Clues by Sam Apr 25, 2026 Answer – Full Solution Explained
A1
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astronaut
B1
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astronaut
C1
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astronaut
D1
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astronaut
A2
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astronaut
B2
👨🚀
astronaut
C2
👩🚀
astronaut
D2
👨🚀
astronaut
A3
👩🚀
astronaut
B3
👨🚀
astronaut
C3
👨🚀
astronaut
D3
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astronaut
A4
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astronaut
B4
👩🚀
astronaut
C4
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astronaut
D4
👨🚀
astronaut
A5
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astronaut
B5
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astronaut
C5
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astronaut
D5
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astronaut
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 Apr 25, 2026 — a Hard solved in 15 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Andre (A1), Helen (C2), Ruth (B4), Saga (C4), Vera (B5) and Xia (D5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.D5 · Xia → CRIMINAL
Orion’s clue says that Xia is one of the exactly 2 criminals in row 5. Since Xia is explicitly named as one of those criminals, no extra deduction about the rest of the row is needed. So Xia must be criminal.
02.D1 · Dave → INNOCENT, D2 · Isaac → INNOCENT
Xia’s clue says the people above Nick contain exactly 2 innocents. Right now there are 0 known innocents in that group, and the only people left there are Dave and Isaac. Since those two spots must supply all 2 innocents required by the clue, both of them have to be innocent. So Dave and Isaac must be innocent.
03.C3 · Luigi → INNOCENT
Dave’s clue says that Luigi is one of the exactly 2 innocents below Chloe. That directly identifies Luigi as an innocent. So Luigi must be innocent.
04.C4 · Saga → CRIMINAL, A5 · Umar → INNOCENT
Row 5 must contain exactly 2 criminals, and Xia is already one of them. The people below Chloe must contain exactly 2 innocents, and Luigi is already one of them. If Saga were innocent while Umar were criminal, then the remaining people involved here, Helen, Vera, and Wanda, would have to satisfy those clue requirements at the same time, but they cannot. So that opposite pairing is ruled out. That makes Saga criminal and Umar innocent.
05.B4 · Ruth → CRIMINAL, B5 · Vera → CRIMINAL, C5 · Wanda → INNOCENT
Row 4 must contain exactly one innocent who has an innocent directly below them, and A4 Orion is the only known innocent in row 4. If Ruth were innocent, Vera were innocent, and Wanda were criminal, then Ruth, Vera, and Wanda together with Tom would have to satisfy that row 4 requirement and the other two clues at the same time, but they cannot. So that tested combination is impossible. This leaves the opposite identities for those three people. Ruth, Vera, and Wanda must be Ruth criminal, Vera criminal, and Wanda innocent.
06.C2 · Helen → CRIMINAL
Dave’s clue says Luigi is one of exactly 2 innocents below Chloe. Below Chloe are Helen, Luigi, Saga, and Wanda, and among them Luigi and Wanda are already innocent while Saga is criminal. If Helen were innocent too, there would be 3 innocents below Chloe, which conflicts with the clue’s exact total of 2. So Helen must be criminal.
07.A2 · Franco → INNOCENT
Ruth’s clue says Franco is one of the exactly 11 innocents on the edges. That directly identifies Franco as an innocent. So Franco must be innocent.
08.B1 · Betsy → INNOCENT
The corner clue says exactly one corner innocent has a criminal directly to the right. Umar already gives that one counted case, because Umar is innocent and Vera is directly to Umar's right and is criminal. The only other direct-right neighbor that could still affect this clue is Betsy, to the right of Andre. If Betsy were criminal, that would create another counted corner case and make the total more than one. So Betsy must be innocent.
09.C1 · Chloe → INNOCENT
Ruth's clue says there are exactly 11 innocents on the edges. Among edge cells, 7 are already known innocents and 2 are already known criminals, so the 5 unknown edge people must contain exactly 4 innocents and 1 criminal. That one edge criminal is accounted for by Andre, Joyce, Nick, and Tom, so Chloe is not in the set that can supply the single remaining edge criminal. Therefore Chloe at C1 must be innocent.
10.D4 · Tom → INNOCENT
Ruth's clue says there are exactly 11 innocents on the edge. Since 8 edge people are already known to be innocent and there are 4 unknown edge people, that leaves room for exactly 1 criminal among Andre, Joyce, Nick, and Tom. Chloe's clue accounts for that one edge criminal among Andre, Joyce, and Nick, so Tom cannot be the criminal in that edge group. That makes Tom innocent.
11.D3 · Nick → INNOCENT
Ruth’s clue says there are exactly 11 innocents on the edge. Among the edge cells, there are already 9 known innocents and 2 known criminals, leaving A1 Andre, A3 Joyce, and D3 Nick as the only unknown edge people. That means exactly one of those three unknown edge people can be criminal. The one criminal among them must be Andre or Joyce, so Nick cannot be that criminal. So Nick must be innocent.
12.B3 · Kumar → INNOCENT
Ruth's clue says the edge cells contain exactly 11 innocents. There are already 10 known innocents on the edge, so the only remaining edge innocents must be Andre and Joyce. Franco's clue says Franco and Ghani have exactly 3 innocent neighbors in common. In their shared neighbor group, Betsy is already innocent, and Andre and Joyce are the other two innocents fixed by the edge clue. That makes the three innocent people in the smaller shared part Andre, Betsy, and Joyce, so the remaining person in the larger shared group, Kumar, must be innocent. So Kumar must be innocent.
13.B2 · Ghani → INNOCENT
Ruth's clue says Franco is one of exactly 11 innocents on the edges. There are already 10 known innocents on the edge, and the only unknown edge people are Andre and Joyce, so those two edge places have to account for the last edge innocent. Nick's clue says Franco's neighbors contain an odd number of criminals. Among Franco's neighbors there are no known criminals, and the unknown neighbors are Andre, Ghani, and Joyce. If Ghani were a criminal, then Andre and Joyce would still have to fit the edge-innocent requirement at the same time, and that clashes with the odd-criminal requirement for Franco's neighborhood. So Ghani must be innocent.
14.A1 · Andre → CRIMINAL
Ghani’s clue says rows 1 and 2 must have the same number of innocents. Row 2 already has 3 innocents, and row 1 also already has 3 known innocents. If Andre were innocent too, row 1 would rise to 4 innocents while row 2 would stay at 3, so the clue would fail. So Andre must be criminal.
15.A3 · Joyce → INNOCENT
Ruth’s clue says Franco is one of exactly 11 innocents on the edges. Among the edge cells, 10 people are already known to be innocent, and Joyce is the only edge person whose status is still unknown. If Joyce were criminal, the edge would have only those 10 innocents, not 11. So Joyce must be innocent.