Clues by Sam Apr 16, 2026 Answer – Full Solution Explained
A1
👩💼
clerk
B1
👩💼
clerk
C1
👩🔧
mech
D1
👨🔧
mech
A2
🕵️♂️
sleuth
B2
👨💼
clerk
C2
👨🔧
mech
D2
👨🍳
cook
A3
👮♂️
cop
B3
🕵️♂️
sleuth
C3
👷♀️
builder
D3
👨🍳
cook
A4
👮♀️
cop
B4
👮♀️
cop
C4
👨🍳
cook
D4
💂♂️
guard
A5
🕵️♀️
sleuth
B5
👷♀️
builder
C5
💂♀️
guard
D5
😬
elephant
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 — 9 criminals.
Clues by Sam answer for Apr 16, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 9 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Jason (D2), Kumar (A3), Megan (C3), Olive (A4), Petra (B4), Salil (D4), Tina (A5), Vicky (C5) and Zoe (D5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.A4 · Olive → CRIMINAL
Salil is at D4, so the people to the left of Salil in row 4 are Olive at A4, Petra at B4, and Rob at C4. Gary’s clue says there are exactly two criminals among those three, and among those two criminals, exactly one is a neighbor of Megan at C3. Petra and Rob are both neighbors of Megan (they are diagonally and directly below-left of her), so they cannot both be the two criminals; at most one of them can be a criminal. Since there must still be two criminals to the left of Salil, the other criminal has to be Olive, who is not Megan’s neighbor. Therefore, we can determine that A4 Olive is CRIMINAL.
02.B4 · Petra → CRIMINAL, A5 · Tina → CRIMINAL
Olive at A4 says that she and Uma at B5 have no innocent neighbors in common. The only people who are neighbors of both Olive and Uma are Petra at B4 (next to both of them) and Tina at A5 (also next to both of them). Since none of their common neighbors can be innocent, Petra and Tina cannot be innocent. Therefore, we can determine that B4 Petra is CRIMINAL and A5 Tina is CRIMINAL.
03.C4 · Rob → INNOCENT
Salil is at D4, so the people to his left in the same row are Olive at A4, Petra at B4, and Rob at C4. Gary’s clue talks about “the 2 criminals to the left of Salil,” which means there are exactly two criminals among those three left-of-Salil positions. Since Olive and Petra are already known criminals, they must be those two, so Rob cannot also be a criminal. Therefore, we can determine that C4 Rob is INNOCENT.
04.C3 · Megan → CRIMINAL
Salil is at D4, and his neighbors are Nick at D3, Rob at C4, Megan at C3, Vicky at C5, and Zoe at D5. Tina’s clue says that among Salil’s neighbors there are exactly 3 criminals total, and exactly 2 of those criminals are on edge spaces. Rob is already known to be innocent, and among the remaining four neighbors only Megan at C3 is not on an edge (Nick, Vicky, and Zoe are all on the edge). Since the third criminal cannot be on an edge, Megan must be that third criminal. Therefore, we can determine that C3 Megan is CRIMINAL.
05.C5 · Vicky → CRIMINAL
Salil is at D4, so the only people in column D who neighbor him are Nick at D3 and Zoe at D5. Rob’s clue says exactly 1 innocent in column D is neighboring Salil, so exactly one of Nick and Zoe is innocent and the other is a criminal. Tina’s clue says Salil has exactly 3 criminal neighbors and exactly 2 of those criminals are on the edges; since Megan at C3 is already a criminal neighbor of Salil and C3 is not on an edge, the other two criminal neighbors must be on edge spaces among C5, D3, and D5. Because only one of D3 and D5 can be a criminal (from Rob’s clue), C5 must be the second edge criminal neighbor. Therefore, we can determine that C5 Vicky is CRIMINAL.
06.D4 · Salil → CRIMINAL
Salil is at D4, and the only people in column D who neighbor Salil are Nick at D3 and Zoe at D5. Rob’s clue says exactly 1 innocent in column D is neighboring Salil, so exactly one of Nick and Zoe is innocent and the other is criminal. Megan’s clue says there are exactly 3 criminals in column D, and only 1 of those criminals is Donna’s neighbor; since Donna at C1 only neighbors Floyd at D1 and Jason at D2 in column D, that means exactly one of Floyd and Jason is criminal, so the other two criminals in column D must be among Nick, Salil, and Zoe. Because Nick and Zoe can contribute only one criminal between them, Salil must be the second criminal in that group. Therefore, we can determine that D4 Salil is CRIMINAL.
07.B1 · Bonnie → INNOCENT
Donna is at C1, so her neighbors are B1, B2, C2, D1, and D2, and the only neighbors she has in column D are D1 and D2. Megan’s clue says that, out of the three criminals somewhere in column D, exactly one of them is Donna’s neighbor, so exactly one of D1 and D2 is a criminal. Vicky’s clue says that, out of the seven criminals on the edges, exactly one is Donna’s neighbor, and Donna’s edge-neighbors are exactly B1, D1, and D2, so among those three squares there can be only one criminal total. Since D1 and D2 already contribute exactly one criminal between them, B1 cannot be a criminal and must be innocent. Therefore, we can determine that B1 Bonnie is INNOCENT.
08.C1 · Donna → INNOCENT
Donna is at C1, and her edge neighbors are B1, D1, and D2. Vicky says there are exactly 7 criminals on the edges, and exactly 1 of those 7 edge-criminals is Donna’s neighbor; since B1 (Bonnie) is already INNOCENT, that forces the one edge-criminal neighbor to be either D1 (Floyd) or D2 (Jason), so at least one of D1 or D2 must be an edge criminal. Now use Salil’s clue that column A has more criminals than column C. If Donna were a CRIMINAL, then column C would already have Megan, Vicky, and Donna as criminals, so the only way for column A to have strictly more is for both unknown people in column A (A1 and A3) to be criminals and for C2 to be innocent. That would make the edge criminals exactly A1, A3, C1 (Donna), A4, A5, C5, and D4, which already fills all 7 edge-criminal slots and leaves no room for D1 or D2 to be an edge criminal neighbor of Donna, even though Vicky’s clue requires one. Therefore, we can determine that C1 (Donna) is INNOCENT.
09.B5 · Uma → INNOCENT
Vicky’s clue says “Only 1 of the 7 criminals on the edges is Donna’s neighbor,” which tells us there are exactly 7 criminals on the edge squares. We already have four edge criminals fixed: A4 (Olive), D4 (Salil), A5 (Tina), and C5 (Vicky), so among the seven edge people whose status is still unknown (A1, A3, D1, D2, D3, B5, D5), exactly three must be criminals. Megan’s clue says there are exactly three criminals in column D, and since D4 is already one of them, exactly two of D1, D2, D3, and D5 are criminals; that uses up two of those three remaining edge-criminal slots, so among A1, A3, and B5 there can be only one criminal in total. Salil’s clue says column A has more criminals than column C, and column C already has at least two criminals (C3 Megan and C5 Vicky), so column A must have at least three criminals; with A4 (Olive) and A5 (Tina) already criminals, that forces at least one of A1 or A3 to be a criminal, which means B5 cannot be the one criminal among A1, A3, and B5. Therefore, we can determine that B5 (Uma) is INNOCENT.
10.B3 · Logan → INNOCENT
Rob is at C4, so his neighbors are B3 Logan, C3 Megan, D3 Nick, B4 Petra, D4 Salil, B5 Uma, C5 Vicky, and D5 Zoe. Bonnie’s clue says there is an odd number of innocents among these neighbors; since Uma is already innocent and Megan, Petra, Salil, and Vicky are already criminals, this means the number of innocents among the three unknown neighbors Logan, Nick, and Zoe must be even. Rob’s clue says exactly 1 innocent in column D is neighboring Salil, and the only column D neighbors of Salil at D4 are Nick at D3 and Zoe at D5, so exactly one of Nick and Zoe is innocent. That already contributes 1 innocent among Nick and Zoe, so to make the total number of innocents among Logan, Nick, and Zoe even, Logan must also be innocent. Therefore, we can determine that B3 Logan is INNOCENT.
11.C2 · Igor → INNOCENT
Donna is at C1, and Vicky’s clue tells us there are exactly 7 criminals on the edge in total, with only 1 of those edge-criminals being a neighbor of Donna. Donna’s only edge-neighbors are D1 and D2 (since B1 is already innocent), so exactly one of D1 and D2 is an edge-criminal. Rob’s clue says exactly 1 innocent in column D neighbors Salil at D4, and the only column-D neighbors of Salil are D3 and D5, so exactly one of D3 and D5 is an edge-criminal as well; that makes exactly 2 edge-criminals among D1, D2, D3, and D5. Since we already have 4 known edge-criminals (Olive at A4, Salil at D4, Tina at A5, and Vicky at C5), the 7 total edge-criminals mean there is exactly 1 more edge-criminal left among A1 and A3, so column A has exactly 3 criminals in total (A4, A5, and one of A1/A3). Salil’s clue says column A has more criminals than column C, and column C already has exactly 2 known criminals (Megan at C3 and Vicky at C5), so column C cannot have a third criminal; that forces Igor at C2 to be innocent. Therefore, we can determine that C2 Igor is INNOCENT.
12.B2 · Henry → INNOCENT
Donna is at C1, so her neighbors are Bonnie at B1, Henry at B2, Igor at C2, Floyd at D1, and Jason at D2. Megan’s clue says that among the three criminals in column D, exactly one is Donna’s neighbor, so exactly one of Floyd or Jason is a criminal. Logan’s clue says Donna has an odd number of criminal neighbors; since Bonnie and Igor are already innocent, the only possible criminal neighbors are Henry plus whichever one of Floyd or Jason is the criminal, giving either 1 or 2 criminals total. Because the total must be odd, it has to be 1, so Henry cannot be a criminal. Therefore, we can determine that B2 Henry is INNOCENT.
13.A1 · Anna → INNOCENT, D1 · Floyd → INNOCENT
Row 1 contains Anna at A1, Bonnie at B1, Donna at C1, and Floyd at D1. Henry’s clue says there are exactly 4 innocents in row 1, and since a row has only 4 people, that means everyone in row 1 must be innocent. Bonnie and Donna already match that, so the remaining two people in row 1, Anna and Floyd, must also be innocent. Therefore, we can determine that A1 Anna is INNOCENT and D1 Floyd is INNOCENT.
14.D2 · Jason → CRIMINAL
Donna is at C1, so the only people in column D who are her neighbors are Floyd at D1 and Jason at D2. Megan’s clue says that out of the three criminals in column D, exactly one of them is Donna’s neighbor. Since Floyd is already known to be INNOCENT, he cannot be that one criminal neighbor, so the only way for the clue to be true is for Jason at D2 to be a CRIMINAL. Therefore, we can determine that D2 Jason is CRIMINAL.
15.A3 · Kumar → CRIMINAL
In column C, the only criminals are Megan at C3 and Vicky at C5, so column C has exactly 2 criminals. Salil’s clue says column A has more criminals than column C, so column A must have at least 3 criminals. In column A we already have Olive at A4 and Tina at A5 as criminals, while Anna at A1 and Gary at A2 are innocents, leaving only Kumar at A3 as the remaining place where a third criminal can be. Therefore, we can determine that A3 Kumar is CRIMINAL.
16.D3 · Nick → INNOCENT
Megan is at C3, so her neighbors are the eight surrounding squares, including diagonals. Among those neighbors, the ones that are on the board’s edge are the three people in column D next to her: D2 Jason, D3 Nick, and D4 Salil. Igor’s clue says that on those edge neighbors, Megan has exactly one innocent neighbor; since Jason and Salil are already known to be criminals, the only way to have exactly one innocent there is for Nick to be innocent. Therefore, we can determine that D3 Nick is INNOCENT.
17.D5 · Zoe → CRIMINAL
Salil is at D4, and the only people in column D who neighbor Salil are Nick at D3 and Zoe at D5, since they are directly above and directly below him. Rob’s clue says that exactly 1 innocent in column D is neighboring Salil. Nick is already known to be innocent, so he already accounts for that one innocent neighbor from column D. That means Zoe cannot also be innocent, so she must be a criminal. Therefore, we can determine that D5 Zoe is CRIMINAL.