Clues by Sam Jul 19, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👨💻
coder
C1
👮♀️
cop
D1
👨🎤
singer
A2
👩💼
clerk
B2
👨⚕️
doctor
C2
👨✈️
pilot
D2
👨✈️
pilot
A3
👩🎤
singer
B3
👨💻
coder
C3
👩💻
coder
D3
👷♀️
builder
A4
👩💼
clerk
B4
👨💼
clerk
C4
👩⚖️
judge
D4
👨⚕️
doctor
A5
👩✈️
pilot
B5
👩⚖️
judge
C5
👨🎤
singer
D5
👷♂️
builder
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 — 16 criminals.
Clues by Sam answer for Jul 19, 2026 — a Brutal solved in 18 steps
Today's Clues by Sam puzzle is rated Brutal and resolves with 16 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Chad (B1), Dana (C1), Eric (D1), Flora (A2), Gabe (B2), Hal (C2), Isaac (D2), Joy (A3), Lucy (C3), Mary (D3), Olivia (A4), Peter (B4), Ruby (C4), Steve (D4), Wally (C5) and Xavi (D5); the remaining 4 suspects are innocent.
The deduction chain, in plain English
01.C5 · Wally → CRIMINAL
Tina’s clue says that the two criminals in row 5 must be connected. In row 5, Tina is already innocent, so the only people who could be those two criminals are Uma, Wally, and Xavi. If Wally were innocent, then the two criminals would have to be Uma and Xavi. But with Wally between them, those two would not be connected, which breaks Tina’s clue. So Wally must be criminal.
02.B5 · Uma → INNOCENT
Wally’s clue says the people strictly between Tina and Xavi contain exactly one innocent. Among those people, there are currently no known innocents, and the only person there whose identity is still unknown is Uma. So the one innocent required between Tina and Xavi has to be Uma. That makes Uma innocent.
03.D5 · Xavi → CRIMINAL
Tina’s clue says that the two criminals in row 5 are connected. In row 5, Tina and Uma are innocent and Wally is already a criminal, so the only person left who could be the second criminal is Xavi. If Xavi were innocent, row 5 would not have both criminals connected as the clue requires. So Xavi must be criminal.
04.B1 · Chad → CRIMINAL
Xavi’s clue says Chad is one of the exactly 3 criminals in row 1. Since Chad is explicitly included among those criminals, Chad at B1 must be criminal.
05.C1 · Dana → CRIMINAL
Xavi says row 1 has exactly three criminals, and Chad is one of them. Since Chad is the only known criminal there, two of Betty, Dana, and Eric must also be criminals. Uma says all criminals in row 1 are connected, so those three criminals must make one continuous block. The only possible three-person blocks are Betty, Chad, and Dana, or Chad, Dana, and Eric. In both cases Dana is included. So Dana must be criminal.
06.A1 · Betty → INNOCENT
Dana’s clue says Gabe has exactly 2 innocent neighbors, and exactly 1 of those innocent neighbors is in row 1. Among Gabe’s neighbors who are in row 1, Chad and Dana are already criminals, so the row 1 part of the clue still needs 1 innocent there. The only unknown person left in that row 1 neighbor group is Betty. So Betty must be innocent.
07.D1 · Eric → CRIMINAL
Xavi’s clue says Chad is one of exactly 3 criminals in row 1. In row 1, Betty is innocent, Chad and Dana are already criminals, and the only person there not yet identified is Eric. If Eric were innocent, row 1 would have only 2 criminals, which conflicts with the clue saying there are exactly 3. So Eric must be criminal.
08.D4 · Steve → CRIMINAL
Eric’s clue says Steve is one of Lucy’s 7 criminal neighbors. That directly places Steve among the criminal neighbors counted around Lucy, so Steve must be criminal.
09.B4 · Peter → CRIMINAL
Steve’s clue says that Peter is one of Joy’s four criminal neighbors. That directly identifies Peter as a criminal. So Peter must be criminal.
10.C2 · Hal → CRIMINAL
Gabe’s clue says his neighbors contain exactly 2 innocents in total, and among Gabe’s row 1 neighbors there is already exactly 1 innocent: Betty. That means the one other innocent among Gabe’s neighbors has to come from the non-row-1 neighbors A2 Flora, A3 Joy, B3 Kumar, or C3 Lucy. Hal is also a non-row-1 neighbor of Gabe, but he is not in that listed group for the remaining innocent, so he cannot be that innocent. So Hal must be criminal.
11.A2 · Flora → CRIMINAL
Gabe’s clue says he has exactly 2 innocent neighbors, and among Gabe’s neighbors the only innocent already identified is Betty. That means there is exactly 1 more innocent among A2 Flora, A3 Joy, B3 Kumar, and C3 Lucy. Dana’s clue also says only 1 of Gabe’s 2 innocent neighbors is in row 1. Betty is in row 1 and is already innocent, so the other innocent cannot be in row 1. Flora is in row 1, so she cannot be that remaining innocent. That makes Flora at A2 criminal.
12.D3 · Mary → CRIMINAL
Gabe’s clue says his neighbors contain exactly two innocents, and exactly one of those innocents is in row 1. In that group, Betty is already the one innocent in row 1, so Gabe’s second neighboring innocent has to be outside row 1. Flora’s clue says row 5 is the only row with exactly two innocents. If Mary were innocent, then the remaining people touched by these clues would have to fit both that row rule and Gabe’s neighbor rule at the same time, but they cannot. So Mary must be criminal.
13.B2 · Gabe → CRIMINAL, C3 · Lucy → CRIMINAL
Hal’s clue says Hal has exactly one innocent neighbor. Among Hal’s unknown neighbors, that one innocent must come from Isaac or Kumar, so it cannot be Gabe or Lucy. That rules out Gabe and Lucy as innocents. So Gabe and Lucy must be criminal.
14.C4 · Ruby → CRIMINAL
Hal’s neighbors contain exactly one innocent, and among those neighbors the only people not yet identified are Isaac and Kumar. Lucy’s neighbors also contain exactly one innocent, and that larger group includes those same possible innocent spots plus Ruby. Since the one innocent allowed in Lucy’s neighbor group must already come from Isaac or Kumar, Ruby cannot be innocent. So Ruby must be criminal.
15.A4 · Olivia → CRIMINAL
Gabe’s clue says his neighbors contain exactly 2 innocents, and exactly 1 of those innocents is in row 1. Since Betty is already that one row-1 innocent among Gabe’s neighbors, the other innocent there has to come from Joy or Kumar. Peter’s clue says his neighbors contain an odd number of innocents. Peter’s neighbors already include 2 known innocents, Tina and Uma, so among Joy, Kumar, and Olivia there must be an odd number of innocents. If Olivia were innocent, then Joy and Kumar would still have to provide Gabe’s one non-row-1 innocent, while Joy, Kumar, and Olivia would also have to make Peter’s total odd, and those requirements conflict. So Olivia cannot be innocent. That makes Olivia criminal.
16.B3 · Kumar → INNOCENT
Steve’s clue says Peter is among Joy’s exactly 4 criminal neighbors. Joy’s neighbors are Flora, Gabe, Kumar, Olivia, and Peter, and among them Flora, Gabe, Olivia, and Peter are already known criminals. That already fills all 4 criminal-neighbor spots from the clue, so Kumar cannot also be a criminal. So Kumar must be innocent.
17.A3 · Joy → CRIMINAL
Gabe’s neighbors contain exactly 2 innocents in total, and exactly 1 of those innocents is in row 1. In Gabe’s row-1 neighbors, A1 Betty is already that 1 innocent. That means among Gabe’s neighbors not in row 1, there is room for exactly 1 innocent, and B3 Kumar is already that innocent. So A3 Joy cannot be innocent. That makes Joy criminal.
18.D2 · Isaac → CRIMINAL
Lucy’s neighbors contain exactly 1 innocent. In that group, Kumar is already the 1 innocent, while Gabe, Hal, Mary, Peter, Ruby, and Steve are criminal, leaving only Isaac unidentified there. Since the single innocent in Lucy’s neighboring group is already accounted for, Isaac cannot be innocent. So Isaac must be criminal.