Clues by Sam Mar 25, 2026 Answer – Full Solution Explained
A1
🕵️♀️
sleuth
B1
👩⚖️
judge
C1
👨🔧
mech
D1
👩🌾
farmer
A2
👨⚕️
doctor
B2
👨🎤
singer
C2
👨🎤
singer
D2
👨🔧
mech
A3
🕵️♂️
sleuth
B3
👩💻
coder
C3
👨🌾
farmer
D3
👨⚖️
judge
A4
👷♂️
builder
B4
👷♂️
builder
C4
👷♀️
builder
D4
👩💻
coder
A5
👮♀️
cop
B5
👮♀️
cop
C5
👨⚕️
doctor
D5
👩💻
coder
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 Mar 25, 2026 — a Medium solved in 15 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 16 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Bonnie (B1), Diane (D1), Erwin (A2), Gabe (B2), Hank (C2), Isaac (D2), Klay (A3), Laura (B3), Martin (C3), Peter (A4), Stella (C4), Tina (D4), Uma (A5), Vera (B5) and Will (C5); the remaining 4 suspects are innocent.
The deduction chain, in plain English
01.A3 · Klay → CRIMINAL
In row 3, the edge positions are A3 and D3. Noah at D3 is already known to be innocent, so there is already one innocent in row 3 who is on an edge. Noah’s clue says there is exactly one innocent in row 3 on the edges, so A3 cannot also be innocent. Therefore, we can determine that A3 (Klay) is CRIMINAL.
02.C4 · Stella → CRIMINAL
Klay’s clue explicitly says, “Stella is one of 4 criminals in column C,” which directly states that Stella is a criminal. Therefore, we can determine that C4 Stella is CRIMINAL.
03.C5 · Will → CRIMINAL
In column C there are Carl at C1, Hank at C2, Martin at C3, Stella at C4, and Will at C5. Klay’s clue says Stella is one of 4 criminals in column C, so there is exactly 1 innocent in that column. Stella’s clue says an odd number of the innocents in column C are Gabe’s neighbors, and the only people in column C who are neighbors of Gabe at B2 are C1, C2, and C3. Since there is only one innocent in column C, for that count to be odd the lone innocent must be among C1–C3, which means Will at C5 cannot be the innocent in column C. Therefore, we can determine that C5 (Will) is CRIMINAL.
04.A2 · Erwin → CRIMINAL
Will’s clue explicitly says that Erwin is one of Laura’s criminal neighbors, which directly assigns Erwin the status CRIMINAL. Therefore, we can determine that A2 Erwin is CRIMINAL.
05.D1 · Diane → CRIMINAL, D2 · Isaac → CRIMINAL
The people above Tina are exactly the three people in column D above D4: Diane at D1, Isaac at D2, and Noah at D3. Erwin’s clue says that among those people, there is only one innocent in total. Since Noah is already known to be innocent, he must be that one innocent above Tina, so Diane and Isaac cannot be innocent. Therefore, we can determine that D1 Diane is CRIMINAL and D2 Isaac is CRIMINAL.
06.D5 · Xia → INNOCENT
Tina is at D4, and her edge neighbors are D3 (Noah), C5 (Will), and D5 (Xia). Diane’s clue says there are exactly three innocents on the entire edge, and exactly two of those edge-innocents are Tina’s neighbors. Since Noah at D3 is already an innocent and Will at C5 is already a criminal, the only remaining way for Tina to have two edge-innocent neighbors is for Xia at D5 to be the second one. Therefore, we can determine that D5 Xia is INNOCENT.
07.D4 · Tina → CRIMINAL
In column D, Diane at D1 and Isaac at D2 are already known criminals, while Noah at D3 and Xia at D5 are already known innocents. Xia’s clue says each column has at least 3 criminals, so column D must contain a third criminal somewhere. Since the only remaining undetermined person in column D is Tina at D4, she must be that third criminal. Therefore, we can determine that D4 Tina is CRIMINAL.
08.A4 · Peter → CRIMINAL, A5 · Uma → CRIMINAL, B5 · Vera → CRIMINAL
Tina at D4 has five neighbors: C3, D3, C4, C5, and D5. Diane’s clue says there are exactly three innocents on the edges in total, and exactly two of those edge-innocents are Tina’s neighbors; since D3 and D5 are already known to be innocents and are both Tina’s neighbors, they must be the two mentioned, leaving only one more edge-innocent anywhere else. Tina’s own clue says row 1 has an odd number of innocents, and since Diane at D1 is a criminal, that means A1, B1, and C1 contain either one innocent or three innocents; it cannot be three, because that would already create more than three edge-innocents when combined with D3 and D5. So exactly one of A1, B1, and C1 is the one remaining edge-innocent, which means no other edge positions can be innocent, including A4, A5, and B5. Therefore, we can determine that A4 (Peter) is CRIMINAL, A5 (Uma) is CRIMINAL, and B5 (Vera) is CRIMINAL.
09.A1 · Anna → CRIMINAL
In column A, we already know Erwin at A2, Klay at A3, Peter at A4, and Uma at A5 are criminals, so column A has at least 4 criminals. Klay’s clue says Stella is one of 4 criminals in column C, which fixes column C at exactly 4 criminals total. Uma’s clue says column A has more criminals than any other column, so column A must have more than 4 criminals, meaning it must have all 5 people as criminals. Therefore, we can determine that A1 Anna is CRIMINAL.
10.B2 · Gabe → CRIMINAL, C2 · Hank → CRIMINAL
Row 5 already has three criminals (Uma at A5, Vera at B5, and Will at C5), because only Xia at D5 is innocent. Anna’s clue says row 2 has more criminals than row 5, so row 2 must have at least four criminals. In row 2, Erwin at A2 and Isaac at D2 are already criminals, so the only way for row 2 to reach four criminals is for both remaining spots, Gabe at B2 and Hank at C2, to be criminals as well. Therefore, we can determine that B2 Gabe is CRIMINAL and C2 Hank is CRIMINAL.
11.B3 · Laura → CRIMINAL
Laura is at B3, so her eight neighbors are Erwin (A2), Gabe (B2), Hank (C2), Klay (A3), Martin (C3), Peter (A4), Ryan (B4), and Stella (C4). Will’s clue says Erwin is one of Laura’s 7 criminal neighbors, so among those eight neighbors exactly seven are criminals; since six of them are already known criminals (Erwin, Gabe, Hank, Klay, Peter, Stella), exactly one of Martin and Ryan must be criminal and the other must be innocent. Stella is at C4, and her neighbors include Noah (D3) and Xia (D5), who are already two innocents, plus the three not-yet-fixed people Laura (B3), Martin (C3), and Ryan (B4). Hank’s clue says Stella has an odd number of innocent neighbors, so the number of innocents among Laura, Martin, and Ryan must be odd; but Martin and Ryan together contribute exactly one innocent, so Laura must contribute zero innocents, meaning she is a criminal. Therefore, we can determine that B3 · Laura is CRIMINAL.
12.C3 · Martin → CRIMINAL
In row 5, Uma at A5, Vera at B5, and Will at C5 are criminals, while Xia at D5 is innocent, so row 5 has exactly three criminals. Laura’s clue says rows 3 and 5 have an equal number of criminals, so row 3 must also have exactly three criminals. In row 3, Klay at A3 and Laura at B3 are already criminals and Noah at D3 is innocent, so the only way for row 3 to reach three criminals is for Martin at C3 to be a criminal. Therefore, we can determine that C3 Martin is CRIMINAL.
13.C1 · Carl → INNOCENT
In column C, the people are Carl at C1, Hank at C2, Martin at C3, Stella at C4, and Will at C5. Klay’s clue says that Stella is one of exactly 4 criminals in column C, so column C must contain exactly 4 criminals total. Since Hank, Martin, Stella, and Will are already confirmed criminals, the only way for column C to have exactly 4 criminals is for Carl to be the lone innocent there. Therefore, we can determine that C1 Carl is INNOCENT.
14.B1 · Bonnie → CRIMINAL
The clue from Diane says “Exactly 2 of the 3 innocents on the edges are Tina’s neighbors,” which only makes sense if there are exactly three innocents on edge spaces in total. Right now we already know three edge-positioned innocents: Carl at C1, Noah at D3, and Xia at D5. Bonnie is also on an edge space (B1), so if Bonnie were innocent there would be four innocents on the edges, and Diane’s clue would not be able to truthfully refer to “the 3 innocents on the edges.” Therefore, we can determine that B1 Bonnie is CRIMINAL.
15.B4 · Ryan → INNOCENT
Laura is at B3, and she has exactly eight neighbors: Erwin at A2, Gabe at B2, Hank at C2, Klay at A3, Martin at C3, Peter at A4, Ryan at B4, and Stella at C4. Will’s clue says that Erwin is one of Laura’s 7 criminal neighbors, which means that among Laura’s eight neighbors, exactly seven are criminals. In that neighbor set, Erwin, Gabe, Hank, Klay, Martin, Peter, and Stella are already known to be criminals, which fills all seven criminal-neighbor slots. That leaves Ryan as the only neighbor who cannot be a criminal. Therefore, we can determine that B4 Ryan is INNOCENT.