Clues by Sam Aug 11, 2026 Answer – Full Solution Explained
A1
👩⚕️
doctor
B1
👩🔧
mech
C1
👨💻
coder
D1
👨💻
coder
A2
😬
raccoon
B2
👩🔧
mech
C2
👨💻
coder
D2
👨🌾
farmer
A3
😬
fish
B3
😬
lamb
C3
👮♂️
cop
D3
👮♂️
cop
A4
👩🌾
farmer
B4
👩⚕️
doctor
C4
👩💼
clerk
D4
😬
fish
A5
👩💼
clerk
B5
🕵️♂️
sleuth
C5
🕵️♂️
sleuth
D5
🕵️♀️
sleuth
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 — 6 criminals.
Clues by Sam answer for Aug 11, 2026 — a Medium solved in 18 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Erwin (A2), Flora (B2), Hank (D2), Ruth (A5) and Thor (B5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.D2 · Hank → CRIMINAL
Nancy’s clue says Hank is one of the exactly 3 criminals in row 2. Since Hank is explicitly included among those criminals, his identity is fixed by the clue itself. So Hank must be criminal.
02.B2 · Flora → CRIMINAL
Hank's clue says the two innocents below Bonnie are connected. Nancy is already a known innocent among the people below Bonnie, so she must be one of those two. That means the other innocent has to be next to Nancy, which in that column leaves only Janet or Thor. Flora is not next to Nancy there, so Flora cannot be that other innocent. So Flora must be criminal.
03.D5 · Xia → INNOCENT
Flora’s clue says that the only criminal to the right of Ruth is also to the left of Xia. Among the people to the right of Ruth, Xia is the one who is not to the left of Xia. That means anyone to the right of Ruth who is outside the “left of Xia” group cannot be the criminal mentioned in the clue. So Xia must be innocent.
04.A5 · Ruth → CRIMINAL
Xia says there is exactly one innocent to her left, so among Ruth, Thor, and Will there is one innocent. Flora says the only criminal to the right of Ruth is to the left of Xia, and the people who satisfy both of those descriptions are Thor and Will. That means the one innocent to Xia's left has to be one of Thor or Will, not Ruth. So Ruth must be criminal.
05.C1 · Clyde → INNOCENT
Ruth’s clue says Bonnie and Gabe have exactly one innocent neighbor in common. Their shared neighbors are only Clyde and Flora, and Flora is already criminal. So that one innocent in the shared pair has to be the only person left there, Clyde. That makes Clyde innocent.
06.A3 · Isaac → INNOCENT, D4 · Pam → INNOCENT
Clyde’s clue says the fish and the sleuths must contain the same number of innocents. Right now the fish have no known innocents, while the sleuths already have one known innocent, Xia. If Isaac and Pam were both criminals, then the fish would have no innocent people at all, so Thor and Will would have to satisfy these clue facts in a way that is impossible. So Isaac at A3 and Pam at D4 must be innocent.
07.D1 · Daniel → INNOCENT
Pam’s clue says exactly one of the three coders has a criminal directly to the right. Gabe already fits that, because Hank is directly to Gabe’s right and Hank is criminal. Clyde also has someone directly to his right, and that person is Daniel, so if Daniel were criminal then Clyde would fit the clue too. That would make two coders fitting a clue that allows only one, so Daniel at D1 must be innocent.
08.C4 · Olivia → INNOCENT
Daniel’s clue says the number of innocents in row 4 who neighbor Will is odd. That group is exactly Nancy, Olivia, and Pam. Nancy and Pam are already known innocents, so there are already 2 innocents in that group. If Olivia were criminal, the total would stay 2, which is even, not odd. So Olivia must be innocent.
09.A1 · Anna → CRIMINAL
Bonnie’s neighbors must contain an odd number of criminals, and they already include 1 known criminal, Flora. So the unknown neighbors there, Anna, Erwin, and Gabe, must add an even number of criminals. Row 2 has exactly 3 criminals. Since Flora and Hank are already the 2 known criminals in that row, Erwin and Gabe must contain exactly 1 more criminal between them. That gives Bonnie’s unknown neighbors in row 2 exactly 1 criminal from Erwin and Gabe, which is odd, so Anna has to be the additional criminal to make the total added by Anna, Erwin, and Gabe even. So Anna must be criminal.
10.D3 · Logan → INNOCENT
Above Pam there must be more innocents than criminals, so that group needs at least 2 innocents and can have at most 1 criminal. In that group, there is already 1 known criminal, Hank, and the only person there whose status is not fixed yet is Logan. That means no remaining person above Pam can be a criminal. So Logan at D3 must be innocent.
11.B1 · Bonnie → INNOCENT
Column B has exactly 3 innocents, and exactly 2 of those innocents are Mary's neighbors. That means exactly 1 innocent in column B is not a neighbor of Mary. The people in column B who are not Mary's neighbors are Bonnie and Flora. Flora is already a criminal, so there are no known innocents in that pair yet, and the one required innocent there has to be the only unknown left. So Bonnie must be innocent.
12.A4 · Mary → INNOCENT
Row 1 has 1 criminal, and row 4 currently has 0. Bonnie's clue says row 1 must have more criminals than row 4. If Mary were a criminal, then row 4 would also have 1 criminal, so row 1 would not have more criminals than row 4. So Mary must be innocent.
13.C3 · Keith → INNOCENT
Mary’s clue says that the two criminals neighboring Janet are also neighbors of Bonnie. Among Janet’s neighbors, Keith is in the group that is not neighboring Bonnie, along with Isaac, Mary, Nancy, and Olivia. Since every criminal neighboring Janet has to be in Bonnie’s neighboring group, nobody outside that group can be a criminal. So Keith must be innocent.
14.A2 · Erwin → CRIMINAL
Column A already has Anna and Ruth as criminals, and Erwin is the only unknown left in that column. Keith’s clue says column A has more criminals than any other column. If Erwin were innocent, column A would stay at 2 criminals. But then the remaining people tied to these clues, namely Gabe, Janet, Thor, and Will, would have to satisfy that clue and Hank’s clue about the innocents below Bonnie at the same time, and that cannot be done. So Erwin at A2 must be criminal.
15.C2 · Gabe → INNOCENT
Nancy’s clue says Hank is one of exactly 3 criminals in row 2. In row 2, Erwin, Flora, and Hank are already the 3 known criminals, and the only other person there is Gabe. If Gabe were criminal too, row 2 would have more than 3 criminals, which conflicts with the clue. So Gabe must be innocent.
16.B3 · Janet → INNOCENT
Gabe’s clue says all innocents in row 3 must be connected in one orthogonally connected block. In row 3, Isaac is innocent, Keith is innocent, and Logan is innocent, while Janet is the only person there not yet identified. If Janet were criminal, then the innocents in that row would be Isaac on one side and Keith and Logan on the other, so they would not form one connected block. So Janet at B3 must be innocent.
17.B5 · Thor → CRIMINAL
Below Bonnie are Flora, Janet, Nancy, and Thor. Hank’s clue says there are exactly two innocents there, and those two innocents must be connected. Janet and Nancy are already innocents in that group. If Thor were also innocent, then the people below Bonnie would not fit the clue’s requirement of exactly two innocents there being connected. So Thor must be criminal.
18.C5 · Will → INNOCENT
Flora says the only criminal to the right of Ruth is to the left of Xia. To the right of Ruth, Thor is already a known criminal, Xia is a known innocent, and Will is the only person there whose identity was not yet fixed. If Will were also criminal, then there would be two criminals to the right of Ruth, Thor and Will. That contradicts Flora’s statement that there is only one criminal there. So Will must be innocent.