Clues by Sam Jul 01, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👩💼
clerk
C1
👩🎤
singer
D1
👩🏫
teacher
A2
👨✈️
pilot
B2
👩💼
clerk
C2
👨🎤
singer
D2
👨🎤
singer
A3
👨🏫
teacher
B3
👨🍳
cook
C3
👮♂️
cop
D3
👮♂️
cop
A4
👩🍳
cook
B4
👩⚕️
doctor
C4
👨💼
clerk
D4
👨⚕️
doctor
A5
👩⚖️
judge
B5
👩🏫
teacher
C5
👩⚖️
judge
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 — 6 criminals.
Clues by Sam answer for Jul 01, 2026 — a Medium solved in 14 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Jason (C2), Kevin (D2), Martin (B3), Noah (C3), Penny (A4) and Uma (A5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.A1 · Amy → INNOCENT, C1 · Claire → INNOCENT
Luigi’s clue says there are exactly 2 innocents in the overlap between row 1 and Bunty’s neighbors. That overlap contains only Amy and Claire. Since those 2 innocent spots in that shared group still have to be filled, and Amy and Claire are the only people there, both of them have to be the innocents. So Amy and Claire must be innocent.
02.B4 · Quita → INNOCENT, C4 · Rohan → INNOCENT
Amy's clue says the people strictly between Penny and Tyler contain exactly 2 innocents. Right now there are 0 known innocents there, and the only people left in that group are Quita and Rohan. Since those two spots must supply all 2 innocents, both of them have to be innocent. So Quita and Rohan must be innocent.
03.B1 · Bunty → INNOCENT
Luigi’s neighbors in column B are Hope and Martin, and exactly 1 of them is a criminal. The larger group, the people who are both in column B and neighboring Gabe, is Bunty, Hope, and Martin, and that group also contains exactly 1 criminal. So Hope and Martin already account for the only criminal allowed in that larger group. The only extra person in the larger group is Bunty, so Bunty cannot be a criminal. That makes Bunty innocent.
04.B5 · Vicky → INNOCENT
Claire’s clue fixes the column-B neighbors of Gabe at exactly 2 innocents, and in that group the only candidates besides Bunty are Hope and Martin. Bunty is already innocent, so Hope and Martin together contribute exactly one more innocent there. Bunty’s clue says there are exactly 3 innocents below Bunty, namely among Hope, Martin, Quita, and Vicky. Since Quita is already innocent and Hope plus Martin contribute exactly one innocent, the third innocent below Bunty has to be the only person left outside that smaller group, Vicky. So Vicky must be innocent.
05.C2 · Jason → CRIMINAL
Claire’s clue makes the shared group of column B people who neighbor Gabe contain exactly 2 innocents. Since Bunty is already the 1 known innocent in that group, the two unknown people there, Hope and Martin, contribute exactly 1 more innocent. Among Noah’s neighbors who are not on the edge, there are exactly 3 innocents in total. That group is Hope, Jason, Martin, Quita, and Rohan, and Quita and Rohan are already innocent, so Hope, Jason, and Martin must contain exactly 1 innocent. But Hope and Martin already account for that 1 innocent, so Jason cannot be innocent. So Jason must be criminal.
06.A5 · Uma → CRIMINAL
Jason's clue says there are exactly 3 criminals on the edge, and exactly 1 of those edge criminals is Penny's neighbor. Among the edge people who are Penny's neighbors, Luigi and Vicky are already innocent. That means this group still needs 1 criminal, and the only unknown person left in it is Uma. So Uma must be criminal.
07.C5 · Wanda → INNOCENT, D5 · Zulu → INNOCENT, D1 · Evie → INNOCENT
Jason’s clue says there are exactly 3 criminals on the edge, and Uma is already one of them, so the other 2 edge criminals must be chosen from the remaining edge unknowns. From the linked restrictions in this step, those 2 must come from Gabe, Kevin, Oscar, Penny, and Tyler. That leaves Evie, Wanda, and Zulu unable to be either of those 2 edge criminals. So Wanda, Zulu, and Evie must be innocent.
08.C3 · Noah → CRIMINAL
Vicky’s clue fixes Noah’s neighborhood at exactly 3 criminals, with exactly 1 of those edge neighbors being an edge criminal among Kevin, Oscar, and Tyler. Quita’s clue also says column C has more criminals than column D, and right now column C already has Jason as 1 known criminal while column D has none. If Noah were innocent, then Hope, Kevin, Martin, Oscar, and Tyler would have to satisfy both of those clue restrictions at the same time, and they cannot. So Noah cannot be innocent. That makes Noah criminal.
09.D4 · Tyler → INNOCENT
Noah’s edge neighbors are D2 Kevin, D3 Oscar, and D4 Tyler. The two clues fix Kevin and Oscar as the two innocents in the shared group of people above Zulu and neighboring Jason, and among Noah’s edge neighbors there must be exactly two innocents in that same overlapping part. That leaves the only person in Noah’s edge-neighbor group who is not in that smaller shared group, namely Tyler, to account for the remaining innocent required there. So Tyler must be innocent.
10.B2 · Hope → INNOCENT, B3 · Martin → CRIMINAL
Luigi’s clue says his neighbors contain exactly 2 criminals, and exactly 1 of those criminals is in column B. Among Luigi’s neighbors, the column B people are Hope, Martin, and Quita, and Quita is innocent, so the one criminal in column B would have to be Hope or Martin. Now test the opposite pairing: Hope criminal and Martin innocent. Martin’s clue says his neighbors contain an odd number of innocents, and Martin’s neighbors already have 3 known innocents, with only Gabe, Hope, and Penny still unknown there. That opposite pairing makes those remaining people unable to satisfy Martin’s odd-innocent clue while also fitting Luigi’s two-criminals clue with exactly one of them in column B. So Hope must be innocent and Martin must be criminal.
11.D3 · Oscar → INNOCENT
Martin’s clue says every row has at least 2 innocents, so in a 4-person row there can be at most 2 criminals. In row 3, Martin and Noah are already known criminals. The only person in that row whose status was still unknown is Oscar, so row 3 cannot take another criminal there. So Oscar must be innocent.
12.D2 · Kevin → CRIMINAL
Noah’s neighboring edge cells are D2 Kevin, D3 Oscar, and D4 Tyler. The clue says exactly 1 of Noah’s 3 criminal neighbors is on the edge, and among those three edge neighbors, Oscar and Tyler are innocent. So the one edge criminal neighbor still has to be Kevin. That makes Kevin criminal.
13.A2 · Gabe → INNOCENT
Martin’s clue says each row has at least 2 innocents. That means row 2 can have at most 2 criminals. Row 2 already has 2 known criminals, Jason and Kevin, and the only person there not yet identified is Gabe. So Gabe at A2 must be innocent.
14.A4 · Penny → CRIMINAL
Rohan's clue says Luigi has exactly 2 criminal neighbors, and exactly 1 of those criminals is in column B. Among Luigi's neighbors in column B, the only criminal is Martin, so the other criminal neighbor must be outside column B. Luigi's neighbors outside column B are Gabe and Penny, and Gabe is innocent. So Penny must be criminal.