/sci/ - Science & Math

>>5404907
> > Can you keep a secret?
> [yes]
So can I.

>>5404905
Can you keep a secret?

>>5404878
Actually several of my namefags shows up in there, along with some trips I totally forgot about.

>>5404878
> dat TN5
*sigh*... I miss that dude.

> tfw no one remembers or cares about you

>>4026593
on convergence issues:
http://en.wikipedia.org/wiki/Newton_fractal

>>4026489
Ignore the other comments. I misunderstood what you were asking. I thought you were providing the root, not the coefficients. (In CF-mode, my mind immediately interprets all lists as continued fractions.)

There is a way to basically use newton's method to pump out continued fractions. I never implemented it, but Gosper described it and it was pretty clever as you fed the continued fraction back into the process so it never bothered to try calculating beyond good values. (Why calculate the first approximation beyond the first term?) It may be that they use some facts about convergence of newton's method to ensure they avoid oscillations. Additionally, convergence depends somewhat critically on the fact that there are no roots in multiplicity.

That's the only guess I have.

>>4026518
Actually, sheesh, there's an infinite number of functions anyway. Weird.

>>4026489
If there is not a unique root, then there are an infinite number of functions, no?

>>4026444
Wrote my own cf arithmetic module. It was great fun. Then I never used it. Such is the life of the eternal toolmaker.

oh look it's the sam-harris-morality-is-objective-guy-again

where you been faggot?

Please forgive me, but every entry shows <span class="math">C_n^{n-1}=1[/spoiler]. Maybe this will clear up my misunderstanding. Here are the possible covers of 5 of length 4:
((4 3 2 1) (5 3 2 1) (5 4 2 1) (5 4 3 1) (5 4 3 2))
There are at least 2 Ys here, not 1.

Maybe this will reveal what I'm misunderstanding.

Man I am fucking up something, because that seems totally impossible to me.

You want to cover (1 ... 6) with Ys composed of 2 element sets. Here are the disjoint two-element sets.
((2 1) (3 1) (4 1) (3 2) (5 1) (4 2) (6 1) (5 2) (4 3) (6 2) (5 3) (6 3) (5 4) (6 4) (6 5))
Because each set has two distinct elements and there are 6 elements to cover you will need at least 3 sets e.g. (1 2) (3 4) (5 6) to cover. This implies you could get away with 5 Ys (there's 15 sets there) but you can't, because elements are repeated since there was no requirement that these are recursively disjoint.

So I must be misunderstanding some criterion.

>>3996888
You're saying <span class="math">C_6^2[/spoiler] is 5 but that seems impossible, can you show me what the five Ys are?

>>3996809
Can you elaborate some of this table, because something seems off

>>3996765
If I understand your suggestion, this isn't right, because when you're maximizing the number of covers you want to minimize the number of times each element occurs in a set. You don't want to include the element 3 any more than you have to, otherwise you may lose a cover (since the requirement is the covers are disjoint). So you can't just "fix" an element.

But I may have misunderstood.

>>3996451
> Civility and politeness are extremely useful tools in communication with people who are more wrong than right, but of very little use with people who are vastly more right than wrong. This counter-intuitive observation comes directly from the fact that we simply do not need civil and polite ways of telling people that they are right about something. So the people who have most to gain from civility and politeness are people who know they are and intend to /stay/ wrong while they force everybody else hold their tongues.
Erik Naggum
http://genius.cat-v.org/erik-naggum/punishers-and-moralists

>>3996274
To generate all the subsets of length 2 from 1..5 is all partitions of length 2 from 3 to 9.

To generate all the subsets of length 3 from 1...5 is all partitions of length 3 from (1+2+3)=6 to (5+4+3)=12.

Since all the partitions of 5 are in the partitions of 6 (since 6=5+1 is the head of the partition tree) then your covers are just showing this relationship.

It may not be the right way to tackle the problem, due to partitions not always being sets. But it's definitely going to generate the only sets you can use for your cover.

pic related on partitions. Notice there's a double length=3 filter step to remove non-sets.

This looks like a way around partition relationships. I mean ordinary partitions of integers, if there is some other obscure meaning (which there usually is). I'm sorry I can't invest a ton of time into this right now due to work, but the first thing that jumps out at me with this problem can be reduced to a problem with counting partitions of some length without repeats (cuz not all partitions of length n are n-element sets due to repeats).

E.g., take all partitions 6 of length 3 which are sets. There's just 1, which is (1 2 3). 7 also has just one, (1 2 4). 8 has (1 2 5) and (1 3 4). 9 has (1 2 6) and (1 3 5). Note that (1 2 6) won't appear in your example because 6 is too large.

Obviously generating partitions is insanity, it's a horribly tree-recursive procedure, but it may help with a combinatorial argument.

Is this one of those Markov Chain text generators?

> writing a string to a character
> program doesn't work
introduce a temporary
> hey it works!
ouch

You've got to declare your input as a fixed character array to use scanf("%ns... or you've got to use getchar().

If it works as you've written it, it is only luck. This guy is 100% right. >>3992009

Of interest possibly:
http://scienceblogs.com/principles/2010/09/_httpksjtrackermitedu20100907e.php

http://blogs.discovermagazine.com/cosmicvariance/2010/10/18/the-fine-structure-constant-is-probably-
constant/

Division by Three... fuck yeah Conway.

A little nighttime reading.