User blog:Syst3ms/Chained Array Notation (second revision)

Hey there, I'm back again. This isn't actually the second revision at all, but i wanted to sum up all of my changes in a new blog.

NCAN :
If you thought CAN looked a lot like HAN, well you are going to be even more correct, since i'm changing rule 4 :

4) \(a[0,\ldots,0,[\cdots[c+1]\cdots],\#] = a[0,\ldots,[0,\ldots,0,[\cdots[0]\cdots],\#],[\cdots[c]\cdots],\#]\)

This subtle change makes NCAN converge towards \(\varphi(\omega,0)\) much faster.

HDCAN :
There is only one thing changed with NCAN : when you nest an array, you look outwards for the innermost set of square brackets.

Meaning that \([0{0,1}1] = [0{[0{0,0}1],0},1]

MDCAN :
Okay, a few changes ahead here :

First, separator arrays don't work the same anymore :
 * They use curly brackets
 * \(\{n@\} = \underbrace{@@\cdots@@}_{n+1}\)

Stage 8 : Hyperspace (HCAN) :
Here is where we get to the massive changes.

First, metadimensional arrays are allowed inside pipe subscripts. So the slash in old ECAN is now \(|_{0,1}\) in HCAN. Note that pipe subscripts have implicit square brackets, meaning that \(|_{0,1} = |_n\).

With that, the limit of pipes is \(a \mapsto |_{0a1}\). And that's when we introduce the slash.

Then we continue onwards in the same way, with 2-slash, 3-slash, (0,1)-slash, until \(a \mapsto /_{0a1}\). Let this be the backslash.

But here we see a pattern. And to formalize it, let's define separator hyperspace.

The first separator in separator hyperspace is the pipe. And on the first row lie all of the pipes. On the second row, you can find all of the slashes. On the third one, the backslashes, etc :

Separator hyperspace :

Row 1 : \(|, |_2, \ldots, |_{0,1}, \ldots\)

Row 2 : \(/, /_2, \ldots, /_{0,1}, \ldots\)

Row 3 : \(\backslash, \backslash_2, \ldots, \backslash_{0,1}, \ldots\)

etc.

So diagonalize over hyperspace, let's use hyperspace arrays, denoted by floor brackets \(\lfloor\rfloor\).

Let's reformulate separator hyperspace in terms of hyperspace arrays :

Row 1 : \(\lfloor0\rfloor, \lfloor0\rfloor_2, \ldots, \lfloor0\rfloor_{0,1}, \ldots\)

Row 2 : \(\lfloor1\rfloor, \lfloor1\rfloor_2, \ldots, \lfloor1\rfloor_{0,1}, \ldots\)

Row 3 : \(\lfloor2\rfloor, \lfloor2\rfloor_2, \ldots, \lfloor2\rfloor_{0,1}, \ldots\)

etc.

As you can see, this makes formalizing much easier. I'm still not going to do it here.

But this is only one entry in hyperspace arrays.

The next hyperspace array, \(\lfloor0,1\rfloor\) is well beyond all of that, it is the limit of \(|, /, \backslash, \ldots\). A way to visualize it is to imagine it being on the second plane of hyperspace. Let's call it the tilde \(\sim\). It is the first planar separator.

We can, again, create a new plane starting at the tilde, and it goes like this :

Row 1 : \(\lfloor0,1\rfloor = \sim, \lfloor0,1\rfloor_2 = \sim_2, \ldots, \lfloor0,1\rfloor_{0,1} = \sim_{0,1}, \ldots\)

Row 2 : \(\lfloor1,1\rfloor, \lfloor1,1\rfloor_2, \ldots, \lfloor1,1\rfloor_{0,1}, \ldots\)

Row 3 : \(\lfloor2,1\rfloor, \lfloor2,1\rfloor_2, \ldots, \lfloor2,1\rfloor_{0,1}, \ldots\)

etc.

To diagonalize over that, we have \(\lfloor0,2\rfloor\).

And then, the limit of that is \(\lfloor0,0,1\rfloor\), the first cubic separator.

Then we can have : Then \(\lfloor0{1}1\rfloor\) is beyond all dimensions.
 * \(\lfloor0,0,2\rfloor\) : the second cubic separator
 * \(\lfloor1,0,2\rfloor\) : the separator on the second row of the first plane of the third realm
 * \(\lfloor0,0,0,1\rfloor\) : the first tesseractic separator
 * \(\lfloor0,0,0,2\rfloor\) : the second tesseractic separator, or the first separator on the third flune
 * \(\lfloor0,0,0,0,1\rfloor\) : the first penteractic separator
 * \(\lfloor3,4,3,5\rfloor\) : the fourth separator on the fifth plane in the fourth realm inside the sixth flune

The limit of hyperspace is \(a \mapsto \(\lfloor0a1\rfloor\).