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Honestants and Swindlecants IV.

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15 replies to this topic

#11 coffee.lady



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Posted 05 November 2007 - 12:59 AM

Another possibilty???

#1 is a swindlecant (easy)

#2 is a swindlecant (being female)

#3 is an honestant (being the only honest man)

I know, it's not indicated in the original and may violate the spirit in which it was given, but isn't it logically possible?

However, if #2 were a swindlecant, #3 would have to be the honestant AND the woman in order to make #2's statement false.
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#12 logic seeker

logic seeker


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Posted 08 January 2008 - 10:40 PM

Of course the second aboriginia is a Honestant and A and C are defintley swindlecants if the second one was lieing that would mean that the first one was telling the truth witch would mean hes a Honestant.

1 Swindlecant

2 Honestant

3 Swindlecant
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#13 MAD


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Posted 19 January 2008 - 12:40 AM

However, if #2 were a swindlecant, #3 would have to be the honestant AND the woman in order to make #2's statement false.

You need to read the two posts following the one you quoted. Then you would see that the following is what was meant:

#1 swindlecant
#2 honestant (female)
#3 honestant (male)

alternately the solution could be:

#1 swindlecant
#2 honestant (male)
#3 honestant (female)

Including the original given solution, there are 3 possibilities. To avoid this ambiguity, the problem statement could be re-written to say that all three are male.

Edited by MAD, 19 January 2008 - 12:42 AM.

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#14 Lawjyk



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Posted 29 July 2008 - 10:29 PM

I think the gringo lying... :P
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#15 grottylittlewanker


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Posted 22 November 2008 - 07:16 PM

<!-- s:?: --><!-- s:?: --> <!-- s:?: --><!-- s:?: --> <!-- s:?: --><!-- s:?: --> how the heck do you use algebra(I'm guessing that's what it is because of the variables for swindlecant and honestant) to solve a logic puzzle? Not all of us are rocket scientists you know.

its not math, its Symbolic logic.

I took a college course on it, it was called Symbolic Logic, but the teacher called it predicate logic...

although the form I know of it is different than what he used.

The version I know:
'^' = and
'v' = or
'~' = not
'->' (an arrow) = if...then
'<->' = if and only if
then an upside down A is 'for every'
and a backwards E is 'there exists'

I know another standard where:
'+' = or
'*' = and
and the rest are the same.

it helps to keep things organized and there are specific rules and such that makes sure you actually prove it in every case. I actually have a program that checks to make sure you prove things correctly using this syntax that I got for free with the book I needed for that class.
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#16 Erdumas



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Posted 14 February 2011 - 08:52 AM

Another possibility: When aborigine A says 'we are all Swindlecants', it is possible 'we' also refers to the gringo. Clearly A is still a swindlecant. But now, B, in saying 'just one of us is an honest man' would also be referring to the four of them. Since the gringo is neither an Honestant nor a Swindlecant, B need not be telling the truth (though he could be). We'd end up with

A-> Swindlecant
B-> Swindlecant
C-> Swindlecant
Gringo -> Neither

We could also have

A-> Swindlecant
B-> Honestant
C-> Swindlecant
Gringo -> Neither
(that is, the old solution is still a solution, but no longer the only one). Of course, if 'honest man' doesn't exclusively mean 'Honestant', then it could also refer to the gringo (supposing he is, by and large, and honest man (person, if you care)), in which case C could be an Honestant, and B would still be lying!
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