Clues by Sam Mar 09, 2026 Answer – Full Solution Explained
A1
💂♂️
guard
B1
👩⚖️
judge
C1
👩⚖️
judge
D1
👩💼
clerk
A2
👩💻
coder
B2
👨⚕️
doctor
C2
🕵️♂️
sleuth
D2
👨🏫
teacher
A3
👨💻
coder
B3
👷♂️
builder
C3
🕵️♀️
sleuth
D3
👨⚕️
doctor
A4
👨🎨
painter
B4
👩🎨
painter
C4
👨⚕️
doctor
D4
👨💼
clerk
A5
💂♀️
guard
B5
👩🏫
teacher
C5
👷♀️
builder
D5
👷♀️
builder
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 — 14 criminals.
Clues by Sam answer for Mar 09, 2026 — a Easy solved in 15 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 14 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Barnie (A1), Chloe (B1), Diane (C1), Evie (D1), Flora (A2), Henry (C2), Kumar (B3), Lucy (C3), Mark (D3), Noah (A4), Tyler (D4), Vicky (A5), Wanda (B5) and Zara (D5); the remaining 6 suspects are innocent.
The deduction chain, in plain English
01.D4 · Tyler → CRIMINAL, D5 · Zara → CRIMINAL, B5 · Wanda → CRIMINAL
Scott is at C4 and Xia is at C5, so the people who are neighbors of both of them are exactly Pam at B4, Tyler at D4, Wanda at B5, and Zara at D5. Pam’s clue says Scott and Xia have only one innocent neighbor in common, meaning exactly one of those four shared neighbors is innocent. Since Pam is already known to be innocent, she must be that one innocent shared neighbor. That forces the other shared neighbors, Tyler, Wanda, and Zara, to all be criminals. Therefore, we can determine that D4 Tyler is CRIMINAL, D5 Zara is CRIMINAL, and B5 Wanda is CRIMINAL.
02.C1 · Diane → CRIMINAL, C2 · Henry → CRIMINAL
Scott is at C4, so the people above Scott are C3 Lucy, C2 Henry, and C1 Diane. Chloe is at B1, and among those three, only Diane at C1 and Henry at C2 are neighbors of Chloe (they are diagonally and directly adjacent to B1, respectively), while Lucy at C3 is too far away to be a neighbor. Wanda’s clue says that none of the people above Scott who neighbor Chloe are innocents, so Diane and Henry cannot be innocent. Therefore, we can determine that C1 Diane is CRIMINAL and C2 Henry is CRIMINAL.
03.C5 · Xia → INNOCENT
Vicky is at A5 and Zara is at D5, so the people strictly between them in row 5 are Wanda at B5 and Xia at C5. Henry’s clue says there is only one innocent in between Vicky and Zara, meaning exactly one of Wanda and Xia is innocent. Since Wanda is already known to be a criminal, the only way to have exactly one innocent between Vicky and Zara is for Xia to be the innocent. Therefore, we can determine that C5 Xia is INNOCENT.
04.C4 · Scott → INNOCENT
Xia’s clue says that Scott is one of Mark’s 2 innocent neighbors, which means Scott is an innocent neighbor of Mark. Therefore, we can determine that C4 Scott is INNOCENT.
05.A4 · Noah → CRIMINAL
In row 4, the people are Noah at A4, Pam at B4, Scott at C4, and Tyler at D4. Scott’s clue says each row has at least 2 criminals, so row 4 must contain at least two criminals. Tyler is already a criminal, and both Pam and Scott are already confirmed innocents, so they cannot be the second criminal. That leaves only Noah as the only possible person who can make row 4 reach at least two criminals. Therefore, we can determine that A4 Noah is CRIMINAL.
06.D2 · Ike → INNOCENT
Row 2 is A2 Flora, B2 Gary, C2 Henry, and D2 Ike, and Barnie at A1 is a neighbor only of A2 and B2 in that row (not C2 or D2). Noah’s clue says there are exactly two innocents in row 2, and exactly one of those two innocents is a neighbor of Barnie. Since Henry at C2 is already a criminal, the two row 2 innocents must come from A2, B2, and D2. If Ike at D2 were a criminal, then the only way to have two innocents in row 2 would be A2 and B2, but then both row 2 innocents would be Barnie’s neighbors, which would make the number equal to two instead of one. Therefore, we can determine that D2 Ike is INNOCENT.
07.C3 · Lucy → CRIMINAL
Mark is at D3, so his neighbors are C2 (Henry), D2 (Ike), C3 (Lucy), C4 (Scott), and D4 (Tyler). Xia’s clue says that Mark has exactly 2 innocent neighbors, and Scott is one of them. Since Ike is already known to be innocent, Scott and Ike fill up Mark’s two innocent-neighbor slots. That means Lucy, as another neighbor of Mark at C3, cannot be innocent and must be criminal. Therefore, we can determine that C3 (Lucy) is CRIMINAL.
08.A5 · Vicky → CRIMINAL
Row 4 has Pam and Scott as innocents, so there are exactly 2 innocents in row 4. Ike’s clue says there are more innocents in row 4 than in row 5, so row 5 must have fewer than 2 innocents. Row 5 already has Xia as an innocent, so it cannot have any other innocent besides Xia. Therefore, we can determine that A5 Vicky is CRIMINAL.
09.B1 · Chloe → CRIMINAL
Look at the four corner people: A1 Barnie, D1 Evie, A5 Vicky, and D5 Zara. Zara at D5 already has exactly one criminal neighbor, because among her neighbors only Tyler at D4 is a criminal while Scott at C4 and Xia at C5 are innocents, so D5 definitely fits “exactly one criminal neighbor.” Vicky’s clue says only one corner person fits that description, so Barnie at A1 cannot also have exactly one criminal neighbor. Barnie’s neighbors are A2 Flora, B2 Gary, and B1 Chloe, and Noah’s clue forces row 2 to have exactly two innocents, meaning exactly one of Flora and Gary is a criminal (since Ike is innocent and Henry is criminal). That guarantees Barnie has exactly one criminal neighbor among A2 and B2, so the only way for Barnie not to have exactly one criminal neighbor overall is for Chloe at B1 to be a criminal as well, making Barnie have at least two criminal neighbors. Therefore, we can determine that B1 Chloe is CRIMINAL.
10.A1 · Barnie → CRIMINAL, D1 · Evie → CRIMINAL
In row 1, Chloe at B1 and Diane at C1 are already known criminals, while Barnie at A1 and Evie at D1 are the only unknowns in that row. Chloe’s clue says row 1 has more criminals than any other row, and row 5 already has three criminals (Vicky, Wanda, and Zara). To have more than three criminals, row 1 must have four criminals, so both of the unknown spots in row 1 must be criminals. Therefore, we can determine that A1 Barnie is CRIMINAL and D1 Evie is CRIMINAL.
11.A3 · John → INNOCENT
Flora is at A2, and her neighbors are Barnie at A1, Chloe at B1, Gary at B2, John at A3, and Kumar at B3. The people in between Chloe at B1 and Wanda at B5 are exactly the ones in column B on rows 2–4: Gary, Kumar, and Pam. Among Flora’s neighbors, the only ones who are in that “in between” group are Gary and Kumar, and the clue also says Flora has exactly two innocent neighbors with only one of them in between Chloe and Wanda. Since Barnie and Chloe are already criminals, the innocent neighbor that is not “in between” must be John at A3. Therefore, we can determine that A3 John is INNOCENT.
12.B3 · Kumar → CRIMINAL
Row 2 contains Flora at A2, Gary at B2, Henry at C2, and Ike at D2. Noah’s clue says there are exactly two innocents in row 2, and since Ike is already one of them and Henry is already a criminal, exactly one of Flora or Gary is innocent and the other is a criminal. John at A3 has five neighbors: A2, B2, B3, A4, and B4; among these, Noah at A4 is a criminal and Pam at B4 is innocent, so the odd-or-even question depends only on A2, B2, and B3. John’s clue says he has an odd number of criminal neighbors, and since A2 and B2 contribute exactly one criminal between them, B3 must also be a criminal to make the total among A2, B2, and B3 even, keeping the overall neighbor-criminal count odd. Therefore, we can determine that B3 Kumar is CRIMINAL.
13.B2 · Gary → INNOCENT
Flora is at A2, so her neighbors are Barnie at A1, Chloe at B1, Gary at B2, John at A3, and Kumar at B3. Barnie, Chloe, and Kumar are already known criminals, and John is already known innocent. Barnie’s clue says that Flora has exactly two innocent neighbors, so the only remaining neighbor who can be the second innocent is Gary. Therefore, we can determine that B2 Gary is INNOCENT.
14.A2 · Flora → CRIMINAL
In row 2, the people are Flora at A2, Gary at B2, Henry at C2, and Ike at D2. Scott’s clue says that each row must contain at least 2 criminals. In row 2 we already know Henry is a criminal, while Gary and Ike are innocents, so the only way for row 2 to reach at least 2 criminals is for Flora to be the second criminal. Therefore, we can determine that A2 Flora is CRIMINAL.
15.D3 · Mark → CRIMINAL
Flora at A2 says there are 14 criminals in total on the 4×5 board, which means there must be exactly 6 innocents overall. Right now we already have exactly 6 people marked INNOCENT: B2 Gary, D2 Ike, A3 John, B4 Pam, C4 Scott, and C5 Xia. Since that already uses up all 6 innocents allowed, everyone else on the board must be a criminal, including Mark at D3. Therefore, we can determine that D3 Mark is CRIMINAL.