Clues by Sam Oct 07, 2026 Answer – Full Solution Explained
A1
🕵️♂️
sleuth
B1
👩🔧
mech
C1
👨🔧
mech
D1
👮♂️
cop
A2
👮♂️
cop
B2
👩🎨
painter
C2
👨🎨
painter
D2
👨🎨
painter
A3
👨💼
clerk
B3
👩⚖️
judge
C3
👩💼
clerk
D3
👮♀️
cop
A4
🕵️♂️
sleuth
B4
👩✈️
pilot
C4
👨✈️
pilot
D4
👨✈️
pilot
A5
👩🎤
singer
B5
👩🎤
singer
C5
👩⚖️
judge
D5
👩⚖️
judge
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 — 13 criminals.
Clues by Sam answer for Oct 07, 2026 — a Tricky solved in 17 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 13 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Aaron (A1), Gabe (A2), Hazel (B2), Isaac (C2), Jerry (D2), Nancy (C3), Olivia (D3), Rose (B4), Scott (C4), Thor (D4), Wanda (B5), Xia (C5) and Zoe (D5); the remaining 7 suspects are innocent.
The deduction chain, in plain English
01.C4 · Scott → CRIMINAL, C5 · Xia → CRIMINAL
Debra’s clue says there are no innocents in the overlap between the people below Eric and Wanda’s neighbors. That shared group is exactly Scott and Xia, and there are already 0 known innocents in it. So Scott and Xia cannot be innocent. That makes Scott and Xia criminal.
02.A1 · Aaron → CRIMINAL
Scott’s clue says Aaron is one of the exactly 2 criminals in column A. That directly identifies Aaron as a criminal. So Aaron must be criminal.
03.A4 · Phil → INNOCENT
If Phil were criminal, then column A would already have its two criminals: Aaron and Phil, because Scott says Aaron is one of exactly two criminals in column A. That means Gabe, Larry, and Vera would all have to be innocent. But then the cops would already include one innocent, Gabe, while the sleuths would include no innocent at all, since Aaron would be criminal and Phil was assumed criminal. Xia's clue says the number of innocent cops and innocent sleuths must be the same, so that cannot happen. So Phil at A4 must be innocent.
04.B5 · Wanda → CRIMINAL
Aaron’s clue fixes column A at exactly 2 criminals, and Phil’s clue says column A must end with more innocents than any other column. Aaron’s other clue also requires an odd number of innocents among the column B neighbors of Nancy, namely Hazel, Mary, and Rose. If Wanda were innocent, then column B would already have Debra and Wanda as innocents, while Hazel, Mary, and Rose would still have to meet that odd-neighbor requirement. At the same time, column A would still have to stay at exactly 2 criminals and also finish with more innocents than every other column. Those requirements conflict if Wanda is innocent. So Wanda must be criminal.
05.A3 · Larry → INNOCENT
Gabe’s neighbors must contain exactly 3 innocents. Among those neighbors, Aaron is already criminal and Debra is already innocent, so among Hazel, Larry, and Mary there is room for exactly 1 criminal. In that same group, the only people who can supply that 1 criminal are Hazel and Mary, so Larry cannot be the criminal there. So Larry must be innocent.
06.B3 · Mary → INNOCENT
Larry’s clue says exactly 2 of Hazel’s innocent neighbors are above Wanda. Among Hazel’s neighbors who are above Wanda, Debra is already known to be innocent, and the only other person there is Mary. Since that group still needs one more innocent to reach the required total of 2, Mary has to be that innocent. So Mary must be innocent.
07.B2 · Hazel → CRIMINAL
Wanda’s clue says Gabe has exactly 3 innocent neighbors. Gabe’s neighbors already include 3 known innocents, and the only neighbor there whose identity is still unknown is Hazel. Since all 3 innocent spots are already accounted for, Hazel cannot be innocent. So Hazel must be criminal.
08.B4 · Rose → CRIMINAL
Aaron’s clue is about the people who are both in column B and neighbors of Nancy. That shared group is Hazel, Mary, and Rose. The clue says an odd number of those people are innocent. Hazel is already criminal and Mary is already innocent, so there is currently exactly one innocent in that group. If Rose were innocent too, the group would have two innocents, which is even, not odd. So Rose must be criminal.
09.C3 · Nancy → CRIMINAL
Isaac has exactly 4 innocent neighbors in total, and exactly 1 of those innocent neighbors is also a neighbor of Rose. The only people who are both neighbors of Isaac and neighbors of Rose are Mary and Nancy. Mary is already known to be innocent, so that one allowed innocent in the shared group is already used up. That means Nancy cannot be innocent, so Nancy must be criminal.
10.A2 · Gabe → CRIMINAL
If Gabe were innocent, then Hazel’s clue would already have its two innocents above Wanda as Debra and Mary, so the other innocent neighbor Hazel still needs cannot be Eric. That makes Eric criminal, and since Hazel’s neighbors must contain exactly four innocents in total, Isaac must be the remaining innocent there. But then the cop and sleuth clue gives the cops the same number of innocents as the sleuths, so with one innocent sleuth, the innocent cop would already be Gabe. That leaves Floyd and Olivia as criminals. Now Isaac’s clue says exactly one of Isaac’s four innocent neighbors is Rose’s neighbor, and the only such person there is Mary, so Isaac still needs three innocent neighbors who are not Rose’s neighbors. Among Isaac’s unknown neighbors, Eric, Floyd, and Olivia are already criminal, so that cannot be met. So Gabe at A2 must be criminal.
11.A5 · Vera → INNOCENT
Scott’s clue says Aaron is one of exactly 2 criminals in column A. In column A, the two criminals are already Aaron and Gabe, while Larry and Phil are innocent and Vera is the only person there not yet identified. Since the two criminal spots in that column are already filled, Vera cannot be a criminal. So Vera must be innocent.
12.D4 · Thor → CRIMINAL, C1 · Eric → INNOCENT
Isaac’s clue says his neighbors contain exactly 4 innocents, and exactly 1 of those innocents is also a neighbor of Rose. Among the people who are both Isaac’s neighbors and Rose’s neighbors, Mary is innocent and Nancy is criminal, so Mary is already that one innocent. That means the other innocents around Isaac have to come from the remaining Isaac-neighbor group, which includes Eric, Floyd, Jerry, and Olivia. Phil’s clue fixes column A at 3 innocents, more than any other column, and Mary’s clue says the people above Zoe contain an odd number of innocents. If Thor were innocent and Eric were criminal, then the remaining people involved here could not satisfy all of those requirements at the same time. So Thor must be criminal and Eric must be innocent.
13.C2 · Isaac → CRIMINAL
Larry’s clue says Hazel has exactly 4 innocent neighbors, and exactly 2 of those innocent neighbors are above Wanda. Those 2 are already Debra and Mary. That means the innocent neighbors of Hazel who are not above Wanda must account for exactly 2 more innocents. In that not-above-Wanda group, Eric and Larry are already innocent, so that quota is full. Isaac is the only unknown person left in Hazel’s neighbors, and he is in that not-above-Wanda group, so he cannot also be innocent. So Isaac must be criminal.
14.D2 · Jerry → CRIMINAL
Isaac’s clue says his neighbors contain exactly 4 innocents, and the only innocent among the Isaac-neighbors who is also a neighbor of Rose is Mary. If Jerry were innocent, then with Debra, Eric, and Mary already innocent, Isaac’s four innocents would be fully used up. That would leave Floyd and Olivia not innocent. But Xia’s clue says the number of innocent cops equals the number of innocent sleuths, and there is already 1 innocent sleuth, Phil, while no cop is known innocent. So Floyd and Olivia would both have to be criminal, leaving 0 innocent cops, which clashes with Xia’s clue. So Jerry at D2 must be criminal.
15.D5 · Zoe → CRIMINAL
Isaac’s neighbors must contain exactly 4 innocents, and 3 of those are already known: Debra, Eric, and Mary. Rose’s clue also says exactly 1 innocent among Isaac’s neighbors is Rose’s neighbor, and Mary is already that 1. So the remaining innocent among Isaac’s neighbors has to come from the two unknown Isaac-neighbors in column D, Floyd or Olivia, which means exactly one of Floyd and Olivia is innocent. Thor’s clue says column D has an odd number of innocents. If Zoe were innocent too, then column D would contain Zoe plus exactly one of Floyd and Olivia as innocents, making 2 innocents there, which is even, not odd. So Zoe must be criminal.
16.D3 · Olivia → CRIMINAL
Zoe’s clue says all innocents in row 3 must be connected as one orthogonally connected block. In row 3, Larry and Mary are already innocent, Nancy is criminal, and if Olivia were innocent then the innocents in that row would be Larry, Mary, and Olivia. But with Nancy criminal between Mary and Olivia, those row 3 innocents would not form one connected block, so that cannot happen. So Olivia must be criminal.
17.D1 · Floyd → INNOCENT
Isaac’s neighbors must contain exactly 4 innocents in total, and Rose’s clue says exactly 1 of those innocents is also Rose’s neighbor. That means the other 3 innocent neighbors of Isaac must be people who are not neighbors of Rose. Among Isaac’s neighbors who are not neighbors of Rose, Debra and Eric are already known innocents, while Hazel, Jerry, and Olivia are criminals. That group still needs 1 more innocent, and the only unknown person left in it is Floyd. So Floyd must be innocent.