The situation is as follows:

At a semiconductor laboratory in Hsinchu a security guard caught five technicians accessing a high level security area reserved for the most trusted scientists working in a new integrated circuit for an upcoming computer. However the security film is not very clear and the security team can only be sure that four out of five technicians have the access key card to enter the chipset room.

During interrogation the security team deduces that two of the technicians are lying and the other are telling the truth.

The answers given by the laboratory technicians were as follows:

Audrey: Gwendolyn does not have an access credential.

Dorothy: I was entrusted an access key.

Marina: Hannah has an access key.

Gwendolyn: Audrey is lying.

Hannah: Dorothy is telling the truth.

Based on this information, which of the technicians does not have access to the chipset room?

I'm stuck at the very beginning. All I could find is this looks like a Knights and Knaves logic problem.

Since there are five individuals the number of possible combinations would be 25=32. 32 combinations seems too big to make case-by-case a practical method.

I need help simplifying this problem to find a solution.

I'm not very knowledgeable with this type of problem. It would help me a lot to visualize what's going on if the proposed solution would include some sort of table or grid so I could identify the concluded result.

  • $\begingroup$ Two technicians are lying. There are $\binom{5}{2}=10$ possible combinations to consider. $\endgroup$ – Daniel Mathias Feb 9 '19 at 13:41
  • $\begingroup$ Similar question on Math StackExchange: math.stackexchange.com/questions/3106076/…> $\endgroup$ – Krad Cigol Feb 9 '19 at 15:45
  • $\begingroup$ @DanielMathias In this kind of situations then should I use $\binom{\textrm{number of people}}{2}$ for two possible answers being one $T$ or $F$ and this always stick?. Sorry. I am not very savvy with the use of combinatorics, hence I assumed that the number of choices would be a power of $2$. In short does the equation I proposed is always valid? $\endgroup$ – Chris Steinbeck Bell Feb 10 '19 at 12:59
  • $\begingroup$ $\binom{5}{2}=\binom{\text{number of people}}{\text{number of liars}}$ $\endgroup$ – Daniel Mathias Feb 10 '19 at 13:49
  • $\begingroup$ @DanielMathias Thanks for the clarification. I thought that it was due the possibilities. I'll take into consideration next time I see these kinds of problems. $\endgroup$ – Chris Steinbeck Bell Feb 11 '19 at 17:15

I think the idea here is that you can rule certain combinations out through logic.

For example,

Gwendolyn says Audrey is lying - this means that either Gwendolyn is telling the truth (and Audrey is indeed lying) or Gwendolyn is lying (and Audrey is not). Either way, Gwendolyn has an access key.

Since we know exactly one of

Gwendolyn and Audrey is lying

we know no more than one of the other three is lying.

Hannah is confirming Dorothy's statement, which is only possible if they're both telling the truth OR if they're both lying. Since the latter is impossible, we know they're both truthful, leaving Marina as the only other person who could be lying.

This gives us the solution that


does not have an access key.

  • $\begingroup$ I think adding these blank spaces to the answers don't actually help much to understand what it is intended to be said. I mean hiding the answer. My intention was to know if there was a table that could be built upon these statements so a conclution to be made, but yes it appears that Hannah was the one without the access key. $\endgroup$ – Chris Steinbeck Bell Feb 9 '19 at 12:26
  • 1
    $\begingroup$ The spoiler blocks are pretty much a courtesy thing for anyone who would want to solve the puzzle themselves, so that they do not immediately have the answer thrown in their face. $\endgroup$ – Braegh Feb 9 '19 at 13:05
  • $\begingroup$ @Chris Steinbeck Bell Spoilers are there to prevent answerers from mistakenly seeing the other answers. They still have to look at the past answers to avoid just writing a duplicate answer, but with spoilers they can take a look only when they want to. $\endgroup$ – Nautilus Feb 9 '19 at 13:05
  • $\begingroup$ @Braegh I'm sorry. I am new to this stack community. I am not sure if the standard practice is to enforce the use of spoilers blanking, but for me as a newcomer is kind of difficult to use for learning as it was intended in my question, therefore the displaying of the steps is not easily for me to follow. But if such is the practice here. I'll have to adapt then, although I voice my view that I politely disagree with this. $\endgroup$ – Chris Steinbeck Bell Feb 10 '19 at 13:20

The solution to the particular problem:

Either Audrey or Gwendolyn is lying, and both Dorothy and Hannah are either lying or telling the truth. Since there are only two liars, the latter two must be telling the truth, so Dorothy has an access key. Marina is one of the liars, which means everyone but Hannah have access. Gwendolyn is telling the truth and Audrey is lying.

How to visualize similar problems with grids:

Like in "Einstein's puzzles", draw grids where columns and rows represent a certain group of results. In this case, we have 5 people, lying/truth telling and having/not having access states, so it makes 3 groups of results. Draw 3*2/2=3 grids and gray/cross out what's impossible in the relevant grid.

In general, beware of:

- People saying one or more others are lying or telling the truth. If the person refers to only another, their statements have the same truth value when he/she says he/she is telling the truth, and the opposite if he/she accuses him of lying. If the person refers to multiple people, it becomes more complicated. Then you can try to find a result coming up regardless if the original person is lying or telling the truth.

- Re: people just talking about the main issue (who's the murderer etc.), there's a limit to the number of liars and truth tellers because of the amount of murderers, knaves or something. People saying or implying the exact same thing are both liars or truth tellers, and the opposing statements have opposite truth values.

  • $\begingroup$ Thanks! But just for starters, can you give me a hand with drawing a table for this particular problem?. I want to learn how to do the same in similar situations, but at the moment I can't find an example of how to do it by my own, hence I would really need help with this. In your comment you mentioned about Einstein's puzzles which I'm not familiar with, but I'll look into it during my studies. You said about drawing grids. But I don't know how to cross the results, therefore can you help me with this for this problem?. Sorry if it sounds redundant but I'd like to learn!. :) $\endgroup$ – Chris Steinbeck Bell Feb 10 '19 at 13:23

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