Murder

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Joseph Bolas
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Murder

Post by Joseph Bolas »

Here is another puzzle from the PerplexCity cards :D.

There's been a murder at the Liar's Club, and as interim sheriff, you are called upon to investigate. Members of the Liar's Club always lie, and there are at least three members, and at least three non-members (who are always truth-tellers), among the seven people involved (the six suspects and the murdered man). Everyone involved knows who is a member and who isn't and even who the killer is, but no one wants to get involved for fear of becoming the next victim. Along with the suspects' statements is a written statement from Hart, the murder victim, which he wrote as he was dying.

The case is definitely not suicide, and it is up to you to determine, from the seven statements, who is a member and who isn't, and who the killer is.

Archer: "Davis is a member, and he would agree that Flint would agree that Hart and Edgar are both members>"
Brown: "Hart would agree that Clark would agree that Archer and Davis are both non-members."
Clark: "Davis and Brown would agree that Edgar would agree that the murderer is not a member."
Davis: "Hart's membership status is the same as the murderer's."
Edgar: "Clark would agree that Brown and HArt would agree that Flint is a member."
Flint: "Brown would agree that Archer and Clark do not have the same membership status."
Hart: "Archer killed me!"
Allan Harmer
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Re: Murder

Post by Allan Harmer »

Joseph - Here are the votes of the Harmer Jury:
Members = Flint, Brown, Davis, Edgar.
Non members = Hart, Archer, Clark.
Murderer = Archer.

It seems too obvious but that's what we came up with,

Allan
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Joseph Bolas
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Re: Murder

Post by Joseph Bolas »

aagharmer wrote:Joseph - Here are the votes of the Harmer Jury:
Members = Flint, Brown, Davis, Edgar.
Non members = Hart, Archer, Clark.
Murderer = Archer.

It seems too obvious but that's what we came up with,

Allan
I've actually misplaced this puzzles answer :oops: so I will have to find it, but I don't think that this is right.
Howard Somerset
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Re: Murder

Post by Howard Somerset »

I've PMd my attempt to Joseph, in order to give others a chance to have a go too.

I agree with Joseph that Allan's solution is wrong. I don't believe that Archer's or Flint's statements are consistent with Allan's solution.
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Joseph Bolas
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Re: Murder

Post by Joseph Bolas »

Howard Somerset wrote:I've PMd my attempt to Joseph, in order to give others a chance to have a go too.

I agree with Joseph that Allan's solution is wrong. I don't believe that Archer's or Flint's statements are consistent with Allan's solution.
Thank you for your PM. I'll let you know via PM if right or wrong :).
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Joseph Bolas
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Re: Murder

Post by Joseph Bolas »

It's been almost a week since this puzzle was first posted and the only person to get this right was Howard, so if Howard would like to explain his method of solving the puzzle and what the answer was please :)
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Re: Murder

Post by Howard Somerset »

Joseph Bolas wrote:It's been almost a week since this puzzle was first posted and the only person to get this right was Howard, so if Howard would like to explain his method of solving the puzzle and what the answer was please :)
Joesph's invited me to explain my method, and to give the answer.

I'll go part way, and show how I approached it. Then others might like to follow on using my method, and have another go at solving the puzzle.

Most of the statements made by the seven people are quite complicated, and can certainly be difficult to unravel if they're examined in detail. My way to approach these liars and truth-tellers puzzles is to realise that if you have a chain of any length, the statement:

X1 says that X2 says that X3 says that ... says that Xn says that Y is a fact.

simply boils down to

If Y is true, then there is an even number of liars in the chain, and if Y is false, there is an odd number of liars in the chain.

It says nothing more, and nothing less, until you know a little more about who are liars and who are truth-tellers.

This puzzle contains many such chains of length 3.

My start point was the simplest chain - the first thing said by Archer. Archer is saying that Davies is a liar. This means that of Archer and Davies, one is a liar and one a truth-teller, but we don't know which.

We do now know, however, that the fact given at the end of Brown's statement is false, and so using the reasoning I gave above, either Brown, Hart and Clark are all liars, or else just one of them is a liar and the other two are truth-tellers, because an odd number of them are liars.

We can now continue in this way, gathering more information, and eliminating a lot, before reaching a solution.

There is an alternative approach, which I abhor, and there is a lot of correspondence in various puzzle forums as to the legitimacy of such an approach. That is to use that fact that it is known that there is a unique solution to the problem in order to find that solution. Personally, I regard the task of solving any puzzle to include the requirement to prove, rather than assume, that the solution is unique.

However, to use that alternative approach here, we can think about what would be the situation if Hart was a liar. We would know that Archer didn't kill him. But how could the puzzle then lead to a unique solution? No other statements say anything at all about identifying who is, or who isn't, the murderer. The only possibility of a unique solution, if Hart was a liar, then is for there to be just three liars, Hart, Archer, and the murderer. So this alternative approach could start with us having eliminated all possibilities except:
(a) Hart is a truth teller and Archer is the murderer
(b) Hart, Archer and the murderer are the only three liars.

I'll leave it there, in case others want to have another go. I'm off to the Lake District again Monday to Wednesday next week, and will also be away on Thursday, but I'll try to look in on Wednesday evening to see how many have solved it now.
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Joseph Bolas
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Re: Murder

Post by Joseph Bolas »

Well explained Howard :D

Not sure though if anyone else is going to have a stab at the puzzle.
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