Clues by Sam Jan 31, 2026 Answer – Full Solution Explained
A1
👩💼
clerk
B1
🕵️♀️
sleuth
C1
👩⚖️
judge
D1
👨⚖️
judge
A2
👷♀️
builder
B2
👨🏫
teacher
C2
👩🏫
teacher
D2
👨💻
coder
A3
👨🏫
teacher
B3
👩🍳
cook
C3
👷♀️
builder
D3
👨🍳
cook
A4
👨💼
clerk
B4
👮♂️
cop
C4
🕵️♀️
sleuth
D4
🕵️♀️
sleuth
A5
👨💼
clerk
B5
👮♂️
cop
C5
👨🍳
cook
D5
👮♀️
cop
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 — 10 criminals.
Clues by Sam answer for Jan 31, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 10 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chloe (B1), Erwin (D1), Gary (B2), Helen (C2), Nancy (C3), Rob (A4), Tina (C4), Uma (D4), Will (B5) and Zara (D5); the remaining 10 suspects are innocent.
The deduction chain, in plain English
01.A3 · Luigi → INNOCENT, D3 · Peter → INNOCENT
Row 3 contains four people: A3 Luigi, B3 Mary, C3 Nancy, and D3 Peter. Debra’s clue says that there are exactly three innocents in row 3, and exactly two of those three are on the edges. In row 3, the only edge positions are A3 and D3, so the two edge-innocents in row 3 must be Luigi at A3 and Peter at D3. Therefore, we can determine that A3 Luigi is INNOCENT and D3 Peter is INNOCENT.
02.D4 · Uma → CRIMINAL
Uma is at D4, so the people to the left of Uma are Rob at A4, Sam at B4, and Tina at C4. Peter’s clue says there is only one innocent to the left of Uma, so exactly one of Rob, Sam, and Tina is innocent. Luigi’s clue says row 4 has exactly one innocent in total, so that one innocent in row 4 must be the same person among Rob, Sam, and Tina, leaving no room for Uma to be innocent. Therefore, we can determine that D4 Uma is CRIMINAL.
03.A4 · Rob → CRIMINAL
The clue from Uma says that exactly 2 of the 3 teachers have a criminal directly below them. On this board, one of the teachers is on the bottom row, so that teacher cannot have anyone directly below them at all, which means the other two teachers must be the two that do have criminals directly below. One of those teachers is directly above Uma, and since Uma at D4 is already known to be a criminal, that teacher already satisfies one of the two required “criminal directly below” cases; the other teacher is Luigi at A3, so the person directly below Luigi, Rob at A4, must also be a criminal to reach the total of exactly two. Therefore, we can determine that A4 Rob is CRIMINAL.
04.D2 · Kumar → INNOCENT
Kumar is at D2, so the people to his left in row 2 are Flora (A2), Gary (B2), and Helen (C2). Rob’s clue says there is only one innocent to the left of Kumar, so among Flora, Gary, and Helen, exactly one is innocent. Luigi’s clue says row 4 is the only row with exactly one innocent, so row 2 cannot have exactly one innocent. Since row 2 already has exactly one innocent among the three left of Kumar, Kumar must also be innocent so that row 2 has more than one innocent in total. Therefore, we can determine that D2 Kumar is INNOCENT.
05.A1 · Barb → INNOCENT
The three sleuths are Chloe at B1, Mary at B3, and Sam at B4, and Kumar’s clue says that exactly 2 of these 3 sleuths have an innocent directly to their left. Mary already has Luigi at A3 directly to her left, and Luigi is known to be innocent, so Mary is definitely one of the two. Sam has Rob at A4 directly to his left, and Rob is known to be criminal, so Sam must be the one sleuth who does not have an innocent directly to the left. That forces Chloe to be the second sleuth who does have an innocent directly to the left, so Barb at A1 must be innocent. Therefore, we can determine that A1 Barb is INNOCENT.
06.C2 · Helen → CRIMINAL
In row 2, Kumar is at D2, so the people to the left of Kumar are Flora at A2, Gary at B2, and Helen at C2. Rob’s clue says there is only one innocent among those three people. Barb’s clue says there is only one innocent to the left of Helen, which in this case means only one innocent among Flora and Gary. Since Flora and Gary already account for the single innocent allowed among Flora, Gary, and Helen, Helen cannot also be innocent. Therefore, we can determine that C2 · Helen is CRIMINAL.
07.B4 · Sam → INNOCENT
Helen’s clue says that Sam is one of Tina’s 4 innocent neighbors, which directly states that Sam is innocent. Therefore, we can determine that B4 Sam is INNOCENT.
08.C4 · Tina → CRIMINAL
Uma is at D4, so the people to the left of Uma in the same row are Rob at A4, Sam at B4, and Tina at C4. Peter’s clue says there is only one innocent among those people. Since Sam is already known to be innocent, Tina cannot also be innocent or there would be more than one innocent to Uma’s left. Therefore, we can determine that C4 Tina is CRIMINAL.
09.B1 · Chloe → CRIMINAL, D1 · Erwin → CRIMINAL
Kumar is at D2, so the people to the left of Kumar are Flora at A2, Gary at B2, and Helen at C2. Rob’s clue says there is only one innocent among those three, and since Helen is already known to be a criminal, exactly one of Flora or Gary is innocent, meaning row 2 has exactly two innocents in total (Kumar plus that one). Tina’s clue says rows 1 and 2 have an equal number of innocents, so row 1 must also have exactly two innocents; but Barb at A1 and Debra at C1 are already innocent, so Chloe at B1 and Erwin at D1 cannot be innocent. Therefore, we can determine that B1 Chloe is CRIMINAL and D1 Erwin is CRIMINAL.
10.A5 · Vince → INNOCENT
Look at Tina at C4: her neighbors include B3, C3, D3, Sam at B4, Uma at D4, and B5, C5, D5 in row 5. Helen’s clue says Tina has exactly 4 innocent neighbors and Sam is one of them; since Sam and D3 are already known innocents and Uma is a known criminal, that means exactly 2 of {B3, C3, B5, C5, D5} must be innocent. Debra’s clue says row 3 has exactly 3 innocents with exactly 2 of them on the edges, and since A3 and D3 are already edge innocents, exactly one of B3 or C3 is innocent; that uses up one of those two needed innocents, so exactly one of {B5, C5, D5} is innocent. If Vince at A5 were not innocent, then row 5 would have exactly one innocent in total, but Luigi’s clue says row 4 is the only row with exactly one innocent, so row 5 cannot have exactly one innocent. Therefore, we can determine that A5 Vince is INNOCENT.
11.B2 · Gary → CRIMINAL
Debra’s clue says there are exactly 3 innocents in row 3, and exactly 2 of those are on the edges; since A3 (Luigi) and D3 (Peter) are already innocent and both on the edge, exactly one of B3 (Mary) or C3 (Nancy) is innocent and the other is criminal. For Kumar at D2, his neighbors are C1 (innocent), D3 (innocent), D1 (criminal), C2 (criminal), and C3, so Kumar has exactly 3 innocent neighbors precisely when C3 is innocent. For Flora at A2, her neighbors are A1 (innocent), A3 (innocent), B1 (criminal), B2, and B3, so Flora has exactly 3 innocent neighbors precisely when B2 is innocent and B3 is criminal. But whenever B3 is criminal, the row 3 fact forces C3 to be innocent, which makes Kumar have exactly 3 innocent neighbors; Vince’s clue says only one person in row 2 can have exactly 3 innocent neighbors, so Flora cannot also have 3 in that situation, meaning B2 cannot be innocent. Therefore, we can determine that B2 · Gary is CRIMINAL.
12.A2 · Flora → INNOCENT
Luigi says row 4 is the only row with exactly one innocent, and row 4 already fits that because Sam is innocent while Rob, Tina, and Uma are criminals. That means every other row must have either no innocents or at least two innocents. In row 2, Kumar is already an innocent, so row 2 cannot have zero innocents and must instead have at least two. Since Gary and Helen are criminals, the only way for row 2 to reach at least two innocents is for Flora to be innocent. Therefore, we can determine that A2 (Flora) is INNOCENT.
13.D5 · Zara → CRIMINAL
Tina at C4 has eight neighbors, and Helen’s clue says Tina has exactly four innocent neighbors and that Sam at B4 is one of them. Since Peter at D3 is also already known innocent, Tina still needs exactly two more innocent neighbors among Mary (B3), Nancy (C3), Will (B5), Xavi (C5), and Zara (D5). Sam at B4 has neighbors including Luigi (A3) and Vince (A5), who are already innocent, and Flora’s clue says Sam has exactly four innocent neighbors in total, with only one of those innocents in column B. That forces exactly two of {Mary, Nancy, Will, Xavi} to be innocent, and because only one innocent can be in column B, one of {Mary, Will} is innocent and the other needed innocent must be in column C, meaning exactly one of {Nancy, Xavi} is innocent. So the two additional innocents Tina needs are already used up among {Mary, Nancy, Will, Xavi}, leaving no room for Zara at D5 to be innocent among Tina’s neighbors. Therefore, we can determine that D5 Zara is CRIMINAL.
14.C5 · Xavi → INNOCENT
Zara at D5 says that everyone has at least one innocent neighbor, so Zara herself must have at least one innocent neighbor among the people next to D5. The only neighbors of D5 are Tina at C4, Uma at D4, and Xavi at C5, and Tina and Uma are already known to be criminals. That leaves Xavi as the only way for Zara to have an innocent neighbor. Therefore, we can determine that C5 (Xavi) is INNOCENT.
15.C3 · Nancy → CRIMINAL
Sam is at B4, so his neighbors are A3 Luigi, B3 Mary, C3 Nancy, A4 Rob, C4 Tina, A5 Vince, B5 Will, and C5 Xavi. Flora’s clue says Sam has exactly four innocent neighbors in total, and exactly one of those innocent neighbors is in column B. We already know three of Sam’s neighbors are innocent (Luigi at A3, Vince at A5, and Xavi at C5), and none of those are in column B, so the one remaining innocent neighbor must be either Mary at B3 or Will at B5. That means Nancy at C3 cannot be the fourth innocent neighbor, so she must be criminal. Therefore, we can determine that C3 Nancy is CRIMINAL.
16.B3 · Mary → INNOCENT
In row 3, the edge positions are A3 and D3, and both Luigi at A3 and Peter at D3 are already known to be innocent. Debra’s clue says there are exactly three innocents in row 3, and exactly two of those innocents are on the edges. Since the two edge spots are already occupied by innocents, the third innocent in row 3 must be one of the non-edge positions, B3 or C3. But C3 (Nancy) is already known to be a criminal, so the only remaining place for the third innocent is B3. Therefore, we can determine that B3 (Mary) is INNOCENT.
17.B5 · Will → CRIMINAL
Tina is at C4, so her eight neighbors are Mary (B3), Nancy (C3), Peter (D3), Sam (B4), Uma (D4), Will (B5), Xavi (C5), and Zara (D5). Helen’s clue says that Tina has exactly 4 innocent neighbors, and Sam is one of those innocents. We can already see four known innocents among Tina’s neighbors: Mary, Peter, Sam, and Xavi, so that uses up all four innocent-neighbor slots. That means Will cannot be innocent, so he must be a criminal. Therefore, we can determine that B5 Will is CRIMINAL.