Clues by Sam Oct 25, 2026 Answer – Full Solution Explained
A1
👩💻
coder
B1
👨💻
coder
C1
👩🎨
painter
D1
👨🎨
painter
A2
👩🍳
cook
B2
👨💻
coder
C2
👨🔧
mech
D2
👩🎨
painter
A3
👨🍳
cook
B3
🕵️♀️
sleuth
C3
👩🔧
mech
D3
👨🔬
scientist
A4
👩🍳
cook
B4
🕵️♂️
sleuth
C4
👩⚖️
judge
D4
👩🔬
scientist
A5
👮♂️
cop
B5
👮♀️
cop
C5
👨⚖️
judge
D5
👨🔬
scientist
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 — 5 criminals.
Clues by Sam answer for Oct 25, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 5 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Diane (C1), Jason (C2), Kay (D2), Ruth (A4) and Vlad (A5); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.B1 · Clyde → INNOCENT
Betty’s clue directly says that Clyde is one of the 10 innocents on the edges. Since Clyde is explicitly included among those edge innocents, Clyde at B1 must be innocent.
02.B5 · Wilma → INNOCENT
Clyde's clue says there is exactly 1 innocent in the overlap between the people to the right of Vlad and Ruth's neighbors. That shared group contains only Wilma. Since that group still needs its 1 innocent, Wilma has to be the one who fills it. So Wilma must be innocent.
03.C1 · Diane → CRIMINAL, C2 · Jason → CRIMINAL
Wilma’s clue says exactly 2 painters have a criminal directly to their left. Diane cannot be one of those painters, because Clyde is directly to Diane’s left and Clyde is innocent. That leaves Eric and Kay as the only painters who can make up the 2 required cases, so the people directly to their left must both be criminals. Eric’s left neighbor is Diane, and Kay’s left neighbor is Jason. So Diane and Jason must be criminal.
04.D1 · Eric → INNOCENT
Diane’s clue says row 2 is the only row with more than one criminal. That means row 2 must have at least two criminals, and every other row can have at most one. Row 1 already has one known criminal, Diane at C1. Eric is the only unknown person left in row 1, so if Eric were a criminal, row 1 would also have more than one criminal, which would break the clue. So Eric must be innocent.
05.C4 · Tina → INNOCENT, D4 · Ulrika → INNOCENT
Diane’s clue says row 2 is the only row with more than one criminal, so row 4 can have at most one criminal. Jason’s clue says the people to the left of Tina, Ruth and Steve, contain exactly one innocent. If Tina and Ulrika were both criminals, then row 4 would already have two criminals in Tina and Ulrika, which breaks Diane’s clue because row 4 would then also have more than one criminal. That combination cannot fit the clues. So Tina and Ulrika must be innocent.
06.D5 · Zed → INNOCENT
Assume Zed were criminal. Then row 5 would already have one criminal, so Diane's clue means row 5 can have at most that one criminal while row 2 must have more than one. Tina's clue says Steve has exactly two criminal neighbors, and exactly one of those neighboring criminals is in row 5. Among Steve's row 5 neighbors, Wilma is already innocent, so the only possible row 5 neighboring criminals are Vlad and Yusuf. That clue therefore fixes the row 5 neighbor count at exactly one, which makes Vlad innocent and Yusuf innocent here. But then Steve would have no criminal neighbor in row 5 at all, contradicting Tina's clue that exactly one of his criminal neighbors is in row 5. So Zed at D5 must be innocent.
07.C5 · Yusuf → INNOCENT
Jason’s clue says there is exactly one innocent to the left of Tina, so among Ruth and Steve, exactly one must be innocent. Zed’s clue says exactly one corner person has exactly one criminal neighbor. Eric already has 2 known criminal neighbors, so Eric cannot be that corner person. If Yusuf were criminal, then Zed would have exactly 1 criminal neighbor, while Betty and Vlad would still depend on Gemma and Henry, and on Ruth and Steve, with Ruth and Steve also having to fit Jason’s “exactly one innocent” clue. Those requirements cannot all be satisfied together. So Yusuf cannot be criminal. That makes Yusuf innocent.
08.A5 · Vlad → CRIMINAL
Tina’s clue says Steve has exactly 2 criminal neighbors, and exactly 1 of those criminal neighbors is in row 5. Among Steve’s neighbors in row 5, Wilma and Yusuf are already innocent, so that row-5 part still needs 1 criminal. The only unknown person left there is Vlad. So Vlad must be criminal.
09.D3 · Pip → INNOCENT
If Pip were a criminal, then row 3 would already have its one criminal, because Diane’s clue says row 2 is the only row with more than one criminal. That makes Melvin, Nicole, and then Olga innocent, and Steve’s clue then fixes Ruth as Steve’s second criminal neighbor alongside Vlad. Betty’s edge clue gives exactly 10 innocents on the edge, so with Melvin innocent and Pip criminal, Gemma and then Kay also have to be innocent. Row 2 still has to be the only row with more than one criminal, so Henry must then be the second criminal in row 2. Now Zed’s corner clue fails. Betty has exactly one criminal neighbor, Eric has two, Zed has none, and Vlad also has exactly one criminal neighbor if Steve is innocent, so Steve has to be criminal to stop there being two such corners. But then row 4 would have two criminals, Ruth and Steve, which contradicts Diane’s clue that only row 2 has more than one criminal. So Pip at D3 must be innocent.
10.C3 · Olga → INNOCENT
Pip’s clue says exactly 3 edge people have a criminal directly to their left. Those 3 are already accounted for by Eric at D1, Kay at D2, and Wilma at B5. The only other unresolved left-neighbor that could add another case to this clue is Olga at C3, for Pip at D3. If Olga were criminal, that would make a fourth such edge person, which would contradict the clue. So Olga must be innocent.
11.A3 · Melvin → INNOCENT
Olga’s clue says that below Betty there are exactly two innocents, and those two innocents must be connected. The people below Betty are Gemma, Melvin, Ruth, and Vlad, and Vlad is already criminal. If Melvin were criminal, then the two innocents below Betty would have to be Gemma and Ruth. But with Melvin between them and Vlad criminal at the bottom, Gemma and Ruth would not be connected as the clue requires. So Melvin must be innocent.
12.D2 · Kay → CRIMINAL
Betty says there are exactly 10 innocents on the edge, and 9 edge innocents are already known. If Kay were innocent, the edge would need one more innocent among the remaining edge people, so Gemma could not be innocent and would have to be criminal. Olga says both innocents below Betty are connected. Below Betty, Melvin is already innocent and Vlad is criminal, so with Gemma criminal, Ruth also has to be criminal, leaving Melvin as the only innocent there instead of two connected innocents. That clashes with Olga's clue, so Kay cannot be innocent. So Kay at D2 must be criminal.
13.B4 · Steve → INNOCENT
If Steve were criminal, then A5 Vlad would have exactly one criminal neighbor in the corners clue, because A5 currently has 0 known criminal neighbors and its only unknown neighbors are Ruth and Steve. But D5 Zed definitely has 0 criminal neighbors, and D1 Eric definitely has 3, so the only corner that could possibly be the one with exactly 1 criminal neighbor is A1 Betty or A5 Vlad. At the same time, column A needs exactly 2 criminals who have an innocent directly to the right, and the only known criminal in column A is Vlad. With Steve criminal, the remaining unknown people A2 Gemma, B2 Henry, B3 Nicole, and Ruth would have to meet that clue as well as the row and corner clues, and they cannot do so consistently. So Steve must be innocent.
14.A4 · Ruth → CRIMINAL
Jason’s clue says there is exactly one innocent to the left of Tina. Among the people to the left of Tina, there is already one known innocent, Steve, and the only other person there whose identity was not yet known is Ruth. Since that group already has its full allotment of one innocent, Ruth cannot be innocent. So Ruth must be criminal.
15.A2 · Gemma → INNOCENT
Olga’s clue says that the two innocents below Betty are connected. Below Betty, the four people are Gemma, Melvin, Ruth, and Vlad, and among them Melvin is already innocent while Ruth and Vlad are criminals. That leaves Gemma as the only person who can be the second innocent needed for that pair below Betty. So Gemma must be innocent.
16.B3 · Nicole → INNOCENT
Tina’s clue says that Steve has exactly 2 criminal neighbors, and exactly 1 of those criminals is in row 5. Among Steve’s neighbors in row 5, Vlad is already that 1 criminal. That means the other criminal neighbor must be outside row 5, and Ruth is already a known criminal there. So there is no room for any other non-row-5 neighbor of Steve to be criminal. That makes Nicole innocent.
17.B2 · Henry → INNOCENT
The clue says exactly one corner person has exactly one criminal neighbor. Eric already has 3 criminal neighbors, Zed has 0, and Vlad has exactly 1, so Vlad is the one corner who fits the clue. Betty currently has 0 known criminal neighbors, and the only unknown neighbor affecting that count is Henry. If Henry were criminal, Betty would also have exactly 1 criminal neighbor, which would give a second matching corner and break the clue. So Henry at B2 must be innocent.