Clues by Sam Apr 15, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👷♀️
builder
C1
👮♀️
cop
D1
👷♂️
builder
A2
💂♀️
guard
B2
👨🌾
farmer
C2
👨🌾
farmer
D2
👷♂️
builder
A3
🕵️♂️
sleuth
B3
👨🍳
cook
C3
👩🌾
farmer
D3
👩🍳
cook
A4
💂♀️
guard
B4
👮♀️
cop
C4
🕵️♀️
sleuth
D4
👨🍳
cook
A5
💂♂️
guard
B5
🕵️♂️
sleuth
C5
👩💼
clerk
D5
👨💼
clerk
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 — 8 criminals.
Clues by Sam answer for Apr 15, 2026 — a Tricky solved in 19 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chloe (C1), Jerry (A3), Kevin (B3), Nicole (D3), Olga (A4), Tyler (D4), Vince (B5) and Wanda (C5); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.D4 · Tyler → CRIMINAL
Stella’s clue explicitly says that Tyler is one of the two criminals located above Xavi. Since all clues are truthful, that directly fixes Tyler’s status as a criminal. Therefore, we can determine that D4 Tyler is CRIMINAL.
02.D1 · Denis → INNOCENT
Look at column D, where Xavi is at D5 and Tyler is at D4. Stella’s clue says that above Xavi there are exactly two criminals, and Tyler is one of them, so among D1–D4 there is exactly one other criminal besides Tyler. Tyler’s clue says all criminals in column D are connected, which in a single column means the criminals must form one uninterrupted vertical block with no innocent between them. Since Tyler is at D4, the only way for there to be exactly one more criminal and still have them connected is for that other criminal to be directly above him at D3, which means D1 cannot be a criminal. Therefore, we can determine that D1 Denis is INNOCENT.
03.B5 · Vince → CRIMINAL
Denis’s clue says that Vince is one of Umar’s 2 criminal neighbors, which means Vince is a neighbor of Umar and is criminal. So Vince’s identity is fixed by the clue itself. Therefore, we can determine that B5 Vince is CRIMINAL.
04.C5 · Wanda → CRIMINAL
Vince at B5 says he has exactly two neighboring criminals, and exactly one of those two is below Chloe. Being below Chloe means being somewhere under Chloe in column C, and among Vince’s neighbors the only person in column C is Wanda at C5 (since Stella at C4 is already known INNOCENT). That means the one neighboring criminal who is below Chloe must be Wanda. Therefore, we can determine that C5 Wanda is CRIMINAL.
05.D3 · Nicole → CRIMINAL
In column D, the people above Xavi at D5 are Denis at D1, Ivan at D2, Nicole at D3, and Tyler at D4. Stella’s clue says there are exactly two criminals among those four, and Tyler is one of them; since Denis is already INNOCENT, the second criminal must be either Ivan or Nicole. Tyler also says all criminals in column D are connected, so any criminal above Tyler must be directly linked to him without an innocent breaking the chain, which forces D3 to be a criminal to connect Tyler at D4 to the second criminal above Xavi. Therefore, we can determine that D3 Nicole is CRIMINAL.
06.D2 · Ivan → INNOCENT
The clue talks about the people above Xavi in column D: Denis at D1, Ivan at D2, Nicole at D3, and Tyler at D4. It says Tyler is one of exactly two criminals above Xavi, so there must be exactly two criminals among those four people, including Tyler. Since Nicole is already known to be a criminal and Tyler is a criminal, those are the two criminals the clue allows above Xavi. That leaves Denis and Ivan as not criminals, so Ivan must be innocent. Therefore, we can determine that D2 Ivan is INNOCENT.
07.A3 · Jerry → CRIMINAL
Hank is at C2, and his row 3 neighbors are B3, C3, and D3. Wanda says Hank has exactly three criminal neighbors in total, and exactly two of those three are in row 3; since Nicole at D3 is already a criminal, that forces exactly one of B3 and C3 to be a criminal so that row 3 contributes exactly two criminal neighbors. Nicole also says there is an odd number of criminals in row 3; with Nicole already one criminal, the number of criminals among A3, B3, and C3 must be even, but B3 and C3 already contribute exactly one criminal, so A3 must be the second to make it even. Therefore, we can determine that A3 Jerry is CRIMINAL.
08.A5 · Umar → INNOCENT
Umar is at A5, and his only neighbors are Olga at A4, Ruth at B4, and Vince at B5. Denis’s clue says Vince is one of Umar’s 2 criminal neighbors, so Umar has exactly two criminal neighbors, meaning exactly one of Olga and Ruth is also a criminal. Vince’s clue says he has exactly two criminal neighbors and only one of them is below Chloe; among Vince’s neighbors, the only person who is below Chloe and could be a criminal is Wanda at C5, so Wanda must be one of Vince’s two criminal neighbors and the other criminal neighbor must be exactly one of Olga, Umar, or Ruth. Since we already know exactly one of Olga and Ruth is a criminal, that leaves no room for Umar to also be a criminal. Therefore, we can determine that A5 Umar is INNOCENT.
09.B2 · Gary → INNOCENT
Hank is at C2, so his neighbors include B1, C1, B2, and also D3 (Nicole) among others. Wanda’s clue says Hank has exactly three criminal neighbors in total, with exactly two of those three in row 3, so the one remaining criminal neighbor not in row 3 must be one of B1, C1, or B2 (since D1 and D2 are already innocent). Jerry’s clue says exactly two of the edge criminals are neighbors of Hank, and Nicole at D3 is already an edge criminal who neighbors Hank, so among Hank’s other edge-neighbors only one of B1 or C1 can be criminal. That means the single criminal among {B1, C1, B2} must be in {B1, C1}, leaving Gary at B2 unable to be that criminal. Therefore, we can determine that B2 Gary is INNOCENT.
10.D5 · Xavi → INNOCENT
On the edge we already have five known criminals: Jerry at A3, Nicole at D3, Tyler at D4, Vince at B5, and Wanda at C5. Jerry’s clue says there are exactly seven criminals on the edges in total, so only two more edge people can still be criminals. The same clue also says exactly two of those seven edge criminals are Hank’s neighbors; since Nicole at D3 is already an edge criminal neighbor of Hank, exactly one of B1 or C1 must be an edge criminal, using up one of the two remaining edge-criminal slots. Ivan’s clue says each column has at least 2 criminals, and column A currently has only Jerry as a criminal and Umar as innocent, so at least one of A1, A2, or A4 must be a criminal, which uses up the other remaining edge-criminal slot because every position in column A is on the edge. That leaves no remaining edge-criminal slot for D5, so Xavi cannot be a criminal on the edge. Therefore, we can determine that D5 Xavi is INNOCENT.
11.A2 · Evie → INNOCENT
Jerry is at A3 and Kevin is at B3, so the people who are neighbors of both of them are A2 (Evie), B2 (Gary), A4 (Olga), and B4 (Ruth). Xavi’s clue says that among these common neighbors, exactly 3 are innocent, so since Gary at B2 is already innocent, only 1 of Evie, Olga, and Ruth can be a criminal. Denis’s clue says Umar has exactly 2 criminal neighbors and that Vince is one of them; Umar’s only neighbors are A4, B4, and B5, and since B5 (Vince) is already criminal, exactly 1 of Olga (A4) and Ruth (B4) must also be criminal. That already uses up the single allowed criminal among Evie, Olga, and Ruth, so Evie cannot be criminal. Therefore, we can determine that A2 (Evie) is INNOCENT.
12.B3 · Kevin → CRIMINAL
Umar at A5 has only three neighbors: Olga at A4, Ruth at B4, and Vince at B5. Denis says Vince is one of Umar’s 2 criminal neighbors, so besides Vince, exactly one of Olga or Ruth is also a criminal and the other is an innocent. Jerry at A3 has five neighbors: Evie at A2 and Gary at B2 (both innocents), plus Olga at A4, Kevin at B3, and Ruth at B4, and Umar says the number of innocents among these five neighbors is odd. Since Olga and Ruth contribute exactly one innocent between them, the total number of innocent neighbors of Jerry is odd only if Kevin is a criminal. Therefore, we can determine that B3 Kevin is CRIMINAL.
13.C3 · Layla → INNOCENT
Nicole is in row 3, which contains Jerry at A3, Kevin at B3, Layla at C3, and Nicole at D3. Her clue says the total number of criminals in that row is odd. Jerry, Kevin, and Nicole are already known criminals, so row 3 already has 3 criminals, which is odd. To keep the total odd, Layla cannot be a criminal, so she must be innocent. Therefore, we can determine that C3 Layla is INNOCENT.
14.C1 · Chloe → CRIMINAL
Ivan is at D2, and his neighbors are C1, D1, C2, C3, and D3. Among those neighbors, the ones that are on the edge of the board are C1, D1, and D3. Layla’s clue says exactly 2 of those edge-neighbors are criminal, and we already know D1 (Denis) is innocent while D3 (Nicole) is criminal, so C1 must be the second criminal to make the total exactly 2. Therefore, we can determine that C1 Chloe is CRIMINAL.
15.B1 · Betsy → INNOCENT
Hank is at C2, so his neighbors are B1 Betsy, C1 Chloe, D1 Denis, B2 Gary, D2 Ivan, B3 Kevin, C3 Layla, and D3 Nicole. Wanda’s clue says that Hank has exactly three criminal neighbors in total, and exactly two of those criminals are in row 3. We already know Chloe (C1) is a criminal, and Kevin (B3) and Nicole (D3) are criminals, which makes exactly three criminal neighbors with exactly two of them in row 3, so none of Hank’s other neighbors can be criminals. Therefore, we can determine that B1 Betsy is INNOCENT.
16.C2 · Hank → INNOCENT
Hank is at C2, and the only currently unknown people on the edge are Anna at A1 and Olga at A4. Jerry’s clue refers to “the 7 criminals on the edges”, and we can already see six known edge criminals (Chloe C1, Jerry A3, Nicole D3, Tyler D4, Vince B5, and Wanda C5), so exactly one of Anna or Olga must be a criminal. Gary’s clue says columns A and C have the same number of criminals; column A then has Jerry plus exactly one of Anna or Olga, so it has exactly 2 criminals, meaning column C must also have exactly 2 criminals. Column C already has Chloe and Wanda as criminals, so Hank cannot be a criminal. Therefore, we can determine that C2 Hank is INNOCENT.
17.A1 · Anna → INNOCENT
The total-criminals clue says there are 8 criminals altogether, and we can already count 7 criminals for sure: Chloe at C1, Jerry at A3, Kevin at B3, Nicole at D3, Tyler at D4, Vince at B5, and Wanda at C5. That means exactly one of the three unknown people (Anna at A1, Olga at A4, and Ruth at B4) can be a criminal. Denis’s clue says Vince is one of Umar’s 2 criminal neighbors; Umar is at A5, and his only neighbors are Olga at A4, Ruth at B4, and Vince at B5, so besides Vince exactly one of Olga or Ruth must also be a criminal. Since the only remaining criminal must be in {A4, B4}, Anna at A1 cannot be that last criminal. Therefore, we can determine that A1 Anna is INNOCENT.
18.A4 · Olga → CRIMINAL
Look at column A: Anna at A1 is innocent, Evie at A2 is innocent, Jerry at A3 is criminal, Olga at A4 is unknown, and Umar at A5 is innocent. Ivan’s clue says each column has at least 2 criminals, so column A must contain at least two criminals. Since Jerry is the only already-known criminal in column A and every other spot in that column is already confirmed innocent except Olga, Olga must be the second criminal in column A. Therefore, we can determine that A4 Olga is CRIMINAL.
19.B4 · Ruth → INNOCENT
Umar is at A5, and his neighbors are Olga at A4, Ruth at B4, and Vince at B5. Denis’s clue says that Umar has exactly 2 criminal neighbors, and Vince is one of those two. Since Vince is already a criminal and Olga is also already a criminal, those are the two criminal neighbors Umar is allowed to have. That leaves Ruth as Umar’s remaining neighbor, so Ruth cannot be a criminal and must be innocent. Therefore, we can determine that B4 Ruth is INNOCENT.