Clues by Sam May 04, 2026 Answer – Full Solution Explained
A1
🕵️♀️
sleuth
B1
👨⚖️
judge
C1
👩🎤
singer
D1
👮♀️
cop
A2
💂♂️
guard
B2
👩⚖️
judge
C2
💂♂️
guard
D2
👮♀️
cop
A3
👮♂️
cop
B3
💂♀️
guard
C3
👨🎨
painter
D3
👩⚖️
judge
A4
👨🏫
teacher
B4
👨🏫
teacher
C4
👩🎤
singer
D4
👨🏫
teacher
A5
🕵️♂️
sleuth
B5
👨🍳
cook
C5
👩🍳
cook
D5
👩🎨
painter
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 May 04, 2026 — a Easy solved in 15 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 15 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alice (A1), Brian (B1), Debra (C1), Ellie (D1), Gary (A2), Habiba (B2), Isaac (C2), Joyce (D2), Kyle (A3), Phil (B4), Tyler (D4), Umar (A5), Vince (B5), Wanda (C5) and Zara (D5); the remaining 5 suspects are innocent.
The deduction chain, in plain English
01.C1 · Debra → CRIMINAL, C2 · Isaac → CRIMINAL
Martin says there are no innocents above him, so the people above him must contain exactly 0 innocents. That group already has all 0 innocents it is allowed to have. The only people there who are not yet identified are Debra and Isaac, so neither of them can be innocent. So Debra and Isaac must be criminal.
02.C4 · Sue → INNOCENT, C5 · Wanda → CRIMINAL
Isaac’s clue says the two innocents in column C must be connected. In column C, Debra and Isaac are already criminals, and Martin is already innocent, so the only undecided spots there are Sue and Wanda. If Sue were criminal and Wanda were innocent, then the two innocents in column C would be Martin and Wanda, and that does not fit the clue’s requirement for the two innocents to be connected. So that opposite assignment cannot be right. This leaves Sue as innocent and Wanda as criminal.
03.B1 · Brian → CRIMINAL
Above Phil, there are exactly two criminals: Brian, Habiba, and Linda. The clue also says exactly one of those criminals is Kyle's neighbor, and among the people above Phil, Kyle's neighbors are Habiba and Linda. So the other criminal above Phil has to be someone above Phil who is not Kyle's neighbor. Among the people above Phil, the only person who is not Kyle's neighbor is Brian, so Brian must be criminal.
04.A3 · Kyle → CRIMINAL
Brian's clue says there is exactly 1 innocent among the people who are both in row 3 and neighboring Linda. That shared group is only Kyle and Martin. Martin is already known to be innocent, so the one innocent allowed by the clue is already accounted for. Kyle at A3 must be criminal.
05.A1 · Alice → CRIMINAL
Above Phil, there must be exactly 2 criminals, and exactly 1 of those criminals is Kyle's neighbor. The people involved there are Brian, Habiba, and Linda, with Brian already known to be a criminal, so Habiba and Linda have to fit Wanda's restriction. Gary's neighbors must contain an odd number of innocents. Right now Gary has 0 known innocents among his neighbors, and the only unknown neighbors are Alice, Habiba, and Linda. If Alice were innocent, then Habiba and Linda would have to satisfy Gary's odd-innocent requirement while also fitting the restriction from above Phil, and those facts clash. So Alice must be criminal.
06.A5 · Umar → CRIMINAL, B5 · Vince → CRIMINAL
Alice’s clue says exactly 2 edge people have an innocent directly to their left. That total is already accounted for by Nancy and Tyler. The only other direct-left neighbor cases still relevant to this clue are Umar and Vince, so if either Umar or Vince were innocent, that would create more than 2 such edge cases. So Umar and Vince must be criminal.
07.B4 · Phil → CRIMINAL
Above Phil, there are exactly two criminals in total, and exactly one of those criminals is a neighbor of Kyle. In that group above Phil, the only possible Kyle-neighboring people are Habiba and Linda. Also, an odd number of the people who are both below Brian and neighbors of Kyle are innocent, and that shared group is Habiba, Linda, and Phil. If Phil were innocent, then Habiba and Linda would have to satisfy both clues at the same time, but they cannot. So Phil must be criminal.
08.D3 · Nancy → INNOCENT
Phil's clue says Nancy is one of Sue's three innocent neighbors. That directly identifies Nancy as innocent. So Nancy must be innocent.
09.D2 · Joyce → CRIMINAL
Nancy's clue says exactly 2 of the edge criminals are neighbors of Ellie. Among the edge cells that neighbor Ellie, C1 Debra is already a criminal, and the only other unknown person there is D2 Joyce. Since that neighboring edge group still needs 1 more criminal to reach the required total of 2, Joyce has to be that criminal. So Joyce must be criminal.
10.B3 · Linda → INNOCENT
Joyce says there is exactly one criminal in row 3. Row 3 already has that one criminal: Kyle. The only person in row 3 whose status was still unknown is Linda, so she cannot also be a criminal. So Linda must be innocent.
11.B2 · Habiba → CRIMINAL
Above Phil, there must be exactly 2 criminals in total. Among the people above Phil, exactly 1 of those criminals is a neighbor of Kyle, and the only people there who are Kyle's neighbors are Habiba and Linda. Linda is innocent, so that required criminal neighbor above Phil cannot be Linda and must be the only unknown left in that group, Habiba. So Habiba must be criminal.
12.D4 · Tyler → CRIMINAL, D5 · Zara → CRIMINAL
Phil’s clue says Nancy is one of Sue’s exactly 3 innocent neighbors. In Sue’s neighbor group, the three innocents are already Linda, Martin, and Nancy. That fills all 3 innocent spots around Sue, so Tyler and Zara cannot also be innocent neighbors there. So Tyler and Zara must be criminal.
13.D1 · Ellie → CRIMINAL
Habiba’s clue says rows 1 and 5 must have the same number of innocents. Right now both rows have 0 known innocents, and row 5 has no unknown people at all, so row 5’s innocent count stays at 0. If Ellie were innocent, row 1 would have 1 innocent while row 5 would still have 0, which conflicts with the clue. So Ellie must be criminal.
14.A4 · Oscar → INNOCENT
Ellie’s clue says row 4 has exactly 2 innocents. In row 4, there is already 1 known innocent, and Oscar is the only person there whose status is still unknown. So row 4 still needs exactly 1 more innocent, and that remaining innocent has to be Oscar. That makes Oscar innocent.
15.A2 · Gary → CRIMINAL
Nancy’s clue says there are exactly 12 criminals on the edge, and exactly 2 of those edge criminals are Ellie's neighbors. Those 2 are already accounted for by Debra and Joyce, so the edge people who are not neighboring Ellie must contain exactly 10 criminals. Among the edge people who are not neighboring Ellie, there are already 9 known criminals: Alice, Brian, Ellie, Kyle, Tyler, Umar, Vince, Wanda, and Zara. The only unknown person in that same group is Gary, so he has to fill the last criminal spot. So Gary must be criminal.