Shadowed Philosopher

Turing complete, to some reasonable approximation thereof. Also, I believe that if you did simply plug in infinite memory (and infinite registers, and...you need a lot of infinite things, actually) to a real processor, you would get something that is actually, theoretically Turing complete.
Incidentally, has anyone here ever looked into the theory of large numbers? By 'large', it is meant 'incomprehensibly, irrepresentably huge', as in 'you cannot write this out if you fill the entire universe with digits at one per Planck volume'. The usual example is Graham's number, which is derived from a simple doubly recursive function similar or identical to the Ackermann function, and is so, insanely huge that the number of universes it would take to write out its decimal representation cannot, itself, be represented in this universe. As far as I know, that's the biggest one that's actually mathematically significant in any way; then you get to things which are just silly, like the XKCD number and the various derived large numbers.
Edited to fix link.

edited 14th Jun '11 2:46:57 PM by alethiophile

Shinigan (Naruto fanfic)

DUMB

[1]
Not really of interest to me, though...

__*__(edit: hate this parser)edited 13th Jun '11 10:17:36 PM by Tzetze

Shadowed Philosopher

Theoretically, a computer with enough swap space should be able to handle numbers up to (2^the number of bits in the address space, plus one

__*__). Which is always huge (even for a 32-bit PC it's, what, 2^(2^32*8), which is about 2^34350000000).edited 14th Jun '11 2:50:29 PM by alethiophile

Shinigan (Naruto fanfic)

Is that cake frosting?

I think that with some care, you could actually handle even bigger numbers, by keeping them in mass storage and performing the arithmetic operations

*without*bringing the whole of them into memory at once. This would be slower, of course, but it is certainly possible. But in any case, even if this number is representable there definitely exist numbers which are too big to be represented by any computer — heck, just take 2 to that number you mentioned and you're already there.edited 23rd Jun '11 11:00:35 PM by Carciofus

But they seem to

know where they are going, the ones who walk away from Omelas.

know where they are going, the ones who walk away from Omelas.

DUMB

In one given schema, you mean. 0 = zero, 1 = Ackermann(23,23), 2 = 235!!!!!, 3 = Graham's number, etc.
That reminds me, Kolmogorov complexity and the related definition of "random sequence" are pretty damn interesting.

edited 23rd Jun '11 11:34:59 PM by Tzetze

^{57}dRoy8th Jul 2011 06:31:43 AM from The Happy Place , Relationship Status: A cockroach, nothing can kill it.

Perpetually clueless

I can't do math. Any math related discussion tends to explode my brains. Yet I understand some of the charm of value and see its worth. I really want to be good in math, if only so that I can understand some of the equations in wikipedia, like equation of force and all. If only I could do math, my life would have been a lot easier...I actually gave up math and gets barely enough score to not to fail.
In short: I suck at math and am jealous of anyone who isn't.

Mother of god...You turned one of the hardest and best Champions into an absolute joke. - Zelenal

I love math but I already forgot most of the things I was supposed to remember. Ah well.
Mostly I derive things on the spot until muscle memory just sets in. More fun that way. I never hated math so I don't really know understand the math-haters' feelings.

Your Lord and Master

Math + Tropes = Win.

A Sadist RP-er

Me too!

__==__Watchman of the Apocalypse

Ok math people I have a question I need help answering.
part 1

I have a math/ problem I need to run by you guys. This should be fairly simple and only slightly tricky. When asked to see figures refer to this image◊ A builder using 1x1x1 blocks is building a step pyramid. His first layer (Figure 1) is two squares one inside the other that form the shell of the structure. The outer most layer (black) is 68 blocks on a side. The inner most square grey is 66 blocks on a side. In order to build the pyramid, the next layers outer square will be built on top of the last layers inner square (see figure 2) and a new inner square placed inside that square. See figure two. See Figure 2(The black is the very first outer square, the blue is built on top of the grey layer, the red forms the next layer to built on top of. Other then the two squares the rest of the layer will be empty. The builder needs to build in this manner up to layer 50. At layer 50 the layer will be filled in completely to build the top of the pyramid. How many total individual blocks does the builder need to achieve his goal? The total of all the layers + 317 for interior constructions should give you the final answer.

Part 2 What I have figured out thus far is this. You find the perimeter of both squares for each layer until the 50th layer. The 50th layer will be the area of that layer. To find the area each new layers sides shrink by two. You add up all the perimeter totals for the layers except 50 which is an area. Then add the extra bits for interior structure. I know each layers total for the two perimeters shrinks by a predictable amount. Ie. The first layers total is 536. The next layer is 528. It shrinks by 8 every time. But I have no way to cut the very long process of writing out then calculating that for 49 layers. The 50th layer I think is 19x19 but I think I might be off on that.

What I need is given all this information how many total units will be needed to complete that structure.

I have a math/ problem I need to run by you guys. This should be fairly simple and only slightly tricky. When asked to see figures refer to this image◊ A builder using 1x1x1 blocks is building a step pyramid. His first layer (Figure 1) is two squares one inside the other that form the shell of the structure. The outer most layer (black) is 68 blocks on a side. The inner most square grey is 66 blocks on a side. In order to build the pyramid, the next layers outer square will be built on top of the last layers inner square (see figure 2) and a new inner square placed inside that square. See figure two. See Figure 2(The black is the very first outer square, the blue is built on top of the grey layer, the red forms the next layer to built on top of. Other then the two squares the rest of the layer will be empty. The builder needs to build in this manner up to layer 50. At layer 50 the layer will be filled in completely to build the top of the pyramid. How many total individual blocks does the builder need to achieve his goal? The total of all the layers + 317 for interior constructions should give you the final answer.

Part 2 What I have figured out thus far is this. You find the perimeter of both squares for each layer until the 50th layer. The 50th layer will be the area of that layer. To find the area each new layers sides shrink by two. You add up all the perimeter totals for the layers except 50 which is an area. Then add the extra bits for interior structure. I know each layers total for the two perimeters shrinks by a predictable amount. Ie. The first layers total is 536. The next layer is 528. It shrinks by 8 every time. But I have no way to cut the very long process of writing out then calculating that for 49 layers. The 50th layer I think is 19x19 but I think I might be off on that.

What I need is given all this information how many total units will be needed to complete that structure.

edited 12th Aug '11 4:32:15 PM by TuefelHundenIV

"Who watches the watchmen?"

I see the Awesomeness.

Hm let's see. Total number for n layer is going to be 8*(68-2(n-1))+4 yes? If you just need that count summed, stick it in a calculator/excel.
Wait, the fiftieth layer should be negative somehow. It's shrinks on one side by two for each layer, you'd need a minimum of one hundred starting width to reach a fiftieth layer, and it would have to be a corner stack that breaks the even pattern.
Is the correct number fifteen or five or what? Or is it fifty wide?

edited 12th Aug '11 6:24:24 AM by Deboss

DUMB

Yes, the pyramid only has thirty-four layers.
If you built a 68-on-a-side pyramid all the way up like that, you'd need 68²+66²+...+2² = 54740 blocks, though.

edited 12th Aug '11 10:04:33 AM by Tzetze

Watchman of the Apocalypse

Hmm something is not quite right with the data then. I know the structure is supposed to creat a flat top for the peak not keep going to a point and beyond. :/ I need to dig this up again and double check first layers measurements.
Charts and graphs ho.
or actually I need check the triangle formula I had to use to get the sides ...grrr that has to be where I messed up.
Crap to get 17/18/19 (whichever of these three it will be) on a side for the 50th layer it has to be over 106 on a side for the first outer most layer.
Let me dig it up after I get home. I can't access the rest of the information I need at work.

edited 12th Aug '11 5:10:10 PM by TuefelHundenIV

"Who watches the watchmen?"

I see the Awesomeness.

Tzetze, it's not a filled area. It's a perimeter two blocks thick.
Are you sure it's not the point where it's fifty blocks per side?

名無しさん

Studying it as part of a double-degree at university.

^{67}dRoy13th Aug 2011 12:54:41 AM from The Happy Place , Relationship Status: A cockroach, nothing can kill it.

Perpetually clueless

Ooh, I wish I could do that. For some reason I think it would help me a lot with my novel project.

Mother of god...You turned one of the hardest and best Champions into an absolute joke. - Zelenal

Watchman of the Apocalypse

Hmm it is 68 for the outermost wall on a side. Then 66 for the inner most. Which means the outer layers decrease by two as they go up. Yet when I make the structure it some how works out. This is very confusing.

edited 13th Aug '11 3:03:16 AM by TuefelHundenIV

"Who watches the watchmen?"

Hooray for math!

I see the Awesomeness.

Wait, I messed up. It should be 67, not 68. The outer perimeter is 4x67, not 4x68.
Upon further thought, you can actually average the two together. So the first layer will have 8*66 blocks. The second layer should have 8*64. So the proper formula for "blocks per layer 'n'" is 8*(68-2n)=blocks.

Not all there

I guess I'll just sign here. According to all my exams and all, I'm useless at maths, which I agree, but I love the visual side of it. There's just something amazing about the fact these patterns, shapes, and... things. To be honest, the only reason I can say that is because of Processing. Can anyone direct me to some stuff on the maths side of things? Fractals, anything that can be applied visually really. I can make things that look nice, and use a fair amount of calculations, but I have no idea what to call the things it's doing inside.

edited 15th Aug '11 10:21:21 AM by Rial

Not all there

Yes. That's one thing I don't like about it; vauge name. I prefer proce55ing.

edited 15th Aug '11 10:52:58 AM by Rial

DUMB

Well, if you want to learn about fractals, I suppose the obvious start is Mandelbrot... do you know how imaginary numbers work?

I prefer proce55ing.

But that's 1337, and 1337 is so very nineties!
edited 15th Aug '11 10:54:00 AM by Tzetze

Not all there

Not really, no... Basic explanation, maybe? I think they can't be converted into fractions, or is that my wild and uncreative imagination?
I u53 1337. 1337speak 15 c00l.