Clues by Sam Oct 23, 2026 Answer – Full Solution Explained
A1
👨💻
coder
B1
👷♂️
builder
C1
👩⚖️
judge
D1
👩🎨
painter
A2
👨💼
clerk
B2
👷♂️
builder
C2
👩⚖️
judge
D2
👨⚖️
judge
A3
👨💼
clerk
B3
👷♂️
builder
C3
👩🎨
painter
D3
👮♂️
cop
A4
👨💻
coder
B4
👨🎨
painter
C4
👩🍳
cook
D4
👨💻
coder
A5
👩💻
coder
B5
👨🍳
cook
C5
👩🍳
cook
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 — 6 criminals.
Clues by Sam answer for Oct 23, 2026 — a Hard solved in 16 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 Berat (B1), John (A3), Linda (C3), Noah (A4), Sam (D4) and Uma (A5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.A5 · Uma → CRIMINAL
Below Alex there are exactly 3 criminals in total. Exactly 2 of those criminals are Kyle's neighbors, and among the people below Alex those neighbors are Frank, John, and Noah. So the people below Alex who are not Kyle's neighbors must contain exactly 1 criminal. The only person below Alex who is not a neighbor of Kyle is Uma, so Uma must be criminal.
02.A1 · Alex → INNOCENT
Dana’s clue fixes the people below Alex at exactly 3 criminals, and exactly 2 of those criminals must be Kyle’s neighbors. Uma’s clue also requires every column, including column A, to contain at least 2 innocents. If Alex were a criminal, then column A would contain Alex, Frank, John, Noah, and Uma, while still needing at least 2 innocents there and also needing the people below Alex to meet Dana’s exact criminal counts at the same time. Those requirements cannot all be met together. So Alex must be innocent.
03.B4 · Ollie → INNOCENT, B2 · Ghani → INNOCENT
Among the people below Alex who are also Kyle's neighbors, namely Frank, John, and Noah, exactly 2 are criminals. Alex's clue says Kyle has exactly 3 criminal neighbors, and only 1 of those is Mike's neighbor, so exactly 2 of Kyle's criminal neighbors must be outside Mike's neighbors; in this outside-Mike group, the possible people are Frank, Ghani, John, Noah, and Ollie. Frank, John, and Noah are already the full set accounting for those 2 criminals in that outside-Mike group. The only other people in that same group are Ghani and Ollie, so neither of them can be criminal. That makes Ollie and Ghani innocent.
04.C3 · Linda → CRIMINAL
Ghani’s clue says Mike has exactly 2 criminal neighbors, and exactly 1 of those criminal neighbors is to Mike’s left. Among Mike’s neighbors, the only person who is to Mike’s left is Linda. So the one criminal neighbor to Mike’s left has to be Linda. That makes Linda criminal.
05.C4 · Ruth → INNOCENT, C2 · Habi → INNOCENT
Alex’s clue says Kyle has exactly 3 criminal neighbors, and exactly 1 of those criminals is also Mike’s neighbor. The people who are both Kyle’s neighbors and Mike’s neighbors are Habi, Linda, and Ruth. Linda is already known to be a criminal, so that one allowed criminal in the shared group is already accounted for. That means the other two people in that shared group cannot be criminals. So Ruth and Habi must be innocent.
06.A3 · John → CRIMINAL, B3 · Kyle → INNOCENT
Dana’s clue says there are exactly 3 criminals below Alex, and exactly 2 of those are Kyle’s neighbors. Since Uma is already a criminal below Alex and is not Kyle’s neighbor, the other two criminals below Alex must come from A2 Frank, A3 John, and A4 Noah, which pushes strongly against John being innocent. Ghani’s clue says Mike has exactly 2 criminal neighbors, and exactly 1 of them is to Mike’s left. Linda is already a criminal to Mike’s left, so the other criminal neighbor of Mike must be on Mike’s right side, among D2 Ike or D4 Sam. Ollie’s clue says exactly one person in row 3 has exactly 4 innocent neighbors. But Kyle already has 4 known innocent neighbors, and Linda also already has 4 known innocent neighbors, so if John were innocent while Kyle were criminal, the remaining unknowns would have to satisfy all three clues at once, and they cannot. So John must be criminal and Kyle must be innocent.
07.D3 · Mike → INNOCENT
Linda is already the one criminal neighbor of Mike who is to Mike's left, and Mike has exactly two criminal neighbors in total. Linda's neighbors also must contain an odd number of innocents, and at the moment that group already has 5 known innocents, with only Ike, Mike, and Sam not yet identified there. If Mike were criminal, then Ike and Sam would have to satisfy both clues at the same time along with Mike, but that cannot be done. So Mike cannot be criminal. Therefore Mike must be innocent.
08.D5 · Zara → INNOCENT
Ollie’s clue says exactly one person in row 3 has exactly 4 innocent neighbors. But Kyle already has 4 known innocent neighbors, while John has only 3 known innocents, Linda already has 6, and Mike has only 2, so that clue fixes the row 3 requirement around Kyle alone. Mike’s clue also requires exactly 4 edge people to have a criminal directly below them. If Zara were criminal, then the remaining people involved here, namely Berat, Emily, Frank, Ike, Noah, Sam, Vic, and Yara, would have to satisfy both clues at once, and they cannot do so. So Zara must be innocent.
09.D1 · Emily → INNOCENT
Mike’s clue fixes his neighboring criminals very tightly: he has exactly 2 criminal neighbors in total, and exactly 1 of those is to Mike’s left. Linda is already that one criminal neighbor to Mike’s left, so the other criminal neighbor has to come from Ike or Sam, and not from anyone else involved here. Above Zara there is already 1 known innocent, Mike, and the total number of innocents above Zara must be odd. If Emily were criminal, then among Emily, Ike, and Sam only Ike and Sam could add innocents above Zara, but those same two people also have to supply exactly one criminal neighbor for Mike. That combination does not fit all the clue requirements at once, so Emily cannot be criminal. So Emily must be innocent.
10.C5 · Yara → INNOCENT
If Yara were criminal, then column C would stay at 3 innocents, so column D also has to stay at 3 innocents to match it. That means neither Ike nor Sam could be innocent, because either one being innocent would raise column D above 3 innocents. But Mike’s clue says his neighbors contain exactly 2 criminals, and exactly 1 of those criminals is to Mike’s left. Linda is already the 1 criminal neighbor to Mike’s left, so the other criminal neighbor must be not to his left. The only unknown neighbors not to Mike’s left are Ike and Sam, so Ike must be criminal and Sam must be criminal, which clashes with column D having to stay at 3 innocents. So Yara at C5 must be innocent.
11.A4 · Noah → CRIMINAL
Yara’s clue directly says that Noah is one of the 5 criminals on the edges. Since the clue explicitly includes him in that criminal group, Noah must be criminal.
12.A2 · Frank → INNOCENT
Below Alex there are exactly three criminals in total. Among the people below Alex who are also Kyle's neighbors, the group is Frank, John, and Noah, and exactly two of those criminals are supposed to be in that neighboring group. But John and Noah are already known criminals in that group, so the two criminal spots there are already filled. That means Frank cannot also be a criminal. So Frank must be innocent.
13.B5 · Vic → INNOCENT
Linda’s clue says every corner person can have at most one criminal neighbor. Uma is a corner person, and she already has one known criminal neighbor: Noah. Vic is another neighbor of Uma, so if Vic were criminal, Uma would have more than one criminal neighbor, which would break the clue. So Vic must be innocent.
14.B1 · Berat → CRIMINAL
Yara’s clue says there are exactly 5 criminals on the edge. On the edge, the known criminals are John, Noah, and Uma, so the remaining edge criminals have to come from the unknown edge people. Ghani’s clue fixes Mike’s neighboring criminals so that the unknown neighbor spots are the ones that account for the remaining edge-criminal places among Mike’s neighboring edge cells, leaving the larger edge group with one more criminal still needed. In the edge cells outside that smaller group, the only person left is Berat. So Berat must be criminal.
15.D2 · Ike → INNOCENT
Emily’s clue says Habi has exactly 6 innocent neighbors. Among Habi’s neighbors, 5 are already known to be innocent, and the only neighbor there whose identity is still unknown is Ike. So the one remaining innocent neighbor required by the clue has to be Ike. That makes Ike innocent.
16.D4 · Sam → CRIMINAL
Mike’s neighbors contain exactly 2 criminals in total. Exactly 1 of those criminals is to Mike’s left, and that one is Linda. So among Mike’s neighbors who are not to his left, there must be exactly 1 criminal. Habi, Ike, and Ruth are innocent, and Sam is the only unknown person in that group, so Sam has to fill that remaining criminal spot. That makes Sam criminal.