Clues by Sam Aug 07, 2026 Answer – Full Solution Explained

01.A2 · Gary → INNOCENT

Karen’s clue says Gary is one of two or more innocents in row 2. That directly includes Gary among the innocents there, and also says he is not the only innocent in that row. So Gary at A2 must be innocent.

02.B1 · Bobby → CRIMINAL, D1 · Evie → CRIMINAL

Karen’s clue says Gary is one of at least two innocents in row 2, and Gary is already innocent, so row 2 must contain at least one more innocent among Hilda, Isaac, and Julie. Gary’s clue says all five of Daniel’s neighbors include exactly one innocent. Among those five neighbors, the only people in row 2 are Hilda, Isaac, and Julie, so that one innocent neighbor has to come from that row-2 group. That means Bobby and Evie cannot be Daniel’s one innocent neighbor. So Bobby and Evie must be criminal.

03.A1 · Amy → CRIMINAL, C1 · Daniel → CRIMINAL

Bobby’s neighbors contain exactly 2 innocents in total. The clue also says exactly 1 of those innocents is a neighbor of Daniel, and among Bobby’s neighbors the people who are Daniel’s neighbors are Hilda and Isaac. That means the other innocent among Bobby’s neighbors has to come from the people there who are not Daniel’s neighbors: Amy, Daniel, and Gary. Gary is already known to be innocent, so that uses up that one innocent spot. Amy and Daniel therefore cannot be innocent. So Amy and Daniel must be criminal.

04.D2 · Julie → CRIMINAL

Bobby’s clue fixes the two unknown neighbors Hilda and Isaac so that exactly one of them is innocent. Daniel’s clue says Daniel has exactly four criminal neighbors, so among Daniel’s five neighbors there is exactly one innocent. Hilda and Isaac are both neighbors of Daniel, so that one innocent in Daniel’s neighbor group is already accounted for within Hilda and Isaac. The only other person in Daniel’s neighbor group outside that pair is Julie, so Julie cannot be innocent. So Julie must be criminal.

05.D5 · Zed → INNOCENT

Bobby says Steve has exactly 6 innocent neighbors, so among Steve's 8 neighbors there must be exactly 2 criminals. Those 2 criminals therefore have to come from Logan, Mary, Nancy, Ruby, Tyler, Wally, and Xavi. That leaves Zed outside the two criminal spots, so Zed must be innocent.

06.B5 · Wally → CRIMINAL

07.D4 · Tyler → INNOCENT

Steve has exactly 6 innocent neighbors, and among Steve's 8 neighbors, Wally is already a known criminal and Zed is already a known innocent. That means the other 6 neighboring spots contain exactly 1 criminal in total: Logan, Mary, Nancy, Ruby, Tyler, and Xavi. The remaining criminal among Steve's unknown neighbors must come from Logan, Mary, Nancy, Ruby, or Xavi, so Tyler cannot be that criminal. So Tyler must be innocent.

08.C4 · Steve → INNOCENT

Evie’s neighbors must contain more criminals than Ruby’s neighbors, and right now Evie already has 2 known criminal neighbors while Ruby has 1. Also, the innocents in row 4 have to form one connected block, and row 4 is Pam, Ruby, Steve, and Tyler, with Tyler already innocent. If Steve were criminal, the remaining people involved here could not satisfy both of those requirements at the same time. That rules out Steve being criminal. So Steve must be innocent.

09.B4 · Ruby → INNOCENT

Row 4 has Pam, Ruby, Steve, and Tyler, and Steve and Tyler are already innocent. Daniel’s clue says all innocents in row 4 must be connected in one orthogonal block. Steve and Tyler already make a connected innocent block at C4 and D4, so if Ruby were criminal, the only way to add any more innocent in row 4 would be Pam at A4, which would be separated from Steve and Tyler. At the same time, Steve’s clue says Vicky’s neighbors have more innocents than Evie’s neighbors. Both sides currently have 0 known innocents, and among Vicky’s neighbors the only candidates to provide innocents are Pam and Ruby. If Ruby were criminal, those clues cannot be satisfied together by Pam and Isaac. So Ruby must be innocent.

10.C5 · Xavi → INNOCENT

Steve must have exactly 6 innocent neighbors, so among his four unknown neighbors there is room for exactly 1 criminal. Ruby also says the people to the right of Karen contain an odd number of criminals, and those people are Logan, Mary, and Nancy, so the 1 criminal among Steve's unknown neighbors has to be in that trio. That leaves Xavi unable to be that criminal. So Xavi must be innocent.

11.A4 · Pam → INNOCENT

12.B3 · Logan → INNOCENT

Steve and Tyler have exactly 3 innocent neighbors in common, and in that shared group Xavi and Zed are already innocent. So among the other shared neighbors, Mary and Nancy together contribute exactly 1 more innocent. Steve has exactly 6 innocent neighbors in total. Ruby, Tyler, Xavi, and Zed already account for 4 of them, so the three unknown neighbors Logan, Mary, and Nancy must contribute exactly 2 innocents. Since Mary and Nancy contribute only 1 of those 2, the remaining one has to be Logan. So Logan must be innocent.

13.A5 · Vicky → INNOCENT

Evie’s neighbors must have more criminals than Ruby’s neighbors. Right now Evie’s neighbors already have 2 known criminals, while Ruby’s neighbors have 1 known criminal, and the unknown people involved in these clues are Isaac, Mary, and Vicky. Column C must contain an odd number of criminals, and it currently has 1 known criminal, with Isaac and Mary the only unknown people there. If Vicky were criminal, then Ruby’s neighbors would rise to 2 criminals, so Isaac and Mary would have to meet the column C odd-count clue while still keeping Evie’s total higher than Ruby’s, and that cannot be done. So Vicky must be innocent.

14.B2 · Hilda → INNOCENT

Steve’s clue says his neighbors contain exactly 6 innocents. Right now 5 of Steve’s neighbors are already known innocent, so among Mary and Nancy exactly one must be innocent and the other must be criminal. Vicky’s clue says Tyler has more criminal neighbors than Karen. At the moment both Tyler’s neighbors and Karen’s neighbors have 0 known criminals, so that comparison depends on Mary and Nancy for Tyler, and on Hilda for Karen. If Hilda were criminal, then Karen’s neighbors would have 1 criminal. But Mary and Nancy cannot both be criminal because Steve needs exactly 6 innocent neighbors, so Tyler’s neighbors cannot reach more than 1 criminal. That would stop Tyler from having more criminal neighbors than Karen. So Hilda must be innocent.

15.C2 · Isaac → CRIMINAL

Bobby’s neighbors contain exactly 2 innocents in total. The clue says exactly 1 of those innocents is also a neighbor of Daniel, and among the people who are neighbors of both Bobby and Daniel, Hilda is already that 1 known innocent. Isaac is the only remaining unknown person in that shared group, so Isaac cannot also be innocent. So Isaac must be criminal.

16.C3 · Mary → CRIMINAL

Logan's clue says column C contains an odd number of criminals. In column C, there are already 2 known criminals, Daniel and Isaac, and the only unknown person left there is Mary. If Mary were innocent, column C would stay at 2 criminals, which is even, not odd. So Mary must be criminal.

17.D3 · Nancy → INNOCENT

Bobby’s clue says Steve’s neighbors contain exactly 6 innocents. Among Steve’s neighbors, 5 are already known to be innocent, and the only neighbor there whose identity is still unknown is Nancy. That means the one remaining innocent in Steve’s neighborhood has to be Nancy. So Nancy must be innocent.