At the last Linux-intro meeting I met an officer of SigLinux
that posed a riddle to me. I'll present the riddle and
explain why I think his solution was flawed:
It went something like:
Sun's CEO wanted to see how smart their engineers were, so he
took them all to a room and gave them black jackets to put on.
He told them that some of them have white Sun logos on the back
of their jackets. When you discover that you have a logo
go to the front of the room and he will check on them
each hour until they have all the logo'ed at the front.
You can not look at the back of your jacket, feel it, use
mirrors or test for texture. You can only see everyone
elses' jacket. How do you determine if you have a logo or not?
Think about this for a minute.
sleep(60);
Ok, I follow the beginning steps of induction here.
Obviously, if you look around and see NO other logos, you
have one right? So you know to go to the front.
Using the same logic, if you see someone else with a logo
AND ASSUME that everyone in the room is smart and knows the
game, you realize that he is not going up because he ALSO
sees a logo, on you; so you go up, and he then knows to go
up after the CEO comes and goes (since the game is not over).
Ok, looks like it's working right?
Well, 3 logos is a little more difficult, but you can assume
that if you see 2 people with logos and nothing is happening,
since you all know they game, they must be seeing 2 logos as well,
meaning you have one, so you go up, then it works with 2.
Ok well at this point the riddler began telling me about how
if the logic works for n (being one logo) and it works for n+1 (2),
then it works for all n. I don't see this being true.
For example, let's say there are 100 jackets with logo, there is
no methodology I can think of that would give these engineers the
hints they would need to determine if they had a logo. It works
with 1,2, and 3, but begins to fail soon after.
I'll also use the Reducto Ad Absurdum method to disprove the
"true for n, true for n+1, true for all n" logic, for some cases
anyway.
n = 1; # one is prime
n+1= 2; # two is prime
therefore:
n = 4; # four is prime
Bwahh haa, four is not prime.
--
You guys see where I am going with this?
If there is someone out there that can prove this works with
say 100 people with logos, I'd love to hear it. It's stumped me
and everyone I have told.
Good luck on finals,
--
Tom Carlile [EMAIL PROTECTED]
-- randomly generated starwars quote:
"Try not. Do. Or do not. There is no try."
-- Yoda
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