Clues by Sam Mar 10, 2026 Answer – Full Solution Explained
A1
👩⚖️
judge
B1
👨🍳
cook
C1
💂♂️
guard
D1
👩💼
clerk
A2
👨⚖️
judge
B2
👷♂️
builder
C2
💂♂️
guard
D2
👨💼
clerk
A3
👩🎨
painter
B3
👩🎨
painter
C3
👩🎨
painter
D3
👷♂️
builder
A4
🕵️♀️
sleuth
B4
👨🌾
farmer
C4
👩🎤
singer
D4
👩🌾
farmer
A5
🕵️♀️
sleuth
B5
👨🍳
cook
C5
👩🎤
singer
D5
👨🎤
singer
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 — 4 criminals.
Clues by Sam answer for Mar 10, 2026 — a Easy solved in 16 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Katie (A3), Lucy (B3), Petra (A4) and Sue (C4); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.C1 · David → INNOCENT, C2 · Ike → INNOCENT
The people above Sue (at C4) are David at C1, Ike at C2, and Nicole at C3. Ellie is at D1, and among those three people above Sue, only David at C1 and Ike at C2 are neighbors of Ellie (C1 is directly left of Ellie, and C2 is diagonally adjacent), while Nicole at C3 is too far away to be a neighbor. Gabe’s clue says that exactly 2 innocents above Sue are neighboring Ellie, so the only two people who can make that count reach exactly 2 must both be innocents. Therefore, we can determine that C1 David is INNOCENT and C2 Ike is INNOCENT.
02.B1 · Barnie → INNOCENT, B2 · Hank → INNOCENT
Gabe is at A2, and the only people who are neighbors of both Gabe and Alice are Barnie at B1 and Hank at B2, because Alice is in the corner at A1 and only touches A2, B1, and B2. Ike’s clue says that exactly 2 of Gabe’s 3 innocent neighbors also neighbor Alice. Since the only possible neighbors that satisfy “neighbor Gabe and also neighbor Alice” are Barnie and Hank, those two must be the “exactly 2” mentioned in the clue. That forces both Barnie and Hank to be innocent. Therefore, we can determine that B1 Barnie is INNOCENT and B2 Hank is INNOCENT.
03.A4 · Petra → CRIMINAL
Barnie’s clue talks about the people below Alice in column A, which are Gabe at A2, Katie at A3, Petra at A4, and Uma at A5. The clue says there are exactly two criminals among them, and that those two criminals are connected, meaning they must be orthogonally adjacent with no gap. Since Gabe at A2 is already known to be innocent, the two criminals must be chosen from A3, A4, and A5, and the only way for two people in that vertical line to be connected is for them to be A3 and A4 together or A4 and A5 together. In both possible connected pairs, Petra at A4 is included, so she must be one of the two criminals. Therefore, we can determine that A4 Petra is CRIMINAL.
04.C3 · Nicole → INNOCENT
Gabe at A2 has five neighbors: Alice at A1, Barnie at B1, Hank at B2, Katie at A3, and Lucy at B3. Ike’s clue says Gabe has exactly three innocent neighbors, and since Barnie and Hank are already known innocent, exactly one of Alice, Katie, and Lucy is innocent, which means the other two are criminals. Those same three people (Alice, Katie, and Lucy) are all neighbors of Hank at B2, so Hank already has exactly two criminal neighbors among them. Petra’s clue says nobody in row 2 can have more than two criminal neighbors, so Nicole at C3 cannot be a criminal neighbor of Hank as well. Therefore, we can determine that C3 Nicole is INNOCENT.
05.C4 · Sue → CRIMINAL
In row 4, Tina is at D4, so the people to the left of Tina are Petra at A4, Ronald at B4, and Sue at C4. Nicole’s clue says that there are exactly 2 criminals to the left of Tina, and that Sue is one of those 2 criminals. Since Petra is already confirmed to be a criminal, the only way for Sue to be “one of the 2 criminals” is for Sue herself to be the other criminal. Therefore, we can determine that Sue is CRIMINAL.
06.B4 · Ronald → INNOCENT
The clue talks about the people to the left of Tina in row 4, which are Petra at A4, Ronald at B4, and Sue at C4. It says that Sue is one of exactly two criminals to the left of Tina, so among those three people there must be exactly two criminals total, and Sue must be one of them. We already know Sue is a criminal and Petra is also a criminal, which already makes the required two criminals to Tina’s left. That means Ronald cannot be a criminal, or there would be three criminals to the left of Tina. Therefore, we can determine that B4 Ronald is INNOCENT.
07.D4 · Tina → INNOCENT
Look at row 4: A4 Petra is a criminal, B4 Ronald is innocent, C4 Sue is a criminal, and D4 Tina is the only one still unknown in that row. Ronald’s clue says that each row must contain at least 2 innocents. Since row 4 currently has only one confirmed innocent (Ronald), Tina must be the second innocent for that row to meet the clue. Therefore, we can determine that D4 Tina is INNOCENT.
08.C5 · Wanda → INNOCENT
Ike is at C2, so the people below him in the same column are Nicole at C3, Sue at C4, and Wanda at C5. Tina’s clue says that among those three people there are exactly 2 innocents. Nicole is already confirmed INNOCENT and Sue is already confirmed CRIMINAL, so only 1 innocent is currently accounted for below Ike. That means Wanda must be the second innocent below Ike. Therefore, we can determine that C5 Wanda is INNOCENT.
09.B5 · Vince → INNOCENT
Uma is at A5, so her only neighbors are Petra at A4, Ronald at B4, and Vince at B5. Wanda’s clue says Uma has exactly 2 innocent neighbors, meaning exactly two of those three people are innocent. Petra is already known to be a criminal, and Ronald is already known to be innocent, so the second innocent neighbor must be Vince. Therefore, we can determine that B5 Vince is INNOCENT.
10.D1 · Ellie → INNOCENT
Ike is at C2, and Ellie at D1 is one of Ike’s neighbors because D1 is diagonally adjacent to C2. Vince’s clue says that Ellie is one of Ike’s 7 innocent neighbors, which directly puts Ellie in the innocent group among Ike’s neighbors. Therefore, we can determine that D1 Ellie is INNOCENT.
11.B3 · Lucy → CRIMINAL
Ellie’s clue says that columns B and C contain the same number of criminals. In column C, everyone is known to be innocent except Sue at C4, who is a criminal, so column C has exactly 1 criminal. Column B already has Barnie, Hank, Ronald, and Vince all confirmed innocent, leaving only Lucy at B3 as the only spot where that one criminal can be. Therefore, we can determine that B3 Lucy is CRIMINAL.
12.D2 · Jason → INNOCENT, D3 · Olof → INNOCENT
Ike is at C2, and his eight neighbors are Barnie at B1, David at C1, Ellie at D1, Hank at B2, Jason at D2, Lucy at B3, Nicole at C3, and Olof at D3. Vince’s clue says that Ellie is one of Ike’s 7 innocent neighbors, so among Ike’s eight neighbors exactly seven are innocent and only one is a criminal. Lucy is already known to be a criminal, so she must be the only criminal in Ike’s neighborhood, which forces every other neighbor of Ike to be innocent, including Jason and Olof. Therefore, we can determine that D2 Jason is INNOCENT and D3 Olof is INNOCENT.
13.D5 · Xavi → INNOCENT
In column B, Barnie at B1, Hank at B2, Ronald at B4, and Vince at B5 are all already INNOCENT, and only Lucy at B3 is a CRIMINAL, so column B has exactly 1 criminal. Jason’s clue says column B has more criminals than column D, so column D must have fewer than 1 criminal, which means it has 0 criminals in total. Since D1, D2, D3, and D4 are already INNOCENT, the only way for column D to have 0 criminals is for Xavi at D5 to also be INNOCENT. Therefore, we can determine that D5 Xavi is INNOCENT.
14.A5 · Uma → INNOCENT
Gabe is at A2, and his neighbors are Alice at A1, Barnie at B1, Hank at B2, Katie at A3, and Lucy at B3. Ike’s clue says Gabe has exactly 3 innocent neighbors, and since Barnie and Hank are already innocent while Lucy is already a criminal, it follows that exactly one of Alice (A1) and Katie (A3) must be a criminal. Xavi’s clue says there are exactly 2 criminals in column A, and we already have Petra at A4 as one criminal, so among A1, A3, and A5 there can be only one more criminal; because that one is already forced to be either A1 or A3, A5 cannot be a criminal. Therefore, we can determine that A5 Uma is INNOCENT.
15.A3 · Katie → CRIMINAL
The clue talks about the people below Alice, meaning the rest of column A: Gabe at A2, Katie at A3, Petra at A4, and Uma at A5. It says “both criminals below Alice,” which means there are exactly two criminals among those four people. Petra at A4 is already known to be a criminal, and Gabe at A2 and Uma at A5 are already known to be innocent, so the only remaining person who can be the second criminal is Katie at A3. Therefore, we can determine that A3 Katie is CRIMINAL.
16.A1 · Alice → INNOCENT
Gabe is at A2, so his neighbors are Alice at A1, Barnie at B1, Hank at B2, Katie at A3, and Lucy at B3. Among these, Barnie and Hank are already known to be innocent, while Katie and Lucy are criminals. Ike’s clue says Gabe has exactly 3 innocent neighbors, so the only way to reach three is for Alice at A1 to be innocent as well; this also fits the rest of the clue because Barnie and Hank both neighbor Alice. Therefore, we can determine that A1 Alice is INNOCENT.