Clues by Sam Apr 01, 2026 Answer – Full Solution Explained
Tricky·Solved
A1
👨🍳
Austin
cook
B1
👩💻
Claire
coder
C1
👩💻
Debra
coder
D1
👩✈️
Emily
pilot
A2
👨🍳
Frank
cook
B2
👨🍳
Gary
cook
C2
👨💻
Henry
coder
D2
👨✈️
John
pilot
A3
👮♂️
Kumar
cop
B3
👩🏫
Lisa
teacher
C3
👩🎨
Megan
painter
D3
👩🎨
Pam
painter
A4
👮♀️
Ruth
cop
B4
👨🔧
Steve
mech
C4
👨🔧
Thor
mech
D4
👩🎤
Uma
singer
A5
👨⚖️
Vince
judge
B5
👩⚖️
Wanda
judge
C5
👩🏫
Xena
teacher
D5
👨🎤
Zed
singer
Final Board State
This puzzle is fully solved.
All characters have been identified as innocent or criminal based on today's clues.
Final Result
Innocent 19Criminal 1Unknown 0
See how each clue leads to the final result
Answer (spoilers)
A quick reference of the final identities. For explanations, see the reasoning above.
▶ Answer list (spoilers)
Innocent · 19
[ A1 ] [ B1 ] [ C1 ] [ D1 ] [ A2 ] [ B2 ] [ C2 ] [ D2 ] [ A3 ] [ B3 ] [ C3 ] [ D3 ] [ A4 ] [ B4 ] [ C4 ] [ D4 ] [ B5 ] [ C5 ] [ D5 ]
Criminal · 1
[ A5 ]
Clues
Raw text reference from the original puzzle
Original clue texts as provided in today's puzzle. No deductions or interpretations are applied here.
▶ Raw clues (original text)
A1 · Austin
"There's an odd number of innocents in column B"
B1 · Claire
"There's been reports that people don't like criminals"
C1 · Debra
"I like where this puzzle is going"
D1 · Emily
"All innocents in row 2 are connected"
A2 · Frank
"Claire and Debra have an equal number of innocent neighbors"
B2 · Gary
"All innocents in column A are connected"
C2 · Henry
"Each column has at least 3 innocents"
D2 · John
"There are at least 2 innocent cooks"
A3 · Kumar
"Exactly 1 innocent in row 5 is in column B"
B3 · Lisa
"There are exactly 2 innocent pilots"
C3 · Megan
"Criminals are so negative, so from now on, puzzles are like this"
D3 · Pam
"Vince and Zed have an equal number of innocent neighbors"
A4 · Ruth
"Seems like I can finally retire!"
B4 · Steve
"All innocents in row 5 are connected"
C4 · Thor
"No one in column D has a criminal neighbor"
D4 · Uma
"This is much better. Best change ever!"
A5 · Vince
"Yeah, right. Nice try. We'll be back tomorrow!"
B5 · Wanda
"Each row has at least 3 innocents"
C5 · Xena
"There's an odd number of innocents neighboring John"
D5 · Zed
"There's an odd number of innocents on the edges"
Answer Explanation
Full reasoning transcript (reference)
This is the full reasoning transcript for today's puzzle. For an interactive walkthrough, use Replay above.
▶ View full transcript (12 steps)
D1 · Emily → INNOCENT, D2 · John → INNOCENT
Because: The only people with the pilot profession on the board are Emily at D1 and John at D2. Lisa’s clue says there are exactly 2 innocent pilots, meaning there must be two pilots who are innocent. Since there are only two pilots total, both of them have to be the innocent pilots Lisa is counting. Therefore, we can determine that D1 Emily is INNOCENT and D2 John is INNOCENT.
Clue:
"There are exactly 2 innocent pilots" — Lisa (B3)
B2 · Gary → INNOCENT, C2 · Henry → INNOCENT
Because: In row 2, John at D2 is already INNOCENT, and Emily’s clue says that every INNOCENT in row 2 must be part of one orthogonally connected chain. John’s clue says there are at least 2 innocent cooks, and the only cooks on the board are A1 (Austin), A2 (Frank), and B2 (Gary), so at least one of the row 2 cooks (A2 or B2) has to be INNOCENT. Any INNOCENT at A2 or B2 must be connected to John at D2 through row 2, which forces C2 (Henry) to be INNOCENT as the necessary link between the left side of the row and D2. Finally, if Gary at B2 were not INNOCENT, then the two innocent cooks would have to be A1 and A2, making A2 INNOCENT, and then row 2 connectivity would require B2 to be INNOCENT to connect A2 to D2 through C2, so Gary at B2 must be INNOCENT. Therefore, we can determine that B2 Gary is INNOCENT, and C2 Henry is INNOCENT.
Clue:
"All innocents in row 2 are connected" — Emily (D1)
"There are at least 2 innocent cooks" — John (D2)
A3 · Kumar → INNOCENT
Because: Column A contains Austin at A1, Frank at A2, Kumar at A3, Ruth at A4, and Vince at A5. Henry’s clue says each column has at least 3 innocents, so column A must contain at least three innocents. Gary’s clue says all innocents in column A are connected, which in a single column means the innocents in column A must form one uninterrupted vertical block with no gap. Any uninterrupted block of three or more spaces in a five-space column must include the middle space A3, so Kumar has to be part of that connected innocent block. Therefore, we can determine that A3 Kumar is INNOCENT.
Clue:
"All innocents in column A are connected" — Gary (B2)
"Each column has at least 3 innocents" — Henry (C2)
B5 · Wanda → INNOCENT
Because: In row 5, the only person who is in column B is Wanda at B5. Kumar’s clue says exactly 1 innocent in row 5 is in column B, so that innocent must be Wanda. Therefore, we can determine that B5 Wanda is INNOCENT.
Clue:
"Exactly 1 innocent in row 5 is in column B" — Kumar (A3)
A2 · Frank → INNOCENT
Because: The only cooks on the board are Austin at A1, Frank at A2, and Gary at B2, and Gary is already known to be innocent. John’s clue says there are at least 2 innocent cooks, so at least one of A1 or A2 must also be an innocent cook. Gary’s clue says all innocents in column A are connected, and since Kumar at A3 is innocent, any innocent in A1 would have to connect to A3 through A2, which means A2 would also have to be innocent. So whether the “extra” innocent cook is A2 directly, or it is A1 (which would force A2 to be innocent anyway), A2 must be innocent. Therefore, we can determine that A2 Frank is INNOCENT.
Clue:
"There are at least 2 innocent cooks" — John (D2)
"All innocents in column A are connected" — Gary (B2)
A1 · Austin → INNOCENT, B1 · Claire → INNOCENT, C1 · Debra → INNOCENT
Because: In row 1, Wanda’s clue “Each row has at least 3 innocents” means that since Emily at D1 is already innocent, at least two of Austin at A1, Claire at B1, and Debra at C1 must be innocent, so there can be at most one criminal among those three. Frank’s clue says Claire and Debra have the same number of innocent neighbors: Claire’s neighbors include Frank, Gary, and Henry who are already innocent, plus Austin and Debra, while Debra’s neighbors include Emily, Gary, Henry, and John who are already innocent, plus Claire. That means Claire’s innocent-neighbor count is 3 plus whether Austin is innocent plus whether Debra is innocent, and Debra’s innocent-neighbor count is 4 plus whether Claire is innocent, so the only way they can be equal is if Austin and Debra are both innocent when Claire is innocent, and if Claire were not innocent then exactly one of Austin or Debra would be innocent. But that “Claire not innocent” outcome would force two criminals among A1, B1, and C1, which row 1 cannot have under Wanda’s clue, so Claire must be innocent, and then Frank’s equality forces both Austin and Debra to be innocent as well. Therefore, we can determine that A1 Austin is INNOCENT, B1 Claire is INNOCENT, and C1 Debra is INNOCENT.
Clue:
"Claire and Debra have an equal number of innocent neighbors" — Frank (A2)
"Each row has at least 3 innocents" — Wanda (B5)
B4 · Steve → INNOCENT
Because: In column B, Claire at B1, Gary at B2, Lisa at B3, and Wanda at B5 are already known to be innocent. That means column B currently has 4 innocents for sure. Austin’s clue says the total number of innocents in column B is odd, so it cannot stay at 4. The only remaining person in column B is Steve at B4, so Steve must also be innocent to make the total 5. Therefore, we can determine that B4 Steve is INNOCENT.
Clue:
"There's an odd number of innocents in column B" — Austin (A1)
C5 · Xena → INNOCENT
Because: In row 5, we already know Wanda at B5 is innocent, and Wanda’s clue says each row has at least 3 innocents, so row 5 needs at least two more innocents among A5, C5, and D5. Even if A5 is innocent, we would still need at least one innocent at C5 or D5 to reach three in the row. Steve’s clue says all innocents in row 5 are connected, and since B5 can only connect to D5 through C5, any innocent on the right side (at C5 or D5) forces C5 itself to be innocent. Therefore, we can determine that C5 (Xena) is INNOCENT.
Clue:
"Each row has at least 3 innocents" — Wanda (B5)
"All innocents in row 5 are connected" — Steve (B4)
C3 · Megan → INNOCENT, D3 · Pam → INNOCENT
Because: Look at John at D2: his neighbors are Debra at C1, Emily at D1, Henry at C2, Megan at C3, and Pam at D3. Debra, Emily, and Henry are already known innocents, so John already has 3 innocent neighbors, which is odd. Xena’s clue says the total number of John’s innocent neighbors is odd, so the pair (Megan and Pam) must add an even number of additional innocents, meaning they are either both criminals or both innocents. Wanda’s clue says each row has at least 3 innocents; in row 3 we already have Kumar and Lisa as innocents, so at least one of Megan or Pam must be innocent, which rules out them both being criminals. Therefore, we can determine that C3 Megan is INNOCENT and D3 Pam is INNOCENT.
Clue:
"Each row has at least 3 innocents" — Wanda (B5)
"There's an odd number of innocents neighboring John" — Xena (C5)
A4 · Ruth → INNOCENT, C4 · Thor → INNOCENT, D4 · Uma → INNOCENT
Because: Row 4 currently has only one known innocent, Steve at B4, and Wanda’s clue says each row has at least 3 innocents, so at least two of Ruth (A4), Thor (C4), and Uma (D4) must be innocent. Pam’s clue compares Vince’s and Zed’s numbers of innocent neighbors: Vince at A5 is adjacent to B5 and B4 (both innocent) and to A4, so Vince has 2 innocent neighbors if Ruth were criminal, but 3 if Ruth is innocent. If Ruth were criminal, then row 4 would force both Thor and Uma to be innocent to reach 3 innocents in that row, which would make Zed at D5 have 3 innocent neighbors (C5, C4, and D4), but Vince would still have only 2, so their counts could not be equal; therefore Ruth must be innocent, making Vince’s innocent-neighbor count 3. For Zed to match 3 innocent neighbors, both Thor and Uma must also be innocent (since C5 is already innocent and Zed’s only other neighbors are C4 and D4). Therefore, we can determine that A4 Ruth is INNOCENT, C4 Thor is INNOCENT, and D4 Uma is INNOCENT.
Clue:
"Each row has at least 3 innocents" — Wanda (B5)
"Vince and Zed have an equal number of innocent neighbors" — Pam (D3)
D5 · Zed → INNOCENT
Because: Thor’s clue says that every person in column D has zero criminal neighbors. That means every neighbor of any column D person must be innocent, because being adjacent to a criminal would break the clue. Uma at D4 is in column D, and Zed at D5 is a neighbor of Uma (directly below). Therefore, we can determine that D5 Zed is INNOCENT.
Clue:
"No one in column D has a criminal neighbor" — Thor (C4)
A5 · Vince → CRIMINAL
Because: The edge positions are all outer spaces: A1 to D1, A5 to D5, plus A2 to A4 and D2 to D4. Every one of those edge people is already known to be INNOCENT except A5 (Vince), whose status is still unknown. If Vince were also INNOCENT, then all 14 edge positions would be innocents, which is an even number, but Zed’s clue says the number of innocents on the edges is odd. So Vince cannot be INNOCENT, meaning he must be the one edge person who is not an innocent. Therefore, we can determine that A5 Vince is CRIMINAL.
Clue:
"There's an odd number of innocents on the edges" — Zed (D5)