Clues by Sam Apr 07, 2026 Answer – Full Solution Explained
A1
👩💻
coder
B1
👩🍳
cook
C1
👨💼
clerk
D1
💂♀️
guard
A2
👩⚖️
judge
B2
💂♂️
guard
C2
🕵️♀️
sleuth
D2
👨💻
coder
A3
💂♀️
guard
B3
👨⚖️
judge
C3
🕵️♂️
sleuth
D3
👨✈️
pilot
A4
👩🎤
singer
B4
👨🎤
singer
C4
🕵️♂️
sleuth
D4
👩🍳
cook
A5
👩🎤
singer
B5
👩💼
clerk
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 — 15 criminals.
Clues by Sam answer for Apr 07, 2026 — a Easy solved in 16 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 15 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alice (A1), Bonnie (B1), Donald (C1), Eve (D1), Freya (A2), Gabe (B2), Hope (C2), Isaac (D2), Joy (A3), Mark (C3), Phil (B4), Tina (A5), Uma (B5), Wally (C5) and Zane (D5); the remaining 5 suspects are innocent.
The deduction chain, in plain English
01.B2 · Gabe → CRIMINAL, A3 · Joy → CRIMINAL
Freya is at A2, so her neighbors are A1, B1, B2, A3, and B3, including the diagonals. Noah’s clue says that any neighbor of Freya who is innocent cannot be a guard, which also means that any guard who neighbors Freya cannot be innocent. Gabe at B2 is a guard and is adjacent to Freya, and Joy at A3 is a guard and is also adjacent to Freya, so neither of them can be innocent. Therefore, we can determine that B2 Gabe is CRIMINAL, A3 Joy is CRIMINAL, and so on.
02.D1 · Eve → CRIMINAL, C2 · Hope → CRIMINAL
Donald is at C1 and Isaac is at D2, so the people who are neighbors of both of them are exactly Eve at D1 and Hope at C2. Joy’s clue says that Donald and Isaac have no innocent neighbors in common, which means none of their shared neighbors can be innocent. Since Eve and Hope are the only shared neighbors, both of them must be criminals. Therefore, we can determine that D1 Eve is CRIMINAL and C2 Hope is CRIMINAL, and so on.
03.B3 · Kevin → INNOCENT
The three guards are Eve at D1, Gabe at B2, and Joy at A3. Eve’s clue says that exactly one of these three guards has a criminal directly to their right. Gabe already has Hope at C2 directly to his right, and Hope is a criminal, while Eve has nobody to her right because she is on the edge. Since Gabe is already the one guard who satisfies the clue, Joy cannot also have a criminal directly to her right, so Kevin at B3 must be innocent. Therefore, we can determine that B3 Kevin is INNOCENT.
04.C3 · Mark → CRIMINAL
Row 3 contains Joy at A3, who is already known to be a criminal, and it also contains Kevin at B3 and Noah at D3, who are both known to be innocent. Kevin’s clue says row 4 is the only row with exactly one criminal, so every other row, including row 3, cannot have exactly one criminal. Since row 3 already has one criminal (Joy), the only way for row 3 to avoid having exactly one criminal is for it to have at least two criminals, which forces Mark at C3 to be a criminal. Therefore, we can determine that C3 Mark is CRIMINAL.
05.C1 · Donald → CRIMINAL
Isaac is at D2, so his neighbors are Donald at C1, Eve at D1, Hope at C2, Mark at C3, and Noah at D3. Mark’s clue says Isaac has only one innocent neighbor, and among those neighbors Noah is already known to be innocent. That means every other neighbor of Isaac must be criminal, including Donald at C1. Therefore, we can determine that C1 Donald is CRIMINAL.
06.D5 · Zane → CRIMINAL
Eve is at D1, so “below Eve” means the four people in column D at D2, D3, D4, and D5. Donald’s clue says there are exactly two innocents among those four, and those two innocents must be connected; with only two people, being connected means they must be directly adjacent up, down, left, or right. Since Noah at D3 is already an innocent, the only way for the two innocents below Eve to be connected is for the other innocent to be at D2 or D4, because those are the only squares directly adjacent to D3. If Zane at D5 were the other innocent, the two innocents would be D3 and D5, which are not adjacent, and making D4 innocent to connect them would create more than two innocents below Eve, breaking “both.” Therefore, we can determine that D5 Zane is CRIMINAL.
07.C4 · Rohan → INNOCENT
Look at the people below Eve in column D: Isaac at D2, Noah at D3, Sofia at D4, and Zane at D5. Donald’s clue says there are exactly two innocents among those four, and since Noah is already INNOCENT and Zane is already CRIMINAL, that forces exactly one of Isaac and Sofia to be INNOCENT. Zane’s clue says Noah has exactly 2 innocent neighbors, and Noah’s only possible innocent neighbors are Isaac, Rohan (at C4), and Sofia because Hope and Mark are CRIMINAL. Since Isaac and Sofia can contribute only one innocent between them, Rohan must be the second innocent neighbor Noah needs. Therefore, we can determine that C4 Rohan is INNOCENT.
08.A1 · Alice → CRIMINAL, B1 · Bonnie → CRIMINAL
Below Eve in column D are Isaac at D2, Noah at D3, Sofia at D4, and Zane at D5, and Donald’s clue says there are exactly two innocents among those people and those two innocents are connected. Since Noah is already an innocent and Zane is already a criminal, that forces exactly one of Isaac or Sofia to be the other innocent, meaning exactly one of them is a criminal. Noah’s neighbors include Hope and Mark (both criminals) plus Isaac and Sofia, so Noah has exactly three criminal neighbors in total. Rohan’s clue says Freya has more criminal neighbors than Noah, so Freya must have at least four criminal neighbors. Freya’s only possible criminal neighbors are Gabe and Joy (already criminals) plus Alice and Bonnie (since Kevin is an innocent), so Alice and Bonnie must both be criminals to reach four. Therefore, we can determine that A1 Alice is CRIMINAL and B1 Bonnie is CRIMINAL.
09.A2 · Freya → CRIMINAL
Gabe is at B2, so his neighbors are A1 Alice, B1 Bonnie, C1 Donald, A2 Freya, C2 Hope, A3 Joy, B3 Kevin, and C3 Mark. Alice’s clue says the total number of criminals among these neighbors is odd. Among the seven neighbors whose status we already know, six are criminals (Alice, Bonnie, Donald, Hope, Joy, and Mark) and Kevin is innocent, so the current criminal count is 6, which is even. To make the number of criminals neighboring Gabe odd, Freya must be a criminal. Therefore, we can determine that A2 Freya is CRIMINAL.
10.C5 · Wally → CRIMINAL
“Both innocents below Eve are connected” refers to the people in column D under Eve at D1, so it is talking about D2, D3, D4, and D5, and it tells us there are exactly two innocents among them. Since Noah at D3 is already an innocent and Zane at D5 is already a criminal, exactly one of Isaac at D2 or Sofia at D4 must be the second innocent, which means column D has exactly two innocents in total. Freya’s clue says only one column has exactly two innocents, so no other column is allowed to have exactly two innocents. Column C already has Rohan at C4 as an innocent and every other spot in column C except Wally at C5 is already a criminal, so Wally cannot be an innocent or column C would have exactly two innocents. Therefore, we can determine that C5 Wally is CRIMINAL.
11.D2 · Isaac → CRIMINAL
The people above Rohan at C4 are the three people in column C on rows 1–3: Donald at C1, Hope at C2, and Mark at C3. Wally’s clue says that among these three, only one has an innocent directly to their right (in column D on the same row). Mark at C3 already fits this, because Noah at D3 is innocent, so Hope at C2 cannot also fit it. That means Isaac at D2 cannot be innocent. Therefore, we can determine that D2 Isaac is CRIMINAL.
12.D4 · Sofia → INNOCENT
Eve is at D1, so “below Eve” refers to the four people in column D underneath her: Isaac at D2, Noah at D3, Sofia at D4, and Zane at D5. The clue says there are exactly two innocents among those four, and those two innocents must form one orthogonally connected group. Noah at D3 is already known to be innocent, while Isaac at D2 and Zane at D5 are already known to be criminals, so the only way to have two innocents below Eve is for Sofia at D4 to be the second innocent. With Noah at D3 and Sofia at D4 directly adjacent, the “connected” requirement is also satisfied. Therefore, we can determine that D4 Sofia is INNOCENT.
13.A5 · Tina → CRIMINAL
Kevin’s clue “Row 4 is the only row with exactly one criminal” forces Row 4 to contain exactly one criminal overall. Since C4 (Rohan) and D4 (Sofia) are already known innocents, that means exactly one of A4 (Olive) and B4 (Phil) is criminal, and the other is innocent. That restriction is completely contained within Row 4 and does not place any requirement on Row 5, so it does not constrain A5 (Tina) at all. Isaac’s text “Does it even count as a crime if it didn’t happen yet?” is not a checkable statement about any positions, counts, neighbors, rows, or columns, so as written it adds no logical restriction either. So with only these two clue texts, Tina’s status is not forced one way or the other; you would need Isaac’s actual rule-style clue (or another clue) to logically pin down A5. Therefore, we cannot determine that A5 (Tina) is CRIMINAL from the information shown here.
14.A4 · Olive → INNOCENT
Look at column A: A1 Alice, A2 Freya, A3 Joy, and A5 Tina are already known to be criminals, and the only person there whose status is not yet fixed is A4 Olive. Tina’s clue says each column has at least one innocent, so column A must contain an innocent somewhere. Since Olive is the only possible spot for an innocent in column A, Olive must be innocent. Therefore, we can determine that A4 Olive is INNOCENT.
15.B4 · Phil → CRIMINAL
Row 4 contains Olive at A4, Phil at B4, Rohan at C4, and Sofia at D4. Kevin’s clue says that row 4 has exactly one criminal (and that no other row has exactly one criminal). Since A4, C4, and D4 are already known to be innocent, the only way for row 4 to have exactly one criminal is for Phil at B4 to be that criminal. Therefore, we can determine that B4 Phil is CRIMINAL.
16.B5 · Uma → CRIMINAL
Look at how many innocents are in each column right now, noting that only Uma at B5 is still unknown. Column D already has exactly two innocents (Noah at D3 and Sofia at D4), while columns A and C each have only one innocent. Freya’s clue says only one column has exactly 2 innocents, so column B cannot also end up with exactly two innocents. If Uma were innocent, column B would have two innocents (Kevin at B3 and Uma at B5), creating a second column with exactly 2 innocents, which the clue forbids. Therefore, we can determine that B5 Uma is CRIMINAL.