Clues by Sam Mar 11, 2026 Answer – Full Solution Explained
A1
👩🍳
cook
B1
👷♀️
builder
C1
👩🌾
farmer
D1
👮♀️
cop
A2
💂♂️
guard
B2
💂♂️
guard
C2
👮♂️
cop
D2
👮♀️
cop
A3
💂♂️
guard
B3
👨💼
clerk
C3
👨🏫
teacher
D3
🕵️♂️
sleuth
A4
👩💼
clerk
B4
👩💼
clerk
C4
👩🏫
teacher
D4
🕵️♀️
sleuth
A5
👩🍳
cook
B5
👨🌾
farmer
C5
👷♂️
builder
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 — 11 criminals.
Clues by Sam answer for Mar 11, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 11 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Anna (A1), Barb (B1), Dana (D1), Gus (C2), Hope (D2), Jason (B3), Klay (C3), Megan (A4), Nicole (B4), Sue (C4) and Vince (B5); the remaining 9 suspects are innocent.
The deduction chain, in plain English
01.C3 · Klay → CRIMINAL, C4 · Sue → CRIMINAL
Dana is at D1, and her only neighbors in column C are C1 (Carol) and C2 (Gus). Wally’s clue says there are exactly three criminals in column C, but only one of those three is Dana’s neighbor, so at most one of C1 and C2 can be a criminal from that set. Since Wally at C5 is already INNOCENT, the other two criminals in column C must come from the remaining non-neighbors of Dana in that column, which are C3 (Klay) and C4 (Sue). Therefore, we can determine that C3 Klay is CRIMINAL and C4 Sue is CRIMINAL.
02.B3 · Jason → CRIMINAL
Nicole is at B4, so her neighbors are A3 Isaac, B3 Jason, C3 Klay, A4 Megan, C4 Sue, A5 Uma, B5 Vince, and C5 Wally. Klay says that exactly 3 of Nicole’s neighbors who are on the edge are innocent; among Nicole’s neighbors, the edge positions are A3, A4, A5, B5, and C5, and since Wally at C5 is already innocent, that forces exactly 2 of A3, A4, A5, and B5 to be innocent, so exactly 2 of them are criminals. Sue says the total number of criminals neighboring Nicole is odd, and since Klay and Sue are already two criminals, the remaining neighbors must contain an odd number of criminals; but the four edge-neighbors A3, A4, A5, and B5 contain exactly 2 criminals (an even number), so Jason at B3 must be the extra criminal to make the total odd. Therefore, we can determine that B3 Jason is CRIMINAL.
03.D2 · Hope → CRIMINAL
Dana is in the top-right corner at D1, so her only neighbors are Carol at C1, Gus at C2, and Hope at D2. Wally’s clue says there are exactly three criminals in column C, and only one of those column-C criminals is Dana’s neighbor; since the only column-C neighbors Dana has are Carol and Gus, this forces that exactly one of Carol and Gus is a criminal and the other is innocent. Jason’s clue says Dana has only one innocent neighbor total, and we already have that one innocent neighbor among Carol and Gus, so Hope cannot also be innocent. Therefore, we can determine that D2 (Hope) is CRIMINAL.
04.B5 · Vince → CRIMINAL
Nicole is at B4, and her neighbors that are on the board edge are Isaac (A3), Megan (A4), Uma (A5), Vince (B5), and Wally (C5). Klay says that exactly 3 of those edge-neighbors are innocent, and since Wally is already known to be innocent, that means exactly 2 of Isaac, Megan, Uma, and Vince are innocent. Hope says that among the people below Anna who neighbor Nicole, exactly 2 are innocent; those people are exactly Isaac, Megan, and Uma, so Isaac, Megan, and Uma already account for the 2 innocents. That leaves no room for Vince to be innocent among the four, so Vince must be criminal. Therefore, we can determine that B5 Vince is CRIMINAL.
05.B2 · Frank → INNOCENT
In column C, we already have two known criminals: Klay at C3 and Sue at C4, and Wally at C5 is known innocent. Wally’s clue says there are exactly three criminals in column C, so the third one must be either Carol at C1 or Gus at C2, which means at least one of Barb’s column-C neighbors (C1 or C2) is a criminal. Vince’s clue says Barb has exactly two criminal neighbors in total, and exactly one of those two is in column A (so one of A1 or A2 is criminal). If Frank at B2 were also a criminal, Barb would have at least three criminal neighbors: Frank at B2, one of A1/A2, and one of C1/C2, which is impossible under Vince’s clue. Therefore, we can determine that B2 Frank is INNOCENT.
06.A3 · Isaac → INNOCENT
Barb is at B1, and the only neighbors of Barb in column A are Anna at A1 and Erwin at A2. Vince’s clue says Barb has exactly two neighboring criminals, and exactly one of those two is in column A, so exactly one of Anna or Erwin is a criminal and the other is innocent. Frank’s clue says there are exactly two innocents above Megan at A4, which means exactly two of Anna (A1), Erwin (A2), and Isaac (A3) are innocent; since Anna and Erwin already include exactly one innocent, Isaac must be the second innocent. Therefore, we can determine that A3 · Isaac is INNOCENT.
07.B4 · Nicole → CRIMINAL
Nicole is at B4, and her edge-neighbors are A3 Isaac, A4 Megan, A5 Uma, B5 Vince, and C5 Wally. Klay says exactly 3 of those edge-neighbors are innocent; since Isaac and Wally are already innocent and Vince is already criminal, that means exactly one of Megan and Uma is innocent. Isaac also says Vince has exactly 2 innocent neighbors, and Vince’s neighbors that could still be innocent are Megan, Nicole, and Uma, with Wally already counting as one innocent neighbor and Sue already known criminal. So among Megan, Nicole, and Uma, exactly one is innocent, and because exactly one of Megan and Uma is that innocent, Nicole cannot be the innocent one. Therefore, we can determine that B4 Nicole is CRIMINAL.
08.B1 · Barb → CRIMINAL
Barb is at B1, and Vince’s clue says Barb has exactly two criminal neighbors, with exactly one of those criminals in column A. Since Barb’s only column A neighbors are Anna at A1 and Erwin at A2, that means exactly one of Anna and Erwin is a criminal, and since Frank at B2 is already innocent, the other criminal neighbor of Barb must come from Carol at C1 or Gus at C2, so exactly one of Carol and Gus is a criminal as well. Nicole’s clue says Frank has exactly five criminal neighbors, and we already see two of them are Jason at B3 and Klay at C3, so among Frank’s other neighboring unknowns (Anna, Barb, Carol, Erwin, Gus) there must be exactly three criminals. But Vince’s clue already limits Anna, Carol, Erwin, and Gus to exactly two criminals in total, so the third criminal among those five must be Barb. Therefore, we can determine that Barb is CRIMINAL.
09.D3 · Logan → INNOCENT
Frank is at B2, and his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Nicole’s clue says Frank has exactly five neighboring criminals, and exactly one of those five is to the left of Hope, which means in row 2 at A2 or C2, so exactly one of Erwin at A2 and Gus at C2 is a criminal. Since Barb, Jason, and Klay are already three neighboring criminals of Frank, that makes row 2 have exactly two criminals in total, namely Hope plus that one of Erwin or Gus. Barb’s clue says rows 2 and 3 have the same number of criminals, so row 3 must also have exactly two criminals, and because Jason and Klay already fill those two spots, Logan cannot be a criminal. Therefore, we can determine that Logan is INNOCENT.
10.D1 · Dana → CRIMINAL
Dana is at D1, so Dana’s neighbors are exactly C1 (Carol), C2 (Gus), and D2 (Hope). Wally’s clue says there are exactly three criminals in column C, and only one of those three is Dana’s neighbor; since the only people in column C who can be Dana’s neighbors are Carol at C1 and Gus at C2, this forces exactly one of Carol and Gus to be a criminal. Logan’s clue says Hope has an odd number of neighboring criminals, and Hope’s neighbors are Carol (C1), Gus (C2), Klay (C3), Dana (D1), and Logan (D3); Klay is already a criminal and Logan is already innocent, so Carol, Gus, and Dana together must contribute an even number of criminals to keep the total odd. Because Carol and Gus contribute exactly one criminal between them, Dana must also be a criminal to make that combined number even. Therefore, we can determine that D1 Dana is CRIMINAL.
11.A4 · Megan → CRIMINAL
Frank is at B2, and the only neighbors of Frank that are to the left of Hope (who is at D2) are Erwin at A2 and Gus at C2. Nicole’s clue says that among the five criminals neighboring Frank, exactly one is to the left of Hope, so exactly one of Erwin and Gus is a criminal. Jason is at B3, and Dana’s clue says Jason has an odd number of criminal neighbors; since Klay, Nicole, and Sue are already three criminal neighbors of Jason, the remaining neighboring spots A2, C2, and A4 must add an even number of criminals. Because A2 and C2 contribute exactly one criminal from Nicole’s clue, A4 must also be a criminal to make that added number even. Therefore, we can determine that A4 Megan is CRIMINAL.
12.A5 · Uma → INNOCENT
Nicole is at B4, and her neighbors are A3, B3, C3, A4, C4, A5, B5, and C5. Among those neighbors, the ones on the edge of the board are A3 Isaac, A4 Megan, A5 Uma, B5 Vince, and C5 Wally. Klay’s clue says exactly 3 of these edge-neighbors are innocent, and we already know Isaac and Wally are innocent while Megan and Vince are criminals, so the only way to reach 3 innocents is for Uma to be the third innocent. Therefore, we can determine that A5 Uma is INNOCENT.
13.D5 · Zed → INNOCENT
Frank is at B2, and his only neighbors that are in row 2 (and therefore “to the left of Hope” at D2) are A2 Erwin and C2 Gus. Nicole’s clue says that among the five criminal neighbors Frank has, exactly one is to the left of Hope, so exactly one of Erwin and Gus is a criminal. That means row 2 can have at most two criminals total: Hope at D2 plus at most one of Erwin or Gus, since Frank is already innocent. Uma’s clue says row 2 has more criminals than row 5, so row 5 cannot have two criminals and must have at most one. Row 5 already has Vince as a criminal, so Zed at D5 cannot be a criminal and must be innocent. Therefore, we can determine that D5 Zed is INNOCENT.
14.D4 · Tina → INNOCENT
In column C, we already know C3 (Klay) and C4 (Sue) are criminals, and Wally’s clue says there are exactly three criminals in column C and only one of those three is Dana’s neighbor. Dana (at D1) has only two neighbors in column C: C1 and C2, so exactly one of C1 and C2 must be a criminal and the other must be innocent. That makes column C have exactly two innocents in total: C5 (Wally) and whichever of C1 or C2 is innocent. Megan’s clue says only one column has exactly two innocents, so no other column is allowed to have exactly two innocents. Column D already has two known innocents, D3 (Logan) and D5 (Zed), so Tina at D4 cannot be a criminal because that would leave column D with exactly two innocents. Therefore, we can determine that D4 Tina is INNOCENT.
15.A2 · Erwin → INNOCENT
Dana is in the corner at D1, and her neighbors include Hope at D2 plus the two column C people next to her, Carol at C1 and Gus at C2. Wally’s clue says that among the three criminals in column C, only one is Dana’s neighbor; since the only column C people who can be Dana’s neighbors are exactly Carol and Gus, this forces exactly one of Carol or Gus to be a criminal. That means Dana has exactly two criminal neighbors in total: Hope for sure, plus exactly one of Carol or Gus. Tina’s clue says only one corner person has exactly two criminal neighbors, so A1 cannot also have exactly two criminal neighbors; but A1 would have exactly two only if Erwin at A2 were a criminal (since Barb at B1 is already a criminal and Frank at B2 is innocent). Therefore, we can determine that A2 Erwin is INNOCENT.
16.A1 · Anna → CRIMINAL
Barb is at B1, so her neighbors are A1, A2, B2, C1, and C2. Vince’s clue says Barb has exactly two neighboring criminals, and exactly one of those two is in column A. The only column A neighbors of Barb are A1 and A2, and A2 (Erwin) is already known to be innocent, so A1 must be the one neighboring criminal in column A. Therefore, we can determine that A1 Anna is CRIMINAL.
17.C1 · Carol → INNOCENT, C2 · Gus → CRIMINAL
Frank is at B2, so his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Nicole’s clue says Frank has exactly five criminal neighbors, and exactly one of those five is to the left of Hope; since Hope is at D2, the only neighbor of Frank that is in Hope’s row and to Hope’s left is Gus at C2 (because A2 is Erwin, who is already innocent). That forces Gus to be the one criminal neighbor to the left of Hope, so Gus must be criminal. With Gus criminal, Frank’s five criminal neighbors are Anna, Barb, Jason, Klay, and Gus, so Carol at C1 cannot be criminal and must be innocent. Therefore, we can determine that C1 Carol is INNOCENT and C2 Gus is CRIMINAL.