Clues by Sam Feb 23, 2026 Answer – Full Solution Explained
A1
👩💻
coder
B1
💂♂️
guard
C1
💂♂️
guard
D1
🕵️♀️
sleuth
A2
👨🏫
teacher
B2
👩🏫
teacher
C2
🕵️♂️
sleuth
D2
🕵️♂️
sleuth
A3
👩🎤
singer
B3
👩🔧
mech
C3
👨🍳
cook
D3
👩🍳
cook
A4
👨🔧
mech
B4
👩🔧
mech
C4
👩🎤
singer
D4
👨💻
coder
A5
👩🎤
singer
B5
👩🍳
cook
C5
👨💻
coder
D5
👨🏫
teacher
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 Feb 23, 2026 — a Easy solved in 16 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Eric (C1), Phil (A4), Saga (C4), Tyler (D4) and Ziad (D5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.B2 · Hazel → INNOCENT, D2 · Jerry → INNOCENT
Gus is at A2, so the people to the right of Gus in row 2 are Hazel at B2, Isaac at C2, and Jerry at D2. Isaac is at C2, and among those people, the only ones who are neighboring Isaac are Hazel (directly left of him) and Jerry (directly right of him). Noah’s clue says that exactly 2 innocents to the right of Gus are neighboring Isaac, and there are exactly two such neighbors available. So Hazel and Jerry must be those two innocents. Therefore, we can determine that B2 · Hazel is INNOCENT and D2 · Jerry is INNOCENT.
02.C2 · Isaac → INNOCENT
Jerry’s clue explicitly says that Isaac is one of Olga’s 3 innocent neighbors, which directly states that Isaac is innocent. Therefore, we can determine that C2 Isaac is INNOCENT.
03.C4 · Saga → CRIMINAL, D4 · Tyler → CRIMINAL
Olga is at D3, so her neighbors are Isaac at C2, Jerry at D2, Noah at C3, Saga at C4, and Tyler at D4. Jerry’s clue says that Isaac is one of Olga’s 3 innocent neighbors, meaning Olga has exactly three innocent neighbors in total. We already know Isaac, Jerry, and Noah are innocent, which fills all three of Olga’s innocent-neighbor slots. That leaves Saga and Tyler as Olga’s remaining neighbors, so neither of them can be innocent. Therefore, we can determine that C4 Saga is CRIMINAL and D4 Tyler is CRIMINAL.
04.A2 · Gus → INNOCENT
Row 2 contains Hazel at B2, Isaac at C2, and Jerry at D2, and all three are already known to be innocent, leaving only Gus at A2 unknown. Isaac’s clue says that row 5 is the only row that has exactly one criminal, so no other row is allowed to have exactly one criminal. If Gus were a criminal, then row 2 would have exactly one criminal (Gus), which the clue forbids. Therefore, we can determine that A2 Gus is INNOCENT.
05.B3 · Lucy → INNOCENT, B5 · Vera → INNOCENT
The three singers are Karen at A3, Saga at C4, and Uma at A5. Gus’s clue says that exactly two of these singers have an innocent directly to their right. Saga’s direct right neighbor is Tyler at D4, and Tyler is a known criminal, so Saga is not one of the singers who has an innocent to the right. That forces the other two singers, Karen and Uma, to be the ones whose right neighbors are innocent, so Lucy at B3 and Vera at B5 must both be innocent. Therefore, we can determine that B3 Lucy is INNOCENT and B5 Vera is INNOCENT.
06.B4 · Ruby → INNOCENT
Phil is at A4, so the people to the right of Phil in the same row are Ruby at B4, Saga at C4, and Tyler at D4. Vera’s clue says there is only one innocent among those people to the right of Phil. Since Saga and Tyler are already known to be criminals, the only remaining person who can be that single innocent is Ruby. Therefore, we can determine that B4 Ruby is INNOCENT.
07.C5 · Wally → INNOCENT
Saga is at C4, and the only people who are both neighbors of Saga and also neighbors of Ziad at D5 are Tyler at D4 and Wally at C5. Ruby’s clue talks about “the 6 innocents neighboring Saga” and says that exactly one of those innocents is a neighbor of Ziad. Tyler cannot be that one because Tyler is already known to be a criminal, so the only way the clue can be true is if Wally is an innocent neighbor of Saga and also Ziad’s neighbor. Therefore, we can determine that C5 Wally is INNOCENT.
08.C1 · Eric → CRIMINAL
Wally is at C5, so the people above him are Eric at C1, Isaac at C2, Noah at C3, and Saga at C4. His clue says there are exactly 2 innocents above him, and Isaac and Noah are already confirmed innocents, making exactly 2. That means neither Eric nor Saga can be innocent, and since Saga is already known to be a criminal, Eric must also be a criminal. Therefore, we can determine that C1 Eric is CRIMINAL.
09.D1 · Freya → INNOCENT
Saga is at C4, and her neighbors include Lucy, Noah, Ruby, Vera, Wally, Olga at D3, and Ziad at D5 (Tyler at D4 is also a neighbor but is a criminal). Ruby’s clue says Saga has exactly 6 innocent neighbors, and we already see five of them for sure (Lucy, Noah, Ruby, Vera, and Wally), so exactly one of Olga (D3) and Ziad (D5) must be innocent. Eric’s clue says there is an odd number of innocents in column D; since Jerry at D2 is already innocent, the other three spots in column D (D1, D3, D5) must contain an even number of innocents. Because D3 and D5 contain exactly one innocent between them, D1 must be innocent to make the total for (D1, D3, D5) equal to two. Therefore, we can determine that D1 Freya is INNOCENT.
10.A4 · Phil → CRIMINAL
The clue talks about the people above Uma, which are the four positions A1, A2, A3, and A4 in column A. It says there are exactly two innocents among those four, and that those two innocents are connected, meaning they must touch orthogonally. Since Gus at A2 is already an innocent, the other innocent above Uma has to be either A1 or A3 so it is directly next to A2; it cannot be A4, because A2 and A4 are not adjacent and having A3 also innocent would make more than two innocents. Therefore, we can determine that A4 Phil is CRIMINAL.
11.A5 · Uma → INNOCENT
Uma is at A5, so “above Uma” refers to A1, A2, A3, and A4 in column A. Freya’s clue says there are exactly two innocents among those four people, and those two innocents are connected; since A4 is already a criminal and A2 is already an innocent, this means there are exactly two innocents total in A1–A4. Phil’s clue says each column has at least 3 innocents, so column A must contain at least three innocents overall; with only two innocents allowed in A1–A4, the third required innocent in column A must be Uma at A5. Therefore, we can determine that A5 Uma is INNOCENT.
12.D5 · Ziad → CRIMINAL
In row 5 we have Uma at A5, Vera at B5, Wally at C5, and Ziad at D5. Isaac’s clue says that row 5 has exactly one criminal. Since Uma, Vera, and Wally are already confirmed innocent, the only person in row 5 who can be that single criminal is Ziad. Therefore, we can determine that D5 Ziad is CRIMINAL.
13.D3 · Olga → INNOCENT
Saga is at C4, and her neighbors are B3 Lucy, C3 Noah, D3 Olga, B4 Ruby, D4 Tyler, B5 Vera, C5 Wally, and D5 Ziad. Ruby’s clue says “the 6 innocents neighboring Saga,” so among those eight neighbors exactly six must be innocent. We already know five of them are innocent (Lucy, Noah, Ruby, Vera, and Wally), and two of them are criminal (Tyler and Ziad), so the only way to reach six innocent neighbors is for Olga at D3 to be innocent as well. Therefore, we can determine that D3 Olga is INNOCENT.
14.A3 · Karen → INNOCENT
Karen is at A3 in row 3, where Lucy at B3, Noah at C3, and Olga at D3 are already known to be innocent. Isaac’s clue says that row 5 is the only row with exactly one criminal, and row 5 already has exactly one criminal (Ziad at D5) with the other three people in that row innocent. That means no other row is allowed to have exactly one criminal, so row 3 cannot end up with exactly one criminal. If Karen were a criminal, row 3 would have exactly one criminal, so Karen must be innocent instead. Therefore, we can determine that A3 Karen is INNOCENT.
15.A1 · Anna → CRIMINAL
Uma is at A5, so the people above her in column A are Phil at A4, Karen at A3, Gus at A2, and Anna at A1. Freya’s clue says there are exactly two innocents above Uma, and those two innocents must be connected. Gus and Karen are already the two innocents above Uma, and they are connected because A2 and A3 are directly adjacent. That means Anna cannot also be an innocent above Uma, so Anna must be a criminal. Therefore, we can determine that A1 Anna is CRIMINAL.
16.B1 · Donald → INNOCENT
Look at the rows and count how many innocents each one has. Row 4 has Phil, Saga, and Tyler as criminals, with only Ruby as innocent, so row 4 already has exactly one innocent. Karen’s clue says only one row on the whole board has exactly one innocent, so no other row can also end up with exactly one innocent. In row 1, we already have Anna and Eric as criminals and Freya as innocent, so if Donald were a criminal then row 1 would also have exactly one innocent, which is not allowed. Therefore, we can determine that B1 Donald is INNOCENT.