Clues by Sam Feb 09, 2026 Answer – Full Solution Explained
A1
👩🏫
teacher
B1
👮♂️
cop
C1
👨🏫
teacher
D1
👨🔧
mech
A2
👷♀️
builder
B2
👮♂️
cop
C2
👩🔧
mech
D2
👨🔧
mech
A3
👷♀️
builder
B3
👩🌾
farmer
C3
👨🎤
singer
D3
👨🌾
farmer
A4
👨💼
clerk
B4
👮♀️
cop
C4
👩🎤
singer
D4
👩🌾
farmer
A5
👩💼
clerk
B5
👩🏫
teacher
C5
👨🎤
singer
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 — 6 criminals.
Clues by Sam answer for Feb 09, 2026 — a Easy solved in 15 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chase (B1), David (C1), Gabe (D1), Ruth (B4), Susan (C4) and Vera (B5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.C2 · Joyce → INNOCENT
Will’s clue explicitly says that “Joyce is one of 2 innocents above me,” which directly states that Joyce is an innocent. Therefore, we can determine that C2 Joyce is INNOCENT.
02.D4 · Tina → INNOCENT
Vera is at B5, so her neighbors in row 4 are only A4 Paul, B4 Ruth, and C4 Susan; D4 Tina is not Vera’s neighbor. Joyce’s clue says there are exactly two innocents in row 4, and exactly one of those two innocents is Vera’s neighbor. Since Tina is the only person in row 4 who is not Vera’s neighbor, the “other” innocent in row 4 must be Tina. Therefore, we can determine that D4 Tina is INNOCENT.
03.C1 · David → CRIMINAL
Look at column C above Will at C5: the people there are David at C1, Joyce at C2, Noah at C3, and Susan at C4. Will’s clue says there are exactly two innocents among those four, and Joyce is one of them, so among David, Noah, and Susan there is exactly one innocent and the other two are criminals. Tina at D4 says exactly two people in column C who neighbor her are innocent; the column C neighbors of D4 are Noah at C3, Susan at C4, and Will at C5, and since Will is already innocent, exactly one of Noah and Susan is innocent. That uses up the single innocent allowed among David, Noah, and Susan, so David cannot be that innocent and must be criminal. Therefore, we can determine that C1 David is CRIMINAL.
04.C3 · Noah → INNOCENT, C4 · Susan → CRIMINAL
The clue talks about the people above Will, meaning the four spots in column C above C5: C1 David, C2 Joyce, C3 Noah, and C4 Susan. It says there are exactly two innocents among those four, and those two innocents must be connected by orthogonal adjacency. Joyce at C2 is already known to be innocent, and David at C1 is already known to be criminal, so the second innocent must be either Noah at C3 or Susan at C4. If Susan at C4 were the second innocent, then the two innocents would be C2 and C4, but they would not be connected unless C3 were also innocent, which would create more than two innocents above Will; so the second innocent must be Noah at C3, and Susan at C4 must be criminal to keep the total at exactly two. Therefore, we can determine that C3 Noah is INNOCENT and C4 Susan is CRIMINAL.
05.D2 · Kumar → INNOCENT
Ivan is at B2, so the people to the right of Ivan in the same row are C2 Joyce and D2 Kumar. Noah’s clue says there are exactly 2 innocents to the right of Ivan, which means both of those right-side positions must be innocent. Since Joyce at C2 is already known to be INNOCENT, the other right-side person, Kumar at D2, must also be INNOCENT to make the total exactly two. Therefore, we can determine that D2 Kumar is INNOCENT.
06.A5 · Uma → INNOCENT
Row 5 has four people: Uma at A5, Vera at B5, Will at C5, and Xavi at D5, and Susan is at C4. Susan’s row-5 neighbors are only the three squares directly above row 5 in her neighborhood: B5, C5, and D5; A5 is the only position in row 5 that is not Susan’s neighbor. Kumar’s clue says there are exactly three innocents in row 5, and exactly two of those three are Susan’s neighbors, so the remaining innocent in row 5 must be the one person who is not Susan’s neighbor, which is Uma at A5. Therefore, we can determine that A5 Uma is INNOCENT.
07.B2 · Ivan → INNOCENT
Kumar is at D2, so the people to the left of Kumar are A2 Hazel, B2 Ivan, and C2 Joyce. David is at C1, and among those three, only Ivan at B2 and Joyce at C2 are neighbors of David (Hazel at A2 is too far away to be a neighbor, even diagonally). Uma’s clue says that exactly two innocents to the left of Kumar are neighboring David, so both of the only possible candidates, Ivan and Joyce, must be innocent. Since Joyce is already confirmed innocent, this forces Ivan to be innocent as well. Therefore, we can determine that B2 Ivan is INNOCENT.
08.B5 · Vera → CRIMINAL
Vera at B5 is adjacent to Paul at A4 and Ruth at B4, and those two are the only row 4 positions that can affect Uma at A5 besides Vera. Joyce’s clue says there are exactly two innocents in row 4, and we already know Tina at D4 is innocent while Susan at C4 is criminal, so exactly one of Paul or Ruth is the second innocent in row 4. That means Uma’s two row 4 neighbors (Paul and Ruth) include exactly one innocent. Ivan’s clue says Uma has an odd number of innocent neighbors, and Uma’s only neighbors are Paul, Ruth, and Vera, so with exactly one innocent already coming from Paul/Ruth, Vera cannot be innocent or the total would become 2, which is even. Therefore, we can determine that B5 Vera is CRIMINAL.
09.D5 · Xavi → INNOCENT
Row 5 has Uma at A5, Vera at B5, Will at C5, and Xavi at D5, and the clue says there are exactly 3 innocents in that row. Since Uma and Will are already known innocents, the only way to reach a total of 3 innocents in row 5 is for Xavi to be the third innocent. This also matches the rest of the clue, because Susan at C4 is a neighbor of C5 and D5 but not of A5, so the two neighbor-innocents can be Will and Xavi. Therefore, we can determine that D5 Xavi is INNOCENT.
10.A3 · Linda → INNOCENT, B3 · Mary → INNOCENT, D3 · Ollie → INNOCENT
Row 5 already has three innocents: Uma at A5, Will at C5, and Xavi at D5, with Vera at B5 being criminal. Xavi’s clue says there are more innocents in row 3 than in row 5, so row 3 must have at least four innocents. Since a row only has four people and Noah at C3 is already innocent, the remaining three people in row 3 (Linda at A3, Mary at B3, and Ollie at D3) must also be innocent to reach four. Therefore, we can determine that A3 Linda is INNOCENT, B3 Mary is INNOCENT, and D3 Ollie is INNOCENT.
11.A2 · Hazel → INNOCENT
Row 3 contains Linda at A3, Mary at B3, Noah at C3, and Ollie at D3, and all four of them are already known to be innocent, so row 3 has 4 innocents. Ollie’s clue says rows 2 and 3 have an equal number of innocents, so row 2 must also have 4 innocents. Since row 2 consists of Hazel at A2, Ivan at B2, Joyce at C2, and Kumar at D2, that forces Hazel to be innocent as well. Therefore, we can determine that A2 Hazel is INNOCENT.
12.A1 · Amy → INNOCENT
Vera is at B5, and her neighbors are A4, B4, C4, A5, and C5. Joyce’s clue says there are exactly two innocents in row 4, and only one of those two is Vera’s neighbor; since Susan at C4 is already a criminal and Tina at D4 is already an innocent, this forces exactly one of A4 or B4 to be an innocent. That means Vera has exactly three innocent neighbors in total: A5 and C5 are innocents, and exactly one of A4 or B4 is an innocent. Hazel’s clue says only one person in column B has exactly three innocent neighbors, so Vera at B5 must be the only one in column B with that count. Chase at B1 would have exactly three innocent neighbors only if Amy at A1 were not an innocent, because B1 already has three known innocent neighbors (A2, B2, and C2) and its other relevant neighbor A1 is the only one that can change that count. Since B1 is not allowed to also have exactly three innocent neighbors, Amy at A1 must be an innocent. Therefore, we can determine that A1 Amy is INNOCENT.
13.B1 · Chase → CRIMINAL, D1 · Gabe → CRIMINAL
Row 1 contains Amy at A1, Chase at B1, David at C1, and Gabe at D1. Amy’s clue says there is only one innocent in row 1, and we already know Amy herself is innocent. That means the other three people in row 1 cannot be innocent, so Chase at B1 and Gabe at D1 must be criminals (and David at C1 already fits as criminal). Therefore, we can determine that B1 Chase is CRIMINAL and D1 Gabe is CRIMINAL.
14.A4 · Paul → INNOCENT
The edge positions are the entire outside ring of the board, which includes A1–D1, A5–D5, A2–A4, and D2–D4. On those edges, we already have exactly nine confirmed innocents: A1, A2, A3, D2, D3, D4, A5, C5, and D5, and every other edge position besides A4 is already confirmed criminal or innocent. Gabe’s clue says there are at least 10 innocents on the edges, so the only remaining edge person who can raise the edge-innocent total from 9 to at least 10 is Paul at A4. Therefore, we can determine that A4 Paul is INNOCENT.
15.B4 · Ruth → CRIMINAL
Row 4 currently has Paul at A4 marked INNOCENT and Tina at D4 marked INNOCENT, while Ruth at B4 is the only unknown in that row. Joyce’s clue says “Only 1 of the 2 innocents in row 4 is Vera’s neighbor,” which fixes that there are exactly two innocents in row 4 total. Since Paul and Tina are already the two innocents, Ruth cannot also be an innocent, so her status must be CRIMINAL. Therefore, we can determine that B4 Ruth is CRIMINAL.