Clues by Sam Aug 02, 2026 Answer – Full Solution Explained
A1
👩⚖️
judge
B1
👩🎤
singer
C1
👩🎤
singer
D1
👩🎤
singer
A2
👩⚖️
judge
B2
👩⚖️
judge
C2
💂♀️
guard
D2
💂♀️
guard
A3
👩💻
coder
B3
👩💻
coder
C3
🕵️♀️
sleuth
D3
💂♀️
guard
A4
👩💻
coder
B4
👩🍳
cook
C4
🕵️♀️
sleuth
D4
👩⚕️
doctor
A5
👩🍳
cook
B5
👩🍳
cook
C5
👩⚕️
doctor
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 Aug 02, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 12 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Celia (C1), Debra (D1), Frida (B2), Ina (D2), Laura (B3), Marianne (C3), Quita (B4), Rose (C4), Uma (A5), Vera (B5), Wanda (C5) and Zara (D5); the remaining 8 suspects are innocent.
The deduction chain, in plain English
01.C1 · Celia → CRIMINAL
Barb’s clue says Debra has exactly 2 criminal neighbors, and exactly 1 of those criminals is in row 1. Among Debra’s neighbors, the only person in row 1 is Celia. So the row 1 criminal neighbor required by the clue has to be Celia. That makes Celia criminal.
02.A2 · Emma → INNOCENT
Celia’s clue says Emma is one of Frida’s exactly five innocent neighbors. That directly identifies Emma’s status, regardless of the other neighbors. So Emma must be innocent.
03.D1 · Debra → CRIMINAL, D2 · Ina → CRIMINAL
Emma's clue says there are exactly 2 criminals above Olga. In that group, there are currently 0 known criminals, and the only people there whose identities are still unknown are Debra and Ina. Since those two spots must supply all 2 criminals required by the clue, both of them have to be criminals. So Debra and Ina must be criminal.
04.C2 · Hanna → INNOCENT
Barb’s clue says Debra has exactly 2 criminal neighbors, and exactly 1 of those criminals is in row 1. The only neighbor of Debra who is in row 1 is Celia, and Celia is already a criminal, so that row 1 criminal is fully accounted for. That leaves Debra’s neighbors outside row 1, C2 Hanna and D2 Ina, to contain exactly 1 criminal in total, and Ina is already that criminal. So Hanna must be innocent.
05.A1 · Anna → INNOCENT
Frida has exactly 5 innocent neighbors, so among Frida's 8 neighbors there must be exactly 2 criminals. One of those is already known to be Celia, which means the other criminal neighbors must come from the four unknown neighbors around Frida. But Anna is not among the three row 3 neighbors Katie, Laura, and Marianne. So the 2 criminal people in that unknown part have to be Katie, Laura, and Marianne, not Anna. That makes Anna innocent.
06.B3 · Laura → CRIMINAL
Frida’s neighbors must contain exactly 5 innocents, and 4 of those are already known: Anna, Barb, Emma, and Hanna. So among the three unknown neighbors, Katie, Laura, and Marianne, exactly one is innocent. Debra’s clue says only one criminal in column C has an innocent directly to the left. In this part of the board, that one innocent among Katie, Laura, and Marianne has to come from Katie or Marianne, not Laura. That means Laura cannot be the innocent one in Frida’s remaining neighbor spots. So Laura must be criminal.
07.B5 · Vera → CRIMINAL
Anna’s clue fixes Frida’s neighborhood at exactly five innocents, and Debra’s clue also has to be satisfied in column C. At the same time, Celia’s cook clue says exactly one cook has exactly six criminal neighbors, and the cooks are Quita, Uma, and Vera. If Vera were innocent, then Frida, Katie, Marianne, Pam, Quita, Rose, Uma, and Wanda would have to satisfy all three of those requirements at once. But that combination cannot be made to fit: it conflicts with the needed innocent count around Frida, the column C condition, and the rule that exactly one cook has six criminal neighbors. So Vera must be criminal.
08.C4 · Rose → CRIMINAL
Frida’s clue fixes her neighborhood at exactly five innocents, and four of those are already Anna, Barb, Emma, and Hanna, so only Katie or Marianne can supply the fifth innocent there. At the same time, the cook clue says exactly one cook has exactly six criminal neighbors, and Laura’s clue says an odd number of the edge neighbors of Vera, namely Pam, Uma, and Wanda, are innocent. If Rose were innocent, then Katie, Marianne, Pam, Quita, Uma, and Wanda would have to meet all of those requirements together. But with Rose innocent, those same people cannot satisfy Frida’s count, the cook condition, and Vera’s edge-neighbor oddness all at once. So Rose cannot be innocent. That makes Rose criminal.
09.B4 · Quita → CRIMINAL
Debra’s clue says exactly one criminal in column C has an innocent directly to the left. In column C, the known criminals are Celia and Rose. If Quita were innocent, then Rose would be a criminal in column C with an innocent directly to the left. But Celia is also a criminal in column C with an innocent directly to the left, so that would give more than one such criminal and conflict with the clue. So Quita must be criminal.
10.A4 · Pam → INNOCENT
Frida’s clue fixes her neighborhood at exactly 5 innocents, and that neighborhood already contains 4 known innocents: Anna, Barb, Emma, and Hanna. Rose’s clue also says that in column B, exactly one person has exactly 4 innocent neighbors, with Barb, Frida, Laura, Quita, and Vera being the relevant people to check. If Pam were criminal, then the remaining unknown people named here, including Frida, Katie, Marianne, Uma, and Wanda, would have to satisfy both of those clues at the same time, but they cannot. That makes the criminal option for Pam impossible. So Pam must be innocent.
11.A5 · Uma → CRIMINAL, C5 · Wanda → CRIMINAL
Among the cooks, exactly one person has exactly 6 criminal neighbors. Uma already has 2 known criminal neighbors and no unknown neighbors, so she cannot be the cook with 6. Vera has 2 known criminal neighbors, and even if both unknown neighbors around her were criminal, she would only reach 4, so she cannot be that cook either. That leaves Quita as the only cook who can be the one with exactly 6 criminal neighbors. Quita already has 3 known criminal neighbors, so to reach 6, the unknown neighbors around her that matter here must be criminal rather than innocent. So Uma and Wanda must be criminal.
12.B2 · Frida → CRIMINAL
Frida’s clue says her neighbors contain exactly 5 innocents, and among those neighbors we already know Anna, Barb, Emma, and Hanna are innocent. So exactly one of Katie and Marianne can be innocent. Pam’s clue says there are exactly 8 innocents on the whole board. There are already 5 known innocents, so among Frida, Katie, Marianne, Olga, Sara, and Zara, exactly 3 must be innocent. Ina’s clue says column D has an odd number of criminals. With Debra and Ina already criminal there, the three unknown people in column D must contribute an odd number of criminals, so among Olga, Sara, and Zara there must be either 1 or 3 criminals, which means either 2 or 0 innocents in column D. If Frida were innocent, then Katie, Marianne, Olga, Sara, and Zara would need to supply exactly 2 more innocents in total, but Katie and Marianne can supply only 1 innocent, while Olga, Sara, and Zara can supply only 2 or 0. That cannot make the required total. So Frida must be criminal.
13.D4 · Sara → INNOCENT
Pam says there are 8 innocents in total, and 5 innocents are already known, so the remaining unknown people must supply 3 more innocents among Katie, Marianne, Olga, Sara, and Zara. Celia’s clue says Frida has exactly 5 innocent neighbors; Frida already has 4 known innocent neighbors, so exactly one of Katie and Marianne must be innocent. If Sara were criminal, then the other three needed innocents would have to be Katie or Marianne, Olga, and Zara, which makes Zara innocent. But Uma’s clue says row 4 has more innocents than row 5. Row 4 already has 1 known innocent and row 5 has 0; if Sara were criminal and Zara were innocent, both rows would have 1 innocent, not more in row 4. So Sara at D4 must be innocent.
14.D5 · Zara → CRIMINAL
Pam’s clue says there are exactly 8 innocents on the whole board. There are already 6 known innocents, so among Katie, Marianne, Olga, and Zara, exactly 2 can be innocent. Sara’s clue says row 3 has more innocents than row 5. Right now both rows have 0 known innocents, so that comparison depends only on Katie, Marianne, Olga, and Zara. If Zara were innocent, then row 5 would already have 1 innocent, so Katie, Marianne, and Olga would have to supply more innocents in row 3 while also fitting Pam’s total of exactly 8 innocents, and that cannot be done. So Zara must be criminal.
15.D3 · Olga → INNOCENT
Ina’s clue says column D has an odd number of criminals. In that column, there are already 3 known criminals: Debra, Ina, and Zara, and the only person there not yet identified is Olga. If Olga were criminal too, that would make 4 criminals in column D, which is even, not odd. So Olga cannot be criminal. That makes Olga innocent.
16.A3 · Katie → INNOCENT
Zara’s clue says there are exactly 4 innocents in column A. In that column, Anna, Emma, and Pam are already known to be innocent, so column A currently has 3 known innocents. Katie is the only unknown person left in column A, and the column still needs 1 more innocent to reach 4. So Katie must be innocent.
17.C3 · Marianne → CRIMINAL
Celia’s clue says Emma is one of Frida’s exactly 5 innocent neighbors. Among Frida’s neighbors, the five innocents are already Anna, Barb, Emma, Hanna, and Katie, while Celia and Laura are criminals and Marianne is the only unknown there. If Marianne were innocent too, Frida would have 6 innocent neighbors, which clashes with the clue. So Marianne at C3 must be criminal.