Clues by Sam Oct 10, 2026 Answer – Full Solution Explained
A1
👩🎤
singer
B1
👩⚖️
judge
C1
👩⚖️
judge
D1
👩🎤
singer
A2
👩⚖️
judge
B2
👷♂️
builder
C2
👷♂️
builder
D2
👷♂️
builder
A3
👨⚕️
doctor
B3
👩⚕️
doctor
C3
👩⚕️
doctor
D3
🕵️♂️
sleuth
A4
👩🌾
farmer
B4
💂♀️
guard
C4
👨🌾
farmer
D4
🕵️♀️
sleuth
A5
👨🌾
farmer
B5
🕵️♂️
sleuth
C5
💂♂️
guard
D5
💂♂️
guard
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 — 6 criminals.
Clues by Sam answer for Oct 10, 2026 — a Hard solved in 15 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Carol (C1), Jose (D2), Lucy (B3), Noah (D3), Terry (C4) and Uma (D4); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.A3 · Kumar → INNOCENT, A5 · Vince → INNOCENT
Zach’s clue says Olive has exactly 4 innocent neighbors, and exactly 2 of those innocents are in column A. Among Olive’s neighbors, the only people in column A are Kumar and Vince. Since those 2 column A innocent spots still have to be filled, Kumar and Vince must be the ones who fill them. So Kumar and Vince must be innocent.
02.B4 · Susan → INNOCENT
Olive’s neighbors contain exactly four innocents, and Kumar and Vince are already known innocents there. That leaves exactly one criminal among the three unknown neighbors: Lucy, Susan, and Will. Vince’s clue says there are more criminal doctors than criminal guards. Among those three neighbors, Lucy is a doctor and Susan is a guard, so the one criminal in that group has to come from Lucy or Will, not Susan. So Susan must be innocent.
03.B1 · Betsy → INNOCENT, B2 · Frank → INNOCENT
Olive’s clue leaves exactly 1 innocent among the neighboring column B people, namely Lucy and Will. Susan is already the other innocent in that neighboring column B group, so column B’s four innocents must be Susan, one of Lucy or Will, and two more people. Susan’s clue says column B contains exactly 4 innocents in total. The only people in column B outside that neighboring group are Betsy and Frank, so they have to provide those two remaining innocents. That makes Betsy and Frank innocent.
04.D4 · Uma → CRIMINAL
Betsy’s clue says that Uma is one of Terry’s three criminal neighbors. That directly identifies Uma as a criminal. So Uma must be criminal.
05.D2 · Jose → CRIMINAL
Uma’s clue says that Jose is one of Mary’s 5 criminal neighbors. That directly places Jose among the neighbors who are criminal. So Jose must be criminal.
06.C5 · Xavi → INNOCENT
If Xavi were criminal, then the guard profession would already have 1 criminal. Vince’s clue says there are more criminal doctors than criminal guards, so the doctors would then need more than 1 criminal, which means both doctor unknowns, Lucy and Mary, would have to be criminals. But Betsy’s clue says Uma is one of Terry’s exactly 3 criminal neighbors. With Lucy and Mary criminal, Uma criminal, and Xavi also criminal, Terry’s neighbors would already include four criminals: Lucy, Mary, Uma, and Xavi. That clashes with Terry having exactly 3 criminal neighbors. So Xavi must be innocent.
07.B3 · Lucy → CRIMINAL, A2 · Eve → INNOCENT
Frank’s clue says he has exactly 6 innocent neighbors, and exactly 1 of those innocent neighbors is also a neighbor of Kumar. The only Frank-neighbors who are also Kumar-neighbors are Eve and Lucy, so exactly one of Eve and Lucy is innocent. Vince’s clue says there are more criminal doctors than criminal guards, and there are no criminal guards at all, so at least one doctor must be criminal. The only doctors not already known innocent are Lucy and Mary. Uma’s clue says Mary has exactly 5 criminal neighbors, including Jose, and with the test case Lucy innocent and Eve criminal, the remaining people involved cannot satisfy all of these requirements together. So the tested opposite pairing fails, which leaves Lucy as criminal and Eve as innocent.
08.B5 · Will → INNOCENT
Olive’s neighbors contain exactly 4 innocents, and exactly 2 of those innocents are in column A. Those 2 are already Kumar and Vince, so the other 2 innocents among Olive’s neighbors must be people not in column A. Among Olive’s neighbors who are not in column A, Lucy is criminal, Susan is already innocent, and Will is unknown. Since that group needs 2 innocents total and only Susan is already one of them, the remaining innocent spot has to be Will. So Will must be innocent.
09.C3 · Mary → INNOCENT, D3 · Noah → CRIMINAL
Frank’s clue fixes his neighborhood at exactly 6 innocents, with exactly 1 of those innocent people being a neighbor of Kumar. Mary’s clue also fixes her neighborhood at exactly 5 criminals, and Jose is one of them; that neighborhood is Frank, Henry, Jose, Lucy, Noah, Susan, Terry, and Uma. Now test the opposite pair: Mary criminal and Noah innocent. Then the remaining people involved here, Amy, Carol, Henry, and Terry, would have to make all of those counts true at the same time, but they cannot. So that opposite pair is impossible. That makes Mary innocent and Noah criminal.
10.A4 · Olive → INNOCENT
Lucy’s clue says her neighbors contain an odd number of innocents. Right now Lucy already has 5 known innocent neighbors, and the unknown neighbors are Henry, Olive, and Terry. Uma’s clue says Mary has exactly 5 criminal neighbors, and among Mary’s neighbors there are already 4 known criminals, with only Henry and Terry still unknown there. So Henry and Terry have to account for the last criminal spot in Mary’s neighborhood. If Olive were criminal, then Henry and Terry would have to satisfy both clues in a way that cannot be done at once. So Olive cannot be criminal. That makes Olive innocent.
11.A1 · Amy → INNOCENT
Frank’s neighbors must contain exactly 6 innocents. Among Frank’s neighbors in column C, exactly 2 are innocent, and that group is Carol, Henry, and Mary, with Mary already one known innocent. That accounts for the two innocent spots in Frank’s neighboring column C group, so the remaining innocent needed to reach 6 among Frank’s neighbors has to come from the only neighbor outside that column C group named here: Amy. So Amy must be innocent.
12.D1 · Diane → INNOCENT
Frank’s clue fixes his neighborhood at exactly 6 innocents, and Jose’s clue says Jose has an odd number of innocent neighbors. Around Jose, Mary is already a known innocent, while Carol, Diane, and Henry are the unknown neighbors involved here. If Diane were criminal, then Carol and Henry would have to be the ones that make both clues work at the same time, but they cannot do that. That rules out Diane being criminal. So Diane must be innocent.
13.C4 · Terry → CRIMINAL
Frank’s neighbors must contain exactly 6 innocents, and among those innocents exactly 1 is also a neighbor of Kumar. In column C, there are already 2 known innocents, and the total number of innocents there has to be odd, so Carol, Henry, and Terry have to fit that requirement. If Terry were innocent, then Carol and Henry would be the other people left to satisfy both Kumar’s clue about Frank’s neighbors and Diane’s clue about column C, but those facts cannot all be met at once. So Terry cannot be innocent. That makes Terry criminal.
14.C2 · Henry → INNOCENT
Frank says Lucy has an odd number of innocent neighbors. Lucy’s neighbors already include 6 known innocents, and the only unknown among Lucy’s neighbors is Henry. Since 6 is even, the count can be odd only if Henry is also innocent. So Henry must be innocent.
15.C1 · Carol → CRIMINAL
Xavi says Jose has an odd number of innocent neighbors. Jose’s neighbors already include three known innocents: Diane, Henry, and Mary, and the only neighbor there whose identity is not yet known is Carol. Since the innocent-neighbor total must stay odd, Carol cannot also be innocent. So Carol must be criminal.