Clues by Sam Feb 11, 2026 Answer – Full Solution Explained
A1
👩🍳
cook
B1
👩⚖️
judge
C1
👨🔧
mech
D1
💂♀️
guard
A2
👨⚕️
doctor
B2
👨🍳
cook
C2
🕵️♂️
sleuth
D2
👩⚖️
judge
A3
👨⚕️
doctor
B3
👷♂️
builder
C3
👷♀️
builder
D3
👨🔧
mech
A4
👨🏫
teacher
B4
👨💼
clerk
C4
🕵️♀️
sleuth
D4
💂♀️
guard
A5
👩🏫
teacher
B5
👩💼
clerk
C5
🕵️♂️
sleuth
D5
👩🍳
cook
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 11, 2026 — a Medium solved in 16 steps
Today's Clues by Sam puzzle is rated Medium and resolves with 5 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Betty (B1), Erwin (A2), Kyle (A3), Olof (A4) and Tina (C4); the remaining 15 suspects are innocent.
The deduction chain, in plain English
01.C2 · Gabe → INNOCENT, D2 · Hilda → INNOCENT
Floyd is at B2, so the people to the right of Floyd in the same row are Gabe at C2 and Hilda at D2. Vera’s clue says there are exactly 2 innocents to the right of Floyd. Since there are only two people to the right of Floyd at all, both of them must be those two innocents. Therefore, we can determine that C2 Gabe is INNOCENT and D2 Hilda is INNOCENT.
02.B4 · Steve → INNOCENT
The people above Wanda in column B are Betty (B1), Floyd (B2), Larry (B3), and Steve (B4). Erwin is at A2, so among those four, the only one who is not Erwin’s neighbor is Steve, because Betty, Floyd, and Larry all touch Erwin diagonally or directly. Gabe’s clue says there are three innocents above Wanda, and exactly two of those three are Erwin’s neighbors, so the remaining one innocent must be someone above Wanda who is not Erwin’s neighbor. Since Steve is the only person above Wanda who is not Erwin’s neighbor, Steve must be that remaining innocent. Therefore, we can determine that B4 Steve is INNOCENT.
03.C4 · Tina → CRIMINAL
Uma is at D4, so the people to her left are Olof at A4, Steve at B4, and Tina at C4. Megan is at C3, and among those three, the ones who neighbor Megan are Steve (B4) and Tina (C4). Hilda’s clue says that exactly one innocent to the left of Uma is neighboring Megan, and since Steve is already known to be innocent and he neighbors Megan, he must be that one. That means Tina cannot be innocent, so she must be the other status. Therefore, we can determine that C4 Tina is CRIMINAL.
04.B2 · Floyd → INNOCENT
The people above Wanda in column B are Betty at B1, Floyd at B2, Larry at B3, and Steve at B4. Gabe says there are exactly three innocents among those four, and exactly two of those three innocents are neighbors of Erwin; since Steve at B4 is the only one of the four who is not a neighbor of Erwin, Steve must be one of the innocents, and the other two innocents must be among B1, B2, and B3, leaving exactly one criminal among B1, B2, and B3. Tina says that among Floyd’s five innocent neighbors, only one is above Steve; the only neighbors of Floyd that are above Steve are Betty at B1 and Larry at B3, so exactly one of B1 and B3 is innocent and the other is criminal. That places the single criminal among B1, B2, and B3 on B1 or B3, so Floyd at B2 cannot be that criminal and must be innocent. Therefore, we can determine that B2 Floyd is INNOCENT.
05.B5 · Wanda → INNOCENT
Above Wanda at B5 are the four people in column B: Betty at B1, Floyd at B2, Larry at B3, and Steve at B4. Gabe’s clue talks about “the 3 innocents above Wanda,” which means there are exactly three innocents among those four people, so exactly one of Betty and Larry must be a criminal (since Floyd and Steve are already known innocents). Floyd’s clue says there are exactly four innocents in column B, so among the three unknowns in that column (Betty, Larry, and Wanda) exactly two must be innocent. Because Betty and Larry together provide only one innocent, Wanda must be the other innocent. Therefore, we can determine that B5 Wanda is INNOCENT.
06.B3 · Larry → INNOCENT
Wanda’s clue explicitly says that Larry is one of Steve’s innocent neighbors, and everyone’s clues are always true. Therefore, we can determine that B3 Larry is INNOCENT.
07.B1 · Betty → CRIMINAL
Wanda is at B5, so the people above her are B1 Betty, B2 Floyd, B3 Larry, and B4 Steve. We already know Floyd, Larry, and Steve are INNOCENT. Gabe’s clue talks about “the 3 innocents above Wanda,” which means there are exactly three innocents among those four people above her. Since Floyd, Larry, and Steve already account for those three innocents, Betty cannot be innocent. Therefore, we can determine that B1 Betty is CRIMINAL.
08.D3 · Noah → INNOCENT
Floyd is at B2 and Steve is at B4, so their common neighbors are exactly A3 (Kyle), B3 (Larry), and C3 (Megan). Larry at B3 is already innocent, and the clue says Floyd and Steve have 2 innocent neighbors in common, so exactly one of Kyle (A3) and Megan (C3) must also be innocent. That means row 3 already contains exactly two innocents if Noah at D3 is not innocent: Larry plus exactly one of Kyle or Megan. But Betty’s clue says row 4 is the only row with exactly 2 innocents, so row 3 cannot end up with exactly 2 innocents. The only way to raise row 3 above two innocents, given Kyle and Megan cannot both be innocent, is for Noah at D3 to be innocent. Therefore, we can determine that D3 Noah is INNOCENT.
09.A4 · Olof → CRIMINAL
Noah’s clue explicitly states that “Olof is a criminal.” Since all clues are truthful, this directly fixes Olof’s identity without needing any other relationships or counts on the board. Therefore, we can determine that A4 Olof is CRIMINAL.
10.D4 · Uma → INNOCENT
In row 4, we already know A4 Olof is a criminal, B4 Steve is an innocent, and C4 Tina is a criminal, leaving only D4 Uma unknown. Betty’s clue says that row 4 has exactly 2 innocents (and no other row has exactly 2), so row 4 must contain one more innocent besides Steve. Since the only undecided person in row 4 is Uma, she has to be that second innocent. Therefore, we can determine that D4 Uma is INNOCENT.
11.C1 · Chris → INNOCENT, C3 · Megan → INNOCENT
Floyd is at B2, so his neighbors are A1, A2, B1, C1, C2, A3, B3, and C3. Tina’s clue says that Floyd has exactly five innocent neighbors; since Gabe at C2 and Larry at B3 are already innocent and Betty at B1 is criminal, that means exactly three of the five unknown neighbors A1, A2, C1, A3, and C3 must be innocent. Uma’s clue says there are exactly two criminals above Olof, and the three people above Olof are A1, A2, and A3, so only one of A1, A2, and A3 can be innocent. Therefore, to reach the needed three innocents among A1, A2, C1, A3, and C3, both C1 and C3 must be innocent. Therefore, we can determine that C1 Chris is INNOCENT and C3 Megan is INNOCENT.
12.A3 · Kyle → CRIMINAL
Floyd at B2 and Steve at B4 share exactly three common neighbors: Kyle at A3, Larry at B3, and Megan at C3. Larry and Megan are already known to be innocent, so that gives Floyd and Steve two innocent neighbors in common. Larry’s clue says they have 2 innocent neighbors in common in total, so Kyle cannot also be innocent. Therefore, we can determine that A3 Kyle is CRIMINAL.
13.C5 · Xavi → INNOCENT
Steve is at B4, and his eight neighbors are Kyle (A3), Larry (B3), Megan (C3), Olof (A4), Tina (C4), Vera (A5), Wanda (B5), and Xavi (C5). Wanda’s clue says Larry is one of Steve’s 5 innocent neighbors, which means Steve has exactly five innocent neighbors in total. Among Steve’s neighbors we already know four innocents: Larry, Megan, Vera, and Wanda, and we also know three criminals: Kyle, Olof, and Tina. That leaves only Xavi as the only neighbor who can make the innocent-neighbor count reach five. Therefore, we can determine that C5 Xavi is INNOCENT.
14.D1 · Donna → INNOCENT, D5 · Zoe → INNOCENT
The edge clue from Megan says there are exactly 10 innocents on the edges, so since there are 14 edge positions in total, there must be exactly 4 edge criminals. We already have three edge criminals fixed: Betty at B1, Kyle at A3, and Olof at A4, so there is room for only one more edge criminal anywhere else. Tina’s clue talks about “the 5 innocents neighboring Floyd,” and among Floyd’s eight neighbors we already know four innocents (Chris, Gabe, Larry, and Megan) and two criminals (Betty and Kyle), so the only way Floyd can have exactly five innocent neighbors is if exactly one of Alice at A1 and Erwin at A2 is innocent and the other is criminal. That means the one remaining edge criminal required by the edge count must be either Alice at A1 or Erwin at A2, leaving no possibility for Donna at D1 or Zoe at D5 to be criminals. Therefore, we can determine that D1 Donna is INNOCENT and D5 Zoe is INNOCENT.
15.A1 · Alice → INNOCENT
In row 4, we already see exactly two innocents: B4 Steve and D4 Uma, since A4 Olof and C4 Tina are criminals. Betty’s clue says row 4 is the only row with exactly 2 innocents, so no other row is allowed to have exactly two innocents. In row 1, C1 Chris and D1 Donna are already innocent, and B1 Betty is criminal, so if A1 Alice were criminal then row 1 would have exactly two innocents, which the clue forbids. Therefore, we can determine that A1 Alice is INNOCENT.
16.A2 · Erwin → CRIMINAL
Floyd is at B2, so his neighbors are A1, B1, C1, A2, C2, A3, B3, and C3. Tina’s clue says “the 5 innocents neighboring Floyd,” which means Floyd has exactly five innocent neighbors. Among those eight neighbors, A1, C1, C2, B3, and C3 are already known innocents, while B1 and A3 are already known criminals, leaving only A2 as the undecided one; to keep the total number of innocent neighbors at exactly five, A2 cannot be innocent. Therefore, we can determine that A2 (Erwin) is CRIMINAL.