Clues by Sam Aug 21, 2026 Answer – Full Solution Explained
A1
👨🏫
teacher
B1
👩⚖️
judge
C1
🕵️♂️
sleuth
D1
🕵️♂️
sleuth
A2
👩⚕️
doctor
B2
👨⚕️
doctor
C2
👩🏫
teacher
D2
👨🏫
teacher
A3
👮♂️
cop
B3
👩🌾
farmer
C3
👷♂️
builder
D3
👷♂️
builder
A4
👮♀️
cop
B4
👩🌾
farmer
C4
👩⚖️
judge
D4
👩💼
clerk
A5
👩💻
coder
B5
👨💻
coder
C5
👩💻
coder
D5
👨💼
clerk
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 — 4 criminals.
Clues by Sam answer for Aug 21, 2026 — a Tricky solved in 18 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Erwin (C1), Larry (A3), Samira (C4) and Vicky (A5); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.C1 · Erwin → CRIMINAL
Will’s clue says that Erwin is one of the exactly two criminals above Xola. That directly identifies Erwin’s status from the clue itself. So Erwin must be criminal.
02.C5 · Xola → INNOCENT
Will’s clue says that among the people above Xola, exactly two are criminals, and Erwin is one of them. So in Xola’s column, only one of Joy, Oscar, and Samira can be criminal. Erwin’s clue says each column has at least 3 innocents. If Xola were criminal, then column C would already have Erwin and Xola as two criminals, while only one of Joy, Oscar, and Samira could be criminal. That makes the clues conflict in column C. So Xola must be innocent.
03.D3 · Paul → INNOCENT
Column D contains Franco, Kyle, Paul, Tina, and Ziad. Erwin’s clue says every column has at least 3 innocents, and Xola’s clue says all innocents in column D must form one connected block. If Paul were criminal, the remaining people in column D would have to provide at least 3 innocents while still keeping all the innocents in that column as one connected block, and that cannot be done together with these clue requirements. So Paul cannot be criminal. That makes Paul innocent.
04.C2 · Joy → INNOCENT
Ruth’s neighbors in column C contain exactly 1 criminal, and in that group the only possible people are Oscar and Samira. Xola is innocent, so that one column C criminal neighbor of Ruth has to be Oscar or Samira. The people above Xola contain exactly 2 criminals: Erwin, Joy, Oscar, and Samira. Since Erwin is already one criminal, the other criminal above Xola must be that same Oscar-or-Samira person. That uses up both criminals allowed above Xola, leaving no room for Joy to be criminal. So Joy must be innocent.
05.D4 · Tina → INNOCENT
Joy says Xola has exactly 4 innocent neighbors, and among Xola’s five neighbors Will is already known to be innocent. That means the other four neighbors, Ruth, Samira, Tina, and Ziad, contain exactly 1 criminal. Xola also says all innocents in column D are connected, with Paul already the known innocent in that column. In this step, that leaves the one criminal among Ruth, Samira, Tina, and Ziad to come from Ruth, Samira, or Ziad, not Tina. So Tina must be innocent.
06.D2 · Kyle → INNOCENT
Column D must contain at least 4 innocents, and all innocents in that column have to form one connected block. We already know Paul at D3 and Tina at D4 are innocent. The only connected innocent blocks in column D that fit those facts are D1-D4, D2-D5, or D1-D5. In every one of those possible blocks, D2 is included. So Kyle must be innocent.
07.A5 · Vicky → CRIMINAL
Kyle’s clue says there are no innocents in the overlap of column A and row 5. That overlap is just A5, Vicky, and there are already 0 known innocents in that shared group. So the only person there cannot be innocent. Therefore Vicky must be criminal.
08.A4 · Quita → INNOCENT
Erwin’s clue says every column has at least 3 innocents, and Vicky’s clue says the people above Quita contain an odd number of criminals. The people above Quita are Akira, Habi, and Larry, and all three are still unknown, so that odd-criminal requirement has to be met within that group. If Quita were criminal, then column A would have to satisfy both clues at the same time with Akira, Habi, Larry, Quita, and Vicky, but that combination is impossible. So Quita cannot be criminal. That makes Quita innocent.
09.D5 · Ziad → INNOCENT
Row 4 has exactly one criminal, and in that row the only people not yet identified are Ruth and Samira. Xola's neighbors also contain exactly one criminal, and those neighbors include Ruth, Samira, Tina, Will, and Ziad. Since Ziad is the only one in Xola's neighbor group who is not part of the row 4 group, the single criminal allowed among Xola's neighbors must already be the same single criminal coming from Ruth or Samira. That means Ziad at D5 must be innocent.
10.D1 · Franco → INNOCENT
Joy’s clue fixes Xola’s neighbors at exactly 4 innocents, and those neighbors already include 3 known innocents, so Ruth and Samira cannot both be innocent. Vicky’s clue says the three people above Quita contain an odd number of criminals, so Akira, Habi, and Larry must include either 1 or 3 criminals. Ziad’s clue says exactly 2 edge people have a criminal directly above them. If Franco were criminal, then Akira, Celia, Habi, Larry, Ruth, and Samira would have to satisfy all three of those restrictions at the same time, and they cannot. So Franco cannot be criminal. That makes Franco innocent.
11.A1 · Akira → INNOCENT
Franco’s clue says there are more criminal sleuths than criminal teachers. The sleuths already have 1 criminal, and the teachers currently have 0 criminals, with Akira the only teacher whose identity is not yet known. If Akira were criminal, then the teachers would also have 1 criminal. That would make criminal sleuths and criminal teachers equal, not more for the sleuths. So Akira must be innocent.
12.B1 · Celia → INNOCENT
The two clues both lean on the same small group of unknowns: Celia, Habi, and Larry. Above Quita there must be an odd number of criminals, and with 0 known criminals above her, that requirement has to be met by Habi and Larry. Akira’s clue says there is an odd number of innocents among the edge-cell neighbors of Igor, namely Akira, Celia, Erwin, Habi, and Larry. Akira is already innocent and Erwin is already criminal, so Celia, Habi, and Larry have to fit that odd-innocent count while Habi and Larry also fit the odd-criminal count above Quita. If Celia were criminal, then Habi and Larry would have to satisfy both clue requirements by themselves along with that fixed Akira and Erwin setup, and they cannot. So Celia cannot be criminal. That makes Celia innocent.
13.C3 · Oscar → INNOCENT, C4 · Samira → CRIMINAL
Celia’s clue says that the two innocents above Xola are connected. Above Xola, the four people are Erwin, Joy, Oscar, and Samira, and Erwin is already a criminal while Joy is already innocent. If Oscar were criminal and Samira were innocent, then the innocents above Xola would be Joy and Samira, and that does not satisfy the clue’s required connected pair. So Oscar must be innocent and Samira must be criminal.
14.B4 · Ruth → INNOCENT
Joy’s clue says Xola has exactly 4 innocent neighbors. Xola’s neighbors already include 3 known innocents, and the only neighbor there whose identity is still unknown is Ruth. So the one remaining innocent neighbor required by the clue has to be Ruth. That makes Ruth innocent.
15.B2 · Igor → INNOCENT
If Igor were criminal, then Habi, Larry, and Nancy would have to make all three clues work at the same time. Above Quita there must be an odd number of criminals, and the only unknown people there are Habi and Larry. Ruth must have exactly 3 criminal neighbors, with exactly 1 of those criminals in column C, and Samira is already that 1 criminal in column C. Also, exactly one farmer must have exactly 3 criminal neighbors, while Nancy currently has 1 known criminal neighbor and Ruth currently has 2 known criminal neighbors. Those requirements cannot all be met if Igor is criminal. So Igor cannot be criminal. That makes Igor innocent.
16.B3 · Nancy → INNOCENT
Igor’s clue says there are exactly 4 criminals on the whole board. Since 3 criminals are already known, if Nancy were a criminal, that would use up the fourth criminal spot. Then Habi and Larry would both have to be innocent. But Vicky’s clue says the people above Quita contain an odd number of criminals, and above Quita the only unknown people are Habi and Larry, with 0 known criminals there now. If both Habi and Larry were innocent, that group would have 0 criminals, which is not odd. So Nancy must be innocent.
17.A3 · Larry → CRIMINAL
Paul’s clue says Ruth’s neighbors contain exactly 3 criminals, and exactly 1 of those criminals is in column C. Among Ruth’s neighbors in column C, the only criminal is Samira, so the other 2 criminals among Ruth’s neighbors must be outside column C. Ruth’s neighbors outside column C are Larry, Nancy, Quita, Vicky, and Will. That group already has 1 known criminal, Vicky, so it still needs 1 more criminal. The only unknown person left in that group is Larry, so Larry at A3 must be criminal.
18.A2 · Habi → INNOCENT
Vicky’s clue says the people above Quita must contain an odd number of criminals. Above Quita, there is already 1 known criminal there, and the only person in that group whose status is not yet fixed is Habi. If Habi were criminal, that group would no longer fit the clue’s requirement. So Habi must be innocent.