Clues by Sam Jan 17, 2026 Answer – Full Solution Explained
A1
👨🎨
painter
B1
👩🏫
teacher
C1
💂♀️
guard
D1
👨🍳
cook
A2
🕵️♂️
sleuth
B2
👷♂️
builder
C2
👨🍳
cook
D2
👩🍳
cook
A3
🕵️♀️
sleuth
B3
👨⚖️
judge
C3
💂♀️
guard
D3
💂♂️
guard
A4
👩⚖️
judge
B4
👩🏫
teacher
C4
👨✈️
pilot
D4
👷♀️
builder
A5
👩🎨
painter
B5
👨⚖️
judge
C5
👨✈️
pilot
D5
👩🎨
painter
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 — 16 criminals.
Clues by Sam answer for Jan 17, 2026 — a Hard solved in 19 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 16 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chase (A1), Evie (C1), Frank (D1), Ivan (B2), Kay (D2), Laura (A3), Martin (B3), Olof (D3), Petra (A4), Susan (B4), Thor (C4), Uma (D4), Vicky (A5), Will (B5), Xavi (C5) and Zoe (D5); the remaining 4 suspects are innocent.
The deduction chain, in plain English
01.A1 · Chase → CRIMINAL
Dana’s clue explicitly says, “Chase is one of 2 criminals above Petra,” which directly identifies Chase as a criminal. Therefore, we can determine that A1 Chase is CRIMINAL.
02.B4 · Susan → CRIMINAL
Chase’s clue explicitly says “Susan is one of Thor’s 7 criminal neighbors,” and since all statements are true, this directly means Susan is a criminal. Therefore, we can determine that B4 Susan is CRIMINAL.
03.C5 · Xavi → CRIMINAL
Relevant people: Thor at C4 and his neighbors, including Susan at B4 and the row 5 trio B5–C5–D5; row 5 connectivity concerns A5–B5–C5–D5. Chase’s clue says Susan is one of Thor’s seven criminal neighbors, so among Thor’s eight neighbors exactly one is innocent. Besides Susan, Thor’s remaining seven neighbors include B5, C5, and D5, so at most one of those three can be innocent—meaning at least two of B5, C5, and D5 are criminals. Susan’s clue says all criminals in row 5 are connected, so any two or more criminals among B5, C5, and D5 must be contiguous. The only way two or more of these three can be connected is if C5 is one of the criminals (B5 and D5 alone would be disconnected). Therefore, we can determine that C5 · Xavi is CRIMINAL.
04.B5 · Will → CRIMINAL
Relevant people: Thor at C4, Susan at B4, Xavi at C5, Will at B5, and Zoe at D5. Thor has exactly 7 criminal neighbors and Susan is one of them, so among his other neighbors (which include Will and Zoe) at most one can be innocent—meaning at least one of Will or Zoe is criminal. With Xavi already criminal in row 5, that gives row 5 at least two criminals; but Xavi also says row 2 is the only row with exactly two criminals, so row 5 must have three or four. Susan states all criminals in row 5 are connected; with C5 in that connected group and the row needing at least three criminals, the connected block must pass through B5. Therefore, we can determine that B5 (Will) is CRIMINAL.
05.D5 · Zoe → CRIMINAL
Relevant people are around Thor at C4: his eight neighbors are B3, C3, D3, B4 (Susan), D4, B5 (Will), C5 (Xavi), and D5 (Zoe). “Susan is one of Thor’s 7 criminal neighbors” means exactly seven of these eight are criminals; with Susan, Will, and Xavi already criminals, the remaining five—B3, C3, D3, D4, and D5—contain exactly one innocent. “There are more innocents in row 3 than row 5,” and since row 5 already has two criminals (Will and Xavi), if D5 were that one innocent then row 3 could have at most one innocent (only A3), while row 5 would have at least one, so row 3 could not have more than row 5. Hence D5 cannot be the lone innocent in that group and must be a criminal. Therefore, we can determine that D5 · Zoe is CRIMINAL.
06.A5 · Vicky → CRIMINAL
Thor (C4) has eight neighbors: B3, C3, D3, B4, D4, B5, C5, and D5, and the clue says exactly seven of them are criminals, with Susan (B4) among those criminals. Since B4, B5 (Will), C5 (Xavi), and D5 (Zoe) are already known criminals, the single innocent among Thor’s neighbors must be one of B3, C3, D3, or D4, so at least three of those four are criminals; in particular, row 3 (B3, C3, D3, A3) already has at least two criminals among B3/C3/D3. Because row 2 is the only row with exactly two criminals, row 3 cannot have exactly two, so it must have at least three criminals and therefore at most one innocent. Will’s clue says row 3 has more innocents than row 5, so row 5 must have zero innocents, which makes A5 (Vicky) a criminal. Therefore, we can determine that Vicky is CRIMINAL.
07.D3 · Olof → CRIMINAL
Ivan is at B2, and Vicky’s clue says he has exactly 4 innocent neighbors with only 1 of those in row 3, so among Ivan’s row‑3 neighbors A3, B3, and C3, exactly two are criminals. Xavi’s clue says row 2 is the only row with exactly two criminals, so row 3 cannot total exactly two criminals. Row 3 already has two criminals among A3/B3/C3, so to avoid row 3 having exactly two, the remaining spot in that row, D3 (Olof), must also be a criminal. Therefore, we can determine that D3 Olof is CRIMINAL.
08.A4 · Petra → CRIMINAL
Consider column A around Petra at A4: above her are A1 (Chase), A2 (Gabe), and A3 (Laura), and Martin’s column-A neighbors are A2, A3, and A4. Dana says there are exactly 2 criminals above Petra and Chase is one of them, so between A2 and A3 exactly one is criminal and the other is innocent. Olof says Martin has exactly three innocent neighbors and only one of them is in column A, so among A2, A3, and A4 there is exactly one innocent. Since A2/A3 already account for exactly one innocent, A4 cannot be innocent. Therefore, we can determine that A4 · Petra is CRIMINAL.
09.C4 · Thor → CRIMINAL
Xavi is at C5, so the people above him are Evie (C1), John (C2), Nicole (C3), and Thor (C4). Petra says there are exactly two innocents among these four, and Zoe adds that exactly two innocents above Xavi are neighboring Ivan. Ivan is at B2, whose neighbors among those four are C1, C2, and C3, but not C4. Therefore the two innocents must be within C1–C3, so C4 cannot be an innocent. Therefore, we can determine that Thor is CRIMINAL.
10.B3 · Martin → CRIMINAL
The people that matter are Ivan at B2 (and his neighbors), Xavi at C5, Petra at A4, and Martin at B3. From Zoe, the only positions “above Xavi” that neighbor Ivan are C1–C3, and exactly two of those three are innocents; from Vicky, Ivan has exactly four innocent neighbors in total. Since B1 (Dana) is already an innocent neighbor, those two innocents among C1–C3 plus B1 account for three of Ivan’s four innocent neighbors, leaving room for only one more innocent among the remaining neighbors A2, A3, and B3 (A1 is a known criminal). From Dana’s clue about Petra, A1 is one of the two criminals above Petra, so between A2 and A3 exactly one is innocent, which uses up that final innocent slot. Therefore, we can determine that B3 · Martin is CRIMINAL.
11.D2 · Kay → CRIMINAL
Focus on row 2 (A2–D2) and on Martin at B3. Olof says Martin has exactly 3 innocent neighbors; with A4, B4, and C4 already criminals, that leaves exactly 2 criminals among his remaining neighbors A2, B2, C2, A3, and C3. Xavi says row 2 is the only row with exactly 2 criminals, so row 3 cannot have exactly 2; since B3 and D3 are already criminals, at least one of A3 or C3 must also be a criminal. That means among A2, B2, and C2 there is at most one criminal; but row 2 must total exactly two criminals, so those three must contribute one criminal and D2 must be the other. Therefore, we can determine that D2 · Kay is CRIMINAL.
12.C2 · John → INNOCENT
Above Xavi (C5) are C1–C4; among these, only C1, C2, and C3 neighbor Ivan at B2. Zoe’s clue says exactly two of C1, C2, and C3 are innocents. Kay’s clue says John’s column‑C neighbors are C1 and C3, and exactly one of them is innocent. Since C1 and C3 contribute exactly one innocent, the second innocent among C1, C2, and C3 must be C2. Therefore, we can determine that C2 · John is INNOCENT.
13.B2 · Ivan → CRIMINAL
The people “above Xavi” are C1, C2, C3, and C4; among these, only C1, C2, and C3 are neighbors of Ivan at B2. Zoe’s clue says exactly two innocents from that neighbor subset are above Xavi; since C2 (John) is already innocent, exactly one of C1 or C3 is innocent and the other is criminal. That fixes column C to have exactly three criminals: C4 and C5 (already criminal) plus exactly one of C1/C3. John’s clue says only one column has exactly three criminals, so column B cannot also total three. Column B already has three criminals (B3, B4, B5); therefore B2 (Ivan) cannot be innocent and must be the fourth criminal. Therefore, we can determine that B2 Ivan is CRIMINAL.
14.A2 · Gabe → INNOCENT
In row 2, the people are Gabe (A2), Ivan (B2, criminal), John (C2, innocent), and Kay (D2, criminal). The clue says row 2 is the only row with exactly 2 criminals, so row 2 must have exactly two criminals. Ivan and Kay already make two; if Gabe were also a criminal, the row would have three, which would break the clue. Therefore, we can determine that A2 Gabe is INNOCENT.
15.A3 · Laura → CRIMINAL
Above Petra (A4) in column A are Chase (A1), Gabe (A2), and Laura (A3). Dana’s clue says there are exactly two criminals among those above Petra, and Chase is one of them. We already know Chase is a criminal and Gabe is innocent, so the second criminal must be Laura. Therefore, we can determine that A3 · Laura is CRIMINAL.
16.C1 · Evie → CRIMINAL
Ivan (B2) has eight neighbors: among them, Dana (B1), Gabe (A2), and John (C2) are already known innocents, while Laura (A3) and Martin (B3) are criminals. The clue says Ivan has exactly four innocent neighbors, and only one of those is in row 3. Since A3 and B3 are criminals, the single row‑3 innocent must be Nicole (C3), which completes the four innocents (B1, A2, C2, C3). That leaves Evie (C1), the only other unknown neighbor, unable to be innocent. Therefore, we can determine that C1 Evie is CRIMINAL.
17.D1 · Frank → CRIMINAL
We only need to compare rows 2 and 1. The clue says row 2 is the only row with exactly 2 criminals. Row 2 already has exactly two (Ivan and Kay), so every other row must have a different count; row 1 already has two known criminals (Chase and Evie), so to avoid also having exactly two, it must have more than two, which forces Frank to be a criminal. Therefore, we can determine that D1 · Frank is CRIMINAL.
18.C3 · Nicole → INNOCENT
The people “above Xavi” are the four in column C at C1–C4, and among them, only C1, C2, and C3 are neighbors of Ivan at B2. Zoe’s clue says exactly two of those neighbors are innocents. We already know C1 (Evie) is a criminal and C2 (John) is an innocent, so to reach exactly two innocents among C1–C3, C3 must also be an innocent. Therefore, we can determine that C3 · Nicole is INNOCENT.
19.D4 · Uma → CRIMINAL
Thor (C4) has eight neighbors: Martin (B3), Nicole (C3), Olof (D3), Susan (B4), Uma (D4), Will (B5), Xavi (C5), and Zoe (D5). The clue states Thor has exactly seven criminal neighbors, and Susan is one of them. Among these neighbors, six are already criminals (Martin, Olof, Susan, Will, Xavi, Zoe) and one is innocent (Nicole), leaving Uma as the only unknown. To make the total reach seven criminal neighbors, Uma must also be criminal. Therefore, we can determine that D4 Uma is CRIMINAL.