Clues by Sam Jun 05, 2026 Answer – Full Solution Explained
A1
👩🔧
mech
B1
👩🌾
farmer
C1
👨🔧
mech
D1
👨🔧
mech
A2
👨🌾
farmer
B2
👩🌾
farmer
C2
👮♂️
cop
D2
👷♀️
builder
A3
👮♂️
cop
B3
👨🍳
cook
C3
👨🍳
cook
D3
👷♂️
builder
A4
💂♀️
guard
B4
👨💼
clerk
C4
👩🎨
painter
D4
👷♂️
builder
A5
👩🎨
painter
B5
👩💼
clerk
C5
👩💼
clerk
D5
💂♀️
guard
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 — 14 criminals.
Clues by Sam answer for Jun 05, 2026 — a Hard solved in 16 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 14 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Betsy (B1), David (C1), Eli (D1), Gabe (A2), Habiba (B2), Igor (C2), Joyce (D2), Martin (C3), Olof (D3), Tina (C4), Umar (D4), Wanda (B5), Xola (C5) and Zoe (D5); the remaining 6 suspects are innocent.
The deduction chain, in plain English
01.C3 · Martin → CRIMINAL, C4 · Tina → CRIMINAL
Ruth’s clue says Steve has exactly 4 criminal neighbors, and exactly 2 of those criminals also neighbor Olof. Among Steve’s neighbors, the only people who also neighbor Olof are Martin and Tina. Those two spots are exactly the 2 criminal neighbors of Steve that the clue requires, so both of them have to be criminal. So Martin and Tina must be criminal.
02.A3 · Keith → INNOCENT, B3 · Logan → INNOCENT
Steve’s neighbors must contain exactly 4 criminals. Martin and Tina are already known criminals there, so the other 2 criminals among Steve’s neighbors have to come from the remaining candidates named in this step. Those 2 must be Vera, Wanda, and Xola’s group, which means Keith and Logan are not part of Steve’s remaining criminal slots. So Keith and Logan must be innocent.
03.D3 · Olof → CRIMINAL
Keith’s clue says every row has at least 2 criminals, so row 3 can have at most 2 innocents. Row 3 already has 2 known innocents, Keith and Logan. The only person left in row 3 whose status was not fixed is Olof, so Olof cannot also be innocent. So Olof must be criminal.
04.D5 · Zoe → CRIMINAL
Steve’s clue fixes his neighborhood at exactly 4 criminals, and the only two of those criminals who also neighbor Olof are already Martin and Tina. Keith’s clue also requires every row, including rows 4 and 5, to have at least 2 criminals, while Martin’s clue says row 4 must contain more innocents than row 5. If Zoe were innocent, the remaining unknown people named here would have to satisfy all of those requirements at the same time, but they cannot. That rules out Zoe being innocent. So Zoe must be criminal.
05.B1 · Betsy → CRIMINAL
Zoe’s clue directly says that Betsy is one of the three criminals in column B. Since the clue explicitly includes Betsy among the criminals there, Betsy must be criminal.
06.B2 · Habiba → CRIMINAL
Vera’s neighbors must contain exactly 2 innocents. Ruth is already innocent there, so among Steve and Wanda there is exactly 1 innocent. Column B must contain exactly 3 criminals, so with Betsy already criminal and Logan innocent, column B contains exactly 2 innocents in total. Steve and Wanda are in column B, and they already account for the one innocent still needed among Vera’s neighbors, which is enough to fill column B’s remaining innocent places together with Logan. That leaves the only other person in column B, Habiba, unable to be innocent. So Habiba must be criminal.
07.C2 · Igor → CRIMINAL, D2 · Joyce → CRIMINAL
Martin says his neighbors contain exactly 6 criminals. Right now that group has 3 known criminals, and the unknown members there are Igor, Joyce, Steve, and Umar. If Steve and Umar were both criminals, then row 4 would have only Ruth as an innocent, so row 5 would need no innocents to keep row 4 ahead. But Steve’s clue says he has exactly 4 criminal neighbors, and exactly 2 of those criminals also neighbor Olof; those 2 are already Martin and Tina. So the other 2 criminal neighbors of Steve must come from Vera, Wanda, and Xola, which makes Vera, Wanda, and Xola all criminals. That leaves row 5 with 0 innocents, giving the clash used above, so Steve and Umar cannot both be criminals. Since Martin still needs 6 criminal neighbors, Igor and Joyce have to supply the remaining criminal spots there. So Igor and Joyce must be criminals.
08.A1 · Alice → INNOCENT
Below Alice there must be exactly 3 innocents, and Keith and Ruth are already innocent, so that group of four below Alice contains exactly 1 criminal. That whole group is also part of column A. Logan's clue says column A has an odd number of criminals, and the people below Alice already account for that odd criminal count. Alice is the only person in column A outside that group, so she cannot also be a criminal. So Alice must be innocent.
09.C5 · Xola → CRIMINAL
Alice’s clue says there is exactly one innocent to the left of Xola, so among Vera and Wanda there is one innocent. Steve’s neighbor group has exactly four innocents, and in that group Keith, Logan, and Ruth are already innocent. That means the only remaining innocent Steve’s neighbors can have is the one coming from Vera or Wanda. Xola is the only person in Steve’s neighbors outside that left-of-Xola pair, so Xola cannot be innocent. So Xola must be criminal.
10.C1 · David → CRIMINAL
Xola’s clue says there is exactly 1 innocent among the people who are both in row 1 and neighboring Betsy. That shared group is just Alice and David. Alice is already the 1 known innocent in that group, so the remaining person there cannot also be innocent. So David must be criminal.
11.B4 · Steve → INNOCENT
Row 4 must have more innocents than row 5, and right now row 4 has 1 known innocent while row 5 has 0. Column B must also have more innocents than column D, and right now column B has 1 known innocent while column D has 0. Now test Steve as criminal. Then the remaining people involved here, Eli, Gabe, Umar, Vera, and Wanda, would also have to satisfy the clue that there are exactly 3 innocents below Alice, along with the corner clue about exactly one innocent corner person having a criminal directly below them. Those requirements cannot all be met at the same time. So Steve at B4 must be innocent.
12.D4 · Umar → CRIMINAL
Keith’s clue says every row has at least 2 criminals, so in row 4 there can be at most 2 innocents. Row 4 already has exactly 2 known innocents there, Ruth and Steve. That means no remaining person in row 4 can also be innocent. So Umar must be criminal.
13.B5 · Wanda → CRIMINAL
Zoe’s clue says Betsy is one of exactly 3 criminals in column B. In column B, Betsy and Habiba are already criminals, while Logan and Steve are innocent, so the only person left there who can make the total reach 3 is Wanda. That makes Wanda criminal.
14.A5 · Vera → INNOCENT
Steve’s neighbors contain exactly 4 criminals in total. Ruth’s clue says exactly 2 of those criminal neighbors also neighbor Olof, and those two are Martin and Tina. That leaves the other 2 criminal neighbors of Steve to come from the people who do not neighbor Olof, and that group already contains Wanda and Xola as known criminals. So Vera cannot also be a criminal. This leaves Vera as innocent.
15.A2 · Gabe → CRIMINAL
Joyce says there are exactly 3 innocents below Alice. Below Alice, the 3 known innocents are Keith, Ruth, and Vera, and the only other person there is Gabe. Since those 3 already fill the clue’s total of exactly 3 innocents, Gabe cannot be innocent. So Gabe must be criminal.
16.D1 · Eli → CRIMINAL
Habiba’s clue says exactly one innocent corner person has a criminal directly below them. In the corners, Alice and Vera are already known innocents, while Eli is the only unknown corner person. If Eli were innocent, then the innocent corners would be Alice, Vera, and Eli, and that would conflict with the clue’s requirement about the corner innocents. So Eli cannot be innocent. That makes Eli criminal.