Barber's paradox

[Chuck]

Ok Barber's paradox is back. This is the concise version of a paradox proposed by Bertrand Russell in early 20th century.

1. The men in the village are two types: men who do not shave themselves and men who do.
2. Village barber shaves all men who don't shave them selves, and he shaves only those men.
3. Everyone in the village is shaved.

Who shaves the barber?

[z00ey]

..i ll answer later (wanna see some others first..) .

[Cope]

Well. I do remember from the old forum no one ever getting this. So I toss out the answers I have.

woman or everyone shaves themselves(no barber)

I remember this from some Philosophy class in college. Teacher loved to make us think. in the end over-analyzing the question. But if I recall one of those is right.

Cope

[Cypher]

1) the barber has leukeimia

or

2) any other man. doesnt matter if they shave themselves or not. doesnt say they dont shave other men.

[Loon]

mabye but barber doesn't live in the village?

but it's basically self-contradiction although nothing states that one of the other men doesn't also shave somebody else (namley the barber)

mabye that guy just wants to remain unknown because he did a bad job or something lol

anyway the real awnser is that the barber was a woman, which is confused slightly by Chucks version as it states "he" shaves only those men, but that's not quite the correct version

[z00ey]

Cope said:

Well. I do remember from the old forum no one ever getting this. So I toss out the answers I have.

woman or everyone shaves themselves(no barber)

I remember this from some Philosophy class in college. Teacher loved to make us think. in the end over-analyzing the question. But if I recall one of those is right.

Cope

i had it but it was deleted immediatley because the seciton of the board has been cleaned up (at least i think so, one MUST have been right..)

yes yes yes.. *shuts up* ,but the way to the solution was harder than the obvious..

[Draconis]

Both mathematicians and philosophers have been battling with this one... it's a paradox... and creates an endless loop, and many break it down into the simple fact of "The barber shaves himself, only if he does not".
The most logical, yet the most confusing explanation I can come up with was also probably the most obvious, and often the one that remains overlooked... one must disprove the existence of the barber by using the simple fact that he can not exist if he can both shave himself and not shave himself at the same time... thereby it's assumed that the village would not exist either, as one man can not shaves all, and only those men in it who do not shave themselves (as the barber already does not exist).
Simply, this is referred to as reductio ad absurdum.

So... am I close?

[Cypher]

or u could just read loons post

[z00ey]

as far as i got you are very right.. ..one of my solutions is exactly like that:
"there cannot be a barber who lives in a village etc "


the next one is actually the same but you can prove it with maths terms like let s saybr]p stands for all the ppl in the village, o stands for one person out of p and s(o1~o2) means o1 shaves o2,, then for every o out of p the barber b could be described likebr] (s(o~o) => !s(b~o)) && (!s(o~o) => s(b~o))

that s the same like the what we conidered first: barber b shaves all o's who don t shave themselves..
..now we just assume (out of the question itself) that the barber HAS to shave himself, so we also assume o == b , and we get thisbr] (s(b~b) => !s(b~b)) && (!s(b~b) => s(b~b))
..and there we have our paradoxon aka the answer that !barber exists who lioves in etc..


the way that s more philosophically but actually not far away from the other ways of solution is simple to set a theorem valid that can t be doubted by the one who asks too, like: "nothing in this word can "be" a thing and "cannot be" the same thing at same time".. ..then you assume the paradox is solvable and have to say "if the barber doesn t not shave himself he has to shave himself" ..but that can t be possible because of our basic theorem about the thing that cannot "be" and "notbe" at the same time

..geez i hope this makes some sense, in my head it s much clearer, lol

*oh btw.. ..i was into mirc a bit right now, lol.. ..so i used "!" as "not" and used "&&" as "and"

[w0rd]

This is like..

• If you go back in time and kill your grandparents (for some highly strange reason)..
• Your parents would not have been born..
• Therefore you could not have been born..
• How are you existing to go back in time and do that?

..isnt it?

[z00ey]

Cypher said:

or u could just read loons post

..if you hadn t maths test now i wouldn t reply, but: nope.. ..it s a totally diff answer he gave


yep, word, another paradoxon but not that tricky: the main point with the barber is that however you solve it you come to a point where the subject is (or does) something at the same time and in the same sphere when he isn t (or doesn t) the same (thing)..

..the time paradoxon you could solve into a direction that you say because it can t be possible to kill your grandparents, therefore it can t be possible at all to start the journey, so $false.. but it s only one sphere (not the time levels in the example itself, the mathmatically sphere) and one reduction needed there, with the barber (if you wrote it out in formulas from the beginning to the end) you have three diff reductions involved.

[w0rd]

Ahhhh..

I'm gonna re-read this thread tomorrow when my brain is fully active

Right now it feels like it's partly switched off

Btw, your thing doesnt work

[z00ey]

had to run and search the phone while i was modifying, lol.. ..passed me

[Axel]

the next one is actually the same but you can prove it with maths terms like let s saybr]p stands for all the ppl in the village, o stands for one person out of p and s(o1~o2) means o1 shaves o2,, then for every o out of p the barber b could be described likebr](s(o~o) => !s(b~o)) && (!s(o~o) => s(b~o))

that s the same like the what we conidered first: barber b shaves all o's who don t shave themselves..
..now we just assume (out of the question itself) that the barber HAS to shave himself, so we also assume o == b , and we get thisbr](s(b~b) => !s(b~b)) && (!s(b~b) => s(b~b))
..and there we have our paradoxon aka the answer that !barber exists who lioves in etc..

....and i though MDX was confusing!

[hacker]

Well he does program in C++, ASM and probably other languages, which requires a fair bit of maths