Clues by Sam Jun 04, 2026 Answer – Full Solution Explained
A1
👮♂️
cop
B1
👩🏫
teacher
C1
👩🏫
teacher
D1
👨🏫
teacher
A2
👮♂️
cop
B2
👨🎤
singer
C2
👨🎤
singer
D2
👩🎤
singer
A3
🕵️♀️
sleuth
B3
👨🎨
painter
C3
👩🎨
painter
D3
🕵️♂️
sleuth
A4
👨🎨
painter
B4
👩💻
coder
C4
👩💻
coder
D4
👨🔧
mech
A5
👮♀️
cop
B5
👨💻
coder
C5
👩🔧
mech
D5
👩🔧
mech
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 — 6 criminals.
Clues by Sam answer for Jun 04, 2026 — a Tricky solved in 14 steps
Today's Clues by Sam puzzle is rated Tricky and resolves with 6 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Carol (B1), Eve (C1), Luigi (B3), Nick (D3), Olsi (A4) and Terry (D4); the remaining 14 suspects are innocent.
The deduction chain, in plain English
01.B1 · Carol → CRIMINAL, C1 · Eve → CRIMINAL
Banda's clue says the people to the left of Frank contain exactly 2 criminals. In that group, there are currently 0 known criminals, and the only people there whose identities are still unknown are Carol and Eve. Since those two spots are exactly the 2 criminals the clue still requires, both of them have to be criminals. So Carol and Eve must be criminal.
02.C2 · Isaac → INNOCENT
Carol’s clue says that Isaac is one of Janet’s 3 innocent neighbors. That directly identifies Isaac as innocent. So Isaac must be innocent.
03.C3 · Max → INNOCENT
Janet’s neighbors contain exactly 2 criminals. Among those neighbors, the column D neighbors of Janet, Frank and Nick, contain exactly 1 criminal. The only other neighbor outside that smaller group that matters here is Max, because Eve is already the other criminal in Janet’s neighbor group and Isaac is already innocent. So the smaller group already accounts for the one criminal it can have, together with Eve for the two criminals Janet’s neighbors can have, which leaves Max unable to be a criminal. That makes Max innocent.
04.D5 · Xia → INNOCENT
In column D there are exactly 2 criminals, and Isaac says exactly 1 of those criminals is Janet's neighbor. Among the people in column D, Janet's neighbors are only Frank and Nick, so exactly 1 criminal is in Frank and Nick. Max says the 2 criminals above Xia are connected. That means the only possible 2-person criminal block in column D is one of these adjacent pairs above Xia: Frank and Janet, Janet and Nick, or Nick and Terry. None of those connected pairs includes Xia. So Xia at D5 must be innocent.
05.D3 · Nick → CRIMINAL, D1 · Frank → INNOCENT
Above Xia, the four people are Frank, Janet, Nick, and Terry, and Max's clue says exactly two of them are criminals and those two must be connected. Xia's clue also says exactly one mech has an innocent directly above them, which involves Terry with Nick above, Wanda with Ruth above, and Xia with Terry above. If Nick were innocent while Frank were criminal, then Janet, Ruth, Terry, and Wanda would have to satisfy both clues at the same time, and that cannot be done. So Nick must be criminal and Frank must be innocent.
06.B5 · Vince → INNOCENT, B2 · Hank → INNOCENT
Column D must contain exactly 2 criminals, and exactly 1 of those criminals is Janet’s neighbor. Right now Nick is already a criminal in column D and he is one of Janet’s neighbors, so the other criminal in column D has to be someone who is not Janet’s neighbor. Also, columns B and D must contain the same number of innocents. Column D already has 2 known innocents, while column B has none yet. If Vince and Hank were both criminals, then the remaining people involved here, Janet, Luigi, Quita, and Terry, would have to make all of those requirements true at once, and they cannot. So Vince and Hank cannot be criminals. That makes Vince and Hank innocent.
07.A2 · Gary → INNOCENT
Hank’s clue says Gary is one of Luigi’s exactly 7 innocent neighbors. That directly identifies Gary as innocent. So Gary must be innocent.
08.B4 · Quita → INNOCENT, B3 · Luigi → CRIMINAL
Luigi’s clue says he has exactly 7 innocent neighbors, and four of those are already fixed: Gary, Hank, Isaac, and Max. Vince’s clue says exactly one criminal in row 3 has an innocent directly to the right, and with Max innocent and Nick already the known criminal in row 3, the row 3 requirement has to be met alongside Luigi’s neighbor count. If you try the opposite statuses here, with Quita criminal and Luigi innocent, then Karen, Olsi, and Ruth would have to satisfy both of those clues at the same time, and they cannot. That opposite assignment does not fit the clues. So Quita must be innocent and Luigi must be criminal.
09.C5 · Wanda → INNOCENT
Xia’s neighbors are Ruth, Terry, and Wanda, and Quita’s clue says that group contains an odd number of criminals. Ruth and Terry are exactly the mechs with an innocent directly above them, and that exact-count group contains exactly 1 criminal, which is already odd. Since Wanda is the only one of Xia’s neighbors outside that two-person exact-count group, she cannot also be a criminal without making Xia’s neighbor total even instead of odd. So Wanda must be innocent.
10.D2 · Janet → INNOCENT
Wanda’s clue says Janet is one of the 9 innocents on the edges. That directly identifies Janet as an innocent edge person. So Janet must be innocent.
11.D4 · Terry → CRIMINAL
Isaac's clue says column D has exactly 2 criminals, and exactly 1 of those criminals is Janet's neighbor. In column D, the only known criminal neighbor of Janet is Nick, so the other criminal in column D must be someone who is not Janet's neighbor. The people in column D who are not Janet's neighbors are Janet, Terry, and Xia. Janet and Xia are innocent, so the only person there who can fill that remaining criminal spot is Terry. That makes Terry criminal.
12.C4 · Ruth → INNOCENT
The mechs here are Terry, Wanda, and Xia, and the clue says exactly one mech has an innocent directly above them. Terry does not, because Nick above Terry is criminal, and Xia does not, because Terry above Xia is criminal. So the only mech who can satisfy the clue is Wanda, which means the person directly above Wanda has to be innocent. That person is Ruth, so Ruth must be innocent.
13.A5 · Uma → INNOCENT
Luigi's neighbors must contain exactly 7 innocents, and among those neighbors there are already 6 known innocents, so the only remaining innocent there has to be Karen or Olsi. The edge cells must contain exactly 9 innocents, and among all edge cells there are already 7 known innocents. Since Karen and Olsi account for the one additional innocent needed among Luigi's edge-neighbor cells, the only edge-cell person left outside that smaller group who can supply the other needed innocent is Uma. So Uma must be innocent.
14.A4 · Olsi → CRIMINAL, A3 · Karen → INNOCENT
Uma’s clue says exactly one innocent person in row 4 has a criminal directly above them. In row 4, the known innocents are Quita and Ruth, and the only other person there who could be innocent is Olsi. If Olsi were innocent, then row 4 would have to meet Uma’s “only one” condition with Olsi included among the innocents there, and at the same time Karen would have to be criminal. That combination is impossible with the facts for this step, so Olsi cannot be innocent and Karen cannot be criminal. So Olsi must be criminal, and Karen must be innocent.