Clues by Sam Apr 20, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
💂♀️
guard
C1
👷♂️
builder
D1
👩💻
coder
A2
💂♀️
guard
B2
👮♂️
cop
C2
🕵️♂️
sleuth
D2
👷♀️
builder
A3
👮♀️
cop
B3
👩⚖️
judge
C3
🕵️♀️
sleuth
D3
👨💻
coder
A4
👨💼
clerk
B4
👩⚖️
judge
C4
💂♂️
guard
D4
👨🔧
mech
A5
👨💼
clerk
B5
👨💼
clerk
C5
👨🔧
mech
D5
👩🔧
mech
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 — 16 criminals.
Clues by Sam answer for Apr 20, 2026 — a Easy solved in 15 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 16 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alice (A1), Carl (C1), Ellie (D1), Hazel (A2), Ike (B2), Karen (D2), Lisa (A3), Mary (B3), Nicole (C3), Peter (A4), Quita (B4), Umar (D4), Vince (A5), Wally (B5), Xavi (C5) and Zara (D5); the remaining 4 suspects are innocent.
The deduction chain, in plain English
01.C3 · Nicole → CRIMINAL, B3 · Mary → CRIMINAL
Olof is at D3, so the people to his left are A3, B3, and C3. Among those, the ones who neighbor John at C2 are B3 and C3, namely Mary and Nicole. Betsy’s clue says there are no innocents in that group, so both Mary and Nicole must be criminals. Therefore, we can determine that B3 Mary is CRIMINAL and C3 Nicole is CRIMINAL.
02.A1 · Alice → CRIMINAL, C1 · Carl → CRIMINAL, D1 · Ellie → CRIMINAL
In row 1, Betsy at B1 is already known to be innocent. Nicole’s clue says each row has at least 3 criminals, so row 1 must contain at least three criminals among its four people. With B1 not a criminal, the only way for row 1 to still have at least three criminals is for the other three spaces in that row, A1, C1, and D1, all to be criminals. Therefore, we can determine that A1 is CRIMINAL, C1 is CRIMINAL, and D1 is CRIMINAL.
03.D5 · Zara → CRIMINAL
Carl’s clue says that Zara is one of the 4 criminals in column D. Since column D has exactly five people, this directly includes Zara among the criminals in that column. Therefore, we can determine that D5 is CRIMINAL.
04.D4 · Umar → CRIMINAL
Zara is at D5, so her neighbors are C4, C5, and D4. Her clue says she has exactly 2 criminal neighbors, and exactly 1 of those 2 is in column C. Among those three neighbors, the people in column C are C4 and C5, while D4 is not in column C. So if exactly one criminal neighbor is in column C and there are two criminal neighbors total, the other criminal neighbor must be D4. Therefore, we can determine that D4 is CRIMINAL.
05.A2 · Hazel → CRIMINAL
Carl is at C1, so his neighbors in row 2 are B2, C2, and D2: Ike, John, and Karen. Umar’s clue says an odd number of those three are innocent. Nicole’s clue says row 2 has at least 3 criminals, and since row 2 has only four people, that means row 2 has either 3 or 4 criminals, so among B2, C2, and D2 there can be at most one innocent. The only odd possibility is exactly one innocent there, which means all three of B2, C2, and D2 cannot all be criminals. So row 2 cannot have 4 criminals, and must have exactly 3 criminals, forcing the remaining person in row 2, A2 Hazel, to be the third criminal. Therefore, we can determine that A2 is CRIMINAL.
06.B2 · Ike → CRIMINAL
Mary is at B3, so the people above her are Betsy at B1 and Ike at B2. Hazel’s clue says there is only one innocent above Mary. Betsy is already known to be innocent, so she is that one innocent, which means Ike cannot also be innocent. Therefore, we can determine that B2 Ike is CRIMINAL.
07.B5 · Wally → CRIMINAL
The corners are A1, D1, A5, and D5, and this clue counts corners that have an innocent immediately to their right. Only the left-side corners, A1 and A5, can even have someone directly to the right, so we only need to check B1 and B5. B1 is already known to be innocent, so A1 does fit the clue. Since Ike says only one corner fits, A5 cannot also have an innocent directly to its right, which means B5 cannot be innocent. Therefore, we can determine that B5 is CRIMINAL.
08.A4 · Peter → CRIMINAL
Peter and Umar are in row 4, with Quita between them at B4 and Sam between them at C4. Wally’s clue says there is only one innocent in between Peter and Umar, so among those two people in between, exactly one is innocent. Since row 4 must have at least 3 criminals from Nicole’s clue, and Umar is already a criminal, Peter also has to be a criminal so that row 4 can reach at least 3 criminals while still allowing exactly one of Quita and Sam to be innocent. Therefore, we can determine that A4 is CRIMINAL.
09.A5 · Vince → CRIMINAL, C5 · Xavi → CRIMINAL
Nicole’s clue says every row has at least 3 criminals. In row 5, B5 and D5 are already criminals, so at least one of the two unknown people there, Vince at A5 or Xavi at C5, must also be a criminal. Peter’s clue says row 3 has more innocents than row 5. In row 3, Mary at B3 and Nicole at C3 are already criminals, so the only possible innocents there are Lisa at A3 and Olof at D3, which means row 3 can have at most 2 innocents. So row 5 must have fewer than 2 innocents, meaning row 5 has at most 1 innocent. Since the only unknowns in row 5 are Vince and Xavi, and row 5 already needs at least 3 criminals, row 5 cannot have even one innocent there: both unknown spots must be criminals. Therefore, we can determine that A5 is CRIMINAL and C5 is CRIMINAL.
10.C4 · Sam → INNOCENT
Zara at D5 says that of her two criminal neighbors, only one is in column C. Zara’s neighbors are C4 Sam, C5 Xavi, and D4 Umar, and the two already known criminals among them are Xavi at C5 and Umar at D4. Since exactly one of Zara’s two criminal neighbors is in column C, the column C one is already Xavi, so Sam at C4 cannot be a criminal neighbor as well. Therefore, we can determine that C4 is INNOCENT.
11.B4 · Quita → CRIMINAL
Nicole’s clue says every row contains at least 3 criminals. In row 4, we already know Peter at A4 and Umar at D4 are criminals, and Sam at C4 is innocent. That means the only way for row 4 to reach at least 3 criminals is for Quita at B4 to be the third criminal. Therefore, we can determine that B4 is CRIMINAL.
12.C2 · John → INNOCENT
Carl’s clue says there are exactly 4 criminals in column D, and since Zara at D5 is one of them, that means among D1, D2, D3, D4, and D5 exactly one person in column D is innocent. We already know D1 is criminal, D4 is criminal, and D5 is criminal, so the only possible innocent in that column must be either D2 or D3. Sam’s clue says Nicole at C3 has an odd number of criminal neighbors. Nicole’s neighbors are B2, C2, D2, B3, D3, B4, C4, and D4. Among these, B3, B4, and D4 are criminals, and C4 is innocent, so the four unknowns B2, C2, D2, and D3 must contain an even number of criminals. But B2 is already criminal, and from Carl’s clue exactly one of D2 and D3 is criminal, so among B2, D2, and D3 there are exactly 2 criminals. That forces C2 to be innocent to keep the total for those four unknowns even. Therefore, we can determine that C2 is INNOCENT.
13.D2 · Karen → CRIMINAL
Nicole’s clue says every row contains at least 3 criminals. In row 2, Hazel at A2 and Ike at B2 are already criminals, and John at C2 is innocent, so Karen at D2 must be the third criminal needed for that row to reach at least 3 criminals. Therefore, we can determine that D2 is CRIMINAL.
14.D3 · Olof → INNOCENT
Column D contains Ellie at D1, Karen at D2, Olof at D3, Umar at D4, and Zara at D5. Carl says Zara is one of 4 criminals in that column, so column D has exactly 4 criminals in total. Since D1, D2, D4, and D5 are already known to be criminals, those are the 4 criminals in column D, leaving D3 as the only non-criminal there. Therefore, we can determine that D3 is INNOCENT.
15.A3 · Lisa → CRIMINAL
Nicole’s clue says every row contains at least 3 criminals. In row 3, Mary at B3 and Nicole at C3 are already criminals, and Olof at D3 is innocent, so the only way for row 3 to still have at least 3 criminals is for Lisa at A3 to be the third criminal. Therefore, we can determine that A3 is CRIMINAL.