Clues by Sam Mar 04, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👩🎨
painter
C1
👩🎨
painter
D1
👨⚖️
judge
A2
👨🌾
farmer
B2
👨🏫
teacher
C2
💂♂️
guard
D2
👨💻
coder
A3
👩🎨
painter
B3
👨🏫
teacher
C3
👮♂️
cop
D3
👩⚖️
judge
A4
👩🌾
farmer
B4
👩🏫
teacher
C4
👩💻
coder
D4
👨⚖️
judge
A5
👩💻
coder
B5
💂♀️
guard
C5
💂♂️
guard
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 — 8 criminals.
Clues by Sam answer for Mar 04, 2026 — a Medium solved in 16 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Denis (D1), Gus (B2), Hal (C2), Klay (B3), Maria (D3), Olive (B4), Vicky (B5) and Zed (D5); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.D2 · Ike → INNOCENT
Nancy’s clue says, “Ike is one of Hal’s 4 innocent neighbors,” which directly states that Ike is an innocent. Therefore, we can determine that D2 Ike is INNOCENT.
02.B2 · Gus → CRIMINAL
In row 2 the people are Erwin at A2, Gus at B2, Hal at C2, and Ike at D2, and we already know Ike is INNOCENT. Ike’s clue says “Both criminals in row 2 are connected,” which means there are exactly two criminals in that row, and they must be orthogonally adjacent with no innocent between them. Since D2 is not a criminal, the only way to place two connected criminals in A2–B2–C2 is either A2 with B2 or B2 with C2; A2 with C2 would not be connected because B2 sits between them. In every connected-two-criminal arrangement that fits the clue, B2 must be one of the criminals. Therefore, we can determine that B2 Gus is CRIMINAL.
03.C1 · Claire → INNOCENT, B1 · Betsy → INNOCENT
Amy is at A1 and Denis is at D1, so the only people strictly between them in that row are Betsy at B1 and Claire at C1. Gus’s clue says there are exactly 2 innocents in between Amy and Denis, and there are exactly two people in between them. That forces both Betsy and Claire to be innocents. Therefore, we can determine that C1 Claire is INNOCENT and B1 Betsy is INNOCENT.
04.A2 · Erwin → INNOCENT
Amy is at A1, so the only people in row 2 who neighbor her are A2 (directly below) and B2 (diagonally down-right). Betsy’s clue says that exactly 1 innocent in row 2 is neighboring Amy, so among A2 and B2, exactly one must be innocent. Since B2 (Gus) is already known to be a criminal, he cannot be that innocent neighbor. That forces A2 (Erwin) to be the one innocent in row 2 who neighbors Amy. Therefore, we can determine that A2 is INNOCENT.
05.C2 · Hal → CRIMINAL
In row 2, A2 (Erwin) and D2 (Ike) are already known to be innocent, and B2 (Gus) is already known to be a criminal. Ike’s clue says there are both criminals in row 2, meaning there are exactly two criminals in that row, and they are connected by orthogonal adjacency within the row. Since B2 is one of them and the only remaining unknown in row 2 is C2 (Hal), the second criminal must be Hal, and B2 and C2 are directly adjacent so they are connected as required. Therefore, we can determine that C2 Hal is CRIMINAL.
06.D1 · Denis → CRIMINAL
Hal is at C2, and his neighbors are Betsy at B1, Claire at C1, Denis at D1, Gus at B2, Ike at D2, Klay at B3, Luigi at C3, and Maria at D3. Nancy’s clue says Ike is one of Hal’s 4 innocent neighbors, so Hal has exactly four innocent neighbors total; since Betsy, Claire, and Ike are already innocent and Gus is already criminal, exactly one of Denis, Klay, Luigi, and Maria can be innocent. Hal’s other clue says there is an odd number of innocents to the right of Joy, meaning among Klay, Luigi, and Maria there must be 1 or 3 innocents, so it cannot be 0. If Denis were the one innocent among Denis, Klay, Luigi, and Maria, then Klay, Luigi, and Maria would all have to be criminal, giving 0 innocents to Joy’s right, which is not odd. Therefore, we can determine that D1 Denis is CRIMINAL.
07.A3 · Joy → INNOCENT
Hal is at C2, and his neighbors are Betsy at B1, Claire at C1, Denis at D1, Gus at B2, Ike at D2, Klay at B3, Luigi at C3, and Maria at D3. Nancy’s clue says Hal has exactly 4 innocent neighbors and Ike is one of them; since Betsy, Claire, and Ike are already known innocents while Denis and Gus are known criminals, exactly one of Klay, Luigi, and Maria must be innocent. Denis’s clue says rows 2 and 3 have the same number of innocents, and row 2 already has exactly 2 innocents (Erwin and Ike), so row 3 must also have exactly 2 innocents. Because only one of Klay, Luigi, and Maria can be innocent, Joy at A3 must be the other innocent in row 3. Therefore, we can determine that A3 Joy is INNOCENT.
08.B3 · Klay → CRIMINAL
Hal is at C2, and his neighbors are Betsy, Claire, Denis, Gus, Ike, Klay, Luigi, and Maria. Nancy’s clue says Ike is one of Hal’s 4 innocent neighbors, and we already see Betsy, Claire, and Ike are innocent while Denis and Gus are criminal, so exactly one of Klay, Luigi, and Maria can be innocent. Joy’s clue about Sam says Sam has 3 innocent neighbors in total, and only 1 of those innocents is in row 3, so among Sam’s row 3 neighbors Luigi (C3) and Maria (D3), exactly one is innocent. Since that already supplies the single allowed innocent from the set {Klay, Luigi, Maria}, Klay cannot be the innocent in that set. Therefore, we can determine that B3 Klay is CRIMINAL.
09.B4 · Olive → CRIMINAL, B5 · Vicky → CRIMINAL
Column B contains Betsy at B1, Gus at B2, Klay at B3, Olive at B4, and Vicky at B5. Klay’s clue says there are exactly 4 criminals in column B, so only 1 person in that column can be innocent. Since Betsy at B1 is already known to be innocent, the other four people in column B must be criminals, which forces both Olive and Vicky to be criminals. Therefore, we can determine that B4 Olive is CRIMINAL and B5 Vicky is CRIMINAL.
10.A1 · Amy → INNOCENT
Row 2 already has exactly two innocents, because Erwin at A2 and Ike at D2 are innocent while Gus at B2 and Hal at C2 are criminal. In row 1, Betsy at B1 and Claire at C1 are innocent and Denis at D1 is criminal, so row 1 currently has two known innocents and Amy at A1 is the only one not yet fixed. Vicky’s clue says there are more innocents in row 1 than in row 2, so row 1 must have more than two innocents. That forces Amy to be the extra innocent in row 1. Therefore, we can determine that A1 Amy is INNOCENT.
11.C3 · Luigi → INNOCENT, C4 · Pam → INNOCENT
The clue says that every person in column B has at most 3 criminal neighbors. Look at Klay at B3, who is in column B: his neighbors include Gus at B2, Hal at C2, and Olive at B4, and all three of those are already known criminals, so Klay already has 3 criminal neighbors. Luigi at C3 and Pam at C4 are also neighbors of Klay, so if either of them were a criminal then Klay would have at least 4 criminal neighbors, which the clue forbids. Therefore, we can determine that C3 Luigi is INNOCENT and C4 Pam is INNOCENT.
12.D3 · Maria → CRIMINAL
Hal is at C2, so his neighbors are Betsy at B1, Claire at C1, Denis at D1, Gus at B2, Ike at D2, Klay at B3, Luigi at C3, and Maria at D3. Nancy’s clue says Ike is one of Hal’s 4 innocent neighbors, meaning Hal has exactly four innocent neighbors in total. We can already see four innocents among Hal’s neighbors: Betsy, Claire, Ike, and Luigi, so the remaining neighbors of Hal must all be criminals. Maria is one of those remaining neighbors, so she must be a criminal. Therefore, we can determine that D3 Maria is CRIMINAL.
13.A5 · Tina → INNOCENT
Sam is at D4, and Sam’s neighbors are C3 Luigi, D3 Maria, C4 Pam, C5 Xavi, and D5 Zed. Joy’s clue says Sam has exactly three innocent neighbors, and only one of those is in row 3; since Luigi (C3) is an innocent neighbor in row 3 and Maria (D3) is criminal, this forces Pam (C4) to be innocent and exactly one of the two row-5 neighbors, Xavi (C5) or Zed (D5), to be innocent. Luigi’s clue says every row has at least two innocents, so in row 5 (A5 Tina, B5 Vicky, C5 Xavi, D5 Zed), with Vicky already criminal and with only one of Xavi or Zed able to be innocent, Tina must be the second innocent needed for that row. Therefore, we can determine that A5 Tina is INNOCENT.
14.D4 · Sam → INNOCENT
Sam is at D4, and his neighbors are C3, D3, C4, C5, and D5. Joy’s clue says there are exactly three innocent neighbors of Sam, and only one of those three is in row 3; among Sam’s row-3 neighbors, C3 is innocent and D3 is criminal, so C3 must be the only innocent neighbor in row 3. Since C4 is also already innocent, that uses up two of the three innocent-neighbor slots, so exactly one of C5 and D5 can be innocent. Tina’s clue says there are 10 innocents on the edges, and right now there are 8 confirmed edge innocents, with only D4, C5, and D5 still undecided on the edge. That means exactly two of {D4, C5, D5} must be innocent, and since exactly one of {C5, D5} is innocent, D4 must be the other innocent. Therefore, we can determine that D4 Sam is INNOCENT.
15.C5 · Xavi → INNOCENT
B3 (Klay) has eight neighbors: A2, B2, C2, A3, C3, A4, B4, and C4, and exactly five of those are already known to be innocent (A2, A3, C3, A4, and C4). Sam’s clue says there is only one person on the whole board who has exactly 5 innocent neighbors, so no one else is allowed to have 5. B4 (Olive) has neighbors A3, B3, C3, A4, C4, A5, B5, and C5, and among the first seven of those, exactly five are already known to be innocent (A3, C3, A4, C4, and A5). That means B4 will have exactly 5 innocent neighbors if and only if C5 is not innocent, because C5 is the only remaining neighbor that can change that count. Since B3 already has exactly 5 innocent neighbors and must be the unique such person, C5 cannot make B4 also equal 5, so C5 must be innocent. Therefore, we can determine that C5 (Xavi) is INNOCENT.
16.D5 · Zed → CRIMINAL
Sam is at D4, and his neighbors are Luigi at C3, Maria at D3, Pam at C4, Xavi at C5, and Zed at D5. Joy’s clue says that among Sam’s neighbors there are exactly three innocents, and only one of those three is in row 3. We can already see three innocents next to Sam: Luigi (row 3), Pam (row 4), and Xavi (row 5), and this set also matches the clue’s “only one in row 3” requirement because only Luigi is in row 3. Since those three slots are already filled by Luigi, Pam, and Xavi, Zed cannot be one of the innocents neighboring Sam. Therefore, we can determine that D5 Zed is CRIMINAL.