Clues by Sam Oct 01, 2026 Answer – Full Solution Explained
A1
👨⚕️
doctor
B1
👨⚕️
doctor
C1
👩⚕️
doctor
D1
👩🎨
painter
A2
👮♀️
cop
B2
👮♂️
cop
C2
👩🔧
mech
D2
👨🎨
painter
A3
🕵️♀️
sleuth
B3
👮♀️
cop
C3
👨🔧
mech
D3
👩🎨
painter
A4
🕵️♀️
sleuth
B4
🕵️♂️
sleuth
C4
👨🎤
singer
D4
👨✈️
pilot
A5
👨🎤
singer
B5
👩🎤
singer
C5
👨✈️
pilot
D5
👩✈️
pilot
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 Oct 01, 2026 — a Tricky solved in 17 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 Chuck (B1), Donna (C1), Eve (D1), Ghani (B2), Hilda (C2), Jose (D2), Kay (A3), Olsi (C3), Penny (D3), Quita (A4), Ronald (B4), Tyler (D4) and Umar (A5); the remaining 7 suspects are innocent.
The deduction chain, in plain English
01.A4 · Quita → CRIMINAL
Xia’s clue says there are no innocents who are both below Bobby and in row 4. The only person in that shared group is Quita, and that group already has 0 innocents counted in it. So Quita cannot be innocent. That makes Quita criminal.
02.A3 · Kay → CRIMINAL
Quita’s clue says Kay is one of Ghani’s five criminal neighbors. That directly identifies Kay as a criminal. So Kay must be criminal.
03.A1 · Bobby → INNOCENT, A2 · Frida → INNOCENT
Kay’s clue says there are exactly 2 innocents above Umar. Among the people above Umar, there are currently 0 known innocents, and the only people there whose identity is still unknown are Bobby and Frida. Since those two innocent spots still have to be filled and Bobby and Frida are the only ones left in that group, both of them have to be the innocents. So Bobby and Frida must be innocent.
04.C2 · Hilda → CRIMINAL
Bobby's clue says column C contains exactly 3 criminals, and exactly 1 of those criminals is to the left of Jose. Among the people in column C who are to the left of Jose, the only person there is Hilda. That left-of-Jose spot in column C still needs that 1 criminal, so Hilda must fill it. So Hilda must be criminal.
05.A5 · Umar → CRIMINAL
Hilda’s clue says every column has at least 3 criminals, so in a 5-person column there can be at most 2 innocents. Column A already has 2 known innocents: Bobby at A1 and Frida at A2. That means no other person in column A can be innocent. So Umar must be criminal.
06.B5 · Vicky → INNOCENT, C5 · Will → INNOCENT
Frida’s clue says row 5 has exactly 3 innocents. In row 5, there is already 1 known innocent, and the only people there whose identity is still unknown are Vicky and Will. That means row 5 still needs exactly 2 more innocents, and those two unknown people are the only places they can go. So Vicky and Will must be innocent.
07.B1 · Chuck → CRIMINAL
Below Chuck, the people who neighbor Olsi are exactly Ghani, Megan, and Ronald. Umar’s clue says an odd number of those people are innocent, and none of them is already known innocent, so at least one of Ghani, Megan, or Ronald must be innocent. But column B must have at least 3 criminals. Since Vicky in column B is already innocent, that column can have at most one more innocent person. If one of Ghani, Megan, or Ronald is that extra innocent, then column B has no room left for Chuck to be innocent as well. So Chuck must be criminal.
08.B2 · Ghani → CRIMINAL
If Ghani were innocent, then column B would need its other three undecided people to supply the criminals required by the clue that each column has at least 3 criminals. That makes Megan and Ronald criminals. But the people strictly between Ghani and Vicky are Megan and Ronald, and that group must contain exactly 1 innocent. If both Megan and Ronald are criminals, there would be no innocent there, which clashes with Chuck's clue. So Ghani at B2 must be criminal.
09.C1 · Donna → CRIMINAL
Ghani’s clue says exactly one corner person has a criminal directly to the left. The known corners do not yet supply that one required case, so there must be one more such corner case. The only remaining direct-left position that can affect this clue is Donna at C1, so Donna has to be the criminal that creates it. So Donna must be criminal.
10.D3 · Penny → CRIMINAL
Row 3 must contain an odd number of criminals, and it already has 1 known criminal, Kay. So the remaining people in row 3, Megan, Olsi, and Penny, must add an even number of criminals. Ghani's neighbors contain exactly 5 criminals, and among those neighbors there are already 4 known criminals. The only unknown neighbors are Megan and Olsi, so exactly 1 of Megan and Olsi is a criminal. That gives row 3 one criminal from Megan and Olsi, which is odd, so Penny has to add one more criminal to make the total added in row 3 even. So Penny must be criminal.
11.D2 · Jose → CRIMINAL
Penny’s clue says Jose is one of Olsi’s 6 criminal neighbors. That directly places Jose among the criminal neighbors named by the clue. So Jose must be criminal.
12.D1 · Eve → CRIMINAL
If Eve were innocent, then row 1 would have 2 innocents while row 3 would need 2 innocents as well, because those two rows must contain the same number of innocents. Quita’s clue says Ghani has exactly 5 criminal neighbors, and Ghani’s neighbors already include 4 known criminals, so the only two unknown neighbors, Megan and Olsi, cannot be criminals. That makes Megan and Olsi both innocent, so row 3 would have only 2 innocents by using both of them, while Ghani’s neighbor count is also fully fixed by that same fact. This clash comes from assuming Eve is innocent, so Eve at D1 must be criminal.
13.C3 · Olsi → CRIMINAL
Olsi’s clue says his neighbors contain exactly 6 criminals, and four of those neighbors are already known criminals: Ghani, Hilda, Jose, and Penny. The remaining neighbors involved here are Megan, Ronald, Scott, and Tyler. Row 4 must have exactly 2 criminals who have a criminal directly above them, and in row 4 the people involved are Quita, Ronald, Scott, and Tyler. Testing Olsi as innocent makes the remaining people named here have to satisfy both Olsi’s neighbor total and that row 4 requirement at the same time, and they cannot do that. So Olsi must be criminal.
14.B3 · Megan → INNOCENT
Quita's clue says Kay is one of the exactly five criminal neighbors of Ghani. Among Ghani's neighbors, the five already known criminals are Chuck, Donna, Hilda, Kay, and Olsi, while Bobby and Frida are innocent and Megan is the only unknown there. Since all five criminal neighbor spots are already filled, Megan cannot be criminal. So Megan must be innocent.
15.B4 · Ronald → CRIMINAL
Hilda’s clue says every column has at least 3 criminals, so in column B there can be at most 2 innocents. Column B already has 2 known innocents there, Chuck and Ghani are criminals, and Ronald at B4 is the only person in that column whose identity was not fixed yet. That means column B cannot fit another innocent. So Ronald must be criminal.
16.C4 · Scott → INNOCENT
Bobby's clue says column C has exactly 3 criminals in total, and exactly 1 of those criminals is to the left of Jose. That one criminal is already identified as Hilda. So among the rest of column C, Donna and Olsi are already the other 2 criminals, while Will is innocent. That leaves no room for Scott to be a criminal, so Scott must be innocent.
17.D4 · Tyler → CRIMINAL
Penny’s clue says that Jose is one of the exactly 6 criminal neighbors of Olsi. Among Olsi’s neighbors, five are already known criminals: Ghani, Hilda, Jose, Penny, and Ronald, while Megan and Scott are innocent. The only neighbor there whose identity is not yet known is Tyler, so Tyler has to be the sixth criminal. So Tyler must be criminal.