Clues by Sam Mar 16, 2026 Answer – Full Solution Explained
A1
👨💻
coder
B1
👨🌾
farmer
C1
👨🍳
cook
D1
👩🏫
teacher
A2
👩💻
coder
B2
👨🎤
singer
C2
👩🍳
cook
D2
👨🌾
farmer
A3
🕵️♀️
sleuth
B3
👩🎤
singer
C3
👩🍳
cook
D3
👨🎨
painter
A4
👨🏫
teacher
B4
👩🏫
teacher
C4
👩✈️
pilot
D4
👨🎨
painter
A5
👩✈️
pilot
B5
👩🌾
farmer
C5
👨✈️
pilot
D5
🕵️♂️
sleuth
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 — 5 criminals.
Clues by Sam answer for Mar 16, 2026 — a Easy solved in 15 steps
Today's Clues by Sam puzzle is rated Easy and resolves with 5 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Emma (D1), Julie (A3), Steve (D4), Wanda (B5) and Zane (D5); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.C2 · Helen → INNOCENT, D2 · Igor → INNOCENT
Gabe is at B2, so the only people to the right of Gabe in the same row are Helen at C2 and Igor at D2. Clyde’s clue says there are exactly 2 innocents to the right of Gabe, and there are only those two people there, so both of them must be innocents. Therefore, we can determine that C2 (Helen) is INNOCENT and D2 (Igor) is INNOCENT.
02.D5 · Zane → CRIMINAL
Below Emma in column D are Igor at D2, Mark at D3, Steve at D4, and Zane at D5. Helen’s clue says there are exactly two innocents among those four, and those two innocents form a single connected group. Since Igor at D2 is already known to be innocent, the only way for exactly one other innocent to be connected with him is for Mark at D3 to be the other innocent (because D3 is the only person directly connected to D2 in that column). That uses up both innocents below Emma, so neither Steve at D4 nor Zane at D5 can be innocent. Therefore, we can determine that D5 Zane is CRIMINAL.
03.D3 · Mark → INNOCENT
The clue talks about the people below Emma in column D: Igor at D2, Mark at D3, Steve at D4, and Zane at D5. It says there are exactly two innocents among those four, and those two innocents must be connected by up/down adjacency. We already know Igor at D2 is INNOCENT and Zane at D5 is CRIMINAL, so the second innocent must be either Mark (D3) or Steve (D4). Since Igor at D2 is only directly connected to D3 and not to D4, the only way for the two innocents to be connected is for D3 to be the other innocent. Therefore, we can determine that D3 Mark is INNOCENT.
04.D4 · Steve → CRIMINAL
Emma is at D1, so the people below her are D2 Igor, D3 Mark, D4 Steve, and D5 Zane, all in column D. Helen’s clue says “Both innocents below Emma are connected,” which means there are exactly two innocents among those four people, and those two innocents must form one connected group by up/down adjacency. We already know Igor at D2 and Mark at D3 are innocents, and they are directly adjacent, so they must be the exactly two innocents the clue is talking about. That leaves no room for Steve at D4 to be innocent. Therefore, we can determine that D4 Steve is CRIMINAL.
05.B2 · Gabe → INNOCENT, B3 · Karen → INNOCENT
Zane’s clue says that both singers are innocent, and the only singers on the board are Gabe at B2 and Karen at B3. Therefore, we can determine that B2 Gabe is INNOCENT and B3 Karen is INNOCENT.
06.B1 · Berat → INNOCENT
Karen’s clue explicitly says that Berat is one of Helen’s innocent neighbors, so Berat’s status is given as innocent by the clue itself. Therefore, we can determine that B1 Berat is INNOCENT.
07.C5 · Xavi → INNOCENT
The three pilots are Ruby at C4, Uma at A5, and Xavi at C5. Berat’s clue says that exactly one of these three pilots has an innocent directly below them. Only Ruby can possibly have someone directly below, because she is the only pilot not on the bottom row: the person directly below Ruby at C4 is Xavi at C5, while Uma and Xavi have nobody directly below them. Since Ruby must be the one pilot who has an innocent directly below, Xavi must be innocent. Therefore, we can determine that C5 Xavi is INNOCENT.
08.B4 · Olivia → INNOCENT
Noah is at A4, so his neighbors are Julie at A3, Karen at B3, Olivia at B4, Uma at A5, and Wanda at B5. The only neighbors of Noah who are above Wanda (in Wanda’s column B) are Karen at B3 and Olivia at B4. Xavi’s clue says that exactly 2 of the 3 innocent neighbors of Noah are above Wanda, so both of those “above Wanda” neighbors must be innocent. Karen is already known to be innocent, so Olivia must be the second innocent above Wanda. Therefore, we can determine that B4 Olivia is INNOCENT.
09.A3 · Julie → CRIMINAL
Noah is at A4, and his neighbors are Julie (A3), Karen (B3), Olivia (B4), Uma (A5), and Wanda (B5). Xavi’s clue says Noah has exactly three innocent neighbors, and exactly two of those are “above Wanda”; since Wanda is at B5, the neighbors above Wanda are in column B, and Karen (B3) and Olivia (B4) are already two innocent neighbors above Wanda, so the third innocent neighbor of Noah must be exactly one of Julie, Uma, or Wanda. Olivia’s clue says each row has at least two innocents, and in row 5 we already have Xavi (C5) innocent and Zane (D5) criminal, so at least one of Uma (A5) or Wanda (B5) must be innocent, meaning Julie cannot be the one innocent among {Julie, Uma, Wanda}. Therefore, we can determine that A3 Julie is CRIMINAL.
10.A4 · Noah → INNOCENT
Noah is at A4, and his neighbors are Julie, Karen, Olivia, Uma, and Wanda. Xavi’s clue says there are exactly three innocent neighbors of Noah, and since Karen and Olivia are already innocent while Julie is already criminal, that forces exactly one of Uma and Wanda to be innocent. Julie’s clue counts the edge-neighbors of Olivia, which are Julie, Noah, Uma, Wanda, and Xavi, and it says an odd number of those edge-neighbors are innocent; with Xavi already innocent and Julie already criminal, that means Noah, Uma, and Wanda must contain an even number of innocents. Since Uma and Wanda together contain exactly one innocent, Noah must be innocent to make the total among Noah, Uma, and Wanda even. Therefore, we can determine that A4 Noah is INNOCENT.
11.D1 · Emma → CRIMINAL
In column D, the people are Emma at D1, Igor at D2, Mark at D3, Steve at D4, and Zane at D5. Noah’s clue says there are exactly 3 criminals in column D. We already know Steve at D4 and Zane at D5 are criminals, and Igor at D2 and Mark at D3 are innocents, so column D currently has exactly 2 criminals and cannot gain any more criminals except through Emma. Therefore, we can determine that D1 Emma is CRIMINAL.
12.C3 · Lucy → INNOCENT
Helen is at C2, so her eight neighbors are Berat (B1), Gabe (B2), Karen (B3), Clyde (C1), Lucy (C3), Emma (D1), Igor (D2), and Mark (D3). Karen’s clue says that Berat is one of Helen’s 7 innocent neighbors, which means exactly 7 of those 8 neighbors are innocent. Since Emma (D1) is already known to be a criminal, she must be the single non-innocent neighbor in Helen’s neighborhood, so every other neighbor of Helen, including Lucy (C3), must be innocent. Therefore, we can determine that C3 Lucy is INNOCENT.
13.A1 · Adam → INNOCENT, C4 · Ruby → INNOCENT
Noah is at A4, and his five neighbors are Julie (A3), Karen (B3), Olivia (B4), Uma (A5), and Wanda (B5). Xavi says Noah has exactly three innocent neighbors, and we already know Karen and Olivia are innocents, so exactly one of Uma and Wanda must also be innocent. That means exactly one of Uma and Wanda is a criminal, so in row 5, Zane is a criminal and there is exactly one more criminal (either Uma or Wanda), making row 5 the one row with exactly two criminals. Emma then says only one row has exactly two criminals, so no other row is allowed to have a second criminal; that forces Adam not to be a criminal in row 1, and Ruby not to be a criminal in row 4. Therefore, we can determine that A1 Adam is INNOCENT and C4 Ruby is INNOCENT.
14.A2 · Flora → INNOCENT
The only edge people whose statuses are still unknown are Flora at A2, Uma at A5, and Wanda at B5. Xavi’s clue about Noah means Noah has exactly three innocent neighbors, and since Karen and Olivia are already innocent neighbors of Noah while Julie is a criminal neighbor, exactly one of Uma and Wanda must be innocent and the other must be a criminal. Adam’s clue says the total number of edge criminals is odd, and since the already-known edge criminals give an even count, the number of criminals among Flora, Uma, and Wanda must be odd; with exactly one criminal among Uma and Wanda, that forces Flora not to be a criminal. Therefore, we can determine that A2 Flora is INNOCENT.
15.A5 · Uma → INNOCENT, B5 · Wanda → CRIMINAL
In column A, the only known criminal is Julie at A3, while A1, A2, and A4 are innocent and A5 (Uma) is the only unknown. In column B, B1 through B4 are all innocent, so B5 (Wanda) is the only place a criminal could be in that column. Flora’s clue says columns A and B have an equal number of criminals, so column B must end up with the same criminal count as column A. If Uma were a criminal, column A would have two criminals but column B could have at most one, so Uma must be innocent, and then Wanda must be the one criminal in column B to match the single criminal in column A. Therefore, we can determine that A5 Uma is INNOCENT and B5 Wanda is CRIMINAL.