Clues by Sam Jan 23, 2026 Answer – Full Solution Explained
A1
👩✈️
pilot
B1
🕵️♂️
sleuth
C1
👩💻
coder
D1
👷♂️
builder
A2
👨🔧
mech
B2
👩🔧
mech
C2
👨💻
coder
D2
👨⚖️
judge
A3
🕵️♀️
sleuth
B3
👩✈️
pilot
C3
👨⚖️
judge
D3
🕵️♂️
sleuth
A4
👩🍳
cook
B4
👩🍳
cook
C4
👨🍳
cook
D4
👩🌾
farmer
A5
👨✈️
pilot
B5
👨🌾
farmer
C5
👷♀️
builder
D5
👩🌾
farmer
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 — 12 criminals.
Clues by Sam answer for Jan 23, 2026 — a Tricky solved in 18 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Barb (A1), Carl (B1), Dana (C1), Ivan (C2), John (D2), Kay (A3), Martin (C3), Olga (A4), Paula (B4), Saga (D4), Vince (A5) and Wally (B5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.C2 · Ivan → CRIMINAL
Nick’s clue explicitly says “Ivan is one of Carl’s 3 criminal neighbors,” which directly states that Ivan is a criminal. Therefore, we can determine that C2 · Ivan is CRIMINAL.
02.B2 · Hope → INNOCENT
Carl (B1) has five neighbors: A1, A2, B2 (Hope), C1, and C2 (Ivan). Nick says Carl has exactly three criminal neighbors and that Ivan is one of them, so among A1, A2, B2, and C1 exactly two are criminals. Ivan says that of all edge innocents, only one is Carl’s neighbor; Carl’s edge neighbors are A1, A2, and C1, so exactly two of these three are criminals. Those two criminals account for the entire “two criminals” among A1/A2/B2/C1, leaving B2 not criminal—so B2 (Hope) must be innocent. Therefore, we can determine that B2 (Hope) is INNOCENT.
03.A2 · Gary → INNOCENT
The relevant people are in row 2: Gary (A2), Hope (B2), Ivan (C2), and John (D2), with Carl at B1. Carl’s neighbors in row 2 are Gary, Hope, and Ivan (John is too far to be a neighbor). The clue says exactly two innocents in row 2 are neighboring Carl; among those three, Hope is already known to be innocent and Ivan is a criminal, so Gary must be the second innocent to make the count exactly two. Therefore, we can determine that A2 · Gary is INNOCENT.
04.A1 · Barb → CRIMINAL, C1 · Dana → CRIMINAL
Carl (B1) has five neighbors: Barb (A1), Dana (C1), Gary (A2), Hope (B2), and Ivan (C2). Nick’s clue says Carl has exactly three criminal neighbors, and Ivan is one of them. Among these five, Gary and Hope are already known to be innocent, and Ivan is already known to be a criminal. To reach a total of three criminal neighbors, the only remaining neighbors—Barb and Dana—must both be criminals. Therefore, we can determine that Barb is CRIMINAL and Dana is CRIMINAL.
05.B5 · Wally → CRIMINAL
The key people are Paula at B4 and row 5. Paula’s edge neighbors are A3, A4, A5, B5, and C5; Dana’s clue says exactly one of these five is innocent, so at least two of A5, B5, and C5 must be criminals. Gary’s clue says that all criminals in row 5 must be in one continuous block with no innocents between them. If at least two of A5, B5, and C5 are criminals and there can be no gap between criminals in that row, the middle spot B5 must be one of the criminals; otherwise A5 and C5 would be split by an innocent at B5. Therefore, we can determine that B5 Wally is CRIMINAL.
06.A3 · Kay → CRIMINAL
Relevant people: Paula at B4 and her edge neighbors A3 (Kay), A4 (Olga), A5 (Vince), B5 (Wally), and C5 (Xena). Paula’s clue says exactly one of her edge neighbors is innocent; since Wally is known criminal, the single innocent must be among A3, A4, A5, and C5. Wally’s clue concerns his edge neighbors A4, A5, and C5: an odd number of them are innocent; combined with Paula’s “only one among A3/A4/A5/C5,” those three can have at most one innocent, so the odd count must be exactly one. That exhausts the single innocent among A3/A4/A5/C5, so A3 (Kay) cannot be the innocent. Therefore, we can determine that A3 Kay is CRIMINAL.
07.A4 · Olga → CRIMINAL
The relevant people are Paula at B4 and her edge neighbors A3 (Kay), A4 (Olga), A5 (Vince), B5 (Wally), and C5 (Xena); Kay and Wally are already criminals. Dana’s clue says Paula has only one innocent neighbor on the edges, so among those five, exactly one is innocent. Kay’s clue says only one of Paula’s three innocent neighbors is in row 5; among her row‑5 neighbors (A5, B5, C5), Wally is criminal, so either A5 or C5 must be that single innocent, which means none of the other edge neighbors—including A4—can be innocent. Therefore, we can determine that A4 Olga is CRIMINAL.
08.C4 · Ryan → INNOCENT
Saga is at D4 and Martin is at C3; we need the neighbors they share. The clue says Saga has four innocent neighbors, and exactly two of those also neighbor Martin. The only people who neighbor both Saga and Martin are Nick (D3) and Ryan (C4). Since Nick is already known to be innocent, Ryan must also be innocent to make those two shared neighbors the required two innocents. Therefore, we can determine that C4 Ryan is INNOCENT.
09.D5 · Zoe → INNOCENT
Relevant people are row 2 (Gary, Hope, Ivan, John) and row 5 (Vince, Wally, Xena, Zoe). Ryan says rows 2 and 5 have the same number of innocents; since row 2 already has exactly two known innocents (Gary and Hope) and Ivan is a criminal, row 2 has either 2 or 3 innocents, so row 5 must also have either 2 or 3 innocents (meaning 2 or 1 criminals). Gary says all criminals in row 5 are connected, and we already know Wally (B5) is a criminal, so if there are two criminals they must be B5 plus one adjacent neighbor (A5 or C5); if there is only one criminal, it’s just B5. In none of these allowed cases can D5 be a criminal, because being criminal at D5 would either be disconnected from B5 or would require C5 also to be criminal, creating three criminals, which row 5 cannot have. Therefore, we can determine that D5 (Zoe) is INNOCENT.
10.D2 · John → CRIMINAL
Paula is at B4, and her edge neighbors are A3 (Kay), A4 (Olga), A5 (Vince), B5 (Wally), and C5 (Xena). Dana’s clue says Paula has only one innocent neighbor on the edges; since Kay, Olga, and Wally are already criminals, exactly one of Vince or Xena is innocent. Row 5 consists of Vince, Wally, Xena, and Zoe; with Wally criminal and Zoe innocent, that makes row 5 have exactly two innocents (Zoe plus exactly one of Vince/Xena). Ryan’s clue says rows 2 and 5 have the same number of innocents, so row 2 must also have exactly two. Row 2 already has Gary and Hope as innocents, so John cannot be innocent. Therefore, we can determine that D2 John is CRIMINAL.
11.B4 · Paula → CRIMINAL
We look at Paula (B4) and the neighbors around her that are on the board’s edge: A3 (Kay), A4 (Olga), A5 (Vince), B5 (Wally), and C5 (Xena). Dana’s clue says Paula has only one innocent neighbor on the edges; since Kay, Olga, and Wally are already criminals, this forces exactly one of Vince (A5) or Xena (C5) to be innocent and the other to be a criminal. Now consider Wally (B5); his neighbors are A4 (Olga, criminal), B4 (Paula), C4 (Ryan, innocent), A5 (Vince), and C5 (Xena). John’s clue says an odd number of Wally’s neighbors are criminals; from above, Olga is criminal and exactly one of Vince/Xena is criminal, which is two criminals so far, so to make the total odd, Paula must also be a criminal. Therefore, we can determine that B4 Paula is CRIMINAL.
12.B1 · Carl → CRIMINAL
The clue directly states that Carl is one of Ivan’s criminal neighbors, which means Carl himself is a criminal. Therefore, we can determine that B1 Carl is CRIMINAL.
13.A5 · Vince → CRIMINAL
Relevant people: Saga (D4), Martin (C3), Xena (C5), and the counts in columns A and C. Olga says Saga has exactly four innocent neighbors, and exactly two of those also neighbor Martin. Saga’s neighbors are C3 (Martin), C4 (Ryan, innocent), C5 (Xena), D3 (Nick, innocent), and D5 (Zoe, innocent), so with three already innocent, exactly one of Martin or Xena must be innocent and the other criminal; that makes column C have exactly three criminals (C1, C2, and exactly one of C3/C5). Carl says column A has more criminals than column C, so column A must have at least four criminals. With A1, A3, and A4 already criminal and A2 innocent, only A5 can provide the fourth. Therefore, we can determine that A5 Vince is CRIMINAL.
14.C5 · Xena → INNOCENT
Paula is at B4; her neighbors that lie on the board’s edge are A3 (Kay), A4 (Olga), A5 (Vince), B5 (Wally), and C5 (Xena). The clue says Paula has only one innocent neighbor among these edge neighbors. Among those five, Kay, Olga, Vince, and Wally are already known criminals, so the only one who can be that single innocent edge neighbor is Xena at C5. Therefore, we can determine that C5 · Xena is INNOCENT.
15.C3 · Martin → CRIMINAL
Relevant people: Saga at D4 and her neighbors—C3 Martin (unknown), C4 Ryan (innocent), D3 Nick (innocent), D5 Zoe (innocent), and C5 Xena (innocent). The clue says Saga has exactly four innocent neighbors, and among those four, exactly two also neighbor Martin. Saga has five neighbors total, and four of them (Ryan, Nick, Zoe, Xena) are already known innocents, so the remaining neighbor, Martin, cannot be innocent. Therefore, we can determine that C3 · Martin is CRIMINAL.
16.B3 · Lisa → INNOCENT
Paula is at B4. The clue says she has exactly 3 innocent neighbors, and among those three, exactly one is in row 5. Her row 5 neighbors are A5 Vince (criminal), B5 Wally (criminal), and C5 Xena (innocent), so Xena is the only row‑5 innocent; another known innocent neighbor is C4 Ryan, making two. Among Paula’s remaining neighbors (A3 Kay, B3 Lisa, C3 Martin, A4 Olga), all are already criminals except Lisa, so Lisa must be the third innocent neighbor to reach the required total of three. Therefore, we can determine that B3 · Lisa is INNOCENT.
17.D1 · Frank → INNOCENT
Ivan is at C2, with neighbors B1 Carl, C1 Dana, D1 Frank, B2 Hope, D2 John, B3 Lisa, C3 Martin, and D3 Nick. The clue says Ivan has exactly four criminal neighbors, and Carl is one of them. Among these neighbors, Carl, Dana, John, and Martin are already criminals, while Hope, Lisa, and Nick are innocents—so the four criminal spots are already filled. No other neighbor can be criminal, which forces Frank to be innocent. Therefore, we can determine that D1 Frank is INNOCENT.
18.D4 · Saga → CRIMINAL
Relevant spots: the edges (outer 14 positions), Carl at B1 and his neighbors (A1, A2, B2, C2, C1), and the unknown edge spot D4. Ivan’s clue says there are exactly 5 innocents on the edges, and only one of those five is Carl’s neighbor. Among Carl’s edge neighbors A1, A2, and C1, only A2 (Gary) is innocent, so he is that one; the five edge innocents are then D1 (Frank), A2 (Gary), D3 (Nick), C5 (Xena), and D5 (Zoe). That already uses up all 5 edge-innocent slots, so D4 cannot also be an edge innocent. Therefore, we can determine that D4 · Saga is CRIMINAL.