Originally posted by royalchickengoing by this post, i'm not sure if i can do it on my limited maths base
if "genius" wishes to earn his nickname, he may attempt the
following. if we
pick a positive integer, c(1) (the n is a subscript), and let:
'''''''''''''''' / c(n)/2 if c(n) = 0 (mod 2)
''''''''''''''''' \ 3c(n)+1 if c(n) = 1 (mod 2)
the problem is to show that for any c(1), there will alwa ...[text shortened]... his
nickname. i'll have the answer to the "dons problem" shortly.
sintubin has a remarkably astute stomach. the gossiping don's can always do it2N-3 isn't optimum: for N = 4 the dons can talk in pairs and then swap around a bit, so they
in 2N-3 moves (i think this is optimum).
Originally posted by royalchickenIf you do maths at Cambridge, you're going to be one of the people who goes to 4th year
genius-i was not entirely fair to you (although, as you are one year my senior,
i feel no need to be particularly gentle). the problem i gave is the Collatz
conjecture-one of the outstanding unsolved problems of mathematics. i have only
made progress with a fairly good knowledge of analysis (calculus et al) and
number theory. i have yet to prove it, though. give it a go-it is infectiously
Originally posted by royalchickenSuppose n dons can communicate all the gossip in k calls. Then if you add another don, all
i am correct in saying that a method for the dons to converse is optimum if
there exists some number n of dons for which there is no more efficient method?
also-just to check that it can always be done in 2n-4, i did prove that this
can be done. i then tried n=4 to 12, but i was curiously unable to work out how
it is done with 13 dons. it must be possible though. can someone tell me how?
Originally posted by royalchickenMy email address is: cdr29ycam.ac.uk (Replace the y with an @, you never know where the
if you (Acolyte) would give me your email address, i could send you the work i
have done on said conjecture. (it would be typographically ungainly in straight
text.) however, i have proved that every starting value will produce a sequence
that TENDS towards one as the number of terms generated increases without bound
(but i have not proven that th ...[text shortened]... oled up in a hovel with gallons of coffee, beating his head against a wall
trying to solve it.)