Clues by Sam Mar 29, 2026 Answer – Full Solution Explained
A1
👮♀️
cop
B1
👷♀️
builder
C1
👨🔬
scientist
D1
👷♂️
builder
A2
🕵️♀️
sleuth
B2
👨🔬
scientist
C2
👩🔧
mech
D2
👩🔧
mech
A3
👨🔬
scientist
B3
💂♀️
guard
C3
👩🔧
mech
D3
👨🌾
farmer
A4
👮♂️
cop
B4
👨🔬
scientist
C4
👩⚕️
doctor
D4
👩⚕️
doctor
A5
💂♂️
guard
B5
💂♀️
guard
C5
👨🌾
farmer
D5
👷♂️
builder
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 — 9 criminals.
Clues by Sam answer for Mar 29, 2026 — a Hard solved in 17 steps
Today's Clues by Sam puzzle is rated Hard and resolves with 9 criminals on a 20-cell, 4-column × 5-row grid. The criminals are Amy (A1), Bonnie (B1), David (D1), Hilda (C2), Katie (D2), Olive (C3), Phil (D3), Vince (A5) and Xavi (C5); the remaining 11 suspects are innocent.
The deduction chain, in plain English
01.C3 · Olive → CRIMINAL, D3 · Phil → CRIMINAL
Uma is at D4, and her neighbors are Olive (C3), Phil (D3), Tina (C4), Xavi (C5), and Zed (D5). Tina’s clue says that among Uma’s neighboring criminals there are exactly three in total, and only one of those three is to the right of Vince. Vince is at A5, so the only neighbors of Uma who are to the right of Vince are Xavi and Zed, meaning exactly one of Xavi or Zed is a criminal neighbor of Uma. Since Tina is already INNOCENT, the three criminal neighbors must come from Olive, Phil, Xavi, and Zed, and with only one criminal among Xavi and Zed, the other two criminals must be Olive and Phil. Therefore, we can determine that C3 Olive is CRIMINAL and D3 Phil is CRIMINAL, and so on.
02.B5 · Wanda → INNOCENT
Vince is at A5, so the people to the right of Vince are Wanda at B5, Xavi at C5, and Zed at D5. Uma is at D4, and the only neighbors of Uma who are also to the right of Vince are Xavi and Zed. Tina’s clue says Uma has exactly three criminal neighbors, and only one of those criminals is to the right of Vince, so exactly one of Xavi and Zed is a criminal. Olive’s clue says there is an odd number of criminals to the right of Vince; since Xavi and Zed already contribute exactly one criminal, Wanda cannot be a criminal or the total would become two, which is even. Therefore, we can determine that B5 Wanda is INNOCENT.
03.A5 · Vince → CRIMINAL
Uma is at D4, and her neighbors are C3 Olive, D3 Phil, C4 Tina, C5 Xavi, and D5 Zed. Tina’s clue says that Uma has exactly three criminal neighbors, and since Olive and Phil are already criminals while Tina is innocent, exactly one of Xavi or Zed must be a criminal to make the total of three. Phil’s clue tells us there are exactly two criminals in row 2, and Wanda’s clue says row 5 has the same number of criminals as row 2, so row 5 also has exactly two criminals. In row 5, Wanda at B5 is innocent, and we already know only one of Xavi (C5) or Zed (D5) can be a criminal, so Vince at A5 must be the second criminal in that row. Therefore, we can determine that A5 Vince is CRIMINAL.
04.D4 · Uma → INNOCENT
Uma’s neighbors are Olive, Phil, Tina, Xavi, and Zed, and Tina tells us Uma has exactly 3 neighboring criminals. Since Olive and Phil are already criminals and Tina is innocent, the third neighboring criminal must be either Xavi or Zed, so exactly one of Xavi and Zed is innocent. Vince’s clue counts the edge-neighbors of Tina that are innocent: among those, Wanda is already one innocent, so the remaining edge-neighbors Uma, Xavi, and Zed must contain an even number of innocents to keep the total odd. Because exactly one of Xavi and Zed is innocent, the only way for Uma, Xavi, and Zed to have an even number of innocents is for Uma to be innocent too. Therefore, we can determine that D4 · Uma is INNOCENT.
05.A2 · Freya → INNOCENT, C2 · Hilda → CRIMINAL
David is at D1, so his only neighbors are C1 (Chase), C2 (Hilda), and D2 (Katie). Uma’s clue says David has only one innocent neighbor, so exactly one of those three is innocent and the other two are criminals. Phil’s clue says there are exactly two criminals in row 2, and those two criminals must be connected, which in a single row means they must be in adjacent squares. If Hilda at C2 were innocent, then D2 would have to be one of David’s two criminal neighbors, but then row 2 would contain a criminal at D2 whose only adjacent row-2 neighbor is C2, so the two row-2 criminals could not be connected. So C2 must be criminal, and since row 2 has exactly two connected criminals total, the other row-2 criminal must be B2 or D2, which means A2 cannot be criminal and is innocent. Therefore, we can determine that A2 (Freya) is INNOCENT, C2 (Hilda) is CRIMINAL, and so on.
06.D1 · David → CRIMINAL
Hilda is at C2, and her eight neighbors are Bonnie B1, Chase C1, David D1, Gary B2, Katie D2, Nicole B3, Olive C3, and Phil D3. Freya’s clue says there are exactly five criminals among those neighbors, and only one of those five is in column B, so the other four criminals must be among the five non-column-B neighbors Chase, David, Katie, Olive, and Phil; since Olive and Phil are already criminals, that means exactly one of Chase, David, and Katie is innocent. Uma’s clue says David has only one innocent neighbor, and David’s neighbors are Chase, Hilda, and Katie; since Hilda is a criminal, exactly one of Chase and Katie is innocent. That uses up the single innocent allowed among Chase, David, and Katie, so David cannot be that innocent and must be criminal. Therefore, we can determine that D1 David is CRIMINAL.
07.B3 · Nicole → INNOCENT
Hilda is at C2, and Freya’s clue says that among Hilda’s neighboring criminals there are exactly 5 in total, with exactly 1 of those in column B; the only column-B neighbors of Hilda are Bonnie (B1), Gary (B2), and Nicole (B3), so exactly one of those three is a criminal. David’s clue about Gary says Gary has exactly 4 neighboring criminals, and exactly 1 of those is to the right of Amy; among Gary’s neighbors, the only people to the right of Amy are Bonnie (B1) and Chase (C1), so exactly one of Bonnie and Chase is a criminal. Whenever Chase is that criminal, then (from Freya’s clue) Katie (D2) must be innocent, and then Phil’s clue that row 2 has exactly two criminals forces Gary to be the second criminal in row 2, so Gary is a criminal; otherwise Bonnie is the criminal. That means at least one of Bonnie or Gary is a criminal, so the “exactly one criminal among Bonnie, Gary, and Nicole” cannot be Nicole. Therefore, we can determine that B3 Nicole is INNOCENT.
08.B4 · Steve → INNOCENT
Hilda is at C2, and Freya’s clue says that among the five criminals neighboring Hilda, exactly one of them is in column B. The only column B neighbors of Hilda are B1 (Bonnie), B2 (Gary), and B3 (Nicole), and since Nicole is already INNOCENT, this forces exactly one of Bonnie or Gary to be a CRIMINAL and the other to be INNOCENT. Nicole’s clue says column A is the only column with exactly 2 criminals, so column B cannot end up with exactly 2 criminals. If Steve at B4 were a CRIMINAL, then column B would have exactly 2 criminals (Steve plus the one forced criminal among Bonnie and Gary), which is not allowed, so Steve must be INNOCENT. Therefore, we can determine that B4 Steve is INNOCENT.
09.A4 · Ronald → INNOCENT
Gary is at B2, and his eight neighbors are A1 Amy, B1 Bonnie, C1 Chase, A2 Freya, C2 Hilda, A3 Larry, B3 Nicole, and C3 Olive. David says Gary has exactly four criminal neighbors, and only one of those criminals is to the right of Amy; among Gary’s neighbors, the only people to the right of Amy (same row as Amy, but in columns B or C) are Bonnie at B1 and Chase at C1, so exactly one of Bonnie or Chase is a criminal. Since Hilda and Olive are already two criminal neighbors of Gary, the other two criminal neighbors must come from Amy, Bonnie, Chase, and Larry, and because only one of Bonnie or Chase can be criminal, at least one of Amy at A1 or Larry at A3 must be criminal. Nicole’s clue says column A is the only column with exactly two criminals, and column A already contains Vince at A5 as a criminal, so the needed additional criminal in column A must be Amy or Larry, leaving no room for Ronald at A4 to be a criminal. Therefore, we can determine that A4 Ronald is INNOCENT.
10.D2 · Katie → CRIMINAL
The edge positions that are still unknown are A1, B1, C1, A3, D2, C5, and D5, and Ronald’s clue says there are exactly 7 criminals on the edges. David’s clue says Gary has exactly 4 criminal neighbors, and exactly 1 of those is to the right of Amy; among Gary’s neighbors, the only people to the right of Amy are B1 and C1, so exactly one of B1 and C1 is criminal, and the other extra criminal neighbor must come from A1 or A3, making exactly 2 criminals among A1, B1, C1, and A3. Tina’s clue says Uma has exactly 3 criminal neighbors and only 1 of them is to the right of Vince; since two of Uma’s neighboring criminals are already C3 and D3, the third must be exactly one of C5 or D5, so exactly 1 of C5 and D5 is an edge criminal. We already have three known edge criminals (D1, D3, and A5), and adding the 2 criminals from A1/B1/C1/A3 and the 1 criminal from C5/D5 gives 6 edge criminals, so the 7th edge criminal must be D2. Therefore, we can determine that D2 (Katie) is CRIMINAL.
11.B2 · Gary → INNOCENT
In row 2, we already know C2 (Hilda) and D2 (Katie) are criminals, and they are directly next to each other, so they are connected. Phil’s clue says “Both criminals in row 2 are connected,” which tells us there are exactly two criminals in that row, and they form a single connected group. Since the two criminals in row 2 are already accounted for as C2 and D2, the remaining people in row 2, A2 (Freya) and B2 (Gary), cannot be criminals. Therefore, we can determine that B2 Gary is INNOCENT.
12.C1 · Chase → INNOCENT
David is at D1, so his only neighbors on the board are Chase at C1 (directly left), Hilda at C2 (down-left), and Katie at D2 (directly below). Uma’s clue says David has only one innocent neighbor. Since Hilda and Katie are both already known to be criminals, neither of them can be that innocent neighbor, so the only possible innocent neighbor is Chase at C1. Therefore, we can determine that C1 Chase is INNOCENT.
13.B1 · Bonnie → CRIMINAL
Hilda is at C2, so her neighbors are B1, C1, D1, B2, D2, B3, C3, and D3. We already know four of those neighbors are criminals: David at D1, Katie at D2, Olive at C3, and Phil at D3, while Chase at C1, Gary at B2, and Nicole at B3 are innocents. Freya’s clue says Hilda has exactly five neighboring criminals, so the only remaining neighbor who can be the fifth criminal is Bonnie at B1; this also matches the clue’s statement that only one of those five criminals is in column B, since B1 is the only column B neighbor not already innocent. Therefore, we can determine that B1 Bonnie is CRIMINAL.
14.C5 · Xavi → CRIMINAL
Look at column C: Chase at C1 is innocent, Hilda at C2 is criminal, Olive at C3 is criminal, Tina at C4 is innocent, and only Xavi at C5 is still unknown. Nicole’s clue says column A is the only column with exactly 2 criminals, so no other column is allowed to end up with exactly 2 criminals. Column C already has exactly 2 criminals (Hilda and Olive), so Xavi cannot be innocent or column C would stay at exactly 2 criminals. Therefore, we can determine that C5 Xavi is CRIMINAL.
15.D5 · Zed → INNOCENT
Uma is at D4, and her neighbors are Olive at C3, Phil at D3, Tina at C4, Xavi at C5, and Zed at D5. Tina’s clue talks about “the 3 criminals neighboring Uma,” so Uma must have exactly three criminal neighbors; we already know Olive, Phil, and Xavi are criminals, which fills all three slots. That forces Zed, the only remaining unknown neighbor of Uma, to not be a criminal, and it also keeps the “only 1 … is to the right of Vince” part true because among those three criminals only Xavi is in row 5 to the right of Vince. Therefore, we can determine that D5 Zed is INNOCENT.
16.A1 · Amy → CRIMINAL
The corners are A1, D1, A5, and D5, and Hilda’s clue says the total number of criminals in these four corner spots is odd. We already know D1 and A5 are criminals, and D5 is innocent, so that gives exactly two criminals among the corners so far, which is even. To make the corner total odd, A1 must be a criminal. Therefore, we can determine that A1 Amy is CRIMINAL.
17.A3 · Larry → INNOCENT
Gary is at B2, so his neighbors are the eight surrounding spaces: A1 Amy, B1 Bonnie, C1 Chase, A2 Freya, C2 Hilda, A3 Larry, B3 Nicole, and C3 Olive. David’s clue says that there are exactly four criminals among Gary’s neighbors, and that only one of those four is positioned to the right of Amy (same row as Amy, in a column to her right). We can already see four criminals in Gary’s neighborhood without using Larry at all: Amy, Bonnie, Hilda, and Olive, and among these only Bonnie is to the right of Amy (Amy is at A1 and Bonnie is at B1). Since the clue allows exactly four neighboring criminals total, Larry cannot be a fifth one. Therefore, we can determine that A3 Larry is INNOCENT.