Clues by Sam Jul 26, 2026 Answer – Full Solution Explained
A1
👷♀️
builder
B1
💂♀️
guard
C1
👨🌾
farmer
D1
👩⚕️
doctor
A2
🕵️♂️
sleuth
B2
🕵️♂️
sleuth
C2
👨🌾
farmer
D2
👨⚕️
doctor
A3
👨🌾
farmer
B3
🕵️♀️
sleuth
C3
👨⚕️
doctor
D3
👩🍳
cook
A4
👮♂️
cop
B4
👨🍳
cook
C4
💂♂️
guard
D4
💂♀️
guard
A5
👮♀️
cop
B5
👮♀️
cop
C5
👷♀️
builder
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 — 10 criminals.
Clues by Sam answer for Jul 26, 2026 — a Hard solved in 18 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 10 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Clyde (C1), Eli (A2), Henry (B2), Klay (D2), Logan (A3), Maria (B3), Nick (C3), Raul (B4), Xena (C5) and Zara (D5); the remaining 10 suspects are innocent.
The deduction chain, in plain English
01.B1 · Barb → INNOCENT, D1 · Dana → INNOCENT
Tina's clue says exactly 2 of the people who are both Clyde's neighbors and in row 1 are innocent. That shared group is only Barb and Dana, and none of those 2 is already known innocent. Since the group still needs 2 innocents and Barb and Dana are the only people there, both of them have to fill those 2 innocent spots. So Barb and Dana must be innocent.
02.B2 · Henry → CRIMINAL
Barb’s clue says exactly one innocent is in the overlap of the people to the right of Eli and Olive’s neighbors, and that overlap is only Ivan and Klay. So among Ivan and Klay, exactly one is innocent. Dana’s clue says an odd number of the people in row 2 who neighbor Clyde are innocent, and that group is Henry, Ivan, and Klay. If Henry were innocent, then Henry plus exactly one of Ivan and Klay would make two innocents in that group, which is even, not odd. So Henry must be criminal.
03.A2 · Eli → CRIMINAL
Barb’s clue says exactly one innocent is both to the right of Eli and neighboring Olive, and that shared group is only Ivan and Klay. Henry’s clue says row 5 is the only row with exactly 2 innocents, so the other rows, including row 2, cannot finish with exactly 2 innocents. If Eli were innocent, then row 2 would have Eli plus exactly one innocent among Ivan and Klay, because Barb’s clue allows exactly one innocent in that Ivan-Klay pair. That would make row 2 have exactly 2 innocents, which clashes with Henry’s clue that only row 5 has exactly 2 innocents. So Eli must be criminal.
04.C4 · Scott → INNOCENT
Eli's clue says Maria has exactly 5 criminal neighbors, and exactly 1 of those criminals is also a neighbor of Uma. Among Maria's neighbors, the only people who are also neighbors of Uma are Phil and Raul, so the other 4 criminal neighbors of Maria must be people who are not neighbors of Uma. Henry and Eli are already 2 known criminals among Maria's neighbors, and neither of them is a neighbor of Uma. That leaves exactly 2 more criminals to come from the remaining Maria-neighbors who are not neighbors of Uma: Ivan, Logan, Nick, and Scott. So Scott cannot be one of those additional criminals. That makes Scott innocent.
05.B3 · Maria → CRIMINAL
Scott’s clue says that Maria is one of Ivan’s 5 criminal neighbors. Since the clue directly places Maria among Ivan’s criminal neighbors, her identity is fixed by that clue. So Maria must be criminal.
06.C3 · Nick → CRIMINAL
Maria has exactly 5 criminal neighbors, and Eli says only 1 of those 5 criminals is also a neighbor of Uma. In the part of Maria's neighborhood that is not neighboring Uma, there is exactly 1 innocent among Ivan, Logan, and Nick. That 1 innocent has to be Ivan or Logan, so Nick cannot be the innocent one there. So Nick must be criminal.
07.B5 · Wanda → INNOCENT
Nick’s clue directly says that Wanda is one of the 8 innocents on the edges. Since Wanda is explicitly included in that innocent edge group, Wanda at B5 must be innocent.
08.A1 · Alice → INNOCENT
Nick says there are exactly 8 innocents on the edge cells, so the edge cells contain exactly 5 criminals in total. Eli is already one of those edge criminals, which means the other 5 edge criminals must come from Clyde, Klay, Logan, Olive, Phil, Uma, Xena, and Zara. That leaves no room for Alice to be one of the edge criminals. So Alice must be innocent.
09.C1 · Clyde → CRIMINAL
Henry must have an odd number of innocent neighbors. He already has 2 known innocents there, so among Clyde, Ivan, and Logan, an odd number must also be innocent. Now test Clyde as innocent. Then the same remaining people involved in these clues, Ivan, Logan, Phil, and Raul, would also have to satisfy Eli's statement that Maria has exactly 5 criminal neighbors and exactly 1 of those criminals is a neighbor of Uma, but that combination cannot be made to fit. So Clyde cannot be innocent. That makes Clyde criminal.
10.C5 · Xena → CRIMINAL
Nick's clue says there are exactly 8 innocents on the edge. Since 5 edge innocents are already known, the 7 unknown edge people contain exactly 3 more innocents. Clyde's clue says columns A and D have the same number of innocents. Column A currently has 1 known innocent and column D has 2 known innocents, so the extra edge innocents have to come from A3 Logan, A4 Phil, A5 Uma, D2 Klay, D3 Olive, and D5 Zara. That leaves C5 Xena outside the only places those 3 remaining edge innocents can be, so Xena must be criminal.
11.D5 · Zara → CRIMINAL
Nick's clue fixes the edge cells at exactly 8 innocents. There are already 5 known innocents on the edge, so the remaining 3 edge innocents must come from the six unknown edge people. From the people considered for those 3 remaining edge innocents, the needed innocents are accounted for by D2 Klay, A3 Logan, D3 Olive, A4 Phil, and A5 Uma, so Zara is not among that set of 3 edge innocents. That makes Zara at D5 criminal.
12.B4 · Raul → CRIMINAL
Scott’s clue fixes Ivan’s neighborhood at exactly 5 criminals. In Ivan’s neighbors, there are already 4 known criminals, so among the two unknown neighbors, Klay and Olive, exactly one of them must be criminal. Alice’s clue says column B has more criminals than column D. Right now column B has 2 known criminals, while column D has 1, and the only unknown in column B is Raul. If Raul were innocent, then column B would stay at 2 criminals, while column D could also reach 2 because Klay and Olive are the remaining people involved here and one of them must be criminal. That would not allow column B to have more criminals than column D, so Raul cannot be innocent. So Raul must be criminal.
13.A5 · Uma → INNOCENT
Henry’s clue says row 5 is the only row with exactly 2 innocents. In row 5, Wanda is already innocent and Xena and Zara are already criminals, so if Uma were criminal then row 5 would have only 1 innocent, not 2. That clashes with the clue’s requirement for row 5. So Uma must be innocent.
14.A4 · Phil → INNOCENT
Henry’s clue says row 5 is the only row with exactly 2 innocents. Row 4 already has exactly 2 known innocents, Scott and Tina, and Phil is the only person left in that row whose identity is not yet known. If Phil were criminal, then row 4 would remain on exactly 2 innocents, which the clue does not allow for any row except row 5. So Phil must be innocent.
15.A3 · Logan → CRIMINAL
Ivan’s neighbors must contain exactly 5 criminals, and four of them are already known: Clyde, Henry, Maria, and Nick. So among the two unknown neighbors, Klay and Olive, exactly one is criminal. The edge cells must contain exactly 8 innocents, so with 14 edge cells that means exactly 6 criminals on the edge. Five edge criminals are already accounted for by Clyde, Eli, Xena, Zara, and the one criminal among Klay and Olive. The only other edge person left to make that total is Logan. So Logan must be criminal.
16.C2 · Ivan → INNOCENT
Maria’s neighbors contain exactly 5 criminals in total. Eli’s clue says exactly 1 of those criminals is also a neighbor of Uma, and among the people who are both Maria’s neighbors and Uma’s neighbors, the only criminal is Raul. That leaves the rest of Maria’s neighbors outside Uma’s neighborhood, and that group already includes 4 known criminals: Eli, Henry, Logan, and Nick. Since those 4 already fill all the remaining criminal spots there, Ivan cannot be a criminal. So Ivan must be innocent.
17.D2 · Klay → CRIMINAL
Barb's clue says there is exactly 1 innocent in the overlap between the people to the right of Eli and Olive's neighbors. That shared group is only Ivan and Klay. Ivan is already known to be innocent, so the one innocent required by the clue is already accounted for. Klay therefore cannot also be innocent, so Klay must be criminal.
18.D3 · Olive → INNOCENT
Scott’s clue says Maria is one of Ivan’s exactly 5 criminal neighbors. Among Ivan’s neighbors, the five already known criminals are Clyde, Henry, Klay, Maria, and Nick, and Olive is the only neighbor there whose identity was not yet known. If Olive were criminal too, Ivan would have 6 criminal neighbors instead of 5, which conflicts with the clue. So Olive must be innocent.