Clues by Sam Apr 08, 2026 Answer – Full Solution Explained
A1
👨💼
clerk
B1
👷♂️
builder
C1
👨💼
clerk
D1
🕵️♂️
sleuth
A2
👩⚖️
judge
B2
👩⚖️
judge
C2
👨🏫
teacher
D2
👨🍳
cook
A3
👩⚖️
judge
B3
👷♀️
builder
C3
👨✈️
pilot
D3
👷♂️
builder
A4
👩🌾
farmer
B4
👩🌾
farmer
C4
🕵️♀️
sleuth
D4
👩🍳
cook
A5
👩🌾
farmer
B5
👩🏫
teacher
C5
👨✈️
pilot
D5
👨✈️
pilot
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 — 13 criminals.
Clues by Sam answer for Apr 08, 2026 — a Medium solved in 17 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 13 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Eric (D1), Hope (B2), Igor (C2), Lisa (A3), Max (B3), Petra (A4), Rose (B4), Tina (C4), Uma (D4), Vera (A5), Wanda (B5), Xavi (C5) and Ziad (D5); the remaining 7 suspects are innocent.
The deduction chain, in plain English
01.B2 · Hope → CRIMINAL
In row 2, the four people are Freya at A2, Hope at B2, Igor at C2, and Kevin at D2, and we already know Kevin at D2 is innocent. Kevin’s clue says there are exactly two criminals in row 2, and those two criminals are connected by orthogonal adjacency within the row, which in a single row means they must be directly next to each other. With D2 not available as a criminal, the only adjacent pair that could make “both criminals” is either A2 with B2 or B2 with C2, and both possibilities include Hope at B2. Therefore, we can determine that B2 (Hope) is CRIMINAL.
02.A1 · Alex → INNOCENT
Freya is at A2, and her neighbors are Alex (A1), Brian (B1), Hope (B2), Lisa (A3), and Max (B3). Hope’s clue tells us that exactly two of those neighbors are innocents, and among those two innocents, exactly one is also a neighbor of Brian. The only neighbors of Freya who are also neighbors of Brian are Alex and Hope, but Hope is already known to be a criminal, so she cannot be the innocent that is Brian’s neighbor. That means Alex must be the one innocent neighbor of Freya who is also Brian’s neighbor. Therefore, we can determine that A1 Alex is INNOCENT.
03.C2 · Igor → CRIMINAL
Nick is at C3, and the people in row 2 who neighbor him are B2 Hope, C2 Igor, and D2 Kevin (since neighbors include diagonals). Alex’s clue says that among those row 2 neighbors of Nick, exactly one is innocent. Kevin at D2 is already confirmed innocent, and Hope at B2 is already confirmed criminal, so Igor at C2 cannot also be innocent or the total would be more than one. Therefore, we can determine that C2 Igor is CRIMINAL.
04.A2 · Freya → INNOCENT
In row 2, we already know Hope at B2 and Igor at C2 are criminals, and Kevin at D2 is innocent while Freya at A2 is unknown. Kevin’s clue says that the both criminals in row 2 are connected, meaning row 2 has exactly two criminals and they must touch each other orthogonally in that row. Since B2 and C2 are already adjacent criminals, they must be the only criminals in row 2, so Freya at A2 cannot be a criminal. Therefore, we can determine that A2 Freya is INNOCENT.
05.D4 · Uma → CRIMINAL
Igor’s clue says that Uma is one of the three criminals in column D, which directly states that Uma is a criminal. Therefore, we can determine that D4 Uma is CRIMINAL.
06.D1 · Eric → CRIMINAL
In column D we have Eric at D1, Kevin at D2 who is innocent, Oscar at D3, Uma at D4 who is criminal, and Ziad at D5. Igor’s clue says there are exactly three criminals in column D, so besides Uma, exactly two of {Eric, Oscar, Ziad} must be criminals. Uma’s clue says there is only one innocent below Kevin, and below Kevin are Oscar, Uma, and Ziad; since Uma is not that innocent, exactly one of {Oscar, Ziad} is innocent and the other is criminal, meaning {Oscar, Ziad} contains exactly one criminal. That leaves Eric as the only way to reach the required two criminals among {Eric, Oscar, Ziad}. Therefore, we can determine that D1 Eric is CRIMINAL.
07.A3 · Lisa → CRIMINAL, A4 · Petra → CRIMINAL, A5 · Vera → CRIMINAL
Look at column A: it contains A1 Alex, A2 Freya, A3 Lisa, A4 Petra, and A5 Vera. Eric’s clue says that each column must have at least 3 criminals. Since Alex at A1 and Freya at A2 are already confirmed INNOCENT, the only way for column A to reach at least 3 criminals is for all three remaining people in that column, Lisa at A3, Petra at A4, and Vera at A5, to be CRIMINAL. Therefore, we can determine that A3 Lisa is CRIMINAL, A4 Petra is CRIMINAL, and A5 Vera is CRIMINAL.
08.B4 · Rose → CRIMINAL
Freya is at A2, and her neighbors are Alex at A1, Brian at B1, Hope at B2, Lisa at A3, and Max at B3. Hope’s clue says Freya has exactly two innocent neighbors, and since Alex is already innocent while Hope and Lisa are already criminals, exactly one of Brian (B1) and Max (B3) must be criminal. Vera’s clue says the number of criminals above Wanda (in column B: B1, B2, B3, B4) is odd, and since Hope at B2 is already one criminal, the total number of criminals among B1, B3, and B4 must be even. Because B1 and B3 contain exactly one criminal, B4 must be criminal to make that group’s total even. Therefore, we can determine that B4 Rose is CRIMINAL.
09.B5 · Wanda → CRIMINAL
Freya is at A2, and her neighbors are Alex (A1), Brian (B1), Hope (B2), Lisa (A3), and Max (B3). Hope’s clue says that Freya has exactly two innocent neighbors, and that exactly one of those two is also a neighbor of Brian. Alex is already an innocent neighbor of Freya, and Alex is also a neighbor of Brian, so the other innocent neighbor of Freya must be someone who is not a neighbor of Brian; among Freya’s remaining neighbors, that forces exactly one of Brian (B1) or Max (B3) to be innocent. Rose’s clue says there are more innocents in column A than in column B, and column A already has exactly two innocents (Alex and Freya), so column B can have at most one innocent total. Since we already need one innocent in column B coming from exactly one of B1 or B3, Wanda at B5 cannot be innocent. Therefore, we can determine that B5 Wanda is CRIMINAL.
10.C3 · Nick → INNOCENT
Wanda’s clue explicitly says that Nick is one of the two innocents in row 3, which directly includes Nick among the innocents for that row. Therefore, we can determine that C3 Nick is INNOCENT.
11.C5 · Xavi → CRIMINAL
Uma is at D4, so her edge neighbors are the three spaces D3 (Oscar), D5 (Ziad), and C5 (Xavi), because those are the neighboring positions that lie on the outer edge of the board. Igor’s clue says there are exactly 3 criminals in column D, and we already have Eric at D1 and Uma at D4 as criminals while Kevin at D2 is innocent, so exactly one of Oscar at D3 and Ziad at D5 must be innocent. Nick’s clue says the number of innocents among Uma’s edge neighbors is odd; since D3 and D5 already contribute exactly one innocent, C5 cannot also be innocent or the total would become 2, which is even. Therefore, we can determine that C5 Xavi is CRIMINAL.
12.C4 · Tina → CRIMINAL
The clue says that every person in row 3 must have 2 or fewer innocent neighbors. Max is at B3 in row 3, and two of Max’s neighbors are already confirmed innocents: Freya at A2 and Nick at C3. Tina at C4 is also a neighbor of Max (diagonally adjacent), so if Tina were innocent, Max would already have at least three innocent neighbors, which the clue forbids. Therefore, we can determine that C4 Tina is CRIMINAL.
13.C1 · Denis → INNOCENT
In column C, we already know that Nick at C3 is innocent, while Igor at C2, Tina at C4, and Xavi at C5 are criminals, leaving only Denis at C1 unknown. Tina’s clue says column B is the only column with exactly one innocent, so no other column is allowed to end up with exactly one innocent. If Denis at C1 were a criminal, then column C would have exactly one innocent (only Nick), which is not allowed. So Denis must be the second innocent in column C. Therefore, we can determine that C1 Denis is INNOCENT.
14.B1 · Brian → INNOCENT
The edge positions already confirmed as innocent are A1 (Alex), C1 (Denis), A2 (Freya), and D2 (Kevin), so we currently have 4 innocents on the edges. Denis’s clue says there are at least 6 innocents on the edges, so we must add at least 2 more edge innocents, and the only edge people not yet decided are B1 (Brian), D3 (Oscar), and D5 (Ziad). Igor’s clue says Uma is one of 3 criminals in column D; since D1 (Eric) and D4 (Uma) are already criminals and D2 (Kevin) is already innocent, exactly one of D3 and D5 can be criminal and the other must be innocent, so D3 and D5 contribute only 1 edge innocent total. That means the second required new edge innocent must be Brian at B1. Therefore, we can determine that B1 Brian is INNOCENT.
15.B3 · Max → CRIMINAL
Freya is at A2, so her neighbors are Alex at A1, Brian at B1, Hope at B2, Lisa at A3, and Max at B3. Hope’s clue says that there are exactly two innocents among Freya’s neighbors, and that only one of those two is a neighbor of Brian. Alex and Brian are already the two known innocents next to Freya (and only Alex is a neighbor of Brian, since Brian cannot be his own neighbor), so there cannot be a third innocent neighbor of Freya. That means Max cannot be innocent, so he must be criminal. Therefore, we can determine that B3 Max is CRIMINAL.
16.D3 · Oscar → INNOCENT
In row 3, the people are Lisa at A3, Max at B3, Nick at C3, and Oscar at D3. Wanda’s clue says that Nick is one of exactly two innocents in row 3, so row 3 must contain exactly two innocents total. Since Lisa and Max are already known to be criminals, the only way for the row to have two innocents is for Oscar to be the second innocent alongside Nick. Therefore, we can determine that D3 Oscar is INNOCENT.
17.D5 · Ziad → CRIMINAL
In column D we have Eric at D1, Kevin at D2, Oscar at D3, Uma at D4, and Ziad at D5. Igor’s clue says that Uma is one of 3 criminals in column D, so there must be exactly three criminals somewhere in that column. Eric and Uma are already confirmed criminals, while Kevin and Oscar are confirmed innocents, so the only remaining place for the third criminal is Ziad at D5. Therefore, we can determine that D5 Ziad is CRIMINAL.