Clues by Sam Mar 19, 2026 Answer – Full Solution Explained
A1
💂♀️
guard
B1
💂♂️
guard
C1
👨⚕️
doctor
D1
👨🍳
cook
A2
💂♂️
guard
B2
👨💻
coder
C2
👩💻
coder
D2
👨💻
coder
A3
👨⚕️
doctor
B3
👩🔧
mech
C3
👩🔧
mech
D3
👨🔧
mech
A4
👨🌾
farmer
B4
👩🍳
cook
C4
👩⚕️
doctor
D4
👮♀️
cop
A5
👩🌾
farmer
B5
👩🌾
farmer
C5
👮♂️
cop
D5
👮♂️
cop
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 — 13 criminals.
Clues by Sam answer for Mar 19, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 13 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Bobby (B1), Erwin (D1), Floyd (A2), Gabe (B2), Hope (C2), Ivan (D2), Layla (B3), Nicole (C3), Olof (D3), Rowan (A4), Tina (C4) and Wanda (B5); the remaining 7 suspects are innocent.
The deduction chain, in plain English
01.C4 · Tina → CRIMINAL
Kevin’s clue explicitly says that Tina is one of Stella’s five criminal neighbors, which directly identifies Tina as a criminal. Therefore, we can determine that C4 Tina is CRIMINAL.
02.C3 · Nicole → CRIMINAL, D3 · Olof → CRIMINAL, B3 · Layla → CRIMINAL
Row 3 contains Kevin at A3, Layla at B3, Nicole at C3, and Olof at D3. Tina’s clue says there is only one innocent in row 3, meaning exactly one of those four people can be innocent. Since Kevin is already confirmed INNOCENT, he must be that one innocent, so the other three people in row 3 cannot be innocent. Therefore, we can determine that B3 Layla is CRIMINAL, C3 Nicole is CRIMINAL, and D3 Olof is CRIMINAL.
03.D2 · Ivan → CRIMINAL
Olof’s clue explicitly says that Ivan is one of the three criminals in column D, which directly includes Ivan among the criminals. Therefore, we can determine that D2 Ivan is CRIMINAL.
04.A4 · Rowan → CRIMINAL, D5 · Zach → INNOCENT
Stella is at B4, and her neighbors are Kevin (A3), Layla (B3), Nicole (C3), Rowan (A4), Tina (C4), Vicky (A5), Wanda (B5), and Xavi (C5). Kevin’s clue says Tina is one of Stella’s 5 criminal neighbors, so among those eight neighbors Stella must have exactly five criminals; since Layla, Nicole, and Tina are already criminals and Kevin is innocent, that forces exactly two more criminals among Rowan, Vicky, Wanda, and Xavi. Ivan’s clue says row 3 has more criminals than row 5, and row 3 already has exactly three criminals (Layla, Nicole, and Olof), so row 5 has at most two criminals. Layla’s clue says row 4 is the only row with exactly two criminals, so row 5 cannot have exactly two criminals; combined with “at most two,” row 5 has at most one criminal. That means Vicky, Wanda, and Xavi together can contain at most one criminal, so to reach the needed two criminals among Rowan, Vicky, Wanda, and Xavi, Rowan must be a criminal and exactly one of Vicky/Wanda/Xavi is a criminal, which also forces Zach (D5) to be innocent so row 5 does not exceed one criminal. Therefore, we can determine that A4 Rowan is CRIMINAL and D5 Zach is INNOCENT.
05.B4 · Stella → INNOCENT, D4 · Uma → INNOCENT
In row 4, we already know that Rowan at A4 is a criminal and Tina at C4 is a criminal. Layla’s clue says that row 4 has exactly 2 criminals, so row 4 cannot contain any additional criminals beyond Rowan and Tina. That forces the two remaining people in row 4, Stella at B4 and Uma at D4, to be innocent. Therefore, we can determine that B4 Stella is INNOCENT and D4 Uma is INNOCENT.
06.D1 · Erwin → CRIMINAL
The clue is about column D, which contains Erwin (D1), Ivan (D2), Olof (D3), Uma (D4), and Zach (D5). Olof says that Ivan is one of 3 criminals in column D, so there must be exactly three criminals somewhere in that column. We already know Ivan and Olof are criminals, and we also know Uma and Zach are innocents, so the only remaining person who can be the third criminal is Erwin at D1. Therefore, we can determine that D1 Erwin is CRIMINAL.
07.A1 · Anna → CRIMINAL
In row 1, the people to the right of Anna are Bobby at B1, Chuck at C1, and Erwin at D1. Nicole’s clue says that exactly one innocent to the right of Anna is neighboring Gabe, and the only people to the right of Anna who neighbor Gabe (at B2) are Bobby and Chuck, so exactly one of Bobby and Chuck must be innocent and the other must be criminal. Since Erwin is already a criminal, that would make row 1 have exactly two criminals unless Anna is also a criminal. Layla’s clue says row 4 is the only row with exactly two criminals, so row 1 cannot have exactly two criminals, which forces Anna to be a criminal. Therefore, we can determine that A1 Anna is CRIMINAL.
08.B2 · Gabe → CRIMINAL
Anna is at A1, so “to the right of Anna” refers to B1, C1, and D1, and we already know D1 (Erwin) is a criminal. Nicole’s clue says there is exactly one innocent to the right of Anna, so among B1 (Bobby) and C1 (Chuck) there must be exactly one innocent and exactly one criminal. Uma’s clue counts the criminals neighboring Hope at C2; Hope already has five known criminal neighbors (D1 Erwin, D2 Ivan, B3 Layla, C3 Nicole, and D3 Olof), so to keep the total odd, the number of criminals among the remaining unknown neighbors B1, C1, and B2 (Gabe) must be even. Since B1 and C1 contribute exactly one criminal between them, Gabe must be a criminal to make that even number. Therefore, we can determine that B2 Gabe is CRIMINAL.
09.A2 · Floyd → CRIMINAL
Stella is at B4, and her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Kevin’s clue says Stella has exactly 5 criminal neighbors, and we already know four of those neighbors are criminals (Layla, Nicole, Rowan, and Tina), so among A5, B5, and C5 exactly one is criminal and the other two are innocents. Nicole’s clue says that exactly one innocent to the right of Anna is neighboring Gabe; since the only positions to the right of Anna that neighbor Gabe are B1 and C1 (and D1 is already a criminal), this means exactly one of B1 and C1 is innocent. Gabe’s clue says there are exactly 6 innocents on the edges, and since the only unknown edge positions are B1, C1, A2, A5, B5, and C5, there must be exactly 3 innocents among these six; but the two innocents among A5/B5/C5 together with the one innocent among B1/C1 already make those 3, leaving no room for A2 to be innocent. Therefore, we can determine that A2 Floyd is CRIMINAL.
10.A5 · Vicky → INNOCENT
Stella is at B4, and her eight neighbors are Kevin (A3), Layla (B3), Nicole (C3), Rowan (A4), Tina (C4), Vicky (A5), Wanda (B5), and Xavi (C5). Kevin’s clue says Tina is one of Stella’s 5 criminal neighbors, and since Layla, Nicole, Rowan, and Tina are already four criminals around Stella, exactly one of Vicky, Wanda, and Xavi must be the fifth criminal neighbor. Floyd’s clue says Stella and Tina have only one innocent neighbor in common; their common neighbors are Layla (B3), Nicole (C3), Wanda (B5), and Xavi (C5), so since Layla and Nicole are criminals, exactly one of Wanda and Xavi is innocent and the other is criminal. That means Wanda and Xavi already contain exactly one criminal, so Vicky cannot be the criminal among Vicky, Wanda, and Xavi. Therefore, we can determine that A5 Vicky is INNOCENT.
11.C2 · Hope → CRIMINAL
To the right of Anna in row 1 are Bobby at B1 and Chuck at C1 (Erwin at D1 is not a neighbor of Gabe). Nicole’s clue says that among the people to the right of Anna who are neighbors of Gabe, there is exactly one innocent, and the only such neighbors are Bobby and Chuck, so exactly one of Bobby and Chuck is innocent. That means row 1 has exactly 1 innocent. Vicky’s clue says row 1 has more innocents than row 2, so row 2 must have 0 innocents, which forces Hope at C2 to be a criminal. Therefore, we can determine that C2 Hope is CRIMINAL.
12.B1 · Bobby → CRIMINAL
The doctors are Chuck at C1, Kevin at A3, and Tina at C4. Hope’s clue says that exactly one doctor has an innocent directly to their left. Tina already has Stella at B4 directly to her left, and Stella is innocent, so Tina is the one doctor that satisfies the clue. That means no other doctor can have an innocent directly to their left, so Chuck cannot have an innocent at B1. Therefore, we can determine that B1 Bobby is CRIMINAL.
13.C1 · Chuck → INNOCENT
Anna is at A1, so the people to her right in the same row are Bobby at B1, Chuck at C1, and Erwin at D1. Nicole’s clue says that among those three, exactly one is innocent, and that specific innocent is neighboring Gabe at B2; of the three to the right of Anna, only Bobby and Chuck neighbor Gabe, since Erwin at D1 is too far away. Erwin is already a criminal, and Bobby is also already a criminal, so neither of them can be the clue’s “exactly 1 innocent.” That forces Chuck at C1 to be the one innocent to the right of Anna who neighbors Gabe. Therefore, we can determine that C1 Chuck is INNOCENT.
14.B5 · Wanda → CRIMINAL
The clue compares the total number of innocent doctors to the total number of innocent farmers, and says the innocent doctors are more. On the board, the innocent doctors are Chuck at C1 and Kevin at A3, so there are exactly two innocent doctors. Vicky at A5 is already an innocent farmer, so there is at least one innocent farmer; if Wanda at B5 were also innocent, there would be two innocent farmers, which would no longer be fewer than the two innocent doctors. So Wanda cannot be an innocent farmer. Therefore, we can determine that B5 Wanda is CRIMINAL.
15.C5 · Xavi → INNOCENT
Stella is at B4, so her neighbors are Kevin (A3), Layla (B3), Nicole (C3), Rowan (A4), Tina (C4), Vicky (A5), Wanda (B5), and Xavi (C5). Kevin’s clue says that Tina is one of Stella’s 5 criminal neighbors, meaning Stella has exactly five criminal neighbors total. Among Stella’s neighbors, Layla, Nicole, Rowan, Tina, and Wanda are already known criminals, which already makes five. That leaves no room for Xavi to be a criminal, so Xavi must be innocent. Therefore, we can determine that C5 Xavi is INNOCENT.