Clues by Sam Oct 04, 2026 Answer – Full Solution Explained
A1
👮♂️
cop
B1
👩🏫
teacher
C1
👩💻
coder
D1
🕵️♀️
sleuth
A2
🕵️♂️
sleuth
B2
👨🎨
painter
C2
👨💼
clerk
D2
👩💼
clerk
A3
👨🌾
farmer
B3
👮♂️
cop
C3
👩💻
coder
D3
👨💻
coder
A4
👩🔧
mech
B4
👩🌾
farmer
C4
👩🏫
teacher
D4
👨🎤
singer
A5
👨🌾
farmer
B5
👩🔧
mech
C5
👨🎤
singer
D5
👩🎨
painter
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 — 6 criminals.
Clues by Sam answer for Oct 04, 2026 — a Hard solved in 18 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Ghani (A2), Ike (C2), Martin (B3), Umar (A5), Vicky (B5) and Wally (C5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.A5 · Umar → CRIMINAL
Quita’s clue says row 5 has exactly 3 criminals, and exactly 2 of those criminals are Ruby’s neighbors. That means exactly 1 criminal in row 5 is not a neighbor of Ruby. In row 5, the only person who is not a neighbor of Ruby is Umar, and that group currently has 0 known criminals, so it still needs that 1 criminal. Since Umar is the only possible person there, Umar must be criminal.
02.B1 · Betsy → INNOCENT
Umar’s clue says Betsy is one of the exactly 3 innocents in column B. Since the clue names Betsy as part of that innocent group, her identity is given directly by the clue. So Betsy must be innocent.
03.A3 · Luigi → INNOCENT
Betsy’s clue says there are exactly two innocents below Aaron, and those two innocents must be connected. Below Aaron are Ghani, Luigi, Paula, and Umar, and Umar is already a criminal. If Luigi were a criminal, then the only way to get the required two innocents below Aaron would be for Ghani and Paula to be the two innocents. But with Luigi between them as a criminal, those two innocents would not be connected, which breaks Betsy’s clue. So Luigi must be innocent.
04.A1 · Aaron → INNOCENT
Column B has exactly 3 innocents, and Betsy and Quita are already 2 of them. Column A must have the same number of innocents as column B, but column A currently has only 1 known innocent, Luigi. If Aaron were criminal, then column A would need all three remaining unknowns there, Aaron, Ghani, and Paula, to supply the missing innocents count without Aaron helping, while the people below Aaron would still have to contain exactly 2 connected innocents among Ghani, Luigi, Paula, and Umar. That combination cannot be made to fit all of the listed facts at once. So Aaron must be innocent.
05.D4 · Tyler → INNOCENT
Quita’s clue fixes the two additional criminals in row 5 among Vicky, Wally, and Xena. Aaron’s clue says that among Ruby’s three criminal neighbors, exactly two are not Ike’s neighbors; in the group of Ruby’s neighbors who are not Ike’s neighbors, those possible criminals are Quita, Tyler, Vicky, Wally, and Xena. But Quita is innocent, and the two criminals required in that group are already exactly the row 5 pair from Vicky, Wally, and Xena. That uses up both criminal places in this larger group, so the only extra person there, Tyler, cannot be a criminal. So Tyler must be innocent.
06.C4 · Ruby → INNOCENT
Tyler’s clue says Ruby is one of Martin’s 6 innocent neighbors. That directly places Ruby among the innocent people in Martin’s neighboring group. So Ruby must be innocent.
07.C3 · Nala → INNOCENT
Assume Nala were criminal. Ruby has exactly 3 criminal neighbors, and exactly 1 of those criminals is also Ike's neighbor; among the shared neighbors of Ruby and Ike, Martin, Nala, and Olsi are the only people in that special group, so with Nala criminal, Martin and then Olsi must be innocent. Column B has exactly 3 innocents, and with Betsy, Quita, and now Martin innocent, Hal must be criminal, which leaves Vicky as Ruby's third criminal neighbor. Martin must have exactly 6 innocent neighbors. With Hal and Nala criminal, that forces Ghani and then Ike to be innocent; then the same clue forces Paula to be innocent too. But below Aaron, the innocents would then be Ghani, Luigi, and Paula, even though Betsy's clue says the innocents below Aaron are exactly 2 connected people. That clash shows the assumption was wrong, so Nala must be innocent.
08.C5 · Wally → CRIMINAL
Column B must contain exactly 3 innocents, and it already has Betsy and Quita as 2 of them. Column C must have the same number of innocents as column B, and it also already has 2 known innocents, Nala and Ruby. If Wally were innocent, then column C would already have its third innocent, so the remaining people named in these clues would have to satisfy the rest of the conditions without adding any more innocents there. But Betsy’s clue also requires exactly 2 criminals among Betsy’s neighbors, with exactly 1 of those criminals being Ellie’s neighbor, and those same remaining people cannot meet all of those requirements at once under that assumption. So Wally at C5 must be criminal.
09.D1 · Ellie → INNOCENT, D2 · Janet → INNOCENT
Diane’s neighbors must contain exactly 4 innocents, and right now Betsy is the only known innocent there, with Ellie, Hal, Ike, and Janet unknown. If Hal and Ike were both innocent, then Martin’s neighbors would already have Luigi, Nala, Quita, Ruby, Hal, and Ike as the 6 innocents, so Ghani would have to be criminal; then among the people below Aaron, the two innocents would have to be connected, which forces Paula to be criminal as well, leaving only Luigi innocent below Aaron and contradicting that clue. So Hal and Ike cannot both be innocent. Since Diane’s neighbors still need 4 innocents total, Hal and Ike are the only pair there that can fail to both be innocent, which means Ellie and Janet have to supply the remaining innocent spots. So Ellie and Janet must be innocent.
10.D3 · Olsi → INNOCENT
Ruby’s clue says exactly one of Betsy’s neighboring criminals is also Ellie’s neighbor, so among the shared people C1 Diane and C2 Ike there is exactly 1 criminal. Ellie’s clue says Janet has exactly 4 innocent neighbors, so among Janet’s neighbors there is exactly 1 criminal in total. Janet’s neighbors are C1 Diane, D1 Ellie, C2 Ike, C3 Nala, and D3 Olsi. Since Diane and Ike already account for the one criminal Janet’s whole neighbor group is allowed to have, the only remaining unknown there, Olsi, cannot be criminal. So Olsi must be innocent.
11.B3 · Martin → CRIMINAL
Ruby has exactly 3 criminal neighbors, and exactly 1 of those criminals is also a neighbor of Ike. Among the people who are neighbors of both Ruby and Ike, Nala and Olsi are already innocent, so that shared group still needs 1 criminal. The only unknown person left in that shared group is Martin. So Martin must be criminal.
12.C1 · Diane → INNOCENT
Olsi says there are exactly 4 criminals on the edge, and 2 edge criminals are already known: Umar and Wally. That means the remaining 2 edge criminals have to come from the unknown edge group. From the step here, those 2 remaining edge criminals are taken from Ghani, Paula, Vicky, and Xena, so Diane is not one of them. That makes Diane innocent.
13.C2 · Ike → CRIMINAL
Ruby’s clue says Betsy has exactly two criminal neighbors, and exactly one of those criminals is also a neighbor of Ellie. Among the people who are neighbors of both Betsy and Ellie, Diane is innocent and the only unknown person left is Ike. So that shared group still needs one criminal, and Ike is the only person who can fill it. That makes Ike criminal.
14.B2 · Hal → INNOCENT
Wally’s clue says Diane has exactly 4 innocent neighbors. Diane’s neighbors already include 3 known innocents, and the only neighbor there whose identity is still unknown is Hal. So Diane’s neighbor set needs 1 more innocent, and Hal is the only person who can fill that spot. That makes Hal innocent.
15.A2 · Ghani → CRIMINAL
Betsy's neighbors must include exactly 2 criminals, and exactly 1 of those criminals is also Ellie's neighbor. Among the people who are neighbors of both Betsy and Ellie, the only criminal is Ike, so the other criminal next to Betsy has to come from Betsy's neighbors who are not Ellie's neighbors. Those people are Aaron, Ghani, and Hal, and Aaron and Hal are innocent. That leaves Ghani as the only person who can fill that remaining criminal spot. So Ghani must be criminal.
16.A4 · Paula → INNOCENT
Betsy's clue says the two innocents below Aaron are connected. Below Aaron, Ghani and Umar are criminals, Luigi is innocent, and Paula is the only person there not yet identified. That means the two innocents below Aaron have to be Luigi and Paula, which gives the connected pair the clue requires. So Paula must be innocent.
17.B5 · Vicky → CRIMINAL
Umar’s clue says Betsy is one of exactly 3 innocents in column B. In column B, the people are Betsy, Hal, Martin, Quita, and Vicky, and Betsy, Hal, and Quita are already the 3 known innocents there. That leaves no room for Vicky to be another innocent in that column. So Vicky must be criminal.
18.D5 · Xena → INNOCENT
Ruby’s neighbors contain exactly 3 criminals in total. Aaron’s clue says exactly 1 of those criminals is also a neighbor of Ike, and among the people who are neighbors of both Ruby and Ike, the only criminal is Martin. So the other 2 criminals among Ruby’s neighbors must come from the remaining group: Quita, Tyler, Vicky, Wally, and Xena. But in that remaining group, Vicky and Wally are already the 2 known criminals. That leaves no room for Xena to be a criminal. So Xena must be innocent.