNOCENT
Janet’s neighbors must contain exactly 3 innocents, and exactly 2 of those innocents are neighbors of Gabe. Debra and Nick already fill those 2 innocent neighbors of Gabe among Janet’s neighbors, so the remaining innocent in Janet’s neighborhood has to come from Eric or Olive, the two people there who are not neighbors of Gabe. Vera’s clue also says there are at least 3 innocents on the edge, but the edge currently has only 1 known innocent, Debra. If Brian were criminal, then the only edge people left for these clues to supply the missing innocents would be Eric and Olive, and that cannot satisfy everything at once. So Brian cannot be criminal. That makes Brian innocent.
14.B2 · Gabe → CRIMINAL
Debra’s clue says everyone has at least 2 criminal neighbors. For Aaron at A1, that means his neighbors can include at most 1 innocent. Aaron’s neighbors already include 1 known innocent, and the only neighbor there whose status is not fixed yet is Gabe at B2. So Gabe at B2 must be criminal.
15.D1 · Eric → CRIMINAL
Zara's clue says the two innocents in row 1 must be connected. In row 1, Brian and Debra are already known to be innocent, Aaron is criminal, and Eric is the only person not yet identified. If Eric were innocent, row 1 would have more than those two innocents, which clashes with the clue that the innocents there are the two that must be connected. So Eric at D1 must be criminal.
16.D3 · Olive → INNOCENT
Janet's neighbors must contain exactly 3 innocents in total, and exactly 2 of those innocents are neighbors of Gabe. The neighbors Janet shares with Gabe are Debra and Nick as innocents, so that uses up those 2 innocent shared-neighbor spots. That means Janet must have exactly 1 innocent neighbor who is not a neighbor of Gabe. Among Janet's neighbors who are not neighbors of Gabe, Eric is criminal and Olive is the only unknown, with no known innocent already there. So Olive has to be the remaining innocent. So Olive must be innocent.
17.B4 · Ruth → CRIMINAL
Lisa’s clue says Nick is one of Sue’s exactly 3 innocent neighbors. Sue’s neighbors already include Lisa, Nick, and Olive as innocent, which fills all 3 innocent neighbor spots. If Ruth were innocent too, Sue would have 4 innocent neighbors instead of 3, which is impossible. So Ruth must be criminal.