Clues by Sam Aug 18, 2026 Answer – Full Solution Explained
A1
👷♂️
builder
B1
🕵️♀️
sleuth
C1
💂♀️
guard
D1
👷♀️
builder
A2
👩🎨
painter
B2
🕵️♀️
sleuth
C2
💂♂️
guard
D2
💂♀️
guard
A3
👨🎨
painter
B3
👩🎨
painter
C3
👨🍳
cook
D3
👨🍳
cook
A4
🕵️♂️
sleuth
B4
👮♂️
cop
C4
👷♂️
builder
D4
👩🌾
farmer
A5
👮♀️
cop
B5
👮♂️
cop
C5
👩🌾
farmer
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 — 15 criminals.
Clues by Sam answer for Aug 18, 2026 — a Medium solved in 15 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 15 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Andre (A1), Bonnie (B1), Chloe (C1), Emily (A2), Igor (A3), Joyce (B3), Klay (C3), Logan (D3), Martin (A4), Ollie (B4), Uma (D4), Vera (A5), Will (B5), Xena (C5) and Zane (D5); the remaining 5 suspects are innocent.
The deduction chain, in plain English
01.C3 · Klay → CRIMINAL, B4 · Ollie → CRIMINAL
Freya’s clue says Joyce has exactly 5 criminal neighbors, and exactly 2 of those criminal neighbors also neighbor Rob. Among Joyce’s neighbors, the only people who also neighbor Rob are Klay and Ollie. Since that shared pair still has to supply those 2 criminal neighbors, both of them have to be criminal. So Klay and Ollie must be criminal.
02.C5 · Xena → CRIMINAL
To the right of Martin, there is exactly one innocent, and the only possible people for that are Rob and Uma, since Ollie there is already criminal. Zane’s neighbors also contain exactly one innocent, and those neighbors are Rob, Uma, and Xena. That means the one innocent among Zane’s neighbors has to come from the shared pair, Rob or Uma. There is no room for any additional innocent in that neighbor group, so Xena cannot be innocent. So Xena must be criminal.
03.B5 · Will → CRIMINAL, D5 · Zane → CRIMINAL
To the right of Martin, the three people are Ollie, Rob, and Uma, and that group contains exactly 1 innocent. Xena’s neighbors are Ollie, Rob, Uma, Will, and Zane, and that larger group also contains exactly 1 innocent. Since Ollie, Rob, and Uma are entirely inside Xena’s neighbor group, that smaller group already accounts for the only innocent Xena’s neighbor group is allowed to have. That means the extra people in Xena’s neighbor group, Will and Zane, cannot be innocent. So Will and Zane must be criminal.
04.A3 · Igor → CRIMINAL
Joyce has exactly 5 criminal neighbors, so among her 8 neighbors there must be exactly 3 innocents. Freya is already one of those innocent neighbors, which leaves exactly 2 innocent spots among Emily, Gary, Igor, Martin, and Rob. Those 2 innocent spots therefore have to come from Emily, Gary, Martin, and Rob, not from Igor. So Igor cannot be innocent. So Igor must be criminal.
05.D2 · Hilda → INNOCENT
In row 2, there must be exactly 3 innocents: Emily, Freya, Gary, and Hilda. Among the row 2 people who are also neighboring Freya, exactly 1 is innocent, and those two people are Emily and Gary. That accounts for only 1 innocent from Emily and Gary, plus Freya as the other already known innocent in row 2. So the remaining row 2 person outside that shared group, Hilda, has to provide the third innocent. That makes Hilda innocent.
06.A1 · Andre → CRIMINAL, C1 · Chloe → CRIMINAL
Among Bonnie’s neighbors, there must be exactly 2 innocents. In the row 2 part of Bonnie’s neighbors, the only candidates are Emily, Freya, and Gary, and Freya is already innocent; with Will’s clue, Emily and Gary contain exactly 1 innocent between them. So that row 2 part already accounts for Bonnie’s full total of 2 innocents. That leaves the other two of Bonnie’s neighbors, Andre and Chloe, unable to be innocent. So Andre and Chloe must be criminal.
07.A5 · Vera → CRIMINAL, B3 · Joyce → CRIMINAL
Uma’s left side contains exactly one innocent, and among those people the only unknowns are Martin and Rob. Ollie’s neighbors also contain exactly one innocent. Since Martin and Rob are inside Ollie’s neighbor group, that one innocent among Ollie’s neighbors must already come from Martin or Rob. That leaves the other Ollie neighbors outside Uma’s left side, namely Joyce and Vera, unable to be innocent. So Vera and Joyce must be criminal.
08.B1 · Bonnie → CRIMINAL
Freya’s clue says exactly one innocent is both in Freya’s neighbors and in row 2, and that shared group is only Emily and Gary. Joyce’s clue says exactly one person in column B has exactly 2 innocent neighbors. If Bonnie were innocent, then Bonnie would already have 2 innocent neighbors only if exactly one of Emily and Gary were innocent, while Freya and Joyce would also be competing for that same “exactly 2 innocent neighbors” condition using those same people, along with Martin and Rob for Joyce and Freya. That makes the column B requirement and Freya’s row 2 neighbor requirement clash when Bonnie is treated as innocent. So Bonnie must be criminal.
09.D1 · Dana → INNOCENT
Andre’s clue says row 1 has more innocents than row 5. Row 5 already has 0 innocents, and row 1 also currently has 0 known innocents; the only person left in row 1 who could change that count is Dana. If Dana were criminal, row 1 would stay at 0 innocents, so it would not have more innocents than row 5. So Dana must be innocent.
10.C2 · Gary → INNOCENT
Logan’s neighbors must contain an odd number of innocents. Since Hilda is already a known innocent there, the unknown neighbors must add an even number of further innocents. To the right of Martin, there must be exactly one innocent, and among those people the only unknowns are Rob and Uma, so Rob and Uma contribute exactly one innocent. But Rob and Uma are also Logan’s neighboring pair from that group, so within Logan’s neighbors they add 1 innocent, which is odd. That means Gary has to be the additional innocent so that Logan’s full neighbor count stays odd. So Gary must be innocent.
11.A2 · Emily → CRIMINAL
Will’s clue is about the people who are both in row 2 and neighboring Freya. That shared group is only Emily and Gary, and the clue says exactly 1 of them is innocent. Gary is already known to be innocent, so that one innocent is already accounted for. Emily therefore cannot also be innocent. So Emily must be criminal.
12.D3 · Logan → CRIMINAL
Martin’s clue says there is exactly one innocent to his right. To Martin’s right, Ollie is already criminal, so the single innocent there has to be either Rob or Uma. Gary’s clue says exactly one person in column D has exactly 3 innocent neighbors. Dana can only have 2, Hilda has 2 and could rise only through Logan, and Uma and Zane are far short unless the same small set of unknowns changes things. Logan is the column D person whose count depends on those same remaining people, Rob and Uma. If Logan were innocent, then Rob and Uma would have to satisfy both clues at once, but they cannot. So Logan cannot be innocent. That makes Logan criminal.
13.D4 · Uma → CRIMINAL
Joyce’s neighbors must contain exactly 5 criminals, and exactly 2 of those criminal neighbors also neighbor Rob. Those 2 are already fixed as Klay and Ollie. So the other criminal neighbors around Joyce have to come from Emily, Igor, Martin, and Rob, with no extra criminal who would change that count. But the whole board must contain 15 criminals, and there are already 13 known criminals, so the three unknown people are tightly constrained by those totals. If Uma were innocent, then Martin and Rob would have to carry the remaining criminal placements while still fitting Joyce’s clue, and that cannot be made to work. So Uma at D4 must be criminal.
14.C4 · Rob → INNOCENT
Klay’s clue says there is exactly one innocent to the right of Martin. Among those people, there are currently no known innocents, and the only person there whose identity is not yet known is Rob. So the one innocent to the right of Martin has to be Rob. That makes Rob innocent.
15.A4 · Martin → CRIMINAL
Freya’s clue says Joyce has exactly 5 criminal neighbors, and exactly 2 of those criminals also neighbor Rob. Those 2 are already identified as Klay and Ollie, so the other 3 criminal neighbors of Joyce must be people who do not neighbor Rob. In Joyce’s neighbors who do not neighbor Rob, Emily and Igor are already criminal, while Freya, Gary, and Rob are innocent. That leaves only Martin to fill the last required criminal spot. So Martin must be criminal.