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#197 - Murder at the Liars Club (black dice)

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#72
#197 - Murder at the Liars Club (black dice) 01/03/2007 18:58:27

mindcandy
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Please use this thread for discussion about this card.
#580
Re: #197 - Murder at the Liars Club (black dice) 03/03/2007 21:56:16

Whiternoise
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Has anyone got a decent method for this one?
#627
#197 - Murder at the Liars Club (black dice) 04/03/2007 10:32:53

Albedo
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This shoud get you started:

Spoiler: (highlight to read)

Consider the two statements that refer directly to the murderer. There are only 4 different combinations of truth/lie for these two statements. Test these possibilities, and bear in mind that the puzzle must be solvable - this will save you some time.
#3885
Re: #197 - Murder at the Liars Club (black dice) 07/04/2007 11:38:25

stamford13
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Good advice.
I'm working at it slowly but surely.
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PM me if you need anything!

#5761
Re: #197 - Murder at the Liars Club (black dice) 13/05/2007 20:14:45

derricks
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Argh! This took me forever! I forgot to take into account one instance of the reality that

Spoiler: (highlight to read)


if one half of an and statement is false, the whole thing is, even if the other is true. I glossed over the and in Archer's statement.


#7295
#197 - Murder at the Liars Club (black dice) 28/06/2007 22:39:17

saintjimmy
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Albedo wrote:

This shoud get you started:

Spoiler: (highlight to read)

Consider the two statements that refer directly to the murderer. There are only 4 different combinations of truth/lie for these two statements. Test these possibilities, and bear in mind that the puzzle must be solvable - this will save you some time.



Nice suggestion, and you should quickly eliminate one of thosebecause

Spoiler: (highlight to read)


If hart and archer are both non-members then Archer is the killer and his statement about Davis being a member is true. But Davis says Hart and the killer have the same status. Which we know to be true.
check out http://www.perplexcitytrades.com/saintjimmy for my trades list. Desperately seeking series one atm. See my thread for the tragic tale. lol
#12061
Re: #197 - Murder at the Liars Club (black dice) 01/11/2008 18:51:03

PJBolas
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I've locked myself out of this puzzle, but before I can access it again, I as wondering if someone could tell me if this is the right solution please:

Spoiler: (highlight to read)

Archer: Member
Brown: Non-member
Clark: Non-member
Davies: Non-member
Edgar: Member (Killer)
Flint: Non-member
Hart: Member
#12063
Re: #197 - Murder at the Liars Club (black dice) 02/11/2008 11:07:10

KingOfWrong
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You'll be able to try that solution in a few hours...

But boolean logic is easy enough to check - it's a classic NP-complete problem (i.e. SAT).

This can be represented in propositional logic by letting A..H be predicates standing for "Archer..Hart is a member of the Liar's club" then "X would agree Y" entails either "(NOT X) AND Y" (i.e. not a member, and the statement is true) or "X AND (NOT Y)" (i.e. a member and the statement is false).

Taking Archer's statement, working back from the end, plugging in your values for the predicates, and reducing, you get:
"... Hart and Edgar are both members" so substatement S1 = (H AND E)

Spoiler: (highlight to read)

= (true AND true)
= true



"... Flint would agree that <S1>" so S2 = ((F AND (NOT S1)) OR ((NOT F) AND S1))

Spoiler: (highlight to read)

= ((false AND (NOT true)) OR ((NOT false) AND true))
= ((false AND false) OR (true AND true))
= (false OR true)
= true



"... he [Davis] would agree that <S2>" so S3 = ((D AND (NOT S2)) OR ((NOT D) AND S2))

Spoiler: (highlight to read)

= ((false AND (NOT true)) OR ((NOT false) AND true))
= ((false AND false) OR (true AND true))
= (false OR true)
= true



"Davis is a member and <S3>" so the entire statement S = (D AND S3)

Spoiler: (highlight to read)

= (false AND true)
= false



Finally, this statement is true if - and only if - Archer is not a member, i.e. ((NOT A) AND S) OR (A AND (NOT S)) must be true.

Spoiler: (highlight to read)

= ((NOT true) AND false) OR (true AND (NOT false))
= ((false AND false) OR (true AND true)
= (false OR true)
= true

Huzzah, those values for A, D, E, F and H are consistent with the first statement!



Yes, that's the most complex one...

There are only about 70 possibilities for the values of A..H, out of a search space of only 128, due to the numbers of members/non-members given in the question.

If you were going to drop this problem into a SAT solver, you could introduce another predicate (say "M") for the murderer's status, and then solve separately for the murderer's identity... or you could introduce 7 more (Ma => Archer is murderer, Mb => Brown is murderer, etc.) and a rather ugly 14 clauses for "Hart's membership status is the same as the murderer's" and another 7 to enforce mutual exclusion on the murderer's identity, which will then - completely mechanically - give you the murderer as well.

Learning formal logic tends to take the fun out of this sort of puzzle
#12067
Re: #197 - Murder at the Liars Club (black dice) 04/11/2008 19:39:14

X9Tim
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I learnt some formal logic 17 years ago.

The trick is to not remember what you learnt...

*lol*

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