Clues by Sam Mar 17, 2026 Answer – Full Solution Explained
A1
👩🔧
mech
B1
👨🌾
farmer
C1
👨🔧
mech
D1
🕵️♀️
sleuth
A2
👩💼
clerk
B2
👩🌾
farmer
C2
🕵️♂️
sleuth
D2
👩⚕️
doctor
A3
👨💻
coder
B3
👩⚕️
doctor
C3
👨💻
coder
D3
💂♂️
guard
A4
💂♂️
guard
B4
👩🍳
cook
C4
👩🍳
cook
D4
👨💼
clerk
A5
👩🎨
painter
B5
👨🎨
painter
C5
👨🍳
cook
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 — 10 criminals.
Clues by Sam answer for Mar 17, 2026 — a Medium solved in 15 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 10 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Emily (A2), Hilda (D2), Joy (B3), Noah (D3), Patrick (A4), Sofia (C4), Terry (D4), Uma (A5) and Wally (C5); the remaining 10 suspects are innocent.
The deduction chain, in plain English
01.A5 · Uma → CRIMINAL
Isaac’s clue says that Uma is one of Vince’s four criminal neighbors, which directly identifies Uma as a criminal. Therefore, we can determine that A5 (Uma) is CRIMINAL.
02.C1 · Chad → INNOCENT, C2 · Gabe → INNOCENT
Donna is at D1 and Hilda is at D2, so the only people who are neighbors of both of them are Chad at C1 and Gabe at C2. Uma’s clue says that Donna and Hilda have exactly 2 innocent neighbors in common, and there are exactly two common neighbors available. That means both Chad and Gabe must be innocent to satisfy the clue. Therefore, we can determine that C1 Chad is INNOCENT and C2 Gabe is INNOCENT.
03.A2 · Emily → CRIMINAL, A4 · Patrick → CRIMINAL, A1 · Anna → CRIMINAL
Column A contains Anna at A1, Emily at A2, Isaac at A3, Patrick at A4, and Uma at A5. Gabe’s clue says there is only one innocent in column A, meaning exactly one of those five people can be innocent. Since Isaac at A3 is already confirmed innocent, he must be that one innocent in column A, so everyone else in column A must not be innocent. Therefore, we can determine that A2 Emily is CRIMINAL, A4 Patrick is CRIMINAL, and A1 Anna is CRIMINAL.
04.B5 · Vince → INNOCENT
Patrick is at A4, so his neighboring squares are A3, B3, B4, A5, and B5, and the only neighbors he has in row 5 are Uma at A5 and Vince at B5. Anna’s clue says that Patrick has exactly three innocent neighbors in total, and exactly one of those three is in row 5. Since Uma is already known to be a criminal, the only way Patrick can have an innocent neighbor in row 5 is for Vince to be innocent. Therefore, we can determine that B5 Vince is INNOCENT.
05.C4 · Sofia → CRIMINAL
The clue talks about the people below Chad in column C, which are Gabe at C2, Mark at C3, Sofia at C4, and Wally at C5. It says there are exactly two criminals among those people, and those two criminals are connected, meaning with only two people they must be directly adjacent up-and-down. The only adjacent pairs available among C3, C4, and C5 are (C3 with C4) or (C4 with C5), so any connected pair of two criminals must include Sofia at C4. Therefore, we can determine that C4 Sofia is CRIMINAL.
06.B2 · Freya → INNOCENT
Patrick is at A4, and his neighbors are A3 Isaac, B3 Joy, B4 Ruth, A5 Uma, and B5 Vince. Anna’s clue says Patrick has exactly three innocent neighbors, and we already know Isaac and Vince are innocents, so exactly one of Joy or Ruth must be innocent and the other must be a criminal. Sofia’s clue says the number of criminals below Berat in column B (Freya at B2, Joy at B3, Ruth at B4, and Vince at B5) is odd; since Vince is innocent and exactly one of Joy or Ruth is a criminal, that count is 1 plus whatever Freya is. For the total to stay odd, Freya cannot be a criminal, so she must be innocent. Therefore, we can determine that B2 Freya is INNOCENT.
07.B1 · Berat → INNOCENT
In column C, Chad (C1) and Gabe (C2) are already innocent, and Sofia (C4) is already a criminal. Vince’s clue says there are exactly two criminals below Chad, so besides Sofia there must be exactly one more criminal among Mark (C3) and Wally (C5), which means the other one is innocent. That makes column C have exactly three innocents in total, so by Freya’s clue it must be the only column with exactly three innocents. Around Patrick (A4), the neighbors include Joy (B3) and Ruth (B4), and Anna’s clue says Patrick has exactly three innocent neighbors; since Isaac (A3) and Vince (B5) are already innocent and Uma (A5) is criminal, exactly one of Joy and Ruth is innocent. With Freya (B2) and Vince (B5) already innocent, column B would have exactly three innocents precisely when Berat (B1) is not innocent, and that is not allowed because column C is the only column that can have exactly three innocents. Therefore, we can determine that B1 · Berat is INNOCENT.
08.D4 · Terry → CRIMINAL
Berat’s clue says “Terry is one of 3 criminals above Xena,” which directly includes Terry in the group of criminals. Since everyone’s clues are true, Terry must be a criminal. Therefore, we can determine that D4 Terry is CRIMINAL.
09.D5 · Xena → INNOCENT
Xena is at D5, so the people above her in column D are Donna at D1, Hilda at D2, Noah at D3, and Terry at D4. Berat’s clue says that Terry is one of exactly 3 criminals above Xena, so among those four people above Xena there must be exactly 3 criminals. Terry’s clue says there are exactly 3 criminals in the entire D column, so if all 3 of them are already above Xena, Xena cannot be a criminal. Therefore, we can determine that D5 Xena is INNOCENT.
10.B4 · Ruth → INNOCENT
Row 4 contains Patrick at A4, Ruth at B4, Sofia at C4, and Terry at D4. Xena’s clue says that each row has at least one innocent person. In row 4, Patrick, Sofia, and Terry are already known to be criminals, so the only way for row 4 to include an innocent is for Ruth to be innocent. Therefore, we can determine that B4 Ruth is INNOCENT.
11.C5 · Wally → CRIMINAL
Vince is at B5, so his neighbors are A4 (Patrick), B4 (Ruth), C4 (Sofia), A5 (Uma), and C5 (Wally). Isaac’s clue says that Uma is one of Vince’s 4 criminal neighbors, which means Vince has exactly four criminal neighbors in total. We already know Patrick, Sofia, and Uma are criminals, and we know Ruth is innocent, so the only remaining neighbor who can make the total reach four criminals is Wally at C5. Therefore, we can determine that C5 Wally is CRIMINAL.
12.B3 · Joy → CRIMINAL
Patrick is at A4, so his neighbors are A3 Isaac, B3 Joy, B4 Ruth, A5 Uma, and B5 Vince. Anna’s clue says that among Patrick’s neighbors there are exactly three innocents, and only one of those three is in row 5. Isaac, Ruth, and Vince are already known innocents, and Vince is the only one of them in row 5, so they must be the three innocent neighbors the clue refers to. That leaves Joy as not part of the three innocents, so Joy cannot be innocent here. Therefore, we can determine that B3 Joy is CRIMINAL.
13.C3 · Mark → INNOCENT
Chad is at C1, so the people “below Chad” are the other spots in column C: C2 Gabe, C3 Mark, C4 Sofia, and C5 Wally. Vince’s clue says “both criminals below Chad,” which means there are exactly two criminals among those four people. Since Sofia at C4 and Wally at C5 are already known to be criminals, they must be the only two criminals below Chad, and the clue’s “connected” part is consistent because C4 and C5 touch orthogonally. That leaves no room for Mark at C3 to be a criminal. Therefore, we can determine that C3 Mark is INNOCENT.
14.D2 · Hilda → CRIMINAL
Row 5 has two innocents, because Vince at B5 and Xena at D5 are innocent while Uma at A5 and Wally at C5 are criminal. Mark’s clue says rows 2 and 5 have the same number of innocents, so row 2 must also have exactly two innocents. In row 2 we already have two innocents, Freya at B2 and Gabe at C2, so Hilda at D2 cannot be innocent without making three. Therefore, we can determine that D2 Hilda is CRIMINAL.
15.D1 · Donna → INNOCENT, D3 · Noah → CRIMINAL
Row 1 currently has two known innocents (Berat at B1 and Chad at C1) and one unknown (Donna at D1), so row 1 can only have either 2 or 3 innocents. Row 3 currently has two known innocents (Isaac at A3 and Mark at C3) and one unknown (Noah at D3), so row 3 can only have either 2 or 3 innocents. Hilda’s clue says row 1 has more innocents than row 3, and the only way for that to be true with these possible counts is for row 1 to have 3 innocents and row 3 to have 2 innocents. That forces Donna to be INNOCENT and Noah to be CRIMINAL. Therefore, we can determine that D1 Donna is INNOCENT and D3 Noah is CRIMINAL.