Clues by Sam Jun 13, 2026 Answer – Full Solution Explained
A1
💂♀️
guard
B1
👩🍳
cook
C1
👩⚕️
doctor
D1
👩⚕️
doctor
A2
👷♀️
builder
B2
👨💼
clerk
C2
👨🍳
cook
D2
👨🍳
cook
A3
👨💼
clerk
B3
👨💼
clerk
C3
👷♀️
builder
D3
👨🌾
farmer
A4
💂♀️
guard
B4
👩🎨
painter
C4
👨✈️
pilot
D4
👩🌾
farmer
A5
👩✈️
pilot
B5
👨🌾
farmer
C5
👨✈️
pilot
D5
👨🎨
painter
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 — 11 criminals.
Clues by Sam answer for Jun 13, 2026 — a Hard solved in 18 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 11 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Alice (A1), Claire (B1), Hank (B2), Ike (C2), Jerry (D2), Kevin (A3), Logan (B3), Oscar (D3), Rob (C4), Stella (D4) and Xavi (C5); the remaining 9 suspects are innocent.
The deduction chain, in plain English
01.D4 · Stella → CRIMINAL, D2 · Jerry → CRIMINAL
Will’s clue says Oscar has exactly 4 criminal neighbors, and exactly 2 of those criminals are in column C. That means the other 2 criminal neighbors of Oscar must be not in column C. Among Oscar’s neighbors, the only people not in column C are Jerry and Stella, and there are no known criminals there yet. So those two have to fill the 2 criminal spots outside column C. That makes Jerry and Stella criminals.
02.C1 · Debra → INNOCENT
Oscar’s neighbors need exactly 2 criminals in column C, and the only neighboring people in column C are Ike, Mary, and Rob. So those two column C criminals must come from Ike, Mary, and Rob. The people above Xavi also contain exactly 2 criminals, and that larger group is Debra, Ike, Mary, and Rob. Since Ike, Mary, and Rob already account for the 2 criminals allowed above Xavi, the only remaining person in that group cannot be a criminal. So Debra must be innocent.
03.D5 · Ziad → INNOCENT
Debra’s clue says that Ziad is one of the exactly 2 innocents in column D. Since the clue directly names Ziad as one of those innocents, Ziad must be innocent.
04.C4 · Rob → CRIMINAL
Oscar’s neighbors contain exactly 4 criminals, and Jerry and Stella are already 2 of them. Will’s clue also says exactly 2 of Oscar’s criminal neighbors are in column C, so among Ike, Mary, and Rob in column C, exactly 2 are criminals and therefore exactly 1 is innocent. That one innocent in Oscar’s neighboring group must be Ike or Mary, not Rob. So Rob at C4 must be criminal.
05.B2 · Hank → CRIMINAL
Rob’s clue says that among Ike’s neighbors, exactly one criminal is also a neighbor of Claire. The only Ike-neighbors who are also Claire-neighbors are Debra and Hank, and Debra is innocent. So that one required criminal in that shared pair cannot be Debra and has to be Hank. That makes Hank criminal.
06.B1 · Claire → CRIMINAL
Ike’s neighbors contain exactly 5 criminals, so among his 8 neighbors there must be exactly 2 innocents. Debra is already one known innocent there, which means the other 2 innocents in that neighbor group are the unknown innocents named in the step. Those 2 innocents must come from D1 Esha, B3 Logan, C3 Mary, and D3 Oscar, so Claire is not one of them. That makes Claire criminal.
07.A3 · Kevin → CRIMINAL
Claire’s clue says Kevin is one of Quita’s 4 criminal neighbors. That directly identifies Kevin as one of the criminals in that neighbor group. So Kevin must be criminal.
08.B4 · Quita → INNOCENT
Kevin’s clue says that Quita is one of his 3 innocent neighbors. That directly identifies Quita as innocent. So Quita must be innocent.
09.A4 · Penny → INNOCENT
Quita has exactly 4 criminal neighbors, and Kevin and Rob are already two of them. That means the remaining 2 criminal neighbors must be chosen from the unknown neighbors. But Logan and Mary are the only people between Kevin and Oscar, and that group must contain exactly 1 innocent, so among Logan and Mary there is exactly 1 criminal. With one of the two remaining criminal-neighbor spots already used by Logan or Mary, the other remaining spot must come from Tina or Xavi. That leaves no criminal-neighbor spot for Penny. So Penny must be innocent.
10.A1 · Alice → CRIMINAL
Claire has exactly 2 innocent neighbors. Among Claire's neighbors, Debra is already one innocent, so there is room for only 1 more innocent among Alice, Flora, and Ike. That remaining innocent cannot be Alice, so the extra innocent must come from Flora and Ike instead. This leaves Alice as criminal.
11.D3 · Oscar → CRIMINAL
Oscar’s clue says his neighbors contain exactly 4 criminals, and exactly 2 of those criminals are in column C. Kevin’s clue fixes Quita as one of the 3 innocents among Kevin’s neighbors, and Alice’s clue says rows 2 and 3 must have the same number of criminals; right now row 2 has 2 known criminals while row 3 has 1. If Oscar were innocent, then A2 Flora, C2 Ike, B3 Logan, and C3 Mary would have to make all of those requirements true at the same time, but they cannot. So Oscar cannot be innocent. So Oscar must be criminal.
12.D1 · Esha → INNOCENT
Debra’s clue says Ziad is one of exactly 2 innocents in column D. In that column, Jerry, Oscar, and Stella are already criminals, and Ziad is already the one known innocent, so Esha is the only person there not yet identified. If Esha were a criminal, column D would still have only 1 innocent, which conflicts with the clue that there are exactly 2. So Esha must be innocent.
13.C3 · Mary → INNOCENT
Quita has exactly 4 criminal neighbors, and Kevin and Rob are already two of them. That means the other two criminal neighbors must be chosen from the four unknown neighbors Logan, Mary, Tina, and Xavi. But the two criminals among those four are accounted for by Logan, Tina, and Xavi. So Mary cannot be one of the two criminal spots in Quita's neighborhood. That makes Mary innocent.
14.C2 · Ike → CRIMINAL
Will’s clue says Oscar has exactly 4 criminal neighbors, and exactly 2 of those criminal neighbors are in column C. Among Oscar’s neighbors in column C, Rob is already a known criminal and Mary is innocent. So that group still needs 1 more criminal, and the only unknown person left there is Ike. That makes Ike criminal.
15.B3 · Logan → CRIMINAL
Ziad’s clue says the people between Kevin and Oscar contain exactly one innocent. Among those people, Mary is already known to be innocent, so that one innocent is already accounted for. The only other person there whose identity was not yet known is Logan, so Logan cannot also be innocent. So Logan must be criminal.
16.A2 · Flora → INNOCENT
Kevin says Quita is one of his 3 innocent neighbors. Among Kevin's neighbors, Penny and Quita are already innocent, Hank and Logan are criminal, and Flora is the only neighbor not yet identified. If Flora were criminal, Kevin would have only those 2 innocent neighbors, Penny and Quita, instead of 3. So Flora must be innocent.
17.A5 · Tina → INNOCENT
Mary’s clue says the corner cells contain exactly 3 innocents. In the corners, 2 are already known to be innocent, and the only corner person not yet identified is Tina at A5. So the corners still need exactly 1 more innocent, and Tina is the only person who can fill that spot. That makes Tina innocent.
18.C5 · Xavi → CRIMINAL
Claire’s clue says Kevin is one of Quita’s exactly 4 criminal neighbors. Among Quita’s neighbors, Kevin, Logan, and Rob are already known criminals, while Mary, Penny, Tina, and Will are known innocents. That gives Quita only 3 known criminal neighbors so far, and the only neighbor not yet identified is Xavi. To reach the required total of 4 criminal neighbors, Xavi at C5 must be criminal.