Clues by Sam Mar 28, 2026 Answer – Full Solution Explained
A1
👩🔬
scientist
B1
👨💻
coder
C1
💂♂️
guard
D1
👨💼
clerk
A2
💂♂️
guard
B2
👨💻
coder
C2
👮♀️
cop
D2
👮♂️
cop
A3
🕵️♀️
sleuth
B3
👩💻
coder
C3
💂♀️
guard
D3
👮♀️
cop
A4
👨💼
clerk
B4
👨🎨
painter
C4
👨🎨
painter
D4
👷♀️
builder
A5
🕵️♀️
sleuth
B5
👩🎨
painter
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 — 8 criminals.
Clues by Sam answer for Mar 28, 2026 — a Hard solved in 15 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 8 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Erwin (A2), Hilda (C2), Ike (D2), Julie (A3), Nicole (D3), Oscar (A4), Vera (A5) and Xavi (C5); the remaining 12 suspects are innocent.
The deduction chain, in plain English
01.B1 · Barnie → INNOCENT, C1 · Chuck → INNOCENT
Alice is at A1 and Donald is at D1, so the only people in between them in that row are Barnie at B1 and Chuck at C1. Raul’s clue says there are exactly 2 innocents in between Alice and Donald, so both Barnie and Chuck must be innocents. Therefore, we can determine that B1 Barnie is INNOCENT and C1 Chuck is INNOCENT.
02.B3 · Kay → INNOCENT
To the right of Julie in row 3 are exactly three people: Kay at B3, Max at C3, and Nicole at D3, and Barnie says an odd number of those three are criminals. Chuck says Hilda and Tina have only one innocent neighbor in common; the only people who are neighbors of both Hilda (C2) and Tina (D4) are Max (C3) and Nicole (D3), so exactly one of Max and Nicole is innocent and the other is a criminal. That means Max and Nicole contribute exactly one criminal among the three to the right of Julie, so for the total number of criminals there to be odd, Kay must be innocent. Therefore, we can determine that B3 Kay is INNOCENT.
03.A3 · Julie → CRIMINAL
Hilda at C2 and Tina at D4 share exactly two common neighbors: Max at C3 and Nicole at D3. Chuck’s clue says Hilda and Tina have only one innocent neighbor in common, so exactly one of Max and Nicole is innocent. That means in row 3, besides Kay (who is already innocent), there is exactly one more innocent among Max and Nicole, so row 3 already has exactly 2 innocents without counting Julie. Kay’s clue says row 4 is the only row with exactly 3 innocents, so row 3 cannot have exactly 3 innocents; therefore Julie cannot be innocent and must be criminal. Therefore, we can determine that A3 Julie is CRIMINAL.
04.A1 · Alice → INNOCENT, D1 · Donald → INNOCENT
Hilda at C2 and Tina at D4 share exactly two common neighbors: Max at C3 and Nicole at D3. Chuck’s clue says Hilda and Tina have only one innocent neighbor in common, so exactly one of Max and Nicole is innocent. Since Kay at B3 is already innocent and Julie at A3 is criminal, that fixes row 3 to have exactly 2 innocents in total. Julie’s clue says row 1 has more innocents than row 3, so row 1 must have at least 3 innocents; but Kay’s clue says row 4 is the only row with exactly 3 innocents, so row 1 cannot have exactly 3 and must have 4. That means both unknowns in row 1, Alice at A1 and Donald at D1, are innocent. Therefore, we can determine that A1 Alice is INNOCENT and D1 Donald is INNOCENT.
05.D4 · Tina → INNOCENT, A5 · Vera → CRIMINAL, C5 · Xavi → CRIMINAL
Wanda at B5 is neighbors with Oscar (A4), Phil (B4), Raul (C4), Vera (A5), and Xavi (C5). Donald’s clue says Wanda has exactly 3 criminal neighbors, and since Raul is already known to be innocent, that means exactly 3 of Oscar, Phil, Vera, and Xavi are criminals. Kay’s clue says row 4 has exactly 3 innocents, so row 4 contains only 1 criminal total, meaning Oscar and Phil cannot both be criminals. Since Oscar and Phil can contribute at most 1 criminal but we need 3 criminals among Oscar, Phil, Vera, and Xavi, Vera and Xavi must both be criminals, and exactly one of Oscar or Phil is the third criminal. With row 4 having only 1 criminal total and that 1 criminal already coming from Oscar or Phil, Tina at D4 must be innocent to keep row 4 at exactly 3 innocents. Therefore, we can determine that D4 Tina is INNOCENT, A5 Vera is CRIMINAL, and C5 Xavi is CRIMINAL.
06.B2 · Gary → INNOCENT
The clue from Vera says that Kay has exactly 4 innocent neighbors. Kay already has one known innocent neighbor, Raul, so among Kay’s other neighbors who are still uncertain (Erwin, Gary, Hilda, Max, Oscar, and Phil), exactly 3 must be innocents. Tina’s clue says Gary has exactly 5 innocent neighbors; since four of Gary’s neighbors are already known innocents (Alice, Barnie, Chuck, and Kay), this forces exactly one of Gary’s uncertain neighbors (Erwin, Hilda, and Max) to be innocent. Kay’s clue about row 4 means row 4 has exactly 3 innocents, and since Raul and Tina are already innocent there, exactly one of Oscar and Phil is innocent, so among (Oscar and Phil) there is exactly one innocent as well. That accounts for exactly two innocents among (Erwin, Hilda, Max, Oscar, Phil), so to reach the required three innocents among (Erwin, Gary, Hilda, Max, Oscar, Phil), Gary must be the remaining innocent. Therefore, we can determine that B2 Gary is INNOCENT.
07.B5 · Wanda → INNOCENT
Gary’s clue says, as a fact, that Wanda is one of the 7 innocents on the edges. Since the clue directly includes Wanda in the edge-innocents group, Wanda must be INNOCENT. Therefore, we can determine that B5 Wanda is INNOCENT.
08.A2 · Erwin → CRIMINAL
Around Gary at B2, the already-known innocent neighbors are Alice at A1, Barnie at B1, Chuck at C1, and Kay at B3, which makes 4 innocent neighbors so far. Tina says Gary has exactly 5 innocent neighbors, so among Erwin at A2, Hilda at C2, and Max at C3 there is exactly 1 innocent and the other 2 are criminals. Barnie says there’s an odd number of criminals to the right of Julie in row 3, and since Kay at B3 is innocent, that means exactly one of Max at C3 and Nicole at D3 is a criminal. Gary also says Wanda is one of 7 innocents on the edges, and we already have 6 edge innocents (A1, B1, C1, D1, D4, and B5), so among the remaining edge unknowns A2, D2, A4, D3, and D5 there is exactly 1 innocent. If D3 is that one remaining edge innocent then A2 cannot be innocent, and if D3 is not that edge innocent then D3 is a criminal, which makes C3 the innocent one, and then A2 cannot be the innocent among A2/C2/C3, so A2 is still a criminal. Therefore, we can determine that A2 Erwin is CRIMINAL.
09.A4 · Oscar → CRIMINAL
Gary at B2 has eight neighbors: A1, B1, C1, A2, C2, A3, B3, and C3, and Tina says exactly five of those neighbors are innocent. Since A1, B1, C1, and B3 are already innocent while A2 and A3 are already criminal, exactly one of C2 and C3 must be criminal. That means column C already has exactly two criminals (Xavi at C5 plus the one criminal among C2 and C3), so Wanda’s clue that columns C and D have an equal number of criminals forces column D to also have exactly two criminals among D2, D3, and D5. Gary’s clue says there are exactly seven innocents on the edges, and we already have six edge innocents (A1, B1, C1, D1, D4, and Wanda at B5), so among the remaining edge unknowns D2, D3, D5, and A4 there is exactly one more innocent; but since D2, D3, and D5 can contain only one innocent, A4 cannot be that innocent and must be criminal. Therefore, we can determine that A4 Oscar is CRIMINAL.
10.B4 · Phil → INNOCENT
Row 4 contains Oscar at A4, Phil at B4, Raul at C4, and Tina at D4. Kay’s clue says that row 4 is the only row with exactly 3 innocents, so row 4 must have exactly 3 innocents in it. In row 4 we already know Raul and Tina are innocents, and Oscar is a criminal, so the only way for the row to reach exactly 3 innocents is for Phil to be innocent as well. Therefore, we can determine that B4 Phil is INNOCENT.
11.D2 · Ike → CRIMINAL
Gary is at B2, and his eight neighbors are Alice, Barnie, Chuck, Erwin, Hilda, Julie, Kay, and Max. Tina’s clue says Gary has exactly 5 innocent neighbors; since Alice, Barnie, Chuck, and Kay are already innocent (4) and Erwin and Julie are criminals, exactly one of Hilda and Max must be innocent. Nicole is at D3, and her neighbors are Hilda, Ike, Max, Raul, and Tina; Oscar’s clue says the number of innocents among these neighbors is odd, and Raul and Tina already contribute 2 innocents, so the number of innocents among Hilda, Ike, and Max must also be odd. Because Hilda and Max contribute exactly 1 innocent between them, Ike must contribute 0 innocents, so Ike must be criminal. Therefore, we can determine that D2 Ike is CRIMINAL.
12.D5 · Zara → INNOCENT
The only people who are neighbors of both Hilda at C2 and Tina at D4 are Max at C3 and Nicole at D3. Chuck’s clue says Hilda and Tina have only one innocent neighbor in common, so exactly one of Max and Nicole is innocent. Ike’s clue says Tina has an odd number of innocent neighbors, and Tina’s neighbors are Max, Nicole, Raul, Xavi, and Zara; among these, Raul is already innocent and Xavi is already criminal. If Tina had only one innocent neighbor, then Max, Nicole, and Zara would all have to be criminal, but that would give Hilda and Tina zero innocent neighbors in common, not one, so Tina must instead have three innocent neighbors. That means two of Max, Nicole, and Zara are innocent, and since only one of Max and Nicole can be innocent, Zara must be innocent to make the second one. Therefore, we can determine that D5 Zara is INNOCENT.
13.D3 · Nicole → CRIMINAL
The “edges” are the 14 outer positions, and Gary’s clue says there are exactly 7 innocents on those edges, with Wanda being one of them. Right now, we can already count 7 edge-innocents: all four people in row 1 (A1, B1, C1, D1), plus Wanda at B5 and Zara at D5, plus Tina at D4. Since the clue allows only 7 edge-innocents in total, every other edge position must be criminal; the only edge person still unknown is Nicole at D3. Therefore, we can determine that D3 Nicole is CRIMINAL.
14.C3 · Max → INNOCENT
Julie is at A3, so the people to the right of Julie are B3 Kay, C3 Max, and D3 Nicole. Barnie’s clue says the number of criminals among those three people is odd. Kay is already INNOCENT and Nicole is already a CRIMINAL, so we currently have exactly one criminal to the right of Julie. If Max were a CRIMINAL too, that would make two criminals there, which would be even, so Max must be INNOCENT. Therefore, we can determine that C3 Max is INNOCENT.
15.C2 · Hilda → CRIMINAL
Gary is at B2, so his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Tina’s clue says Gary has exactly 5 innocent neighbors. Among those neighbors, A1, B1, C1, B3, and C3 are already known to be innocent, which is exactly 5, while A2 and A3 are already known criminals. That means C2 cannot be innocent, or Gary would have 6 innocent neighbors instead of 5. Therefore, we can determine that C2 (Hilda) is CRIMINAL.