Clues by Sam Feb 02, 2026 Answer – Full Solution Explained
A1
👨🌾
farmer
B1
🕵️♀️
sleuth
C1
👩🌾
farmer
D1
👩✈️
pilot
A2
👨🌾
farmer
B2
🕵️♂️
sleuth
C2
👨✈️
pilot
D2
👨✈️
pilot
A3
🕵️♀️
sleuth
B3
👨🎨
painter
C3
👩⚕️
doctor
D3
💂♂️
guard
A4
👨🔧
mech
B4
👨🎨
painter
C4
👩⚕️
doctor
D4
💂♀️
guard
A5
👩🔧
mech
B5
💂♀️
guard
C5
👨🔧
mech
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 — 7 criminals.
Clues by Sam answer for Feb 02, 2026 — a Easy solved in 15 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 7 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Austin (A1), Eric (A2), Floyd (B2), Gary (C2), Uma (A5), Will (C5) and Zoe (D5); the remaining 13 suspects are innocent.
The deduction chain, in plain English
01.D1 · Dana → INNOCENT
Ivan’s clue says that Dana is one of Claire’s 3 innocent neighbors, which directly labels Dana as innocent. Therefore, we can determine that D1 Dana is INNOCENT.
02.B2 · Floyd → CRIMINAL, C2 · Gary → CRIMINAL
Eric is at A2, so the people to the right of Eric in the same row are Floyd at B2, Gary at C2, and Ivan at D2. Dana’s clue says there is only one innocent among those three people. Since Ivan at D2 is already known to be innocent, Floyd and Gary cannot also be innocent or the clue would be broken. Therefore, we can determine that B2 Floyd is CRIMINAL and C2 Gary is CRIMINAL.
03.B1 · Betty → INNOCENT
Claire is at C1, so her neighbors are Betty at B1, Floyd at B2, Gary at C2, Dana at D1, and Ivan at D2. Ivan’s clue says that Claire has exactly 3 innocent neighbors, and Dana is one of those innocents. We already know Dana is innocent and Ivan is also innocent, while Floyd and Gary are criminals, so the only remaining neighbor who can be the third innocent is Betty. Therefore, we can determine that B1 Betty is INNOCENT.
04.C3 · Lisa → INNOCENT, C4 · Susan → INNOCENT
Kumar is at B3, so his neighbors include the three people in column C that touch him: Gary at C2 (diagonal up-right), Lisa at C3 (directly right), and Susan at C4 (diagonal down-right). Betty’s clue says that exactly 2 innocents in column C are neighboring Kumar, so among those three column-C neighbors, exactly two must be innocent. Since Gary at C2 is already known to be a criminal, the only way to still have exactly two innocent column-C neighbors of Kumar is for Lisa at C3 and Susan at C4 to be the two innocents. Therefore, we can determine that C3 Lisa is INNOCENT and C4 Susan is INNOCENT.
05.D3 · Martin → INNOCENT
Lisa’s clue says that Martin is one of the 4 innocents in column D, which directly includes Martin among the innocents. Therefore, we can determine that D3 Martin is INNOCENT.
06.C5 · Will → CRIMINAL
Claire is at C1, so the people below her in column C are Gary at C2, Lisa at C3, Susan at C4, and Will at C5. Martin’s clue says “Both innocents below Claire are connected,” which means there are exactly two innocents somewhere below Claire. We already know Lisa (C3) and Susan (C4) are innocents, and they are connected because they are directly adjacent in the same column, so they must be the two innocents the clue refers to. That leaves no room for Will (C5) to also be innocent, so Will must be a criminal. Therefore, we can determine that C5 Will is CRIMINAL.
07.C1 · Claire → INNOCENT
In column C, we already know that Gary at C2 and Will at C5 are criminals, while Lisa at C3 and Susan at C4 are innocent, leaving only Claire at C1 unknown. Will’s clue says that column C is the only column with exactly 2 criminals, so column C must have exactly 2 criminals in total. Since those two criminals are already Gary and Will, Claire cannot be a criminal. Therefore, we can determine that C1 Claire is INNOCENT.
08.B4 · Rob → INNOCENT, B5 · Vera → INNOCENT
Will is at C5, and his five neighbors are Rob at B4, Susan at C4, Tina at D4, Vera at B5, and Zoe at D5. Claire’s clue says Will has exactly 4 innocent neighbors, so exactly one of those five neighbors is a criminal. Lisa’s clue says there are exactly 4 innocents in column D; since Dana at D1, Ivan at D2, and Martin at D3 are already innocent, exactly one of Tina at D4 and Zoe at D5 must be a criminal. That guarantees Will already has a criminal neighbor among Tina and Zoe, so Rob and Vera cannot be the criminal neighbor and must both be innocent. Therefore, we can determine that B4 Rob is INNOCENT and B5 Vera is INNOCENT.
09.B3 · Kumar → INNOCENT
The clue says that column C is the only column that has exactly 2 criminals. In column C we already have Gary at C2 and Will at C5 as criminals, while Claire, Lisa, and Susan are innocents, so column C does have exactly 2 criminals. In column B, the only criminal currently is Floyd at B2, and Betty, Rob, and Vera are already innocents, so Kumar at B3 is the only person who could make column B reach exactly 2 criminals. Since no other column is allowed to have exactly 2 criminals, Kumar cannot be a criminal. Therefore, we can determine that B3 Kumar is INNOCENT.
10.A1 · Austin → CRIMINAL
Look at column A: Austin is at A1, and the people below him are Eric (A2), Joy (A3), Ollie (A4), and Uma (A5). Rob’s clue says “Both innocents below Austin are connected,” which means exactly two of those four people are innocents, so the other two below Austin must be criminals. If Austin were innocent, then column A would contain exactly those two criminals and no others, so column A would have exactly 2 criminals. Will’s clue says column C is the only column with exactly 2 criminals, so column A cannot have exactly 2 criminals, which forces Austin not to be innocent. Therefore, we can determine that A1 Austin is CRIMINAL.
11.A3 · Joy → INNOCENT
Row 1 already has exactly one criminal, because Austin is a criminal and Betty, Claire, and Dana are all confirmed innocents. Austin’s clue says that only one row on the whole board has exactly one criminal, so no other row is allowed to end up with exactly one criminal. In row 3, Kumar, Lisa, and Martin are all confirmed innocents, so if Joy were a criminal then row 3 would have exactly one criminal, which is not allowed. Therefore, we can determine that A3 Joy is INNOCENT.
12.A5 · Uma → CRIMINAL
The clue talks about the people below Austin in column A, which are Eric at A2, Joy at A3, Ollie at A4, and Uma at A5. It says there are exactly two innocents among those four, and that those two innocents are connected using only up-down-left-right adjacency. Joy at A3 is already known to be innocent, so she must be one of those two innocents. For there to be only two innocents total and for them to be connected, the other innocent has to be directly above or directly below Joy, meaning it must be Eric at A2 or Ollie at A4. If Uma at A5 were the other innocent instead, she would not be connected to Joy unless Ollie at A4 were also innocent, which would create three innocents below Austin and break the clue. Therefore, we can determine that A5 Uma is CRIMINAL.
13.A4 · Ollie → INNOCENT, D4 · Tina → INNOCENT
Row 3 has Joy, Kumar, Lisa, and Martin, and all four of them are already known to be innocent, so row 3 contains 4 innocents. Joy’s clue says rows 3 and 4 have an equal number of innocents, so row 4 must also contain 4 innocents. In row 4, Rob and Susan are already innocent, so the only way for row 4 to reach 4 innocents is for Ollie at A4 and Tina at D4 to also be innocent. Therefore, we can determine that A4 Ollie is INNOCENT and D4 Tina is INNOCENT.
14.D5 · Zoe → CRIMINAL
In column D we have Dana at D1, Ivan at D2, Martin at D3, Tina at D4, and Zoe at D5. Lisa’s clue says that Martin is one of exactly four innocents in column D, meaning there are four innocents total in that column. Since Dana, Ivan, Martin, and Tina are already all confirmed innocents, those four slots fill the clue’s total and leave no room for Zoe to also be innocent. Therefore, we can determine that D5 Zoe is CRIMINAL.
15.A2 · Eric → CRIMINAL
Will’s clue says that column C is the only column that has exactly 2 criminals. In column C, we can already see exactly 2 criminals (Gary at C2 and Will at C5), with everyone else in that column innocent, so the clue’s “exactly 2” condition is satisfied there. That means no other column is allowed to end up with exactly 2 criminals. In column A, we already have two known criminals (Austin at A1 and Uma at A5), so if Eric at A2 were innocent then column A would also have exactly 2 criminals, which is not allowed; Eric must be the third criminal in column A. Therefore, we can determine that A2 Eric is CRIMINAL.