Clues by Sam Oct 21, 2026 Answer – Full Solution Explained
A1
👨🔬
scientist
B1
👨🔬
scientist
C1
👩💻
coder
D1
👨🏫
teacher
A2
💂♂️
guard
B2
💂♂️
guard
C2
😬
cat
D2
😬
cat
A3
👨🌾
farmer
B3
💂♀️
guard
C3
👨💼
clerk
D3
👩💼
clerk
A4
👩🌾
farmer
B4
👩🎤
singer
C4
👨💼
clerk
D4
👨🏫
teacher
A5
🥷
ninja
B5
👨🎤
singer
C5
👨🔬
scientist
D5
👩🏫
teacher
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 Oct 21, 2026 — a Tricky solved in 18 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 11 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Boaz (B1), Daniella (C1), Emil (D1), Gabe (B2), Ipsum (C2), Lorem (D2), Sofia (B4), Vera (A5), Will (B5), Xavi (C5) and Zoe (D5); the remaining 9 suspects are innocent.
The deduction chain, in plain English
01.C5 · Xavi → CRIMINAL
Tyler’s clue says there are exactly 2 criminal scientists, and exactly 1 of those criminal scientists is in row 1. That means the other criminal scientist must be a scientist who is not in row 1. Among the scientists not in row 1, there is only Xavi, and there are no known criminals there yet. So Xavi has to fill that remaining criminal scientist spot. That makes Xavi criminal.
02.A3 · Martin → INNOCENT
Xavi’s clue says that Martin is one of the exactly two innocents below Frank. That identifies Martin directly as an innocent from the clue itself. So Martin must be innocent.
03.C1 · Daniella → CRIMINAL, D1 · Emil → CRIMINAL
Tyler's clue leaves exactly one criminal scientist in row 1, and the only row 1 scientists are Allan and Boaz, so row 1 gets exactly one criminal from that pair. Martin's clue says row 1 has exactly one innocent, so the four people in row 1 must include exactly three criminals. After Allan and Boaz account for one of those three criminals, the remaining two people in row 1 are Daniella and Emil, and they must supply the other two criminals. So Daniella and Emil must be criminal.
04.A2 · Frank → INNOCENT
Below Allan there must be an odd number of innocents. Martin is already one known innocent there, so A2 Frank, A4 Ruby, and A5 Vera must add an even number of further innocents. Below Frank there are exactly 2 innocents, and Martin is one of them, so Ruby and Vera must contain exactly 1 more innocent. That leaves Frank as the person who must supply the extra innocent needed so the total below Allan stays odd. So Frank must be innocent.
05.B3 · Nancy → INNOCENT
Frank’s clue says that among Nancy’s neighbors, exactly 2 criminals are also neighbors of Martin. Those shared people are A2 Frank, B2 Gabe, A4 Ruby, and B4 Sofia. Emil’s clue says Martin has exactly 3 innocent neighbors, so Martin’s neighbor group has exactly 2 criminals in total: A2 Frank, B2 Gabe, B3 Nancy, A4 Ruby, and B4 Sofia. The shared group is part of that larger group, and it already accounts for exactly 2 criminals. The only person in Martin’s neighbor group who is not in that shared group is Nancy, so she cannot be a criminal. So Nancy must be innocent.
06.B5 · Will → CRIMINAL
Below Frank, there are exactly 2 innocents among Martin, Ruby, and Vera, and Martin is one of them. Those same three people are also among Sofia's edge neighbors, and that larger group of Sofia's edge neighbors has exactly 2 innocents total: Martin, Ruby, Vera, Will, and Xavi. Since the smaller group already accounts for both innocents allowed in the larger group, the only extra person there, Will, cannot be innocent. So Will must be criminal.
07.D3 · Quita → INNOCENT
Will’s clue says that the two innocents above Zoe are connected. Above Zoe, Emil is criminal, and if Quita were criminal too, then the only way to get those two innocents would be for Lorem and Umar to be the innocents. But then the people above Zoe would be Emil criminal, Lorem innocent, Quita criminal, and Umar innocent. Those two innocents would be separated by Quita, so they would not be connected, which breaks Will’s clue. So Quita must be innocent.
08.D5 · Zoe → CRIMINAL
Nancy’s clue says there are exactly 8 criminals on the edge, so among the edge people who are not Sofia’s neighbors there must be exactly 5 non-neighbor edge criminals. In that non-neighbor edge group, Daniella and Emil are already known criminals, and Frank and Quita are already known innocents, which leaves exactly 2 innocents to be found among Allan, Boaz, Lorem, Umar, and Zoe. Those 2 innocent spots must be filled by Allan, Boaz, Lorem, and Umar, so Zoe cannot be one of them. That makes Zoe criminal.
09.B2 · Gabe → CRIMINAL
Vera’s neighbors can contain exactly one innocent, and among the people relevant here that means Ruby and Sofia account for the innocent neighbor allowed there. Martin’s neighbors must contain exactly three innocents, and two of those are already Frank and Nancy. That leaves room for only one more innocent in Martin’s neighbor group, and that one has to come from Ruby or Sofia, not from the person outside Vera’s group. So Gabe must be criminal.
10.A1 · Allan → INNOCENT
Below Frank there are exactly 2 innocents, and Martin is already one of them, so that group’s two innocent spots are entirely within Martin, Ruby, and Vera. Column A has exactly 4 innocents total: Frank and Martin are already innocent, and Ruby and Vera are the only other people from that below-Frank group who can add to that count. The only person in column A who is outside that smaller group is Allan, so Allan has to supply the remaining innocent needed in column A. So Allan must be innocent.
11.B1 · Boaz → CRIMINAL
Tyler's clue says there are exactly 2 criminal scientists, and exactly 1 of those criminal scientists is in row 1. Among the scientists, Xavi is already a known criminal and he is not in row 1, so the one criminal scientist who is in row 1 still has to be found among Allan and Boaz. Allan is already innocent, leaving only Boaz. So Boaz must be criminal.
12.D4 · Umar → INNOCENT
Row 4 must contain an odd number of innocents. Since Tyler is already the one known innocent in that row, Ruby, Sofia, and Umar together must add an even number of further innocents. Martin’s clue says his neighbors contain exactly 3 innocents. Frank and Nancy are already innocent there, so among Ruby and Sofia there must be exactly 1 more innocent. So Ruby and Sofia contribute 1 innocent to row 4, and to keep the whole row odd, the remaining person in that row, Umar, must add 1 more innocent. That makes Umar innocent.
13.D2 · Lorem → CRIMINAL
Nancy’s clue says there are exactly 8 criminals on the edge, and exactly 3 of those edge criminals are Sofia’s neighbors. So exactly 5 edge criminals are not neighbors of Sofia. Among the edge people who are not neighboring Sofia, Boaz, Daniella, Emil, and Zoe are already known criminals. In that same non-neighbor edge group, the only person whose identity is still unknown is Lorem. That group needs one more criminal to reach 5, so Lorem at D2 must be criminal.
14.C3 · Olsi → INNOCENT
Sofia’s clues talk about the same small group around her: Olsi, Ruby, and Vera are the only neighbors of Sofia whose identities are still unknown, and among edge cells there must be exactly 8 criminals total, with exactly 3 edge criminals neighboring Sofia. Sofia’s other clue says she has an odd number of innocent neighbors, and she already has 3 known innocent neighbors. If Olsi were criminal, then Ruby and Vera would be the only remaining people available to make both of those clues come out right at the same time, but they cannot do that. So Olsi cannot be criminal. That makes Olsi innocent.
15.C2 · Ipsum → CRIMINAL
Frank’s clue says Nancy has exactly 3 criminal neighbors, and exactly 2 of those criminals also neighbor Martin. That means exactly 1 of Nancy’s criminal neighbors does not neighbor Martin. Among Nancy’s neighbors who do not neighbor Martin, the people are Ipsum, Martin, Olsi, and Tyler. Martin, Olsi, and Tyler are already innocent, and there are no known criminals in that group yet. So the one criminal required in that group has to be Ipsum. That makes Ipsum criminal.
16.B4 · Sofia → CRIMINAL
If Sofia were innocent, then Olsi's clue that there are 11 criminals in total would have to be filled by the two unknown people, Ruby and Vera. That makes Ruby criminal and Vera criminal. Nancy's clue says the edge cells contain exactly 8 criminals, and exactly 3 of those edge criminals are Sofia's neighbors. In the edge cells that are not neighbors of Sofia, there are already 5 known criminals: Boaz, Daniella, Emil, Lorem, and Zoe. So the only way to reach exactly 3 edge criminals who are Sofia's neighbors is for Ruby and Vera to be criminals along with Will and Xavi, which gives 4 such neighbors, not 3. That clashes with Nancy's clue, so Sofia cannot be innocent. So Sofia at B4 must be criminal.
17.A4 · Ruby → INNOCENT
Emil’s clue says Martin has exactly 3 innocent neighbors. Among Martin’s neighbors, 2 are already known to be innocent, and the only neighbor there whose identity is still unknown is Ruby. That means the one remaining innocent neighbor Martin still needs has to be Ruby. So Ruby must be innocent.
18.A5 · Vera → CRIMINAL
Nancy’s clue says exactly 3 of the criminals on the edge are Sofia’s neighbors. Among the edge people who neighbor Sofia, Will and Xavi are already known criminals, while Martin and Ruby are innocent, so that group still needs 1 more criminal to reach 3. The only unknown person left in that edge-neighbor group is Vera. So Vera must be criminal.