Clues by Sam Feb 06, 2026 Answer – Full Solution Explained
A1
👩🍳
cook
B1
💂♂️
guard
C1
👩🔧
mech
D1
👩🎤
singer
A2
👨🍳
cook
B2
💂♀️
guard
C2
👨🔧
mech
D2
💂♂️
guard
A3
👩🎤
singer
B3
👨⚖️
judge
C3
👨⚕️
doctor
D3
👨⚕️
doctor
A4
👷♀️
builder
B4
👷♂️
builder
C4
👨🍳
cook
D4
👨⚕️
doctor
A5
👩⚖️
judge
B5
👩🌾
farmer
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 — 5 criminals.
Clues by Sam answer for Feb 06, 2026 — a Tricky solved in 15 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 5 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Ellie (C1), Gary (A2), Ivan (C2), Katie (A3) and Rob (C4); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.D2 · Jason → INNOCENT
Helen’s clue says that Jason is one of Ellie’s 4 innocent neighbors, which directly states that Jason is innocent. Therefore, we can determine that D2 Jason is INNOCENT.
02.C2 · Ivan → CRIMINAL
Ellie at C1 and Noah at D3 share exactly two common neighbors: Jason at D2 and Ivan at C2. Jason’s clue says that Ellie and Noah have only one innocent neighbor in common, so among those two shared neighbors, exactly one is innocent. Since Jason at D2 is already confirmed INNOCENT, he must be that one innocent common neighbor, which forces Ivan at C2 to be the other one. Therefore, we can determine that C2 Ivan is CRIMINAL.
03.D1 · Freya → INNOCENT, B1 · Chase → INNOCENT
Ellie is at C1, so her neighbors are B1 (Chase), B2 (Helen), C2 (Ivan), D1 (Freya), and D2 (Jason). Helen’s clue says that Jason is one of Ellie’s 4 innocent neighbors, so among these five neighbors Ellie must have exactly four innocents. We already know Helen at B2 and Jason at D2 are innocent, and Ivan at C2 is criminal, so the only way for Ellie to have four innocent neighbors is for both remaining neighbors, Chase at B1 and Freya at D1, to be innocent. Therefore, we can determine that D1 Freya is INNOCENT and B1 Chase is INNOCENT.
04.A2 · Gary → CRIMINAL
Row 2 contains Helen at B2 and Jason at D2, and both are already confirmed innocents. Chase’s clue says row 2 is the only row with exactly 2 innocents, so row 2 must have exactly two innocents total. That means the remaining unknown person in row 2, Gary at A2, cannot be an innocent because that would make three innocents in the row. Therefore, we can determine that A2 Gary is CRIMINAL.
05.D4 · Scott → INNOCENT
Logan is at B3, and his neighbors in row 4 are exactly A4, B4, and C4. Ivan’s clue says Logan has exactly 4 innocent neighbors in total, and exactly 2 of those 4 are in row 4, so exactly two of A4/B4/C4 are innocents. That means row 4 already contains exactly 2 innocents among A4–C4, and the only remaining person in row 4 is D4 Scott. Chase’s clue says row 2 is the only row with exactly 2 innocents, so row 4 cannot end up with exactly 2 innocents; therefore Scott cannot be criminal and must be an additional innocent in row 4. Therefore, we can determine that D4 Scott is INNOCENT.
06.D3 · Noah → INNOCENT
Noah is at D3, so “to the left of Noah” means the three people in A3, B3, and C3. Scott’s clue says there are exactly 2 innocents among those three, so row 3 already contains exactly 2 innocents before counting Noah himself. Chase’s clue says row 2 is the only row with exactly 2 innocents, so row 3 cannot end with exactly 2 innocents in total. That forces Noah to be an innocent so that row 3 has more than 2 innocents altogether. Therefore, we can determine that D3 (Noah) is INNOCENT.
07.B3 · Logan → INNOCENT
Logan is at B3, so his neighboring row 2 people are Gary at A2, Helen at B2, and Ivan at C2, and his neighboring row 4 people are Olive at A4, Phil at B4, and Rob at C4. Ivan’s clue says Logan has exactly 4 innocent neighbors in total, and exactly 2 of those 4 are in row 4, so exactly two of Olive/Phil/Rob are innocent, and the other two innocent neighbors must come from the remaining neighbors. In row 2, only Helen is innocent, so the last needed innocent neighbor must be exactly one of Katie at A3 or Mark at C3, meaning row 3 has Noah plus exactly one of Katie/Mark as innocents before counting Logan. Chase’s clue says row 2 is the only row with exactly 2 innocents, so row 3 cannot have exactly 2 innocents, and the only way to avoid that is for Logan himself to be an innocent as well. Therefore, we can determine that B3 Logan is INNOCENT.
08.A4 · Olive → INNOCENT
Look at the farmers in row 5: Uma at B5 has Phil at B4 directly above, Vera at C5 has Rob at C4 directly above, and Zoe at D5 has Scott at D4 directly above. Logan’s clue says exactly one farmer has a criminal directly above them, and since Scott is already known to be innocent, Zoe cannot be that farmer, so exactly one of Phil or Rob must be a criminal and the other must be innocent. That means row 4 already contains Scott plus exactly one of Phil/Rob as innocents, so row 4 has either 2 innocents (if Olive is criminal) or 3 innocents (if Olive is innocent). Chase’s clue says row 2 is the only row with exactly 2 innocents, so row 4 cannot have exactly 2 innocents, forcing Olive to be the extra innocent. Therefore, we can determine that A4 Olive is INNOCENT.
09.A5 · Tina → INNOCENT, B5 · Uma → INNOCENT, C5 · Vera → INNOCENT, D5 · Zoe → INNOCENT
Row 2 already has exactly two innocents, Helen at B2 and Jason at D2, because Gary at A2 and Ivan at C2 are criminals. Chase’s clue says Row 2 is the only row that can have exactly two innocents, so Rows 1, 3, and 4 cannot stay at two innocents; since each of those rows already has two known innocents, each of them must end up with either three or four innocents. Olive’s clue says Row 5 has more innocents than any other row, so no other row is allowed to reach four innocents, because then Row 5 would need more than four which is impossible on a four-person row. That forces Rows 1, 3, and 4 to have exactly three innocents each, so Row 5 must have four innocents to be strictly higher than them. Since Row 5 contains Tina at A5, Uma at B5, Vera at C5, and Zoe at D5, all four must be innocents. Therefore, we can determine that A5 Tina is INNOCENT, B5 Uma is INNOCENT, C5 Vera is INNOCENT, and D5 Zoe is INNOCENT.
10.A1 · Bonnie → INNOCENT
Chase’s clue says row 2 is the only row with exactly 2 innocents. In row 1, B1 and D1 are already innocent, so row 1 cannot end with only those two; it must have at least one more innocent among A1 and C1. If that extra innocent is C1, then we have at least one innocent mech, and Vera’s clue (“more innocent cooks than innocent mechs”) forces there to be at least two innocent cooks; since A2 is a criminal, the only possible innocent cooks are A1 and C4, so A1 must be innocent. If the extra innocent in row 1 is A1 instead, then A1 is innocent immediately. Therefore, we can determine that A1 (Bonnie) is INNOCENT.
11.C1 · Ellie → CRIMINAL
Row 5 already has four innocents (Tina at A5, Uma at B5, Vera at C5, and Zoe at D5). Olive’s clue says Row 5 has more innocents than any other row, so no other row is allowed to also have four innocents. In Row 1, Bonnie (A1), Chase (B1), and Freya (D1) are already innocent, so if Ellie (C1) were also innocent then Row 1 would have four innocents and would tie Row 5, which is not allowed by the clue. Therefore, we can determine that C1 Ellie is CRIMINAL.
12.B4 · Phil → INNOCENT
The two singers are Freya at D1 and Katie at A3. Freya’s neighbors include Ellie at C1 and Ivan at C2, and both are criminals, so Freya has exactly 2 criminal neighbors. Freya’s clue says only one of the two singers has exactly 2 criminal neighbors, so Katie must not have exactly 2 criminal neighbors. Katie already has Gary at A2 as one criminal neighbor, and the only unknown among her neighbors is Phil at B4, so if Phil were a criminal Katie would also have exactly 2; therefore Phil cannot be a criminal. Therefore, we can determine that B4 Phil is INNOCENT.
13.C4 · Rob → CRIMINAL
Logan is at B3, so his neighbors are A2 Gary, B2 Helen, C2 Ivan, A3 Katie, C3 Mark, A4 Olive, B4 Phil, and C4 Rob. Ivan’s clue says that Logan has exactly 4 innocent neighbors in total, and exactly 2 of those innocents are in row 4. The row 4 neighbors of Logan are Olive at A4, Phil at B4, and Rob at C4, and we already know Olive and Phil are innocent, so row 4 already contains the required two innocent neighbors. That means Rob cannot also be innocent, so Rob must be criminal. Therefore, we can determine that C4 Rob is CRIMINAL.
14.A3 · Katie → CRIMINAL
The edge positions are the 14 outer spaces: all of row 1, all of row 5, plus A2–A4 and D2–D4. Among these edges, everyone is already known to be innocent except C1 and A2 (both criminals) and A3 (Katie), who is the only unknown edge person. Counting the known edge innocents gives 11, so Katie being innocent would make 12 edge innocents, which would be even, but Gary’s clue says the number of innocents on the edges is odd. Therefore, we can determine that A3 Katie is CRIMINAL.
15.C3 · Mark → INNOCENT
Logan is at B3, so his neighbors are A2 Gary, B2 Helen, C2 Ivan, A3 Katie, C3 Mark, A4 Olive, B4 Phil, and C4 Rob. Ivan’s clue says that among Logan’s neighboring innocents there are exactly 4 in total, and exactly 2 of those are in row 4. In row 4 next to Logan, Olive and Phil are already innocent, so they must be the two row-4 innocents, which means there must be exactly 2 more innocent neighbors of Logan outside row 4. Outside row 4, Helen is already innocent and Gary, Ivan, and Katie are already criminals, so the only way to reach two innocents there is for Mark to be innocent. Therefore, we can determine that C3 Mark is INNOCENT.