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Comments:
<0> ekiM: I don't get your humour :P
<1> I don't get either.
<1> both
<0> oh, i get hendrik's
<2> ekiM : finitely many primes ?
<2> ekiM : rather appy some contradiction scheme for twin primes
<2> to prove something useful :-P
<3> there's an infinite number of primes
<4> i'll alert the media
<3> good
<2> Catfive : you may wanna consider to call 911 first, might be urgent
<4> % PrimeQ[911]
<5> Catfive: True
<2> haha
<2> interresting observation
<6> % PrimeQ[31337]
<5> sek: True
<6> nice:p
<7> Anyone familiar with the term Unitarily Irreducible?
<7> applied to a matrix
<7> I have a definition for Irreducible
<7> however I don't have one for unitarily irreducible
<7> Anyone, anyone?
<8> http://64.233.183.104/search?q=cache:NRKuvIowZzMJ:nxzobi.people.wm.edu/numerang.pdf+unitarily+irreducible&hl=en&gl=uk&ct=clnk&cd=1&client=firefox-a
<7> lol
<7> I just printed that out
<7> and am looking at it
<7> thanks anyway flips and rails
<7> I have a matrix analysis text
<9> "There exists an A such that there exists a B such that X" is the same as "there exists an A and a B such that X", right?
<7> I don't know why it's not in here
<8> Its not greatly clear but there are plent of things on google
<8> s/plent/plenty
<7> So if a matrix A is unitarily reducible
<10> ihope : existential quantifiers commute past each other, yes
<7> you can break into this sequence of direct sums
<7> with Unitaries
<7> on the left and right
<7> If it's irreducible
<7> you can't do that?
<9> What happens to stuff like "there exists an A such that for all B, there exists a C such that..." or "for all A, there exists a B such that for all C..."?
<8> Safrole: I think so, its not very clear and I'm not that sure :S
<10> ihope : generally, quantifiers of different kinds does not commute with each other
<10> (ihope : though there's something called skolemizing ..)
<7> lol
<7> He gives another definiton
<7> I should say they
<7> sorry
<7> They give another condition an iff for A being unitarily irreducible
<7> u(A) = n
<7> hmm now to figure out what u(A) is :)
<7> okay I think I have it
<7> so if you can write
<7> A in terms of n matrices that are directly summed
<7> where A is nxn
<7> it is irreducible
<7> I love when definitions make sense
<7> lol
<9> So I'm looking for a list of logical axioms.
<10> ihope : axioms for logic ? or for something, expressed in logic ?
<9> Well...
<9> A list of stuff like the law of contrapositives and... stuff.
<10> for which logic ?
<10> (i ***ume you're after a hilbert presentation)
<9> Well, what is there?
<9> Stuff that just deals with truths and things, that is, without any actual values.
<10> propositional logic, then ?
<9> Yeah, I guess.
<9> Well, I found a list... but thanks for telling me what I was looking for :-)
<11> tower of hanoi question,.. say i start with a radom configuration of disk on the three pegs, is it possible to make more than 2^n - 1 moves to get it sorted. ie. can the minimum number of moves required to sort the disks increase if the disks are not ordered at the beginning?
<9> I don't think so, but I'm not sure.
<11> yeah, a proof would be nice
<9> Well, you're saying that there's no algorithm that can always get the job done in less than 2^n moves.
<9> Or, rather, asking.
<12> It's pretty easy to get the job done in n moves with one arrangement.
<9> And what arrangement is that?
<11> ihope: last disk on the third peg
<12> Actually, 1 move.
<12> Actually, zero moves.
<12> This just keeps getting better
<1> Actually, -1 moves
<11> the question is, weather is possible to require more than 2^n -1 moves for randomly distributed disks
<12> If the disks randomly are positioned at the end position, you're done.
<9> Well, I think we don't have to consider possibilities where the biggest disk is on the destination peg.
<11> Olathe: well i was looking at the "worst" case
<9> That's the same as if that disk didn't exist.
<1> what are the rules exactly? You have three pegs and n discs of different sizes, and an allowed move is to switch a disk which is on top to some other peg but only if the top disk in the peg it goes to is greater than the one you're moving?
<11> correct
<1> ok
<11> hanoitower.info if you want to play a short game to get the feeling
<13> azi : surely you're actually asking "is there ANY configuration that requires more than 2^n-1 moves
<13> I#m not sure what "random" has to do with it
<11> ekiM: correct
<13> ok
<11> i suspect that's not possible, but i can't make a general proof
<12> You are attempting, in the middle of the game, to get all the pegs larger than a certain one onto one peg and all the others on another.
<1> oh and what is the object of the game, i'm missing that :)
<1> to take all disks to the same peg, i ***ume
<1> but to a paricular peg?
<12> "all the discs"
<12> So, you start with a random placement.
<11> koro: the third one
<1> ok
<9> How do you know it's in the middle of the game, and not before it?
<14> please help! I can't solve the equation tan(x/2)-sinx=0
<1> freaka: use the formula sin(x) = 2sin(x/2)cos(x/2)
<12> It takes zero or one move to get the smallest on top of the second smallest. It takes up to four moves to get those on top of the third smallest. Etc.
<12> So, the number of moves can't be larger than it is normally.
<1> end write tan(x/2) as cos(x/2)/sin(x/2), and work it out
<11> Olathe: are you sure it's impossible?
<12> Yes. Look at the subgoals.
<15> don't write tan(x/2) as cos(x/2)/sin(x/2) because that's false.
<12> You want to have the smallest n discs on one peg that's not the starting peg.
<1> yeah well write it as it should be written
<1> i did that on purpose to see if you were paying attention
<12> You start the subgoal with the smallest n - 1 discs in order.
<11> HiLander: why?
<11> HiLander: ah
<12> Either you're immediately done because the n disc is already under the n - 1 disc, or it takes the standard number of moves or less.
<12> So, if each subgoal takes the same amount of moves as normal, or less, you know the entire puzzle does.
<11> Olathe: actually the "worst" case is when all the disks "sits" on the first peg
<12> I thought the arrangement was random, but no larger peg on top of smaller.
<1> Olathe: how does it take the standard number o moves or less?
<11> Olathe: correct
<1> you need to stick a disk under a pile of n disks
<1> not to move your pile
<12> koro : Because you have one pile with the n disc on top.
<12> koro : You have one pile with the n - 1 and lower discs nicely ordered.
<12> koro : This is how it happens normally.
<12> koro : So, it can't take more time than it does normally.
<1> Olathe: you don't want to move your whole pile of n-1 disks to the top of the n disk
<11> you want to say, it can take more time than in the worst case?
<1> you want to move the n disk under the pile of n-1 disks
<1> that's not the usual case
<12> koro : That's the same thing, with different words.
<1> huh?
<12> koro : In one, the n - 1 discs end up on top of the n disc. In the other, the n disc ends up below the n - 1 discs.
<1> yes; and those are two different situations
<1> one is the one you want. the other is not
<4> var - nice one.
<1> in one you end up in a given peg, in the other, you end up in a different peg
<12> I don't see how.
<1> I'm not saying this is wrong
<1> I just don't see how the 'standard procedure' allows you to stick the nth disk under the ordered pile of n-1 disks
<12> You choose the nongoal peg the n - 1 discs aren't on to put the n disc on.
<12> Then, you put the n - 1 discs on it.
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