Clues by Sam Jan 18, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👷♂️
builder
C1
👩✈️
pilot
D1
👨🍳
cook
A2
👨🌾
farmer
B2
👩🎤
singer
C2
👩🎤
singer
D2
👨💻
coder
A3
👨⚖️
judge
B3
👩🍳
cook
C3
👮♂️
cop
D3
👩🏫
teacher
A4
👩🌾
farmer
B4
👷♀️
builder
C4
👨✈️
pilot
D4
👨✈️
pilot
A5
👩🌾
farmer
B5
👩⚖️
judge
C5
👨🏫
teacher
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 — 8 criminals.
Clues by Sam answer for Jan 18, 2026 — a Hard solved in 19 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alice (A1), Eli (A2), Freya (B2), Hazel (C2), Ike (D2), Rose (B4), Steve (C4) and Xavi (D5); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.D3 · Olive → INNOCENT
Terry is confirmed truthful, and his clue says, “Olive is one of 2 innocents below Ike,” which directly states Olive is an innocent. Therefore, we can determine that D3 · Olive is INNOCENT.
02.D5 · Xavi → CRIMINAL
In column D, the people below Ike (D2) are Olive (D3), Terry (D4), and Xavi (D5). The clue says there are exactly two innocents below Ike, and Olive is one of them. Olive and Terry are already known to be innocent, which accounts for both. That leaves no room for Xavi to be innocent. Therefore, we can determine that D5 · Xavi is CRIMINAL.
03.B3 · Karen → INNOCENT
Rose is at B4, so “criminals above Rose” refers to Bruce (B1), Freya (B2), and Karen (B3). Celia is at C1, and the only column-B neighbors of Celia are Bruce (B1) and Freya (B2); Olive’s clue says exactly one of those two is innocent, so among B1 and B2 there is exactly one criminal. Xavi’s clue says the total number of criminals among B1–B3 is odd; with B1–B2 already contributing exactly one criminal, Karen cannot add another or the total would become even. Therefore, we can determine that B3 · Karen is INNOCENT.
04.A2 · Eli → CRIMINAL
Relevant people are Alice at A1, Celia at C1, and Alice’s neighbors Bruce (B1), Eli (A2), and Freya (B2). Olive’s clue says exactly one innocent in column B neighbors Celia; since the only column B neighbors of Celia are Bruce and Freya, exactly one of Bruce/Freya is innocent. Karen’s clue says Alice has an odd number of innocent neighbors; Alice’s neighbors are exactly Bruce, Eli, and Freya. With exactly one innocent among Bruce and Freya, the total around Alice stays odd only if Eli does not add another innocent, so Eli must be a criminal. Therefore, we can determine that A2 (Eli) is CRIMINAL.
05.B5 · Vera → INNOCENT
We focus on column B: Bruce (B1), Freya (B2), Karen (B3), Rose (B4), and Vera (B5). Rose is at B4, and the only people in column B who are her neighbors are directly above and below her—B3 and B5. The clue says exactly 2 of the 3 innocents in column B are Rose’s neighbors, so those two neighbors (B3 and B5) must both be innocents. Since Karen at B3 is already known to be innocent, B5 (Vera) must also be innocent. Therefore, we can determine that B5 Vera is INNOCENT.
06.B4 · Rose → CRIMINAL
Consider column B: B1 Bruce, B2 Freya, B3 Karen (innocent), B4 Rose, and B5 Vera (innocent). Xavi says there’s an odd number of criminals above Rose, i.e., among B1, B2, B3; since B3 is innocent, exactly one of Bruce or Freya is a criminal and the other is innocent. Eli says column B has exactly three innocents, and exactly two of them are Rose’s neighbors; in column B, only B3 and B5 neighbor Rose, and both are already known innocents, so there is exactly one more innocent among B1, B2, and B4. Because exactly one of Bruce/Freya is innocent, that single remaining innocent must be either Bruce or Freya, leaving no room for Rose to be innocent. Therefore, we can determine that Rose (B4) is CRIMINAL.
07.C3 · Larry → INNOCENT
Rose’s clue says “Larry is one of Freya’s 5 innocent neighbors,” and since all statements are true, this directly labels Larry as innocent. Therefore, we can determine that C3 Larry is INNOCENT.
08.A3 · Jose → INNOCENT
Relevant people: Bruce (B1), Freya (B2), Karen (B3), Alice (A1), Jose (A3), Celia (C1), Hazel (C2), Eli (A2), Larry (C3). Xavi says there’s an odd number of criminals above Rose; above her are Bruce, Freya, and Karen, and since Karen is innocent, exactly one of Bruce and Freya is a criminal (they are opposites). Rose says Freya has exactly five innocent neighbors; with Eli criminal and Karen and Larry innocent, this forces exactly three innocents among Alice, Jose, Bruce, Celia, and Hazel. Vera says Bruce has an odd number of innocent neighbors; since Eli isn’t one, that means an odd number of innocents among Alice, Freya, Celia, and Hazel. Because Bruce and Freya are opposites, replacing Freya with Bruce flips odd to even, so among Alice, Bruce, Celia, and Hazel the number of innocents is even. To make the five-person total exactly three (an odd number), Jose must supply the extra innocent. Therefore, we can determine that A3 Jose is INNOCENT.
09.C2 · Hazel → CRIMINAL
Freya sits at B2, whose neighbors are A1, B1, C1, A2, C2, A3, B3, and C3; among these we already know A2 is a criminal while A3, B3, and C3 (Larry) are innocents. Rose’s clue says Freya has exactly 5 innocent neighbors, so among the remaining four spots {A1, B1, C1, C2} there must be exactly 2 innocents. Jose’s clue says the number of criminals to the left of Donald (i.e., among A1, B1, C1) is odd, which means the number of innocents among A1, B1, C1 is even. To total exactly 2 innocents across {A1, B1, C1, C2} while A1–B1–C1 contribute an even amount, those three must supply the full 2 innocents and C2 must supply none, so C2 is a criminal. Therefore, we can determine that C2 Hazel is CRIMINAL.
10.C5 · Wally → INNOCENT
Steve sits at C4; his edge neighbors are D3 (Olive), D4 (Terry), B5 (Vera), C5 (Wally), and D5 (Xavi). Hazel’s clue says exactly four of the innocents on the edges are Steve’s neighbors, which means exactly four of these five edge neighbors must be innocent. Among them, Olive, Terry, and Vera are already confirmed innocent, and Xavi is a criminal, so Wally must be the fourth innocent. Therefore, we can determine that C5 Wally is INNOCENT.
11.A5 · Uma → INNOCENT
Relevant people: the edge positions, rows 1 and 5, and Uma at A5. Known edge criminals are A2 (Eli) and D5 (Xavi). Hazel’s clue fixes the total to exactly 10 innocents on the edges, so there are exactly 4 edge criminals. With A2 and D5 already criminal, only 2 more edge criminals remain among the seven unknown edge spots (which include A5 and all of row 1). Larry’s clue says rows 1 and 5 must have the same number of criminals. Row 5 already has D5 as a criminal; if A5 were also criminal, row 5 would have 2 and row 1 would need 2 as well, demanding 3 criminals among those seven edge spots, but only 2 are available. Therefore, A5 cannot be a criminal and must be innocent. Therefore, we can determine that A5 · Uma is INNOCENT.
12.D1 · Donald → INNOCENT
Focus on Freya at B2 and her neighbors: A1, B1, C1, A2, C2, A3, B3, and C3. Rose says “Larry is one of Freya’s 5 innocent neighbors,” so Freya has exactly 5 innocent neighbors; since Eli (A2) and Hazel (C2) are already criminals, there must be exactly one more criminal among A1, B1, C1, A3, B3, and C3. But A3, B3, and C3 (including Larry) are known innocents, so exactly one of A1/B1/C1 is a criminal. Larry also says rows 1 and 5 have the same number of criminals, and row 5 has exactly one (only Xavi at D5), so row 1 must also have exactly one criminal. That leaves no room for D1 to be a criminal. Therefore, we can determine that D1 Donald is INNOCENT.
13.C1 · Celia → INNOCENT
Freya is at B2, whose neighbors are A1, B1, C1, A2 (Eli, criminal), C2 (Hazel, criminal), A3 (Jose, innocent), B3 (Karen, innocent), and C3 (Larry, innocent). Rose’s clue says Freya has exactly 5 innocent neighbors and Larry is one of them, so with Jose, Karen, and Larry already innocent and Eli and Hazel criminal, exactly two of A1, B1, and C1 must be innocent. Uma’s clue says two of Eli’s edge neighbors are innocent; Eli’s edge neighbors are A1, B1, and A3, and since A3 (Jose) is already innocent, exactly one of A1 or B1 is innocent. With only one innocent among A1 and B1 but two needed among A1, B1, and C1, C1 must supply the second innocent. Therefore, we can determine that C1 Celia is INNOCENT.
14.D2 · Ike → CRIMINAL
In column B above Rose (B4) are Bruce (B1), Freya (B2), and Karen (B3), with Karen already known innocent; also Vera (B5) is innocent. Xavi says there’s an odd number of criminals above Rose, so among Bruce and Freya exactly one is a criminal, meaning exactly one of them is an innocent; thus column B has exactly three innocents (Karen, Vera, and one of Bruce/Freya). Celia says columns B and D have the same number of innocents; column D already has three innocents without Ike (Donald, Olive, Terry), so to keep the counts equal Ike cannot be innocent. Therefore, we can determine that D2 Ike is CRIMINAL.
15.A4 · Penny → INNOCENT
Relevant spots are the three unknown edge positions A1, B1, and A4. Hazel’s clue tells us there are exactly 10 innocents on the edges; from the board we already know 8 edge innocents, so exactly two of A1, B1, and A4 must be innocent. Rose’s clue says Freya has exactly 5 innocent neighbors and Larry is one of them; among Freya’s neighbors we already have four innocents (A3, B3, C1, C3) and two criminals (A2, C2), so exactly one of A1 or B1 is innocent. That leaves one more edge innocent still needed, which must be A4. Therefore, we can determine that A4 · Penny is INNOCENT.
16.C4 · Steve → CRIMINAL
Freya (B2) has exactly five innocent neighbors, and Larry (C3) is one of them. Among Freya’s neighbors we already know four innocents (C1, A3, B3, C3) and two criminals (A2, C2), so to reach five, exactly one of Alice (A1) and Bruce (B1) must be innocent—giving row 1 exactly three innocents with C1 and D1. Ike says row 1 has more innocents than row 4; if Steve (C4) were innocent, row 4 would also have three (A4, C4, D4), which would not be more. Therefore, we can determine that C4 Steve is CRIMINAL.
17.A1 · Alice → CRIMINAL
Celia (C1) is neighbored by only B1 (Bruce) and B2 (Freya) from column B, so Olive’s clue means exactly one of Bruce or Freya is innocent. Steve’s clue says only one person in column A has exactly four innocent neighbors, and Penny (A4) already has exactly four: A3 (Jose), B3 (Karen), B5 (Vera), and A5 (Uma). Eli (A2) always has A3 and B3 as innocent neighbors and, from the first clue, exactly one of B1/B2; he would reach four only if Alice (A1) were also innocent. Since A4 already uniquely has four, A2 must not have four, so Alice cannot be innocent. Therefore, we can determine that A1 Alice is CRIMINAL.
18.B1 · Bruce → INNOCENT
Relevant people: Freya at B2, her neighbors A1, B1, C1, A2, C2, A3, B3, C3, and Larry at C3. The clue says Freya has exactly five innocent neighbors and that Larry is one of them. Among those neighbors we already know A1, A2, and C2 are criminals, while C1, A3, B3, and Larry (C3) are innocents—leaving only Bruce (B1) undecided. To reach the required total of five innocent neighbors for Freya, Bruce must be the fifth innocent. Therefore, we can determine that B1 · Bruce is INNOCENT.
19.B2 · Freya → CRIMINAL
The clue concerns the people above Rose in column B: B1 (Bruce), B2 (Freya), and B3 (Karen). Xavi says there is an odd number of criminals above Rose, and clues are always true. Since Bruce and Karen are already known to be innocent, the only way to have an odd count among those three is for Freya to be a criminal. Therefore, we can determine that B2 Freya is CRIMINAL.