Clues by Sam Oct 17, 2026 Answer – Full Solution Explained
A1
👩⚖️
judge
B1
👮♀️
cop
C1
👨💼
clerk
D1
👨🏫
teacher
A2
👩⚖️
judge
B2
👮♀️
cop
C2
👩💼
clerk
D2
👨🏫
teacher
A3
👨⚖️
judge
B3
👮♂️
cop
C3
👩💻
coder
D3
👩🏫
teacher
A4
👩🔧
mech
B4
👩💻
coder
C4
👨💻
coder
D4
👨✈️
pilot
A5
👨🔧
mech
B5
👨🔧
mech
C5
👩✈️
pilot
D5
👨✈️
pilot
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 — 9 criminals.
Clues by Sam answer for Oct 17, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 9 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Bonnie (B1), Hazel (C2), Isaac (D2), Pam (D3), Sam (C4), Tom (D4), Vince (B5) and Wanda (C5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.C4 · Sam → CRIMINAL
Quita’s clue says the two criminals in row 4 must be connected. If Sam were innocent, then row 4 would be Quita innocent, Ruby unknown, Sam innocent, and Tom unknown. To get exactly two criminals in that row and have them connected, Ruby would have to be a criminal, and then Tom would also have to be a criminal, which leaves the two criminals separated by Sam. That breaks the clue that both criminals in row 4 are connected. So Sam must be criminal.
02.D3 · Pam → CRIMINAL
Sam’s clue says that Pam is one of the exactly 3 criminals above Zach. That directly identifies Pam as a criminal. So Pam must be criminal.
03.D2 · Isaac → CRIMINAL
Pam says all criminals in column D are connected, so the criminals there have to make one unbroken vertical block. Sam’s clue says Pam is one of the three criminals above Zach, so among David, Isaac, Pam, and Tom there must be exactly three criminals including Pam. That means the only possible criminal blocks in column D are David, Isaac, and Pam; or Isaac, Pam, and Tom; or Isaac, Pam, Tom, and Zach. In every one of those connected blocks, Isaac is included. So Isaac must be criminal.
04.B3 · Kumar → INNOCENT
If Kumar were criminal, then Mary’s neighbors would have to contain exactly 5 criminals, with exactly 2 of those criminals in row 2. In row 4, Quita’s clue says the two criminals there must be connected, and with Quita innocent and Sam already criminal, Ruby and Tom cannot be criminals there, so Ruby and Tom are innocent. That makes the assumption impossible. So Kumar must be innocent.
05.C3 · Mary → INNOCENT
Assume Mary were a criminal. Sam has exactly 4 criminal neighbors, and exactly 2 of those criminal neighbors are in row 5, so the other 2 criminal neighbors are not in row 5. With Pam already a known criminal neighbor, that leaves just one more criminal neighbor outside row 5. Quita says the two criminals in row 4 are connected. Since Quita is innocent and Sam is already the known criminal in row 4, row 4 must be Quita innocent, Ruby innocent, Sam criminal, and Tom innocent. That makes both Ruby and Tom innocent, so neither of Sam’s row 4 neighbors can be that one extra criminal neighbor outside row 5. That makes the assumption about Mary clash with these clues, so Mary at C3 must be innocent.
06.C5 · Wanda → CRIMINAL
Row 5’s criminals have to be connected, and among Sam’s row 5 neighbors exactly two are criminals: Vince, Wanda, and Zach contain exactly two criminals. That means the row 5 criminal group has to be one connected block that fits those two facts. The only connected row 5 blocks that work are Vince and Wanda, Umar and Vince and Wanda, or Wanda and Zach. In every one of those possible connected blocks, Wanda is included. So Wanda must be criminal.
07.A5 · Umar → INNOCENT
If Umar were criminal, then row 5 would have Umar and Wanda as criminals. Kumar’s clue says exactly 2 of Sam’s 4 criminal neighbors are in row 5, and among Sam’s row 5 neighbors the only possibilities are Vince, Wanda, and Zach. Since Wanda is already one of those row 5 neighboring criminals, the other two in that group, Vince and Zach, have to be innocent. But Wanda’s clue says every row has at least 2 innocents, and with Umar criminal that would leave row 5 short of that requirement. So Umar must be innocent.
08.A2 · Esha → INNOCENT
If Esha were criminal, then row 2 would need Freya and Hazel to be the two innocents there, because every row has at least 2 innocents. But Isaac's clue says Mary's neighbors contain exactly 5 criminals in total, and exactly 2 of those criminals are in row 2. In Mary's neighboring row-2 group, Isaac is already a criminal, so with Esha excluded from Mary's neighbors, the only way to reach 2 row-2 criminals there would be to make both Freya and Hazel criminals. That clashes with row 2 needing Freya and Hazel to be innocents, so Esha at A2 must be innocent.
09.B1 · Bonnie → CRIMINAL
David’s clue says his neighbors contain exactly one innocent, and among the neighbors of David who are also in Freya’s non-column-A neighbor group, the only possibilities are Chris and Hazel. So Chris and Hazel account for the one innocent allowed there. In Freya’s neighbors who are not in column A, there must be exactly three innocents, and two of those are already Kumar and Mary. That means the remaining one innocent in that group has to come from Chris and Hazel, leaving no innocent slot for anyone else in that same group. The only person left there is Bonnie, so Bonnie must be criminal.
10.D1 · David → INNOCENT
Assume David were criminal. Isaac’s clue says exactly 2 of Mary’s 5 criminal neighbors are in row 2, and since Isaac is already one of them, Freya and Hazel cannot also be criminals in that row. So Freya and Hazel must be innocent. But then Chris in row 1 would already have 3 criminal neighbors: Bonnie, Isaac, and David. Bonnie’s clue says no one in row 1 has more than 3 criminal neighbors, so Chris cannot have any additional criminal neighbor. That clashes with the assumption that David is criminal in this step, so David at D1 must be innocent.
11.D4 · Tom → CRIMINAL
Sam’s clue says that above Zach there are exactly 3 criminals, and Pam is one of them. The people above Zach are David, Isaac, Pam, and Tom. David is innocent, while Isaac and Pam are already criminals, so the third criminal above Zach has to be Tom. So Tom must be criminal.
12.B4 · Ruby → INNOCENT
Mary’s neighbors contain exactly 5 criminals, and exactly 2 of those criminals are in row 2. Outside row 2, Mary’s neighboring people are Kumar, Pam, Ruby, Sam, and Tom. In that group, Pam, Sam, and Tom are already known criminals, so the 3 criminals outside row 2 are already fully accounted for. That means Ruby cannot be a criminal. So Ruby must be innocent.
13.C1 · Chris → INNOCENT, C2 · Hazel → CRIMINAL
Sam’s clue fixes his neighboring criminals at exactly 4, with exactly 2 of them in row 5. Freya’s clue fixes her neighboring criminals at exactly 3, with exactly 1 of them in column A. Ruby’s clue also fixes the edge criminals at exactly 7, with exactly 2 of those edge criminals neighboring Ruby. If Chris were criminal and Hazel were innocent, then Anna, Jason, Vince, and Zach would have to make all three of those exact counts come out correctly at the same time. They cannot do that while also meeting Sam’s row 5 count, Freya’s column A count, and Ruby’s edge-neighbor count together. So the opposite assignment fails, which means Chris must be innocent and Hazel must be criminal.
14.B2 · Freya → INNOCENT
Mary’s neighbors contain exactly 5 criminals in total, and exactly 2 of those criminals are in row 2. Among Mary’s neighbors who are in row 2, Hazel and Isaac are already known criminals. That already fills the clue’s row 2 total of 2 criminals, so the only remaining row 2 neighbor of Mary, Freya, cannot be a criminal. So Freya must be innocent.
15.B5 · Vince → CRIMINAL
If Vince were innocent, then column B would have 4 innocents. Tom’s clue says only one column can have more than 3 innocents, so every other column must stay at 3 or fewer. Umar’s clue says Freya has exactly 3 criminal neighbors, and exactly 1 of those is in column A. Freya already has 2 known criminal neighbors outside column A, so the 1 criminal in column A has to come from Anna and Jason. That makes Anna and Jason both criminal, which would raise column A above 3 innocents as well. Then both column A and column B would have more than 3 innocents, contradicting Tom’s clue. So Vince at B5 must be criminal.
16.D5 · Zach → INNOCENT
Kumar says exactly 2 of Sam's neighboring criminals are in row 5. Among Sam's neighbors in row 5, Vince and Wanda are already known criminals, which already makes 2. That means the only remaining row 5 neighbor of Sam, Zach, cannot also be a criminal. So Zach must be innocent.
17.A1 · Anna → CRIMINAL, A3 · Jason → INNOCENT
Ruby says exactly 2 of the 7 edge criminals are her neighbors. In the edge-neighbor group, Vince and Wanda are already criminals, while Quita and Umar are innocent. If Jason were criminal, then Ruby would have 3 criminal neighbors on the edge there: Jason, Vince, and Wanda, which contradicts Ruby's clue that there are exactly 2. So Jason must be innocent, and since the edge cells must contain exactly 7 criminals in total and there are already 6 known edge criminals, Anna must be the remaining edge criminal. So Anna must be criminal and Jason must be innocent.