# Puzzle to find one good person [duplicate]

There are 15 people who may lie or say the truth(bad people) and 16 people who will always say the truth(good) . Suppose A- bad person , B-good person and C-bad person, when asked about A, B will say that A is a bad person(truth), when asked C about A, it may lie or say the truth. How can we find atleast one good person from the lot.

• Unlimited number of questions ? Jun 24, 2016 at 13:11
• Possible duplicate of Knights and jokers or Faulty computers
– f''
Jun 24, 2016 at 13:17
• Can we only ask yes or no questions? Jun 24, 2016 at 13:21
• @wrangler it does not work, bad people can tell the truth everytime if they want. Jun 24, 2016 at 13:39
• Not a duplicate of "Knights and jokers" because that one asks for all individuals to be correctly identified. Looks very close to being a duplicate of "Faulty computers" except that both have particular numbers of people/computers and the numbers are different; I doubt this makes any difference once the number is bigger than 3 or so, but haven't checked. Jun 24, 2016 at 16:12

• Worst case = 496 questions asked. Jun 24, 2016 at 14:11

I'm not sure if this is optimal

Take $16$ of them and split them into $2$ groups of size $8$, $A$ and $B$
- there will be at least $1$ truth-teller amongst the $16$

Ask all $31$ in turn, "Is the number of truth-tellers in $A$ greater than the number of truth-tellers in $B$?".
until you have $16$ agreeing answers, at which point you know this is the truth.

"Yes" identifies $A;$ "No" identifies $B$
(when equal $B$ is identified, but $B$ will still contain at least $1$ truth-teller)

Split the identified group into $2$ again and repeat, until you have identified a group of size $1$

The worst case is that you ask a truth-teller last in every round and the others all lie at every opportunity, $4\times31=124$ questions.

Alternatively, if we are not limited to "yes" or "no" questions

Label each person $1,2,3,4,\cdots,31$

Ask all $31$ in turn, "Please list all the truth-tellers, including yourself if you are one, by label"

Once you have $16$ responses identifying the same set you have found all your truth-tellers.

The accepted answer to the "Knights and Jokers" question of which this is almost a duplicate guarantees to identify everyone in at most $31+15-1=45$ questions.

The accepted answer to the "Faulty computers" question of which this is almost a duplicate claims, assuming it translates in the obvious way, to guarantee to identify one good person (= working computer) in at most $31-2=29$ questions. I say "claims to" simply because it references an answer to the "Knights and Jokers" question which (1) was not optimal and (2) is alleged in a comment to be incorrect.

... OK, having looked at that second accepted answer, it looks OK to me and doesn't depend on the difference between 16+15 here and 50+49 there. To summarize: we are going to build a chain of people, each of whom vouches for the next and containing more good people than bad. In particular there is at least one good person in the chain. Everyone after a good person in the chain must also be good, so in particular the last person in the chain is good. Here's how we do it: start with an empty chain and a "candidate pool" containing everyone. Now repeat the following. Take the last person in the current chain, or anyone at all if the current chain is empty. Ask them about someone from the candidate pool. If they say "good", put that person on the end of the chain. If they say "bad", remove that person from the candidate pool and the person who said "bad" from the end of the chain. Note that at least one of the people you have removed is bad, so there remain more good people than bad after the removal. Eventually you will run out of candidates, so everyone you haven't removed is in the chain. Done.

How many questions does that require us to ask? The first question that creates a non-empty chain adds 2 to the chain. Other than that, every question either adds 1 to the chain or subtracts 1 from the chain and 2 from the total number of people uneliminated; hence every question reduces (#uneliminated - #chain) by exactly 1; at the start this figure is the total number of people, and at the end all the uneliminated people are in the chain so it's zero. So on the face of it this requires (#people-1) questions. BUT there's an optimization available: as soon as there are more people in the chain than out of it, we can stop because there has to be at least one good person in the chain; in particular, if ever there is only one person out of the chain, we're done. (What if the chain is empty? Then, because there are always more good people than bad, that one person is known to be good.) So we never need more than (#people-2) questions.

OK. Now, really the only thing missing here is that we haven't proved this optimal. So suppose we have $n$ bad people and $n+1$ good people, and we start asking questions "person $i$, is person $j$ good?" and every answer is no. A "no" answer tells you precisely that either $i$ or $j$ is bad, so what we have is: $N=2n+1$ people, just over half of whom are good, and $N-3$ facts of the form "these two people are not both good". Perhaps this cannot be enough to definitely identify a single good person.

That would mean that for each of the $2n+1$ people, we can delete that person from the list and still find a way of marking $n+1$ as good and not violating any of our $2n-2$ facts. That will be true if every graph with $2n$ vertices and $2n-2$ edges has an independent set of size at least $n+1$. The usual lower bound on independent set size comes from Turan's theorem but I think that isn't sharp when, as here, the number of edges is very small. I haven't yet found a proof (or disproof) of the bound we need in this case, but I haven't tried terribly hard.

• Yes, I was thinking about ways to use who answered what to improve upon my strategy. I didn't think to check for the similar problem (of which it looks like this is indeed a sub-problem). Jun 24, 2016 at 16:25
• There is one more optimisation for $N$ computers/people and unknown $j<\frac{N}2$ jokers/faulty computers - if $N>4$ and even we can discard one person from the pool right at the start and use $N−3$ questions. The full solution to a puzzle of the same form was provided by Chris Peikert, Grant Wang, and Abhi Shelat of MIT (I don't know if they were the first to ever solve it this way though). Jun 26, 2016 at 0:30