Clues by Sam Oct 03, 2026 Answer – Full Solution Explained
A1
👩🍳
cook
B1
👨🔬
scientist
C1
👨💼
clerk
D1
👩💼
clerk
A2
👨🍳
cook
B2
👩🍳
cook
C2
👨💻
coder
D2
👨🎤
singer
A3
👨🍳
cook
B3
👨💻
coder
C3
👩💻
coder
D3
👨🎤
singer
A4
👮♂️
cop
B4
👩🎤
singer
C4
👩🍳
cook
D4
👨🔬
scientist
A5
👷♀️
builder
B5
👩🍳
cook
C5
👩🍳
cook
D5
👩🔬
scientist
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 03, 2026 — a Hard solved in 16 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 Freya (D1), Jerry (D2), Kumar (A3), Ruby (B4), Tom (D4) and Xia (C5); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.C2 · Isaac → INNOCENT, B3 · Luigi → INNOCENT
Megan says exactly 2 of Hope's innocent neighbors also neighbor Megan. The only people who fit that shared-neighbor spot are Isaac and Luigi. Since those 2 innocent people are still needed there, both of those remaining shared neighbors have to be innocent. So Isaac and Luigi must be innocent.
02.B1 · Chris → INNOCENT
Hope has exactly 7 innocent neighbors, and the only innocent neighbors of Hope who also neighbor Megan are Isaac and Luigi. So the other 5 innocent neighbors of Hope must all be people who do not neighbor Megan: A1 Betty, B1 Chris, C1 Denis, A2 Gabe, and A3 Kumar. If Chris were criminal, then among those five, Betty, Denis, Gabe, and Kumar would still have to be innocent. That would put Betty and Denis as innocents in row 1 with Chris between them. Isaac says all innocents in row 1 are connected, but Betty and Denis would not form one connected block if Chris were criminal. So Chris at B1 must be innocent.
03.C1 · Denis → INNOCENT
Hope’s clue says exactly 7 of her neighbors are innocent, and exactly 2 of those innocent neighbors also neighbor Megan. Those 2 are already fixed as Isaac and Luigi, so the other 5 of Hope’s innocent neighbors must all come from the neighbors of Hope who are not neighbors of Megan: A1 Betty, B1 Chris, C1 Denis, A2 Gabe, and A3 Kumar. If Denis were criminal, then those 5 would have to be A1 Betty, B1 Chris, A2 Gabe, and A3 Kumar plus one more impossible person, so Betty, Gabe, and Kumar are forced innocent. In row 1 that would leave Betty and Chris as known innocents, and Chris’s clue says row 1 has an odd number of innocents, so Freya would also have to be innocent. But Isaac’s clue says all innocents in row 1 are connected, and with Betty, Chris, and Freya innocent while Denis is criminal, the row 1 innocents would be split. So Denis at C1 must be innocent.
04.A4 · Ollie → INNOCENT, A5 · Vicky → INNOCENT
Hope’s clue fixes her neighborhood at exactly 7 innocents, and the only two of those innocents who also neighbor Megan are Isaac and Luigi. Denis’s clue says columns A and C must have the same number of criminals, while Luigi’s clue says column A must have an odd number of criminals. If Ollie and Vicky were both criminals, then Betty, Gabe, Kumar, Susan, and Xia would have to satisfy all of those requirements at the same time, and they cannot. So Ollie at A4 and Vicky at A5 must be innocent.
05.D5 · Zara → INNOCENT
Vicky’s clue says Zara is one of Tom’s four innocent neighbors. That directly identifies Zara as innocent. So Zara must be innocent.
06.D3 · Nick → INNOCENT
Hope has exactly 7 innocent neighbors, and exactly 2 of those innocent neighbors also neighbor Megan. Those 2 are already fixed as Isaac and Luigi, so none of Hope's other neighbors can be innocent neighbors of Megan. If Nick were criminal, then Tom's clue about having exactly 4 innocent neighbors, with Zara as one of them, makes Susan and Xia innocent. Denis's clue then keeps the criminal counts in columns A and C equal, so Betty, Gabe, and Kumar are innocent as well. But then Hope's neighbors would have Chris, Denis, Isaac, Luigi, Betty, Gabe, and Kumar as 7 innocents, and among Hope's innocent neighbors the ones who also neighbor Megan would be Isaac and Luigi as required. Susan and Xia would also be innocent from Tom's clue, which clashes with the restriction created by Megan's clue under the assumption that Nick is criminal. So Nick at D3 must be innocent.
07.C4 · Susan → INNOCENT
Nick’s clue says Megan has exactly 5 innocent neighbors, and exactly 2 of those innocent neighbors also neighbor Ruby. Among the people who are both Megan’s neighbors and Ruby’s neighbors, B3 Luigi is already known innocent and the only other person there is C4 Susan. Since that group needs 1 more innocent person to reach the required total of 2, Susan has to be that innocent person. So Susan must be innocent.
08.C5 · Xia → CRIMINAL
Vicky’s clue says Zara is one of Tom’s exactly 4 innocent neighbors. Tom’s neighbors are Megan, Nick, Susan, Xia, and Zara, and among them Megan, Nick, Susan, and Zara are already known to be innocent. That already fills all 4 innocent-neighbor spots from the clue, so Xia cannot also be innocent. So Xia must be criminal.
09.D4 · Tom → CRIMINAL, B4 · Ruby → CRIMINAL
Megan’s clue says she has exactly 5 innocent neighbors in total. Four of Megan’s neighbors are already known innocent, so there is room for exactly 1 more innocent among B2 Hope, D2 Jerry, B4 Ruby, and D4 Tom. Nick’s clue says exactly 2 of Megan’s 5 innocent neighbors also neighbor Ruby. Those 2 are already Luigi and Susan, so the one extra innocent in Megan’s neighbor list cannot be someone who also neighbors Ruby. Among those four candidates, Ruby and Tom are the ones in that disallowed group. So Ruby and Tom cannot be that remaining innocent neighbor. That makes Ruby and Tom criminal.
10.D2 · Jerry → CRIMINAL
Hope’s clue says she has exactly 7 innocent neighbors, and the only two of those who also neighbor Megan are already fixed as Isaac and Luigi. Gabe’s clue says Gabe has exactly 4 innocent neighbors, while among Gabe’s neighbors only Chris and Luigi are already known innocent, so A1 Betty, B2 Hope, and A3 Kumar have to supply the rest of that requirement. Ruby’s clue also says row 2 has an odd number of innocents, and row 2 currently has Isaac as the only known innocent, with Gabe, Hope, and Jerry undecided. If Jerry were innocent, then Betty, Gabe, Hope, and Kumar would have to satisfy all of those counts at the same time, but they cannot. So Jerry cannot be innocent. That makes Jerry criminal.
11.A2 · Gabe → INNOCENT
Megan’s neighbors must contain exactly 5 innocents, and 4 of them are already known, so the only unknown there, Hope, has to be innocent. Then row 2 has Isaac and Hope as innocents, so it already contains 2 innocents, and Ruby’s clue says the number of innocents in row 2 is odd. The only other unknown person in row 2 is Gabe, so Gabe has to be innocent to make that row’s innocent count odd. So Gabe must be innocent.
12.B2 · Hope → INNOCENT
Nick’s clue says Megan has exactly 5 innocent neighbors, and exactly 2 of those innocents also neighbor Ruby. Those 2 are already Luigi and Susan, so the other 3 innocent neighbors of Megan must be among Megan’s neighbors who do not neighbor Ruby. That non-Ruby-neighbor group is Hope, Isaac, Jerry, Nick, Ruby, and Tom. In that group, Isaac and Nick are already innocent, while Jerry, Ruby, and Tom are criminal, so one more innocent is still needed there. The only unknown person left in that group is Hope, so Hope at B2 must be innocent.
13.B5 · Wanda → INNOCENT
If Wanda were criminal, then row 5 would have 2 criminals. Jerry’s clue says rows 1 and 5 must have the same number of criminals, so row 1 would also need 2 criminals. Row 1 already has Chris and Denis as innocents, and Chris’s clue says row 1 contains an odd number of innocents. That leaves no unknown people in row 1 who can be innocent, so Betty and Freya would both have to be criminals. But then row 1 would have only 1 known criminal while row 5 has 2, which conflicts with the requirement that the two rows have equal numbers of criminals. So Wanda must be innocent.
14.D1 · Freya → CRIMINAL
Hope’s neighbors must contain exactly 7 innocents, and the only unknown people in that group are Betty and Kumar. Megan’s clue also fixes the pattern inside Hope’s neighborhood: exactly 2 of those innocents are neighbors of Megan, namely Isaac and Luigi. The edge clue says the edge cells contain an odd number of innocents. There are already 8 known innocents on the edge, and the only unknown edge people are Betty, Freya, and Kumar. If Freya were innocent, then Betty and Kumar would be the same remaining people that also have to satisfy Hope’s neighborhood clue, and those facts cannot all hold at once. So Freya must be criminal.
15.A1 · Betty → INNOCENT
Chris’s clue says row 1 has an odd number of innocents. In row 1, there are already 2 known innocents, Chris and Denis, and the only unknown person left there is Betty. Since 2 is even, Betty has to be innocent to make the total number of innocents in row 1 odd. So Betty must be innocent.
16.A3 · Kumar → CRIMINAL
Hope’s neighbors must contain exactly 7 innocents in total. Of those innocents, exactly 2 also neighbor Megan, and those 2 are already Isaac and Luigi. That leaves the other Hope-neighbors who do not neighbor Megan as Betty, Chris, Denis, Gabe, Kumar, and Megan, and among that group Betty, Chris, Denis, Gabe, and Megan are already known innocents. Since the full total of 7 innocent neighbors for Hope is already reached without Kumar, Kumar cannot be innocent. So Kumar must be criminal.