Clues by Sam Aug 24, 2026 Answer – Full Solution Explained
A1
👩⚖️
judge
B1
👩💻
coder
C1
👩⚖️
judge
D1
👮♂️
cop
A2
👩🔧
mech
B2
👨🔧
mech
C2
👮♀️
cop
D2
👨⚖️
judge
A3
👩🎨
painter
B3
👩🔧
mech
C3
👨💻
coder
D3
👨✈️
pilot
A4
👨🎨
painter
B4
👮♂️
cop
C4
🕵️♀️
sleuth
D4
👨✈️
pilot
A5
👷♀️
builder
B5
🕵️♂️
sleuth
C5
👷♂️
builder
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 — 7 criminals.
Clues by Sam answer for Aug 24, 2026 — a Easy solved in 17 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 7 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Donna (B1), Helen (A2), Kevin (D2), Mary (B3), Ryan (B4), Sofia (C4) and Zara (D5); the remaining 13 suspects are innocent.
The deduction chain, in plain English
01.C2 · Jane → INNOCENT
Nick’s clue says Jane is one of Olof’s 3 innocent neighbors. That directly identifies Jane as one of the innocents in that neighbor group. So Jane must be innocent.
02.A2 · Helen → CRIMINAL, D2 · Kevin → CRIMINAL
Jane's clue says there are exactly 2 criminals who are both in row 2 and on the edge. In that shared group, the only people are Helen at A2 and Kevin at D2. Since that group still needs 2 criminals and Helen and Kevin are the only people available there, both of them have to fill those two criminal spots. So Helen and Kevin must be criminal.
03.C5 · Xavi → INNOCENT
Olof’s neighbors still need exactly 1 more innocent, and the only candidates for that are Sofia and Tyler. Zara’s neighbors need exactly 2 innocents among Sofia, Tyler, and Xavi. Since Sofia and Tyler together can contribute only that 1 innocent required around Olof, Zara’s second innocent has to come from the one person in her neighbor group who is not in Olof’s group, which is Xavi. So Xavi must be innocent.
04.C4 · Sofia → CRIMINAL
Xavi’s clue says exactly 1 innocent is both above Xavi and neighboring Ryan. That shared group is only Nick and Sofia, and Nick is already known to be innocent. Since the one innocent in that group is already accounted for by Nick, Sofia cannot also be innocent. So Sofia must be criminal.
05.D4 · Tyler → INNOCENT
Nick’s clue says that Jane is one of Olof’s exactly 3 innocent neighbors. Among Olof’s neighbors, Jane and Nick are already known to be innocent, while Kevin and Sofia are criminal, so the only neighbor left who could supply the third innocent is Tyler. If Tyler were criminal, Olof would have only those 2 innocent neighbors instead of 3. So Tyler must be innocent.
06.C1 · Emily → INNOCENT
Tyler’s clue is about the people who are both in column C and neighbors of Isaac. That shared group is Emily, Jane, and Nick. The clue says an odd number of those people are innocent. Jane and Nick are already innocent, so that group already contains 2 innocents. If Emily were criminal, the group would stay at 2 innocents, which is even, not odd. So Emily must be innocent.
07.B2 · Isaac → INNOCENT
Emily’s clue says there are exactly two innocents in row 2, and those two innocents are connected. In row 2, Helen is criminal, Jane is innocent, and Kevin is criminal, leaving only Isaac unknown. If Isaac were criminal, then Jane would be the only innocent in row 2, so the clue about both innocents in that row being connected could not be true. So Isaac must be innocent.
08.B5 · Wally → INNOCENT
Isaac’s clue says that Wally is innocent. That directly identifies Wally’s status, so Wally at B5 must be innocent.
09.D1 · Floyd → INNOCENT
Wally’s clue says column D has exactly 3 innocents, and exactly 2 of those innocents are Sofia’s neighbors. That means exactly 1 innocent in column D is not a neighbor of Sofia. The people in column D who are not neighbors of Sofia are Floyd and Kevin. Kevin is already criminal, so there are currently no known innocents in that non-neighbor group, and the one innocent required there has to be Floyd. So Floyd must be innocent.
10.A3 · Lucy → INNOCENT
Floyd’s clue says exactly one of the two painters has a criminal directly above them. Lucy already fits that, because Helen is directly above Lucy and Helen is criminal. Peter would also fit the clue if Lucy were criminal, since Lucy is directly above Peter. That would make two painters with a criminal directly above them, which contradicts the clue, so Lucy at A3 must be innocent.
11.B1 · Donna → CRIMINAL, B3 · Mary → CRIMINAL
Above Ryan there are three people: Donna, Isaac, and Mary. Lucy's clue says there are more criminals than innocents above Ryan, so that group must have at least 2 criminals and therefore at most 1 innocent. Isaac is already the 1 known innocent in that group, and the only people there not yet identified are Donna and Mary. That means neither of them can be innocent. So Donna and Mary must be criminal.
12.B4 · Ryan → CRIMINAL
Column B already has 2 known criminals, and Ryan is the only unknown person left in that column. Donna’s clue says column B has more criminals than column A, column C, and column D. Wally’s clue fixes column D at exactly 3 innocents in total, with exactly 2 of those innocents among Sofia’s neighbors in D3, D4, and D5. If Ryan were innocent, then the remaining unknown people tied to these clues, including Betty, Peter, Vicky, Olof, and Zara, would have to satisfy both of those column clues at the same time, and that cannot be done. So Ryan at B4 must be criminal.
13.A5 · Vicky → INNOCENT
Ryan’s clue says Peter and Olof must have the same number of criminal neighbors. Olof’s neighbors are fully known and contain exactly 2 criminals. Peter’s neighbors also already contain exactly 2 known criminals, and the only unknown among Peter’s neighbors is Vicky. If Vicky were criminal, Peter would have 3 criminal neighbors while Olof would still have 2, which breaks the clue. So Vicky must be innocent.
14.A1 · Betty → INNOCENT
Peter’s neighbors are fixed at 2 criminals, and Donna’s neighbors currently have 1 criminal plus Betty. The clue says Peter has more criminal neighbors than Donna. If Betty were a criminal, then Donna’s neighbors would also have 2 criminals. That would make Donna’s criminal-neighbor count equal to Peter’s, not more. So Betty must be innocent.
15.A4 · Peter → INNOCENT
Betty’s clue says that all innocents below her must be connected in one orthogonal chain. Below Betty are Helen at A2, Lucy at A3, Peter at A4, and Vicky at A5, and among those, Lucy and Vicky are already innocent. If Peter were criminal, then the innocents below Betty would be Lucy at A3 and Vicky at A5, and they would not form one connected block. So Peter at A4 must be innocent.
16.D3 · Olof → INNOCENT
Peter’s clue says the number of innocent painters equals the number of innocent pilots. The painters already have 2 innocents, Lucy and Peter, while the pilots currently have only 1 innocent, Tyler. If Olof were criminal, the pilots would stay at just 1 innocent, so they would not match the painters’ 2 innocents. That contradicts Peter’s clue. So Olof must be innocent.
17.D5 · Zara → CRIMINAL
Column D has exactly three innocents in total. Among the people in column D who are Sofia's neighbors, the relevant group is Olof, Tyler, and Zara, and the clue says exactly two of the column D innocents are in that neighboring group. Olof and Tyler are already known innocents, so those two already fill the clue's full count of innocent people from column D who neighbor Sofia. That means Zara cannot also be innocent. So Zara must be criminal.