Today's answer, solved step by step.
A1
👷♂️
builder
B1
👮♀️
cop
C1
🕵️♀️
sleuth
D1
🕵️♀️
sleuth
A2
👨🌾
farmer
B2
👨🌾
farmer
C2
👨🌾
farmer
D2
🕵️♂️
sleuth
A3
👨💻
coder
B3
👨⚕️
doctor
C3
👮♀️
cop
D3
👷♀️
builder
A4
👨🎨
painter
B4
👨🎨
painter
C4
💂♀️
guard
D4
💂♀️
guard
A5
👩🎨
painter
B5
👩⚕️
doctor
C5
👨💻
coder
D5
💂♀️
guard
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 — 4 criminals.
Clues by Sam answer for Sep 20, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 4 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Austin (A1), Barb (B1), Mary (C3) and Xavi (C5); the remaining 16 suspects are innocent.
The deduction chain, in plain English
01.A4 · Phil → INNOCENT, A3 · Jose → INNOCENT
Larry’s clue says the people between Frank and Uma contain exactly 2 innocents. Right now there are 0 known innocents in that group, and the only people left there are Jose and Phil. Since the group still needs 2 innocents and there are exactly 2 people available to fill those spots, both of them have to be innocents. So Phil and Jose must be innocent.
02.A5 · Uma → INNOCENT, B5 · Vera → INNOCENT
Phil’s clue says the people to the left of Xavi contain exactly 2 innocents. Right now there are 0 known innocents in that group, and the only people there whose identities are still unknown are Uma and Vera. Since that group still needs exactly 2 innocents and Uma and Vera are the only two people left to fill those spots, both of them have to be innocent. So Uma and Vera must be innocent.
03.C2 · Hal → INNOCENT
If Hal were criminal, then Uma's clue that exactly 3 people above Xavi are innocent would force the other three above Xavi, Claire, Mary, and Susan, to all be innocent. Then Vera's clue that Claire has exactly 4 innocent neighbors would apply to Claire's remaining neighbors B1 Barb, D1 Eve, B2 Gary, and D2 Igor, so all four of them would also have to be innocent. But then Hal's neighbors would contain Larry plus Claire, Mary, Barb, Eve, Gary, and Igor, which is already 7 innocents, even before considering Olive, while Jose's clue says Hal has exactly 6 innocent neighbors. That cannot happen, so Hal at C2 must be innocent.
04.C4 · Susan → INNOCENT
Gary’s neighbors in column C, namely Claire, Hal, and Mary, contain exactly 1 criminal. Those same people are part of the group above Xavi, and the whole group above Xavi contains exactly 1 criminal total. So the single criminal allowed above Xavi is already accounted for within Claire, Hal, and Mary. The only person above Xavi who is not in that smaller group is Susan, so she cannot be criminal. That makes Susan innocent.
05.D3 · Olive → INNOCENT
Hal has exactly 6 innocent neighbors, so among his 7 unknown neighbors there must be exactly 2 criminals. Those 7 are Barb, Claire, Eve, Gary, Igor, Mary, and Olive. But the 2 criminals must come from Barb, Claire, Eve, Gary, Igor, and Mary, so Olive cannot be one of Hal's criminal neighbors. That makes Olive innocent.
06.B2 · Gary → INNOCENT
Assume Gary were criminal. Claire has exactly 4 innocent neighbors, and Hal is already one of them, so Barb must be innocent; then Eve must be innocent; then Igor must be innocent, because those are the remaining neighbors needed to bring Claire's total to 4 innocents. Gary’s clue says he has exactly 3 criminal neighbors, and exactly 1 of those criminals is in column C. With Hal innocent and Claire, Barb, Eve, and Igor all innocent, the only possible criminal in Gary’s column C neighbor set is Mary, so the other two criminals among Gary’s neighbors have to be Austin and Frank. But Jose has at least 4 innocent neighbors, and he currently has only 2 known innocents, so making Gary criminal while also making Frank criminal conflicts with that clue. So Gary at B2 must be innocent.
07.D2 · Igor → INNOCENT
If Igor were criminal, then Claire’s neighbors would need four innocents, and with Gary and Hal already innocent that would force Barb and Eve to be innocent too. Gary has exactly three criminal neighbors, and exactly one of those criminals is in column C. With Barb innocent, the three criminals around Gary must be Austin, Frank, and Claire, so Mary is innocent and Claire is the one criminal in column C. That gives row 1 exactly two criminals, Austin and Claire. But Gary also says row 1 is the only row with exactly two criminals, while row 2 would also have exactly two criminals, Frank and Igor. That clash means the assumption was wrong, so Igor at D2 must be innocent.
08.A1 · Austin → CRIMINAL
Austin’s neighbors must contain exactly 2 innocents, and Gary is already one of them. So among the two shared neighbors, Barb and Frank, exactly 1 is innocent. Now look at Gary’s neighbors that are not in column C: Austin, Barb, Frank, Jose, and Larry. That group must contain exactly 3 innocents, and Jose and Larry already account for 2 of them, so only 1 innocent is allowed among Austin, Barb, and Frank. But Barb and Frank already supply that 1 allowed innocent. That leaves no room for Austin to be innocent, so Austin must be criminal.
09.C1 · Claire → INNOCENT
Row 1 has exactly 2 criminals, and Austin is already one of them, so row 1 needs exactly 1 more criminal. Claire’s neighbors must contain exactly 4 innocents, and among those neighbors Gary, Hal, and Igor are already innocent, so the only way to reach 4 innocents is for exactly one of Barb and Eve to be criminal. Barb and Eve are both in row 1, so that shared pair already accounts for the one remaining criminal row 1 needs. Claire is also in row 1 but not part of that Barb-or-Eve count, so Claire cannot be criminal. So Claire must be innocent.
10.C3 · Mary → CRIMINAL
Uma’s clue says the people above Xavi contain exactly 3 innocents. Above Xavi, Claire, Hal, and Susan are already known innocents, which already fills all 3 innocent spots. The only person there whose identity was still unknown is Mary, so she cannot also be innocent. So Mary must be criminal.
11.B4 · Ryan → INNOCENT
Claire’s clue says her neighbors contain exactly 4 innocents, and among those neighbors there are already 3 known innocents, with only Barb and Eve still unknown. Susan’s clue also says exactly one column has 5 innocents, and Mary’s clue says column B has more innocents than column C even though both columns currently already have 3 known innocents. Now test Ryan as criminal. Then Barb, Eve, Frank, Tina, Xavi, and Zara would be the remaining people who would have to satisfy all of those requirements at once, but they cannot. So Ryan cannot be criminal. That makes Ryan innocent.
12.D5 · Zara → INNOCENT
Assume Zara were criminal. Then column D would already have 1 criminal, and Gary's clue says column C must have more criminals than column D. But column C has 1 known criminal already, so to have more than column D, Xavi cannot be criminal and must be innocent. That clashes with the clue pattern being used here, so the assumption about Zara fails. So Zara must be innocent.
13.D4 · Tina → INNOCENT
Claire’s clue says her neighbors contain exactly 4 innocents. Right now those neighbors already include 3 known innocents, and the only unknown neighbors are Barb and Eve, so that clue tightly restricts how Barb and Eve can be placed. At the same time, Susan says only one column has exactly 5 innocents, and Ryan says column C has more criminals than column D. Column C already has 1 known criminal, while column D has none, so column D cannot pick up too many criminals without conflicting with that comparison and the one-full-innocent-column condition. If Tina were a criminal, then the remaining people involved here, namely Barb, Eve, Frank, and Xavi, would have to satisfy all of those facts at once, and they cannot. So Tina cannot be a criminal. That makes Tina innocent.
14.C5 · Xavi → CRIMINAL
If Xavi were innocent, column C would have 4 innocents. Column B already has 4 known innocents, so for column B to have more innocents than column C, Barb would have to be innocent too, giving column B 5 innocents. Also, if Xavi were innocent, column C would still have 1 criminal while column D has 0 known criminals, so for column C to have more criminals than column D, Eve would have to be innocent rather than adding a criminal to column D. But then Claire’s neighbors would be Barb, Eve, Gary, Hal, and Igor all innocent, which makes 5 innocent neighbors, not exactly 4. So Xavi must be criminal.
15.A2 · Frank → INNOCENT
If Frank were criminal, then with Vera's clue that there are exactly 4 criminals in total, the four criminals would already be Austin, Mary, Xavi, and Frank. That leaves Barb and Eve as innocent. But Claire must have exactly 4 innocent neighbors. Her neighbors would then be Barb, Eve, Gary, Hal, and Igor, which is 5 innocent neighbors, not 4. So the assumption fails, and Frank at A2 must be innocent.
16.B1 · Barb → CRIMINAL
Hal’s clue says Gary has exactly 3 criminal neighbors, and exactly 1 of those criminals is in column C. Among Gary’s neighbors in column C, Claire and Hal are innocent and Mary is criminal, so that 1 criminal in column C is already Mary. That means the other 2 criminal neighbors of Gary must be outside column C. Outside column C among Gary’s neighbors, Austin is already criminal, Frank, Jose, and Larry are innocent, and Barb is the only person left who could fill the remaining criminal spot. So Barb must be criminal.
17.D1 · Eve → INNOCENT
Vera says Claire has exactly 4 innocent neighbors. Among Claire's neighbors, 3 are already known to be innocent, and the only neighbor there whose identity is still unknown is Eve. So the one remaining innocent neighbor Claire needs has to be Eve. That makes Eve innocent.