Clues by Sam Mar 23, 2026 Answer – Full Solution Explained
A1
👩🎨
painter
B1
👩🏫
teacher
C1
👨🏫
teacher
D1
👨🍳
cook
A2
👩⚕️
doctor
B2
👷♂️
builder
C2
👮♀️
cop
D2
👩⚕️
doctor
A3
👮♂️
cop
B3
👷♂️
builder
C3
👮♂️
cop
D3
👷♂️
builder
A4
👨🍳
cook
B4
👨🔧
mech
C4
👩🍳
cook
D4
👨🔧
mech
A5
👩🎨
painter
B5
👨🔧
mech
C5
👩🌾
farmer
D5
👨🌾
farmer
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 Mar 23, 2026 — a Easy solved in 16 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 Don (D1), Ivan (B2), Nick (C3), Olof (D3), Paul (A4), Sue (C4) and Vince (B5); the remaining 13 suspects are innocent.
The deduction chain, in plain English
01.B5 · Vince → CRIMINAL
The clue from Zane says that among the people in row 5 who are neighbors of Wanda, there is exactly 1 innocent. In row 5, the only neighbors of Wanda (at C5) are Vince (at B5) and Zane (at D5), since Uma is too far away and Wanda doesn’t count as her own neighbor. Because Zane is already known to be innocent, he must be the one innocent neighbor of Wanda in row 5, so Vince cannot be innocent. Therefore, we can determine that B5 Vince is CRIMINAL.
02.C3 · Nick → CRIMINAL, D3 · Olof → CRIMINAL
Katie is at D2 and Thor is at D4, and we look at the people who are neighbors of both of them. The only shared neighbors they have are Nick at C3 (diagonal to both) and Olof at D3 (directly between them in the same column). Vince’s clue says Katie and Thor have no innocent neighbors in common, so any person who is a neighbor of both of them cannot be innocent. That forces both Nick and Olof to be criminals. Therefore, we can determine that C3 Nick is CRIMINAL and D3 Olof is CRIMINAL.
03.C5 · Wanda → INNOCENT, C1 · Chuck → INNOCENT
In column C we have Chuck at C1, Joy at C2, Nick at C3, Sue at C4, and Wanda at C5, and we already know Nick at C3 is a criminal. Olof’s clue says there are exactly two criminals in column C, and those two criminals are connected by a continuous up/down chain. With only two criminals available, the only way for Nick at C3 to be connected to the other criminal is for that other criminal to be directly above or directly below him, at C2 or C4; if the other criminal were at C1 or C5, there would have to be additional criminals in between to form the connection, which would make more than two. That means Chuck at C1 cannot be one of the two criminals, and Wanda at C5 cannot be one of the two criminals. Therefore, we can determine that C5 Wanda is INNOCENT and C1 Chuck is INNOCENT.
04.A3 · Larry → INNOCENT, B3 · Mark → INNOCENT
The clue says that in each row, there are at least 2 innocents. In row 3 (A3 Larry, B3 Mark, C3 Nick, D3 Olof), we already know that Nick at C3 is a criminal and Olof at D3 is a criminal. That means the only people left in row 3 who could supply the required at least 2 innocents are Larry at A3 and Mark at B3, so both of them must be innocents. Therefore, we can determine that A3 Larry is INNOCENT and B3 Mark is INNOCENT.
05.A5 · Uma → INNOCENT
Row 5 contains Uma at A5, Vince at B5, Wanda at C5, and Zane at D5. Larry’s clue says there are exactly 3 innocents in row 5. We already know Wanda and Zane are innocents, and Vince is a criminal, so row 5 currently has 2 innocents and 1 criminal. To make the total exactly 3 innocents in that row, the remaining person, Uma, must be an innocent. Therefore, we can determine that A5 (Uma) is INNOCENT.
06.A1 · Anna → INNOCENT
In row 1 there are Anna at A1, Barb at B1, Chuck at C1, and Don at D1, and Chuck is already known to be innocent. Uma’s clue says there are exactly three innocents in row 1, and among those three innocents only one is a neighbor of Chuck. In row 1, the only neighbors of Chuck are Barb (B1) and Don (D1), while Anna (A1) is not a neighbor of Chuck at all. Since two other people in row 1 must be innocent besides Chuck, and only one of those innocents can be Chuck’s neighbor, the other innocent must be the only non-neighbor in the row, which is Anna at A1. Therefore, we can determine that A1 Anna is INNOCENT.
07.D2 · Katie → INNOCENT
Uma’s clue tells us there are exactly three innocents in row 1, and that among those three, only one is a neighbor of Chuck at C1. The only people in row 1 who neighbor Chuck are B1 and D1, so exactly one of B1 and D1 is innocent and the other is criminal. Anna’s clue says an odd number of edge-innocents neighbor Joy at C2; the edge neighbors of Joy are B1, C1, D1, D2, and D3. We already know C1 is innocent and D3 is criminal, so the parity depends on B1, D1, and D2, and it requires that the number of innocents among B1, D1, and D2 is even. Since exactly one of B1 and D1 is innocent, D2 must also be innocent to make that count even. Therefore, we can determine that D2 Katie is INNOCENT.
08.D4 · Thor → INNOCENT
In column C, we already know Nick at C3 is a criminal and Chuck at C1 and Wanda at C5 are innocents. Olof’s clue says there are exactly two criminals in column C and they are connected, so the second criminal in column C must be directly next to Nick, meaning exactly one of Joy at C2 and Sue at C4 is a criminal. Katie’s clue says Olof has an odd number of innocent neighbors; among Olof’s five neighbors, Katie is already an innocent and Nick is already a criminal, so among the remaining three neighbors (Joy, Sue, and Thor) there must be an even number of innocents. Since Joy and Sue include exactly one innocent and one criminal, Thor must be an innocent to make the total number of innocents among those three even. Therefore, we can determine that D4 Thor is INNOCENT.
09.A2 · Freya → INNOCENT
Anna is at A1 and Paul is at A4, so the only people strictly between them in column A are Freya at A2 and Larry at A3. Thor’s clue says there are exactly 2 innocents in between Anna and Paul, which means both of those in-between people must be innocents. Larry at A3 is already known to be innocent, so the other in-between person, Freya at A2, must also be innocent. Therefore, we can determine that A2 Freya is INNOCENT.
10.D1 · Don → CRIMINAL
In column C, we already know Nick at C3 is a criminal, while Chuck at C1 and Wanda at C5 are innocents. Olof’s clue says there are exactly two criminals in column C and those two are connected, so the second criminal must be either Joy at C2 or Sue at C4, and the other one of those two must be innocent. That means column C has exactly three innocents in total (C1, C5, and whichever of C2 or C4 is not the second criminal). Freya’s clue says columns C and D have the same number of innocents, and column D already has three confirmed innocents (Katie at D2, Thor at D4, and Zane at D5), so Don at D1 cannot be an innocent. Therefore, we can determine that D1 Don is CRIMINAL.
11.B1 · Barb → INNOCENT
Row 1 contains Anna at A1, Barb at B1, Chuck at C1, and Don at D1. Uma’s clue says that there are exactly three innocents in row 1. We already know Anna and Chuck are innocents, and Don is a criminal, so the only way for row 1 to have three innocents is for Barb to be the remaining innocent. Therefore, we can determine that B1 Barb is INNOCENT.
12.A4 · Paul → CRIMINAL
Look at column A: Anna at A1 is innocent, Freya at A2 is innocent, Larry at A3 is innocent, and Uma at A5 is innocent, leaving only Paul at A4 still undecided in that column. Barb’s clue says each column has at least one criminal, so column A must contain a criminal somewhere. Since every other person in column A is already confirmed innocent, the only place that required criminal can be is Paul at A4. Therefore, we can determine that A4 Paul is CRIMINAL.
13.B2 · Ivan → CRIMINAL
Freya at A2 has five neighbors: Anna, Barb, Ivan, Larry, and Mark, and four of those (Anna, Barb, Larry, Mark) are already known to be innocent, so Freya has either 4 innocent neighbors (if Ivan is criminal) or 5 innocent neighbors (if Ivan is innocent). Olof at D3 has five neighbors: Katie and Thor (both innocent), Nick (criminal), and Joy and Sue (unknown), so Olof has either 2, 3, or 4 innocent neighbors depending on Joy and Sue. Paul’s clue says Freya has exactly one more innocent neighbor than Olof, so the only way Ivan could be innocent is if Freya had 5 innocent neighbors, which would force Olof to have 4 innocent neighbors, meaning Joy and Sue would both have to be innocent. Since we already know Joy and Sue cannot both be innocent, Ivan cannot be innocent, and the only remaining possibility is that Ivan is criminal. Therefore, we can determine that B2 Ivan is CRIMINAL.
14.C2 · Joy → INNOCENT
Row 3 already has two innocents: Larry at A3 and Mark at B3, while Nick at C3 and Olof at D3 are criminals, so row 3 has exactly 2 innocents. Row 2 currently has two confirmed innocents, Freya at A2 and Katie at D2, and Ivan at B2 is a criminal, leaving only Joy at C2 undecided. Ivan’s clue says there are more innocents in row 2 than in row 3, so row 2 must have at least 3 innocents to beat row 3’s 2. That is only possible if Joy is an innocent. Therefore, we can determine that C2 Joy is INNOCENT.
15.C4 · Sue → CRIMINAL
In column C, Chuck at C1, Joy at C2, and Wanda at C5 are already known to be innocent, and Nick at C3 is already known to be a criminal, leaving only Sue at C4 unresolved. Olof’s clue says there are both criminals in column C, meaning there are exactly two criminals in that column, and they must be connected. Since Nick is one of those criminals and every other spot in column C is already innocent except Sue, Sue has to be the second criminal, and she is also directly below Nick so they are connected. Therefore, we can determine that C4 Sue is CRIMINAL.
16.B4 · Rob → INNOCENT
Look at row 4: Paul at A4 is a criminal, Rob at B4 is unknown, Sue at C4 is a criminal, and Thor at D4 is an innocent. Wanda’s clue says each row must contain at least 2 innocents. Since row 4 currently has only one known innocent (Thor), and Paul and Sue cannot contribute any innocents, Rob must be the second innocent required for that row. Therefore, we can determine that B4 Rob is INNOCENT.