Clues by Sam Jul 20, 2026 Answer – Full Solution Explained

01.C2 · Gabe → INNOCENT

Salil’s clue says Gabe is one of Hope’s 4 innocent neighbors. Since Gabe is explicitly included among those innocent neighbors, his identity is fixed by the clue itself. So Gabe must be innocent.

02.B2 · Flora → CRIMINAL, D2 · Hope → CRIMINAL

Gabe’s clue means there is exactly 1 innocent among the guards. The guards already have that 1 innocent, because Gabe is known to be innocent. That leaves the only other guards, Flora and Hope, unable to be innocent. So Flora and Hope must be criminal.

03.B1 · Betty → INNOCENT

Flora’s clue says there is exactly 1 innocent among the people who are both in column B and neighboring Adam. That shared group is only Betty and Flora. Flora is already a criminal, so the group still needs 1 innocent, and Betty is the only unknown person left in it. So Betty must be innocent.

04.D1 · Diane → INNOCENT

Betty’s clue says there are more innocent teachers than innocent guards. Right now there is 1 known innocent teacher and 1 known innocent guard, and there are no unknown guards left. If Diane were criminal, the teachers would stay at 1 innocent, which would not be more than the guards’ 1 innocent. So Diane must be innocent.

05.D3 · Mary → INNOCENT

Flora’s clue says exactly 2 of her 5 innocent neighbors are in column C, and those column C neighbors are C1 Chase, C2 Gabe, and C3 Kim. Hope’s clue says Hope has exactly 4 innocent neighbors, and those neighbors are C1 Chase, D1 Diane, C2 Gabe, C3 Kim, and D3 Mary. The column C people here are the shared smaller group inside Hope’s neighbor set, and the only person in Hope’s neighbor set who is outside that smaller group is D3 Mary. Since Hope needs 4 innocent neighbors while the column C part accounts for only 2 innocents, that remaining difference has to come from D3 Mary. So Mary must be innocent.

06.D4 · Ruth → INNOCENT, D5 · Xavi → CRIMINAL

Mary’s clue says there are exactly two innocents below Diane, and those two innocents must be connected. Below Diane are Hope, Mary, Ruth, and Xavi, with Hope already criminal and Mary already innocent. If Ruth were criminal and Xavi were innocent, then the two innocents below Diane would be Mary and Xavi, and that does not satisfy the clue’s requirement for the two innocents below Diane. So Ruth must be innocent and Xavi must be criminal.

07.A1 · Adam → INNOCENT

Xavi’s clue says that Adam is innocent. Since this clue directly identifies Adam’s status, Adam at A1 must be innocent.

08.B3 · Justin → CRIMINAL

Adam's clue is about the people who are both in column B and neighboring Elsa. That shared group is Betty, Flora, and Justin, and the clue says exactly 1 of them is innocent. Betty is already known to be innocent, so that one innocent in the group is already accounted for. That means the remaining unknown person in that group cannot also be innocent. So Justin must be criminal.

09.B4 · Olive → CRIMINAL

Justin’s clue says Olive is one of Tyler’s 3 criminal neighbors. Since the clue directly includes Olive among those criminal neighbors, Olive’s identity is fixed by the clue itself. So Olive must be criminal.

10.A4 · Nicole → CRIMINAL

Justin says Olive is one of Tyler's 3 criminal neighbors, so among Tyler's neighbors there must be exactly 3 criminals. Since Olive is already criminal and Salil is already innocent, the only people left to fill Tyler's remaining neighbor counts are Nicole, Pam, and Wally. Olive also says Chase has more criminal neighbors than Xavi. Chase already has 2 criminal neighbors and no unknown neighbors, so Xavi cannot have 2 criminal neighbors; among Xavi's neighbors, Pam and Wally therefore cannot both be criminal. That means Pam and Wally must include exactly 1 innocent person, so the single innocent among Nicole, Pam, and Wally has to be Pam or Wally, not Nicole. So Nicole must be criminal.

11.B5 · Tyler → INNOCENT

Nicole’s clue says Salil and Chase have the same number of criminal neighbors. Chase’s neighbors already contain exactly 2 criminals, and Salil’s neighbors also already contain exactly 2 known criminals. The only unknown among Salil’s neighbors is Tyler, so if Tyler were criminal, Salil would have 3 criminal neighbors while Chase would still have 2, which breaks the clue. So Tyler must be innocent.

12.C3 · Kim → INNOCENT, C1 · Chase → CRIMINAL

Flora has exactly 5 innocent neighbors, and exactly 2 of those innocents are in column C. In Flora’s neighboring column C spots, Gabe is already innocent, so if Chase were innocent and Kim were criminal, Gabe and Chase would be the two column C innocents. Tyler’s clue says the edge neighbors of Flora contain an odd number of innocents. That edge-neighbor group is Adam, Betty, Chase, Elsa, and Igor, with Adam and Betty already innocent, so making Chase innocent would force Elsa and Igor to fit the rest of the counts at the same time as Flora’s 5-neighbor total and Diane’s column C total, and that cannot be done. So Kim at C3 must be innocent, and Chase at C1 must be criminal.

13.C4 · Pam → CRIMINAL

Flora’s clue fixes her neighborhood at exactly 5 innocents, and exactly 2 of those innocents are in column C. In Flora’s neighboring column C people, Gabe and Kim are already innocent, so those 2 column C innocents are already accounted for. Now test Pam as innocent. Justin’s clue says his neighbors must contain an odd number of innocents, and he currently has 2 known innocents, so among Elsa, Igor, and Pam the total number of innocents would have to make that count odd. But with Pam counted as an innocent while Flora’s column C innocent total is already filled by Gabe and Kim, Elsa and Igor cannot satisfy both clues at once. So Pam cannot be innocent. That makes Pam criminal.

14.C5 · Wally → INNOCENT

Justin’s clue says Olive is one of Tyler’s exactly 3 criminal neighbors. Tyler’s neighbors are Nicole, Olive, Pam, Salil, and Wally, and among them Nicole, Olive, and Pam are already criminal while Salil is innocent. That already fills all 3 criminal neighbor spots named by the clue, so Wally cannot also be criminal. So Wally must be innocent.

15.A2 · Elsa → INNOCENT

16.A3 · Igor → CRIMINAL

Flora’s neighbors contain exactly 5 innocents, and Diane’s clue says exactly 2 of those innocents are in column C. In Flora’s neighboring spaces that are in column C, the two innocents are already Gabe and Kim. That means the neighbors not in column C must contain the other 3 innocents, and that group already has Adam, Betty, and Elsa as known innocents. So Igor cannot also be innocent. That makes Igor criminal.