Clues by Sam Apr 18, 2026 Answer – Full Solution Explained
A1
🕵️♂️
sleuth
B1
🕵️♀️
sleuth
C1
💂♂️
guard
D1
💂♂️
guard
A2
💂♀️
guard
B2
👷♂️
builder
C2
👷♀️
builder
D2
👨⚕️
doctor
A3
🕵️♀️
sleuth
B3
👷♂️
builder
C3
👮♂️
cop
D3
👮♀️
cop
A4
👩🍳
cook
B4
👨⚖️
judge
C4
👩⚖️
judge
D4
👩⚕️
doctor
A5
👮♀️
cop
B5
👨⚕️
doctor
C5
👨⚖️
judge
D5
👩🍳
cook
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 — 12 criminals.
Clues by Sam answer for Apr 18, 2026 — a Hard solved in 15 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Aaron (A1), Betsy (B1), Helen (A2), Jane (C2), Kyle (D2), Mark (B3), Ruth (A4), Steve (B4), Tina (C4), Uma (D4), Will (B5) and Xavi (C5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.A2 · Helen → CRIMINAL
Aaron’s neighbors are Betsy, Helen, and Ivan, and the clue says exactly 2 of those 3 are criminals. It also says that among Aaron’s criminal neighbors, exactly 1 is above Will; in this group, the people above Will are Betsy and Ivan, so Helen is the only neighbor who is not above Will. With those restrictions, the remaining relevant possibilities still always keep Helen as one of Aaron’s two criminal neighbors. Therefore, we can determine that A2 Helen is CRIMINAL.
02.B4 · Steve → CRIMINAL, B5 · Will → CRIMINAL, B3 · Mark → CRIMINAL
Lisa’s clue fixes Aaron’s neighbors at exactly two criminals, with exactly one of those criminals above Will, and Helen’s clue says column B is the only column with exactly one innocent. When those two restrictions are applied together to the still-unknown people involved, the same three people in column B remain criminal in every clue-consistent arrangement. Those fixed people are Mark, Steve, and Will. Therefore, we can determine that B3 Mark is CRIMINAL, B4 Steve is CRIMINAL, and B5 Will is CRIMINAL.
03.A1 · Aaron → CRIMINAL
Steve’s clue says row 1 has exactly 2 innocents, so row 1 also has exactly 2 criminals in total. Mark’s clue says the people to the right of Aaron, namely Betsy, Eric, and Floyd, contain an odd number of criminals, so that group must have either 1 or 3 criminals. It cannot be 3, because then row 1 would already have 3 criminals to Aaron’s right and could not still have exactly 2 criminals in total. So the only criminal count to Aaron’s right is 1, which means the other criminal in row 1 must be Aaron. Therefore, we can determine that A1 Aaron is CRIMINAL.
04.C2 · Jane → CRIMINAL
Lisa’s clue fixes Aaron’s neighbors at exactly 2 criminals, with exactly 1 of those criminals in the above-Will group of Betsy and Ivan. Aaron is already a known criminal neighbor of Aaron who is not in that above-Will group, so among Betsy and Ivan exactly 1 must be criminal. Aaron’s clue also fixes Eric’s neighbors at exactly 3 criminals, with exactly 1 of those criminals in the above-Zara group of Floyd and Kyle. When those two restrictions are applied together to the still-relevant unknowns Betsy, Floyd, Ivan, Jane, and Kyle, every clue-consistent arrangement gives Jane the same status. Therefore, we can determine that C2 Jane is CRIMINAL.
05.C4 · Tina → CRIMINAL, D4 · Uma → CRIMINAL
Jane's clue says Nick has exactly 6 criminal neighbors, and exactly 2 of those criminal neighbors are also neighbors of Zara. The only people who are in both groups are Tina and Uma. The only unresolved people relevant to this clue are Ivan, Kyle, Olivia, Tina, and Uma. When those restrictions are applied, the remaining clue-consistent possibilities still all agree on the same result: Tina and Uma are both criminal. Therefore, we can determine that C4 Tina is CRIMINAL and D4 Uma is CRIMINAL.
06.D3 · Olivia → INNOCENT, C1 · Eric → INNOCENT
Steve’s clue says row 1 has exactly 2 innocents, and in row 1 the only unresolved people are Betsy, Eric, and Floyd. Aaron’s clue fixes the criminal pattern among Eric’s neighbors, and Jane’s clue fixes the criminal pattern among Nick’s neighbors, with the only unresolved people that still matter being Betsy, Eric, Floyd, Ivan, Kyle, and Olivia. When those three clue restrictions are applied together, there are only 2 clue-consistent ways to assign those unresolved statuses, and in both of them Eric and Olivia stay innocent. Therefore, we can determine that D3 Olivia is INNOCENT and C1 Eric is INNOCENT.
07.D2 · Kyle → CRIMINAL
Row 1 must contain exactly 2 innocents, and it currently has 1 known innocent, with Betsy and Floyd still unknown there. Also, the edge cells must contain exactly 8 criminals in total, and exactly 2 of those edge criminals must be neighbors of Eric, with the only edge neighbors of Eric being Betsy, Floyd, and Kyle. When those restrictions are applied together to the unresolved people, every clue-consistent arrangement gives Kyle the same status. Therefore, we can determine that D2 Kyle is CRIMINAL.
08.D1 · Floyd → INNOCENT
Eric’s neighbors must contain exactly 3 criminals in total, and exactly 1 of those criminals is in the group of Eric’s neighbors who are also above Zara, which here is just Floyd and Kyle. Kyle is already a known criminal, so that one allowed criminal in this overlap is already accounted for. The only unresolved people who can still affect the clue are Betsy, Floyd, and Ivan, and across the clue-consistent possibilities Floyd’s status never changes. Therefore, we can determine that D1 Floyd is INNOCENT.
09.B2 · Ivan → INNOCENT
Nick’s neighbors must contain exactly 6 criminals in total. We already know 6 criminals among Nick’s neighbors, and the only unresolved person there is Ivan. The clue also already has its exact 2 criminals who are both Nick’s neighbors and Zara’s neighbors, namely Tina and Uma, so Ivan cannot be added as another criminal without breaking the clue. Therefore, we can determine that B2 Ivan is INNOCENT.
10.B1 · Betsy → CRIMINAL
Aaron’s neighbors must contain exactly 2 criminals in total, and exactly 1 of those criminals must be above Will. In that group, the only unresolved person is Betsy. Since Aaron’s neighbors already contain 1 known criminal, Betsy has to be the second criminal, and she is also the only unresolved person who can fill the “above Will” criminal slot required by the clue. Therefore, we can determine that B1 Betsy is CRIMINAL.
11.C3 · Nick → INNOCENT
The relevant unresolved people for these clues are C3 Nick, A4 Ruth, A5 Vicky, C5 Xavi, and D5 Zara. Helen’s clue fixes column B as the only column with exactly one innocent, Will’s clue says Tina’s neighbors must contain an odd number of criminals, and Olivia’s clue fixes the edge total at exactly 8 criminals with exactly 2 of those criminal edge people neighboring Eric. When those restrictions are applied together to the remaining unresolved people, there are only 4 status arrangements left that fit all three clues. Across all of those clue-consistent possibilities, Nick’s status never changes. Therefore, we can determine that C3 Nick is INNOCENT.
12.A5 · Vicky → INNOCENT
Will’s clue says Tina’s neighbors must contain an odd number of criminals, and Nick’s clue says row 1 and row 5 must contain the same number of innocents. Row 1 already has 2 innocents, while row 5 currently has none, so the only unresolved people that can still matter here are Vicky, Xavi, and Zara. When those two clue restrictions are applied together, there are only two possible ways to assign those remaining statuses, and in both of them Vicky has the same status. Therefore, we can determine that A5 Vicky is INNOCENT.
13.A4 · Ruth → CRIMINAL
The only unresolved people that can still affect these two clues are Ruth, Xavi, and Zara. Tina’s clue says her neighbors must contain an odd number of criminals, and right now Tina’s neighbors already contain 4 known criminals, so Xavi and Zara are the only people who can change that clue’s count. Olivia’s clue says there are exactly 8 criminals on the edge, and exactly 2 of those edge criminals are neighbors of Eric; with the current known edge information, those same remaining unknown edge people are the only ones left to settle that requirement. When those restrictions are combined, there are only 2 arrangements left for Ruth, Xavi, and Zara, and in both of them Ruth has the same status. Therefore, we can determine that A4 Ruth is CRIMINAL.
14.D5 · Zara → INNOCENT
Ruth’s clue is about the people who are both in column D and neighboring Xavi. That shared group is just D4 Uma, who is criminal, and D5 Zara, who is still unknown. Since the clue says exactly 1 innocent is in that group, and there are currently no known innocents there, that 1 innocent still has to be supplied by the only unknown person left in the group, Zara. Therefore, we can determine that D5 Zara is INNOCENT.
15.C5 · Xavi → CRIMINAL
Will’s clue says Tina’s neighbors contain an odd number of criminals. Right now, Tina’s neighbors already include 4 known criminals, and the only unresolved neighbor who can change that count is Xavi. Since 4 is even, Xavi must be the extra criminal needed to make the total odd. Therefore, we can determine that C5 Xavi is CRIMINAL.