Clues by Sam Jan 16, 2026 Answer – Full Solution Explained
A1
👩🎨
painter
B1
👩🏫
teacher
C1
💂♂️
guard
D1
💂♂️
guard
A2
👷♂️
builder
B2
👨🏫
teacher
C2
👷♂️
builder
D2
👨🍳
cook
A3
👨💼
clerk
B3
👨🌾
farmer
C3
👩🌾
farmer
D3
💂♀️
guard
A4
👩💼
clerk
B4
👩🎨
painter
C4
👨🌾
farmer
D4
👨💻
coder
A5
👩🍳
cook
B5
👩💼
clerk
C5
👩💻
coder
D5
👩💻
coder
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 — 4 criminals.
Clues by Sam answer for Jan 16, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chase (C1), Floyd (A2), Wanda (B5) and Zoe (D5); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.C2 · Henry → INNOCENT
The clue directly states that Henry is one of Igor's innocent neighbors, so Henry is innocent. Therefore, we can determine that C2 Henry is INNOCENT.
02.D3 · Megan → INNOCENT
Relevant people are Igor at D2 and his neighbors: C1 (Chase), D1 (Daniel), C2 (Henry), C3 (Lucy), and D3 (Megan). Clue 1 says Igor has exactly 4 innocent neighbors and Henry is one of them; Daniel is also already known to be innocent. Clue 2 says exactly two innocents above Xena (C1, C2, C3, C4) are neighboring Igor; only C1, C2, and C3 neighbor Igor, and since C2 is already innocent, exactly one of C1 or C3 is innocent. That gives Igor three innocents among his neighbors so far (Daniel, Henry, and exactly one of C1/C3), so to reach four, D3 (Megan) must also be innocent. Therefore, we can determine that D3 · Megan is INNOCENT.
03.B1 · Betty → INNOCENT
The three guards are known to be Chase at C1, Daniel at D1, and Megan at D3, so the people directly to their left are, respectively, Betty at B1, Chase at C1, and Lucy at C3. Megan says that exactly two of those three left neighbors are innocent. Daniel’s clue fixes that Igor (D2) has exactly four innocent neighbors; with Daniel (D1), Henry (C2), and Megan (D3) already innocent, only one of Chase (C1) and Lucy (C3) can be innocent. Since exactly one of C1 and C3 is innocent, Betty at B1 must be the second innocent among the three left-of-guard positions. Therefore, we can determine that B1 Betty is INNOCENT.
04.B3 · Kumar → INNOCENT
Relevant people: Henry (C2), Igor (D2), Daniel (D1), Megan (D3), Chase (C1), Lucy (C3), Gus (B2), Betty (B1), and Kumar (B3). Daniel says Henry is one of Igor’s four innocent neighbors; Igor’s five neighbors are Henry, Daniel, Megan, Chase, and Lucy, and since Daniel, Megan, and Henry are already innocent, exactly one of Chase or Lucy is innocent. Betty says exactly three of Gus’s six innocent neighbors also neighbor Henry; the people who neighbor both Gus and Henry are Betty, Chase, Kumar, and Lucy, so exactly three of these four are innocent. With Betty already innocent, exactly two among Chase, Kumar, and Lucy are innocent, and together with “exactly one of Chase or Lucy,” Kumar must be the second innocent. Therefore, we can determine that B3 · Kumar is INNOCENT.
05.A4 · Nicole → INNOCENT, A5 · Uma → INNOCENT
Gus (B2) and Henry (C2) share exactly four common neighbors: B1, C1, B3, and C3; among these, B1 and B3 are already known to be innocent. Betty’s clue says Gus has exactly 6 innocent neighbors and exactly 3 of those also neighbor Henry, so among the four common neighbors exactly three are innocent—therefore exactly one of C1 or C3 is innocent and the other is criminal. Counting Gus’s neighbors then gives four fixed innocents (B1, B3, Henry at C2, and one of C1/C3), so to reach six, exactly two of A1, A2, and A3 must be innocent. Kumar’s clue says column A has exactly four innocents; with A1–A3 contributing exactly two, the remaining two must be A4 and A5. Therefore, we can determine that A4 Nicole is INNOCENT and A5 Uma is INNOCENT.
06.A1 · Alice → INNOCENT
Relevant spots are column A (A1–A5) and Kumar at B3, whose edge neighbors are A2, A3, and A4; we already know A4 and A5 are innocent. Kumar says column A has exactly 4 innocents, and Uma says exactly 2 of the edge innocents are Kumar’s neighbors. Since the only edge neighbors of Kumar are A2, A3, and A4, and A4 is already innocent, exactly one of A2 or A3 can be innocent. Column A needs 4 innocents total; with A4 and A5 already innocent, two of A1, A2, and A3 must be innocent, but only one of A2/A3 can be, so A1 must be the other. Therefore, we can determine that A1 Alice is INNOCENT.
07.D4 · Sam → INNOCENT
The clue explicitly says Sam is one of Raul’s six innocent neighbors, which directly makes Sam an innocent. Therefore, we can determine that D4 · Sam is INNOCENT.
08.B4 · Pam → INNOCENT, C3 · Lucy → INNOCENT
Raul sits at C4 with eight neighbors: B3, C3, D3, B4, D4, B5, C5, and D5. Alice’s clue says Sam is one of Raul’s six innocent neighbors, so exactly two of Raul’s neighbors are criminals; among these eight we already know B3 (Kumar), D3 (Megan), and D4 (Sam) are innocent. Nicole’s clue says row 5 has exactly two criminals; among Raul’s neighbors, the row‑5 trio B5 (Wanda), C5 (Xena), and D5 (Zoe) therefore contain exactly those two criminals. That uses up both criminal spots among Raul’s neighbors, so the remaining unknown neighbors C3 (Lucy) and B4 (Pam) must both be innocent. Therefore, we can determine that B4 Pam is INNOCENT and C3 Lucy is INNOCENT.
09.C1 · Chase → CRIMINAL
Relevant people: Igor at D2 and his neighbors C1 (Chase), D1 (Daniel), C2 (Henry), C3 (Lucy), and D3 (Megan). The clue says Igor has exactly 4 innocent neighbors and Henry is one of them. Among these five neighbors, Daniel, Henry, Lucy, and Megan are already known to be innocent—this makes four. That leaves Chase as the only neighbor who cannot also be innocent. Therefore, we can determine that C1 Chase is CRIMINAL.
10.D2 · Igor → INNOCENT
Kumar (B3) has three edge neighbors: A2, A3, and A4. Uma’s clue says exactly 2 of the 10 edge innocents are Kumar’s neighbors; since A4 is already innocent, exactly one of A2 or A3 is also innocent (the other is criminal). On the edges we already know A1, B1, D1, A4, D3, D4, and A5 are innocent; adding exactly one of A2/A3 makes 8 edge innocents, so we still need exactly 2 more among D2, B5, C5, and D5. Nicole’s clue says row 5 has exactly 2 criminals; with A5 innocent, exactly one of B5, C5, and D5 is innocent. To reach the needed two more edge innocents, D2 must be the second one. Therefore, we can determine that D2 Igor is INNOCENT.
11.C4 · Raul → INNOCENT
Row 4 already has Nicole, Pam, and Sam confirmed innocent, so only Raul could make row 4 have any criminals. Nicole says row 5 is the only row with exactly two criminals, so row 2 cannot have two and thus has at most one. Lucy says row 2 has more criminals than row 4, which can only happen if row 4 has zero criminals (since row 2 cannot reach two). That forces Raul to be innocent. Therefore, we can determine that C4 Raul is INNOCENT.
12.B5 · Wanda → CRIMINAL
Row 5 is the only row with exactly two criminals, and Uma (A5) is already innocent, so exactly one of Wanda (B5), Xena (C5), and Zoe (D5) must be innocent. Raul’s clue says the number of innocent coders must equal the number of innocent guards. Comparing the professions of Wanda, Xena, and Zoe against the innocents already placed outside Row 5, the only way to keep those two counts equal is for the lone innocent in Row 5 to be one of Xena or Zoe; if Wanda were the lone innocent, the innocent coder/guard totals would no longer match. Therefore, we can determine that B5 Wanda is CRIMINAL.
13.B2 · Gus → INNOCENT
Around Gus (B2) the neighbors are A1, B1, C1, A2, C2 (Henry), A3, B3, and C3; of these, A1, B1, C2, B3, and C3 are already innocent, and C1 is criminal. Betty’s clue says Gus has exactly 6 innocent neighbors, so with one criminal already at C1, exactly one of the two unknown neighbors A2 or A3 must be criminal. Wanda’s row-3 clue applied to Kumar (B3) means B3 cannot have more than one criminal neighbor; since B3’s only possible criminal neighbors are A2, A3, and Gus, at most one of those three can be criminal. Combining these, because exactly one of A2 or A3 is criminal, Gus cannot also be criminal. Therefore, we can determine that B2 Gus is INNOCENT.
14.A2 · Floyd → CRIMINAL
We’re comparing rows 2 and 4. Row 4 is fully known and all four (Nicole, Pam, Raul, Sam) are innocent, so row 4 has 0 criminals. In row 2, Gus, Henry, and Igor are confirmed innocent, leaving only Floyd undecided. Lucy’s clue says row 2 must have more criminals than row 4, so row 2 must have at least 1 criminal; the only way to achieve that is for Floyd to be a criminal. Therefore, we can determine that A2 Floyd is CRIMINAL.
15.A3 · Jerry → INNOCENT
Gus (B2) has eight neighbors: A1 Alice, B1 Betty, C1 Chase, A2 Floyd, C2 Henry, A3 Jerry, B3 Kumar, and C3 Lucy; Jerry is the only one among them with unknown status. The clue states Gus has exactly six innocent neighbors (and among those six, exactly three also neighbor Henry). We already know five of these neighbors are innocent—Alice, Betty, Henry, Kumar, and Lucy—and two are criminals—Chase and Floyd—so the sixth innocent must be Jerry. Therefore, we can determine that A3 · Jerry is INNOCENT.
16.D5 · Zoe → CRIMINAL
We look column by column. Columns A, B, and C already contain a criminal (Floyd at A2, Wanda at B5, and Chase at C1), so they satisfy the clue. In column D, Daniel, Igor, Megan, and Sam (D1–D4) are all confirmed innocent, leaving only Zoe (D5) as the sole possible criminal to meet “each column has at least one criminal.” Therefore, we can determine that D5 Zoe is CRIMINAL.
17.C5 · Xena → INNOCENT
Row 5 contains Uma (innocent), Wanda (criminal), Xena (unknown), and Zoe (criminal). The clue says Row 5 is the only row with exactly 2 criminals. Since Wanda and Zoe already make two criminals in that row, Xena cannot also be a criminal without exceeding two. Therefore, we can determine that C5 Xena is INNOCENT.