Clues by Sam Aug 05, 2026 Answer – Full Solution Explained
A1
👮♂️
cop
B1
👩🌾
farmer
C1
👷♂️
builder
D1
👨🌾
farmer
A2
👮♀️
cop
B2
🕵️♂️
sleuth
C2
🕵️♂️
sleuth
D2
👩🍳
cook
A3
🕵️♂️
sleuth
B3
💂♀️
guard
C3
👩🎨
painter
D3
👩🎨
painter
A4
👮♀️
cop
B4
👨✈️
pilot
C4
👨✈️
pilot
D4
💂♀️
guard
A5
💂♀️
guard
B5
👨✈️
pilot
C5
👷♀️
builder
D5
👨🍳
cook
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 Aug 05, 2026 — a Tricky solved in 18 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Ben (A1), Derek (C1), Erwin (D1), Freya (A2), Kevin (A3), Mary (C3), Nancy (D3), Olivia (A4), Peter (B4), Rob (C4), Tina (D4) and Vince (B5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.C4 · Rob → CRIMINAL
Xia’s clue directly says that Rob is one of the three criminals above Xia. Since Rob is explicitly included in that criminal group, Rob must be criminal.
02.C3 · Mary → CRIMINAL, C2 · Ike → INNOCENT
Rob’s clue says there are exactly two criminals below Derek, and those two criminals must be connected. Below Derek are Ike, Mary, Rob, and Xia, with Rob already known to be a criminal and Xia already known to be innocent. If Mary were innocent and Ike were criminal, then the two criminals below Derek would be Ike and Rob, but that assignment conflicts with the clue’s requirement for the criminals below Derek. So Mary must be criminal and Ike must be innocent.
03.C1 · Derek → CRIMINAL
Xia’s clue says that above Xia there are exactly 3 criminals, and Rob is one of them. The people above Xia are Derek, Ike, Mary, and Rob; among them, Mary and Rob are already criminals and Ike is innocent. If Derek were innocent too, then only Mary and Rob would be criminals above Xia, which clashes with Xia’s count of exactly 3 criminals there. So Derek must be criminal.
04.D4 · Tina → CRIMINAL
Below Joy are Nancy, Tina, and Zane, and that group must contain an odd number of innocents. Since none of them is already known innocent, those three must supply an odd number of innocents themselves. Tina’s neighbors include Mary and Rob, who are criminals, and Xia, who is innocent, so Tina’s clue leaves exactly one more innocent among Nancy and Zane. That means Nancy and Zane already contribute 1 innocent below Joy, which is the odd total that clue needs there. If Tina were also innocent, the people below Joy would have 2 innocents instead, which is even. So Tina must be criminal.
05.B5 · Vince → CRIMINAL
Among the edge cells that are neighbors of Rob, Tina is already criminal and Xia is already innocent, so the only places left for the one innocent in that group are Nancy, Vince, and Zane. Mary's clue says the people below Joy contain an odd number of innocents, and with Tina already criminal that one innocent must come from Nancy or Zane. That means the innocent in Rob's edge-neighbor group is already accounted for by Nancy or Zane, so Vince cannot be that innocent. So Vince must be criminal.
06.D1 · Erwin → CRIMINAL
Vince’s clue says that Erwin is one of the exactly 3 criminals in column D. Since Erwin is explicitly included among those criminals, his identity is fixed by the clue itself. So Erwin must be criminal.
07.D2 · Joy → INNOCENT
Vince's clue says column D has exactly 3 criminals. Erwin and Tina are already criminals there, so among Joy, Nancy, and Zane there is room for exactly 1 more criminal. Mary's clue says the people below Joy contain an odd number of innocents. Below Joy, Tina is already a criminal, so the needed innocents must come from Nancy and Zane. That means the one extra criminal in column D has to be among Nancy and Zane, not Joy. So Joy must be innocent.
08.B1 · Carol → INNOCENT
Column A has exactly 4 criminals, and those five people are A1 Ben, A2 Freya, A3 Kevin, A4 Olivia, and A5 Uma. The edge cells that are not neighboring Rob also must contain exactly 6 criminals, and that group is those five column A people plus B1 Carol, C1 Derek, D1 Erwin, D2 Joy, with Derek and Erwin already criminal and Joy innocent. Since the five people in column A account for 4 of those criminals, Derek and Erwin account for the other 2, and that already reaches the full total of 6 criminals allowed in this larger group. The only extra person in that larger group beyond those already counted is Carol, so she cannot be criminal. That makes Carol innocent.
09.A1 · Ben → CRIMINAL
Tina’s clue fixes the edge people who are not Rob’s neighbors so that, among A1 Ben, A2 Freya, A3 Kevin, A4 Olivia, and A5 Uma, there is exactly 1 innocent. Erwin’s clue says the people below Ben contain an odd number of criminals, so among A2 Freya, A3 Kevin, A4 Olivia, and A5 Uma there is exactly 1 innocent. That means the single innocent in the five-person group has to be one of Freya, Kevin, Olivia, or Uma, not Ben. So Ben must be criminal.
10.A2 · Freya → CRIMINAL
Ben’s clue says an odd number of the innocents below him neighbor Peter. Among the people below Ben, the only ones who neighbor Peter are Kevin, Olivia, and Uma, while Freya does not neighbor Peter. Since exactly one innocent in that relevant edge-cell group has to be among Kevin, Olivia, and Uma, Freya cannot be that innocent person. So Freya must be criminal.
11.B3 · Lisa → INNOCENT
Column A must contain exactly 4 criminals, and Ben and Freya already account for 2 of them. So among Kevin, Olivia, and Uma, exactly 2 must be criminals, which means exactly 1 of those three is innocent. Peter’s neighbors must contain an odd number of innocents. Among those neighbors, Xia is already a known innocent, and the other people involved here are Kevin, Lisa, Olivia, and Uma. If Lisa were criminal, then the only possible additional innocents among these neighbors would come from Kevin, Olivia, and Uma, but that group contains exactly 1 innocent, making Peter’s total innocent neighbors 2, which is even, not odd. So Lisa must be innocent.
12.A4 · Olivia → CRIMINAL
Lisa’s clue says exactly 1 cop has an innocent directly below them, and among the people directly below cops the innocent count is exactly 1 across Freya, Kevin, and Uma. Tina’s clue implies that among the edge people who are not neighbors of Rob, there are exactly 3 innocents, namely Carol, Joy, and one from that below-a-cop group. That larger edge group is A1 Ben, B1 Carol, C1 Derek, D1 Erwin, A2 Freya, D2 Joy, A3 Kevin, A4 Olivia, and A5 Uma. Since its 3 innocent places are already accounted for by Carol, Joy, and the one innocent among Freya, Kevin, and Uma, the only remaining person in that group outside the smaller set, Olivia, cannot be innocent. So Olivia must be criminal.
13.B4 · Peter → CRIMINAL
Below Joy, the innocents must total an odd number, and at the moment the only unknown people there are Nancy and Zane. Rob's neighbors also must contain an odd number of innocents, and among those neighbors there are already 2 known innocents, with the only unknowns being Nancy, Peter, and Zane. If Peter were innocent, then the innocence count among Rob's neighbors would depend on Nancy and Zane in a way that has to satisfy Rob's odd-neighbor clue, while Nancy and Zane also have to satisfy Joy's below clue at the same time. Those same two people cannot make both clues work if Peter is innocent. So Peter must be criminal.
14.D3 · Nancy → CRIMINAL
Ike’s clue says the number of innocent painters must equal the number of innocent pilots. Right now the pilots have 0 innocent people, and the painters also have 0 known innocents; Nancy is the only painter whose identity is not yet fixed. If Nancy were innocent, the painters would have 1 innocent while the pilots would still have 0, which breaks Ike’s clue. So Nancy must be criminal.
15.D5 · Zane → INNOCENT
Mary's clue says the people below Joy contain an odd number of innocents. Below Joy, Nancy and Tina are already known criminals, so there are currently 0 known innocents there, and the only person left in that group is Zane. If Zane were criminal, the number of innocents below Joy would stay 0, which is not odd. So Zane must be innocent.
16.A5 · Uma → INNOCENT
Zane’s clue says exactly 2 pilots have a criminal directly to their left. Peter already fits that because Olivia is directly to his left, and Rob already fits that because Peter is directly to his left. Those two already make up the full count of 2. If Uma were criminal, then Vince would also be a pilot with a criminal directly to his left, which would make 3 instead of 2. So Uma must be innocent.
17.A3 · Kevin → CRIMINAL
Tina's clue says there are exactly 9 criminals on the edge, and exactly 3 of those edge criminals are neighbors of Rob. Those 3 are already accounted for by Nancy, Tina, and Vince, so the edge cells that are not neighbors of Rob must contain exactly 6 criminals. Among the edge cells that are not neighbors of Rob, Ben, Derek, Erwin, Freya, and Olivia are already known criminals, which makes 5. Kevin is the only unknown person left in that group, so he has to be the sixth criminal. So Kevin must be criminal.
18.B2 · Hal → INNOCENT
Freya’s clue says Lisa has exactly 2 innocent neighbors. Among Lisa’s neighbors, 1 is already known to be innocent, and the only neighbor there whose identity is still unknown is Hal. So Lisa’s neighbors need exactly 1 more innocent, and Hal is the only person who can fill that spot. That makes Hal innocent.