Clues by Sam Mar 06, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👨✈️
pilot
C1
👩🏫
teacher
D1
👨🏫
teacher
A2
💂♂️
guard
B2
💂♂️
guard
C2
👩✈️
pilot
D2
👩✈️
pilot
A3
💂♀️
guard
B3
👩🌾
farmer
C3
👨🎨
painter
D3
👩🎨
painter
A4
👨🍳
cook
B4
👨🍳
cook
C4
👩🌾
farmer
D4
👩🎨
painter
A5
👩⚕️
doctor
B5
👨⚕️
doctor
C5
👮♂️
cop
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 — 11 criminals.
Clues by Sam answer for Mar 06, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 11 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Hope (C2), Joyce (D2), Katie (A3), Oscar (C3), Pam (D3), Raul (A4), Scott (B4), Uma (D4), Xavi (C5) and Zane (D5); the remaining 9 suspects are innocent.
The deduction chain, in plain English
01.A5 · Vicky → INNOCENT, B5 · Will → INNOCENT
Zane is at D5, so the people to the left of Zane are Xavi at C5, Will at B5, and Vicky at A5. Raul is at A4, and the only people in row 5 who neighbor Raul are Vicky at A5 (directly below) and Will at B5 (diagonally below-right); Xavi at C5 does not neighbor Raul. Ethan’s clue says that exactly two innocents to the left of Zane are neighboring Raul, so those two must be exactly Vicky and Will, and they both must be innocent. Therefore, we can determine that A5 Vicky is INNOCENT and B5 Will is INNOCENT.
02.C3 · Oscar → CRIMINAL
In column C, the people are Claire at C1, Hope at C2, Oscar at C3, Tina at C4, and Xavi at C5. Vicky says there are exactly three criminals in column C, and only one of those three is a neighbor of Will; since Will is at B5, the only people in column C who are his neighbors are Tina (C4) and Xavi (C5), so exactly one of Tina and Xavi is a criminal and the other two criminals in column C must be among Claire, Hope, and Oscar. Will also says there is only one innocent above Oscar, meaning among Claire (C1) and Hope (C2) exactly one is innocent and the other is a criminal, so that gives only one criminal among Claire and Hope. Since we still need two criminals among Claire, Hope, and Oscar, Oscar must be the second one. Therefore, we can determine that C3 Oscar is CRIMINAL.
03.D3 · Pam → CRIMINAL, D5 · Zane → CRIMINAL
Uma at D4 has five neighbors: Oscar at C3, Pam at D3, Tina at C4, Xavi at C5, and Zane at D5. Oscar’s clue says Uma has exactly 4 criminal neighbors, and since Oscar is already a criminal, that forces exactly 3 criminals among the other four neighbors Pam, Tina, Xavi, and Zane. Vicky’s clue says there are 3 criminals in column C and only 1 of those is a neighbor of Will at B5; the only people in column C who neighbor Will are Tina and Xavi, so exactly 1 of Tina and Xavi is a criminal. That means Tina and Xavi contribute exactly 1 criminal to the set {Pam, Tina, Xavi, Zane}, so the remaining 2 criminals required in that set must be Pam and Zane. Therefore, we can determine that D3 Pam is CRIMINAL and D5 Zane is CRIMINAL.
04.A4 · Raul → CRIMINAL
Pam’s clue says that Raul is one of the three criminals in row 4, which directly includes Raul among the criminals. Therefore, we can determine that A4 Raul is CRIMINAL.
05.D4 · Uma → CRIMINAL
In row 4, the people are Raul at A4, Scott at B4, Tina at C4, and Uma at D4. Pam’s clue says Raul is one of 3 criminals in row 4, so row 4 has exactly one innocent. Zane’s clue says there is only one innocent to the left of Uma, and the squares to the left of D4 are A4, B4, and C4; since Raul at A4 is already a criminal, that one innocent must be either Scott at B4 or Tina at C4. That uses up the only innocent allowed in row 4, so Uma cannot be innocent. Therefore, we can determine that D4 Uma is CRIMINAL.
06.A1 · Anna → CRIMINAL
Uma’s clue says that Anna is one of Gus’ 4 criminal neighbors, and Anna at A1 is indeed a neighbor of Gus at B2 (diagonally). Since the clue explicitly labels Anna as a criminal neighbor, Anna must be a criminal. Therefore, we can determine that A1 Anna is CRIMINAL.
07.C2 · Hope → CRIMINAL
Anna’s clue says that Hope is one of Oscar’s 5 criminal neighbors, which directly states that Hope is a criminal neighbor of Oscar. Since all clues are truthful, Hope must be a criminal. Therefore, we can determine that C2 Hope is CRIMINAL.
08.C1 · Claire → INNOCENT
In column C, Hope at C2 and Oscar at C3 are already criminals, and neither of them is a neighbor of Will at B5. Vicky’s clue says there are exactly three criminals in column C in total, and exactly one of those three criminals is Will’s neighbor. That means the third criminal in column C must be someone in column C who is actually adjacent to Will, which can only be Tina at C4 or Xavi at C5, so Claire at C1 cannot be the third criminal. Therefore, we can determine that C1 Claire is INNOCENT.
09.D1 · Daniel → INNOCENT
Gus is at B2, and Uma’s clue says Anna is one of Gus’ 4 criminal neighbors, which fixes that Gus has exactly 4 criminal neighbors. Raul’s clue then says that in row 2, only one person has exactly 4 criminal neighbors, so no one else in row 2 can have exactly 4 criminal neighbors, including Joyce at D2. Joyce’s neighbors are Claire (C1), Daniel (D1), Hope (C2), Oscar (C3), and Pam (D3); among these, Hope, Oscar, and Pam are already criminals and Claire is innocent, so Joyce would have exactly 4 criminal neighbors precisely if Daniel were a criminal. Since Joyce cannot have exactly 4 criminal neighbors, Daniel cannot be a criminal. Therefore, we can determine that D1 Daniel is INNOCENT.
10.B1 · Bruce → INNOCENT
Oscar at C3 has 5 criminal neighbors, and we can already see three of them are Hope, Pam, and Uma, so exactly two of Oscar’s other neighbors (Gus, Joyce, Lucy, Scott, and Tina) must be criminals. Row 4 has exactly 3 criminals, and since Raul and Uma are already criminals there, exactly one of Scott or Tina is criminal, which means the other needed criminal around Oscar must be exactly one of Gus, Joyce, and Lucy. Hope’s neighbors include Oscar and Pam (already two criminals), plus Gus, Joyce, Lucy, and Bruce; with exactly one criminal among Gus/Joyce/Lucy, Hope has 3 criminal neighbors if Bruce is innocent but would have exactly 4 if Bruce were criminal. Gus is stated to have exactly 4 criminal neighbors, and Raul says only one person in row 2 has exactly 4 criminal neighbors, so Hope cannot also have 4, forcing Bruce to be innocent. Therefore, we can determine that B1 Bruce is INNOCENT.
11.B3 · Lucy → INNOCENT
The people who matter here are Oscar at C3 and his neighbors, especially Gus at B2, Joyce at D2, and Lucy at B3. Anna’s clue says Hope is one of Oscar’s 5 criminal neighbors, and since Hope, Pam, and Uma are already known criminals next to Oscar, that means only 2 more of Oscar’s other neighbors can be criminals. Pam’s clue says there are exactly 3 criminals in row 4, and since Raul and Uma are already criminals there, exactly one of Scott (B4) and Tina (C4) is a criminal, so the other remaining criminal neighbor of Oscar must be exactly one person among Gus (B2), Joyce (D2), and Lucy (B3). Bruce’s clue says row 1 has more innocents than row 2; row 1 already has 3 innocents, so row 2 cannot have 3 innocents, which forces at least one of Gus or Joyce to be a criminal, leaving no criminal slot for Lucy in that set of three. Therefore, we can determine that B3 Lucy is INNOCENT.
12.A3 · Katie → CRIMINAL
Gus is at B2, so his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Uma’s clue says Anna is one of Gus’ 4 criminal neighbors, which means Gus has exactly four criminal neighbors in that set and Anna at A1 is one of them. Among those neighbors, Anna at A1, Hope at C2, and Oscar at C3 are already criminals, while Bruce at B1, Claire at C1, Ethan at A2, and Lucy at B3 are innocents, leaving only Katie at A3 undecided. Since Gus must have four criminal neighbors total and only Katie can supply the missing one, Katie has to be a criminal. Therefore, we can determine that A3 Katie is CRIMINAL.
13.B2 · Gus → INNOCENT
Lucy is at B3, so her neighbors are Ethan at A2, Gus at B2, Hope at C2, Katie at A3, Oscar at C3, Raul at A4, Scott at B4, and Tina at C4. Lucy’s clue says the number of innocents among those eight neighbors is odd; right now, the only definite innocent among them is Ethan, while Hope, Katie, Oscar, and Raul are definite criminals. Pam’s clue says Raul is one of 3 criminals in row 4, and since row 4 already has Raul and Uma as criminals, exactly one of Scott and Tina must be innocent and the other must be criminal. That means among Lucy’s unknown neighbors (Gus, Scott, Tina), Scott and Tina contribute exactly one innocent, so to make the total number of innocent neighbors odd, Gus must also be innocent. Therefore, we can determine that B2 Gus is INNOCENT.
14.D2 · Joyce → CRIMINAL
Row 1 has three innocents (Bruce at B1, Claire at C1, and Daniel at D1), since only Anna at A1 is a criminal. Row 2 already has two innocents (Ethan at A2 and Gus at B2), and Hope at C2 is a criminal, leaving only Joyce at D2 unknown. Bruce’s clue says there are more innocents in row 1 than in row 2, so row 2 cannot also have three innocents. That means Joyce cannot be innocent, because that would make row 2’s innocent count equal to row 1’s. Therefore, we can determine that D2 Joyce is CRIMINAL.
15.B4 · Scott → CRIMINAL
Will is at B5, and among the people in column C the only ones who are Will’s neighbors are Tina at C4 and Xavi at C5. Vicky’s clue says there are exactly three criminals in column C, and only one of those criminals is Will’s neighbor; since Hope (C2) and Oscar (C3) are not neighbors of Will, that forces exactly one of Tina (C4) and Xavi (C5) to be a criminal and the other to be an innocent. Scott at B4 has three already-known innocent neighbors (Lucy at B3, Vicky at A5, and Will at B5), and with exactly one of C4 or C5 also being innocent, Scott has exactly four innocent neighbors. Joyce’s clue says only one person in row 4 has exactly four innocent neighbors, so Raul at A4 must not also have four innocent neighbors; Raul would have four innocent neighbors precisely when Scott is innocent, so Scott cannot be innocent. Therefore, we can determine that B4 Scott is CRIMINAL.
16.C4 · Tina → INNOCENT
Row 4 contains Raul at A4, Scott at B4, Tina at C4, and Uma at D4. Pam’s clue says that Raul is one of 3 criminals in row 4, so there must be exactly 3 criminals in that row. We already know Raul, Scott, and Uma are criminals, which makes three criminals accounted for in row 4. That means Tina cannot also be a criminal, or the row would have four criminals instead of three. Therefore, we can determine that C4 Tina is INNOCENT.
17.C5 · Xavi → CRIMINAL
The clue says that in column C there are exactly three criminals in total, and among those three, exactly one is a neighbor of Will at B5. In column C, Hope at C2 and Oscar at C3 are already criminals, while Claire at C1 and Tina at C4 are innocents, so the only way for column C to have three criminals is for Xavi at C5 to be the third one. This also fits the rest of the clue because Will’s neighbors include C5 but do not include C2 or C3, so exactly one of the column C criminals is Will’s neighbor. Therefore, we can determine that C5 Xavi is CRIMINAL.