Clues by Sam Feb 24, 2026 Answer – Full Solution Explained
A1
🕵️♀️
sleuth
B1
💂♂️
guard
C1
💂♂️
guard
D1
💂♀️
guard
A2
👮♀️
cop
B2
👨✈️
pilot
C2
👩🏫
teacher
D2
👨🏫
teacher
A3
👮♂️
cop
B3
👨⚕️
doctor
C3
👨💻
coder
D3
👨💻
coder
A4
🕵️♀️
sleuth
B4
👷♀️
builder
C4
👩💻
coder
D4
👷♀️
builder
A5
🕵️♀️
sleuth
B5
👩✈️
pilot
C5
👮♂️
cop
D5
👨⚕️
doctor
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 Feb 24, 2026 — a Medium solved in 17 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Brian (B1), Clyde (C1), Donna (D1), Hazel (C2), Isaac (D2), Logan (B3), Olive (A4), Tina (D4), Vera (A5), Wanda (B5) and Xavi (C5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.C2 · Hazel → CRIMINAL
In row 2 we have Flora at A2, Gary at B2, Hazel at C2, and Isaac at D2. Flora’s clue says there are exactly two criminals in row 2, and those two criminals are connected, which in a single row means they must be in neighboring squares with no gap between them. Since A2 is already INNOCENT, the two criminals must be chosen from B2, C2, and D2, and the only adjacent pairs available are B2 with C2 or C2 with D2. Either way, C2 is one of the two connected criminals. Therefore, we can determine that C2 Hazel is CRIMINAL.
02.B2 · Gary → INNOCENT
Hazel’s clue explicitly says that Gary is one of her three innocent neighbors. Since all clues are truthful, Gary must be innocent. Therefore, we can determine that B2 Gary is INNOCENT.
03.D2 · Isaac → CRIMINAL
In row 2, A2 (Flora) and B2 (Gary) are already known to be innocent, and C2 (Hazel) is already known to be a criminal, leaving only D2 (Isaac) undecided. Flora’s clue says “Both criminals in row 2 are connected,” which means there are exactly two criminals in that row and they must be adjacent with no innocent between them. Since A2 and B2 are innocent, the only way for C2 to have the second row-2 criminal connected to it is for D2 to be the other criminal. Therefore, we can determine that D2 Isaac is CRIMINAL.
04.A5 · Vera → CRIMINAL
Vera is at A5, so the people to the right of her in row 5 are Wanda at B5, Xavi at C5, and Zach at D5. Gary’s clue says there is only one innocent to the right of Vera, so among Wanda, Xavi, and Zach there is exactly 1 innocent in total. Isaac’s clue says row 5 has an odd number of innocents, so the total innocents in row 5 must be either 1 or 3; since the three people to Vera’s right already contribute exactly 1 innocent, Vera must contribute 0 innocents for the row total to stay odd. Therefore, we can determine that A5 Vera is CRIMINAL.
05.B1 · Brian → CRIMINAL, B3 · Logan → CRIMINAL
Hazel at C2 says that Gary is one of her 3 innocent neighbors, and Hazel’s neighbors are B1 Brian, C1 Clyde, D1 Donna, B2 Gary, D2 Isaac, B3 Logan, C3 Martin, and D3 Noah. Since Gary is already innocent, that leaves exactly 2 innocents among the other six neighbors B1, C1, D1, B3, C3, and D3, so exactly 4 of those six must be criminals. Vera also says Isaac at D2 has exactly 3 criminal neighbors; Hazel is one of Isaac’s neighbors and is already criminal, so among Isaac’s other neighbors C1, D1, C3, and D3 there must be exactly 2 criminals. Those four people are part of Hazel’s six unknown neighbors, so they can contribute only 2 of the 4 criminals Hazel still needs among those six, which forces the remaining two people B1 Brian and B3 Logan to be criminals. Therefore, we can determine that B1 Brian is CRIMINAL and B3 Logan is CRIMINAL.
06.B4 · Paula → INNOCENT
Tina is at D4, so the people to her left in the same row are Olive at A4, Paula at B4, and Susan at C4. Logan’s clue says there are exactly 2 innocents to the left of Tina, so exactly two of Olive, Paula, and Susan are innocents. Brian’s clue says all innocents in row 4 are connected, so those row-4 innocents cannot be split apart by a criminal sitting between them. If Paula were not an innocent, then the two innocents to the left of Tina would have to be Olive and Susan, but they would be separated by Paula at B4 and would not be connected in row 4. Therefore, we can determine that B4 Paula is INNOCENT.
07.B5 · Wanda → CRIMINAL
Look at column B: Brian at B1 is a criminal and Logan at B3 is a criminal, while Gary at B2 and Paula at B4 are innocent. Paula’s clue says each column must contain at least 3 criminals, so column B still needs one more criminal beyond Brian and Logan. The only remaining person in column B who can supply that third criminal is Wanda at B5. Therefore, we can determine that B5 Wanda is CRIMINAL.
08.A4 · Olive → CRIMINAL
Look at Gary at B2: his column A neighbors are A1 (Anna), A2 (Flora), and A3 (Kevin). Wanda’s clue says that among the five criminals neighboring Gary, only one is in column A, and since Flora at A2 is already INNOCENT, that forces exactly one of Anna (A1) or Kevin (A3) to be a CRIMINAL, meaning the other one is INNOCENT. Now use Paula’s clue that each column has at least 3 criminals: in column A we already have Vera (A5) as a CRIMINAL and exactly one of A1/A3 as a CRIMINAL, but A2 is INNOCENT, so without Olive (A4) being a CRIMINAL the column could only reach 2 criminals. Therefore, we can determine that A4 Olive is CRIMINAL.
09.C4 · Susan → INNOCENT
Tina is at D4, so the people to the left of Tina are exactly A4 Olive, B4 Paula, and C4 Susan. Logan’s clue says there are exactly 2 innocents among those three people. We already know Olive is a criminal and Paula is innocent, so Susan must be the second innocent to make the total exactly 2. Therefore, we can determine that C4 Susan is INNOCENT.
10.C3 · Martin → INNOCENT
Hazel at C2 says Gary is one of her 3 innocent neighbors, and Hazel’s neighbors are Brian, Clyde, Donna, Gary, Isaac, Logan, Martin, and Noah. Since Brian, Isaac, and Logan are already criminal and Gary is innocent, exactly two of the remaining four neighbors (Clyde, Donna, Martin, and Noah) must be innocent. Susan’s clue counts the edge neighbors of Isaac at D2, which are Clyde at C1, Donna at D1, and Noah at D3, and says an odd number of them are innocent. That odd number cannot be 3, because then Hazel would already have at least three innocent neighbors besides Gary, so it must be exactly 1, which forces the second innocent among Hazel’s remaining four neighbors to be Martin. Therefore, we can determine that C3 Martin is INNOCENT.
11.C5 · Xavi → CRIMINAL, C1 · Clyde → CRIMINAL
Look at column C: Clyde at C1 is Unknown, Hazel at C2 is CRIMINAL, Martin at C3 is INNOCENT, Susan at C4 is INNOCENT, and Xavi at C5 is Unknown. Paula’s clue says each column must contain at least 3 criminals, so column C must have at least three CRIMINAL people among these five. Since C3 and C4 are already confirmed INNOCENT, the only way for column C to reach at least three criminals is for both remaining unknowns, C1 and C5, to be CRIMINAL. Therefore, we can determine that C5 Xavi is CRIMINAL and C1 Clyde is CRIMINAL.
12.D5 · Zach → INNOCENT
Vera is at A5, so the people to the right of Vera are the other three people in row 5: Wanda at B5, Xavi at C5, and Zach at D5. Gary’s clue says there is only one innocent among those people to the right of Vera. Wanda and Xavi are already known to be criminals, so neither of them can be that one innocent. That leaves only Zach as the single innocent to the right of Vera. Therefore, we can determine that D5 Zach is INNOCENT.
13.A1 · Anna → CRIMINAL
Brian is at B1 and Gary is at B2, so their common neighbors (people who are neighbors of both) are A1, A2, C1, and C2. Clyde’s clue says that among these common neighbors, exactly one is innocent. We already know A2 (Flora) is innocent, and C1 (Clyde) and C2 (Hazel) are criminals, so A1 cannot also be innocent. Therefore, we can determine that A1 Anna is CRIMINAL.
14.A3 · Kevin → INNOCENT
Gary is at B2, and his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Wanda’s clue says that exactly five of those neighbors are criminals, and among those five criminals, exactly one is in column A. We already know five neighboring criminals: Anna at A1, Brian at B1, Clyde at C1, Hazel at C2, and Logan at B3, and only Anna is in column A, so the clue’s count is already fully satisfied. That means Kevin at A3 cannot be another neighboring criminal, so he must be innocent. Therefore, we can determine that A3 Kevin is INNOCENT.
15.D1 · Donna → CRIMINAL
Row 5 already has exactly three criminals: Vera at A5, Wanda at B5, and Xavi at C5, while Zach at D5 is innocent. Anna’s clue says that only one row in the whole board has exactly three criminals, so no other row is allowed to have exactly three. Row 1 currently has three known criminals at A1, B1, and C1, so if Donna at D1 were innocent then Row 1 would also have exactly three criminals, which is not allowed. Donna must therefore be criminal so that Row 1 has four criminals instead of three. Therefore, we can determine that D1 Donna is CRIMINAL.
16.D3 · Noah → INNOCENT
Hazel is at C2, so her neighbors are B1 Brian, C1 Clyde, D1 Donna, B2 Gary, D2 Isaac, B3 Logan, C3 Martin, and D3 Noah. Her clue says that she has exactly 3 innocent neighbors, and that Gary is one of them. We already know Gary is innocent and Martin at C3 is also innocent, and every other neighbor of Hazel besides Noah is already confirmed criminal. That means Noah must be the third innocent neighbor Hazel is talking about. Therefore, we can determine that D3 Noah is INNOCENT.
17.D4 · Tina → CRIMINAL
Look at column D, which contains Donna (D1), Isaac (D2), Noah (D3), Tina (D4), and Zach (D5). Paula’s clue says that each column must have at least 3 criminals. In column D we already know Donna and Isaac are criminals, while Noah and Zach are innocents, so the only remaining way to reach at least 3 criminals in that column is for Tina to be a criminal. Therefore, we can determine that D4 Tina is CRIMINAL.