In 11 Among Us players, some of them are crewmates, who always tell the truth, and imposters, who can either lie or tell the truth. Can you figure out who the imposters are? (There can be any number of imposters.)

Here is what they said.

Red: Purple is a crewmate.

Orange: Exactly two of the players with names beginning with 'B' or 'G' are the same role.

Yellow: Red and Orange are different roles.

Lime: Blue is lying iff he is an imposter.

Green: At least two of the above are crewmates.

Cyan: There is at least one crewmate below me.

Blue: Lime is an imposter iff Maroon is one too.

Purple: Green is telling the truth if he is an imposter.

Pink: Cyan and Brown both told the truth or both lied.

Gray: Lime just told the truth iff Blue did too.

Brown: Yellow and Green are different roles iff Red and Pink are too.

Maroon: Purple and Green both either told the truth or lied.

  • 5
    $\begingroup$ Are you asking if this is solvable because you found it somewhere? In that case you should provide attribution. Or did you create this yourself? $\endgroup$
    – SQLnoob
    Jun 2 at 15:00
  • 1
    $\begingroup$ Did you mean for there to be only one impostor, maybe? $\endgroup$
    – Esther
    Jun 2 at 20:50
  • 1
    $\begingroup$ The definition of "possible" really needs to be clarified. Do you mean "possible to find a solution for" or "possible to solve logically"? Also, please do respond to the earlier comment about who made this puzzle. $\endgroup$
    – bobble
    Jun 2 at 22:35

2 Answers 2


This puzzle

does not appear to be solvable, given the information provided. There are 200 216 permutations of crew and impostors that could fit with the given statements, including the cases where:

- All 12 of them are impostors
- Exactly one of Lime, Blue, Brown or Maroon is a crewmate
- Orange is the only impostor

and 210 other cases in between, where there can be anywhere from 1 to 11 impostors. The only things I can determine for certain are that:

- They are not all crewmates (although it is possible that all 12 statements are true), and
- Maroon is definitely telling the truth.

Proof that they are not all crewmates: If Yellow is a crewmate, then Red and Orange are different roles and therefore one of them must be an impostor.

Proof that Maroon is telling the truth: Purple says "If Green is an impostor, then Green is telling the truth." If Purple is telling the truth, then either Green is an impostor who tells the truth, or he's a crewmate; either way Green is also telling the truth. On the other hand, if Purple is lying, the negation of their statement is "Green is a lying impostor." In other words, Green's statement is true if Purple's is true, and Green's statement is false if Purple's is false. So Maroon's statement is true regardless.

For fun, here's an image illustrating the solutions I found:


Small edit:

I rewrote my logic to clean up the spaghetti I created while I was originally trying to solve the problem, and now come up with 216 possible solutions, instead of 200 before. Doesn't meaningfully change any of the previous conclusions.

  • 1
    $\begingroup$ I would consider this puzzle to be solvable, just not uniquely solvable $\endgroup$
    – xyldke
    Jun 2 at 15:37
  • 6
    $\begingroup$ A puzzle with 200 different solutions isn't much of a puzzle imo. $\endgroup$
    – SQLnoob
    Jun 2 at 15:49

Here's one solution which relies on a problem with the puzzle.

Blue is a crewmate, everyone else is an impostor.

Blues's statement "Lime is an imposter iff Maroon is one too." is the truth, if Lime and Maroon are both impostors.

Everyone else could tell the truth or lie, which means you can have their statements be true or false depending on what you need. If a statement conflicts with your interpretation it's a lie, if it supports your interpretation it's the truth.

In this case the statements are:
Red: LIE (purple is an impostor)
Orange: TRUTH (Gray and Green are the same role, Blue and Brown are not)
Yellow: LIE (Red and Orange are both impostors
Lime: TRUTH OR LIE, since Blue is not an impostor this statement is never evaluated (also the 'only if' is unneccesary since crewmates always tell the truth)
Green: LIE, all above are impostors
Cyan: TRUTH, Blue is a crewmate
Blue: TRUTH (see above)
Purple: LIE, Green is lying, despite being an impostor
Pink: TRUTH or LIE depending on the answer for Brown
Gray: TRUTH or LIE, depending on the answer for Lime
Brown: TRUTH OR LIE, since Red and Pink are the same role, as are Yellow and Green, so the "only if" part is true. The "if" part is unknown, hence the statement can be either true or false.
Maroon: TRUTH, since Purple and Green both lied.

Here's the problem with the puzzle:

This same logic could apply to more people being the sole crewmate.
Maroon, Lime and Brown could all be the sole crewmate and there might be even more solutions.
The big problem is that statements made by impostors are meaningless since either the statement or its negation (or in other words anything) could be true.

  • $\begingroup$ I agree with your general conclusion, although rot13(Cvax pnaabg or n fbyr perjzngr. Vs Cvax vf n Perjzngr, gura Plna vf gryyvat gur gehgu, naq gurersber Oebja vf nyfb gryyvat gur gehgu. Ohg gura bar bs Lryybj, Terra, be Erq zhfg nyfb or n Perjzngr. Nyfb, V qba'g oryvrir vg'f cbffvoyr sbe Znebba'f fgngrzrag gb or n yvr.) $\endgroup$
    – SQLnoob
    Jun 2 at 14:58
  • $\begingroup$ You're right about pink, but I don't think you're right about Maroon: r13(Nf ybat nf Terra vf na vzcbfgbe, Znebba vf nyjnlf gryyvat gur gehgu, ohg vs Terra jrer n perjzngr, gur irenpvgl bs Checyr'f fgngrzrag pna abg or qrgrezvarq.) $\endgroup$
    – xyldke
    Jun 2 at 15:35
  • $\begingroup$ Va guvf pbagrkg V'z nffhzvat gung gb pnyy n fgngrzrag n yvr vf gb fnl vgf artngvba vf gehr. Gur ybtvpny artngvba bs Checyr'f fgngrzrag vf "Terra vf ylvat." Fb vs Checyr vf ylvat, gura Terra vf ylvat. V thrff vg'f n znggre bs frznagvpf jurgure vg'f nccebcevngr gb nccyl ybtvpny bcrengbef va gung jnl ohg gung frrzf yvxr gur zbfg fgenvtugsbejneq jnl gb rinyhngr gur gehgu inyhrf bs gur fgngrzragf. $\endgroup$
    – SQLnoob
    Jun 2 at 15:46

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