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Honestants and Swindlecants IV.

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Honestants and Swindlecants IV. - Back to the Logic Problems

Our gringo was lucky and survived. On his way to the pub he met three aborigines. One made this statement: "We are all Swindlecants." The second one concluded: "Just one of us is an honest man." Who are they?

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Honestants and Swindlecants IV. - solution

The first one must be a swindlecant (otherwise he would bring himself into a liar paradox), and so (knowing that the first one is lying) there must be at least one honestant among them. If the second one is lying, then (as the first one stated) the third one is an honestant, but that would make the second one speak the truth. So the second one is an honestant and C is a swindlecant.

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• 3 months later...

1st - Swindlecant

2nd - Honestant

3rd - Swindlecant

1st one says "We are all Swindlecants". If he was a Honestant, this would be false. Hence, he is a Swindlecant. Since he is a Swindlecant, this statement has to be false, and hence, atleast one of the three is a Honestant. Thus, either one or both of the others are Honestants.

2nd one says "Just one of us is an honest man". If he was a Swindlecant, then, as concluded, from the 1st person's statement, the 3rd person would be the Honestant. However, that would make his statement true, which is not possible. Hence, he is a Honestant, and his statement is true.

From the 2nd person's statement, we can infer that the 3rd person is a Swindlecant.

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• 2 weeks later...

Too easy

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• 1 month later...

In classical logic form:

A: This person is a swindlecant

B: This person is an honestant

X="Ai + Aii + Aiii"

Y="Bi xor Bii xor Biii"

If X, then not Y

If not X, then Y

f Y, then Bii

Ai xor not Ai

If Ai, not X

If not Ai, then X

If X, Ai

If not Ai, Ai (reductio ad absurdum)

----------

Ai

Not X

----------

Y

Bii

If Bii, not Biii

----------

Not Biii

Aiii

So our three conclusions are: Ai, Bii, Aiii or First person is Swindlecant, Second is Honestant, and Third is Swindlecant

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• 1 month later...
In classical logic form:

A: This person is a swindlecant

B: This person is an honestant

X="Ai + Aii + Aiii"

Y="Bi xor Bii xor Biii"

If X, then not Y

If not X, then Y

f Y, then Bii

Ai xor not Ai

If Ai, not X

If not Ai, then X

If X, Ai

If not Ai, Ai (reductio ad absurdum)

----------

Ai

Not X

----------

Y

Bii

If Bii, not Biii

----------

Not Biii

Aiii

So our three conclusions are: Ai, Bii, Aiii or First person is Swindlecant, Second is Honestant, and Third is Swindlecant

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.

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the first one is lying because he said "we are all swindlecants" and then the second said that one of them was an honest man. therefore if the first is lying he is a swindlecant and the second is an honestant and the third would have to be a swindlecant.

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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?

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Great observation, slmo! But... if gender is a factor, then:

(1) B and C must be Honestants; and

(2) either one of them could answer "one Honest man" truthfully

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(1) B and C must be Honestants; and

(2) either one of them could answer "one Honest man" truthfully

Thanks...ur right, my bad, B is honest

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this one was pretty easy.

it is impossible for aborigine that said "We are all Swindlecants." to be an honestant. so he is a swindlecant. he is lieing about all of them being swindlecants. so the one that said "Just one of us is an honest man." is an honestant.

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• 3 weeks later...
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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• 2 months later...

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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• 2 weeks later...
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.

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• 6 months later...

I think the gringo lying...

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• 3 months later...
<!-- 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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• 2 years later...

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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