Clues by Sam Apr 27, 2026 Answer – Full Solution Explained
A1
👩🍳
cook
B1
👩🔧
mech
C1
🕵️♂️
sleuth
D1
👩🎤
singer
A2
👨🍳
cook
B2
🕵️♂️
sleuth
C2
👩💻
coder
D2
👨🎤
singer
A3
🕵️♂️
sleuth
B3
👨🔧
mech
C3
👨💻
coder
D3
👩🏫
teacher
A4
👩🎨
painter
B4
👩🏫
teacher
C4
👨🎨
painter
D4
👩🎨
painter
A5
👩🌾
farmer
B5
👩🌾
farmer
C5
👨🏫
teacher
D5
👨🎤
singer
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 — 11 criminals.
Clues by Sam answer for Apr 27, 2026 — a Easy solved in 16 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 11 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Evie (D1), Hazel (C2), Igor (D2), John (A3), Mark (C3), Olive (D3), Ruth (B4), Salil (C4), Uma (A5) and Vicky (B5); the remaining 9 suspects are innocent.
The deduction chain, in plain English
01.B5 · Vicky → CRIMINAL
Gary’s clue says that the two innocents below Claire are connected. Among the people below Claire, Gary is already known to be one of those innocents. That means the other innocent in that group has to be next to Gary, so it can only be Larry, and Vicky is not next to Gary in that line. So Vicky cannot be the second innocent there. That makes Vicky criminal.
02.B3 · Larry → INNOCENT
Below Claire, the clue says there are exactly two innocents, and those two must be connected. In that group we already have Gary as innocent and Vicky as criminal, leaving Larry and Ruth as the only unknowns there. If Larry were criminal, then Ruth would be the only possible second innocent below Claire. That would make the two innocents Gary and Ruth, which does not satisfy the clue’s requirement for both innocents below Claire to be connected. So Larry must be innocent.
03.B4 · Ruth → CRIMINAL
Gary’s clue says that exactly two innocents below Claire are connected. Below Claire, Gary and Larry are already known innocents, Vicky is a known criminal, and Ruth is the only unknown person there. If Ruth were innocent, then there would be three innocents below Claire: Gary, Larry, and Ruth. That conflicts with the clue’s requirement that the innocents below Claire are both innocents, meaning exactly two. So Ruth must be criminal.
04.C5 · Xavi → INNOCENT, D5 · Zane → INNOCENT
Vicky's clue says exactly 2 of the people who are both Salil's neighbors and in row 5 are innocent. That shared group is B5 Vicky, C5 Xavi, and D5 Zane. Vicky is already criminal, so the group currently has no known innocents, and the only people left there who could fill those 2 innocent spots are Xavi and Zane. So Xavi and Zane must be innocent.
05.B1 · Claire → INNOCENT
Zane’s clue says column B is the only column with exactly 3 innocents. In column B, Gary and Larry are already innocent, Ruth and Vicky are already criminal, and Claire is the only unknown person left there. If Claire were criminal, then column B would stay at only 2 innocents, which conflicts with the clue that column B has exactly 3 innocents. So Claire must be innocent.
06.D4 · Tina → INNOCENT
Claire’s clue says exactly 1 painter has an innocent directly to the right. Petra cannot be that painter, because the person directly to her right is Ruth, and Ruth is criminal. That leaves the remaining painter case to be supplied by the other painter-with-right-neighbor possibility, which is Salil with Tina directly to his right. For Salil to be the one painter counted by the clue, Tina has to be innocent. So Tina must be innocent.
07.A5 · Uma → CRIMINAL
Tina’s clue fixes the four people above Uma as exactly 2 innocents, with exactly 1 of those 2 being Ruth’s neighbor, and the only above-Uma neighbors of Ruth are John and Petra. Zane’s clue also fixes column B as the only column with exactly 3 innocents. If Uma were innocent, then Anna, Frank, John, Petra, David, Hazel, Evie, Igor, Mark, Olive, and Salil would have to satisfy both of those clue requirements at the same time, and that cannot be done. So Uma cannot be innocent. So Uma must be criminal.
08.A4 · Petra → INNOCENT
Tina’s clue says there are exactly 2 innocents above Uma, and exactly 1 of those innocents is Ruth’s neighbor. Among the people above Uma who are Ruth’s neighbors, the only possibilities are John and Petra, and that group contains exactly 1 criminal. That means the criminal in that pair has to be John, so Petra cannot be the criminal there. So Petra must be innocent.
09.A3 · John → CRIMINAL
Above Uma, there are exactly 2 innocents in total. Exactly 1 of those innocents is a neighbor of Ruth, and among the people above Uma who are Ruth's neighbors, Petra is already that 1 known innocent. The only other person in that same overlapping group is John, so John cannot be innocent. So John must be criminal.
10.C3 · Mark → CRIMINAL, C4 · Salil → CRIMINAL
Petra’s clue says the people below Hazel contain exactly one innocent. That group already has its one innocent, Xavi. The only other people left in that group are Mark and Salil, so neither of them can be innocent. So Mark and Salil must be criminal.
11.D3 · Olive → CRIMINAL
Mark’s clue says the teachers contain an odd number of innocents. Among the teachers, Xavi is already known innocent and Ruth is already known criminal, so Olive is the only teacher whose status is still undecided. If Olive were innocent, that would make 2 innocent teachers in total, which is even, not odd. So Olive at D3 must be criminal.
12.D1 · Evie → CRIMINAL, D2 · Igor → CRIMINAL
Column B already has exactly 3 innocents, so no other column can also have exactly 3 innocents. Olive’s clue also says every column has at least 2 criminals. If Evie and Igor were both innocent, then column D would have Evie, Igor, Tina, and Zane as innocents, leaving only Olive as a criminal there. That breaks the requirement that each column has at least 2 criminals, and it also clashes with the fact that column B is the only column singled out by the 3-innocent clue. So Evie and Igor must be criminal.
13.C2 · Hazel → CRIMINAL
Igor’s clue says the people below David contain exactly 3 criminals. In that group, 2 are already known criminals. The only unknown person left in that same group is Hazel, so the group’s one remaining criminal has to be Hazel. So Hazel must be criminal.
14.C1 · David → INNOCENT
Above Uma there must be exactly 2 innocents, and exactly 1 of those innocents is Ruth's neighbor. In that group, Petra is already an innocent and she is Ruth's neighbor, so the second innocent above Uma has to be one of Anna or Frank, not both. The edge cells must contain an odd number of innocents. There are already 5 known innocents on the edge, and the only unknown edge people are Anna, Frank, and David. If David were criminal, then only Anna and Frank could change that edge-innocent total, but the condition above Uma allows exactly one of Anna and Frank to be innocent, which would add exactly 1 more edge innocent and make the total 6, not odd. So David must be innocent.
15.A1 · Anna → CRIMINAL
David’s clue says rows 1 and 5 have the same number of criminals. Row 5 already has 2 criminals, while row 1 currently has only 1 known criminal, with Anna the only unknown person left in that row. If Anna were innocent, row 1 would stay at 1 criminal, which would not match row 5’s 2. So Anna must be criminal.
16.A2 · Frank → INNOCENT
Above Uma there must be exactly 2 innocents in total. Exactly 1 of those innocents is a neighbor of Ruth, and among the people above Uma who are Ruth's neighbors, Petra is the only innocent there. So the other innocent above Uma has to be among the people above Uma who are not Ruth's neighbors, which are Anna and Frank. Anna is criminal, so the remaining innocent spot there has to be Frank. So Frank must be innocent.