corrora & //apploc.dek

(d)irect
0000//// +-(O, A) +-((A, O) +-A, O))    unit 1
0000\\\\ +-(A, O) +-(A, A) +-(O, A).     unit 2

(s)tatic
0000//// +-(O, A +-A, O) +-(A, O)
0000\\\\ +-(A, O) +-(A, A) +-(O, A).

 

Step 1: Determine Order for Direct. (x) cancel, (y) is next selection.

 

+-((A, O) +-A, O))::units:: x(predetermined) with +-(A, A)

 

+A, Ox
-A, Ox
-A, A
+A, A

 

(x)|A, O|, y|A, A|, y|O, A|, x|O, A|

 

+A, Ox -- these eventually cancel
-A, Ox -- eventually cancel, label (x)
-A, A  -- THIS IS THE BANK
+A, A  -- WE HAVE |A, A| = +- A, A.

 

|A, A| + (x)|A, O| + ""bank""|A, A|""bank""

-A, Ay -- y is secondary term
+A, Ay -- y is also the second term selected
+A, O(x) -- this (prediction)
-A, O(x) -- happens when canceling

 

Prediction: We can either cancel out all the units or try to raise the
bank.

 

Step 2: Determine which is next to invoke after cancel. (x)|O, A|

 

(y)|O, A| + |O, A| + (x)|A, O|::

""|A, O| + |A, A| + (x)|O, A| - (x)|O, A|"" -- don't mind the quotes

 

+O, A(x)
-O, A(x)
-O, A(x)
+O, A(x)
-A, O(y) -- notice these terms aren't identical
+A, O(y) -- also notice that (x)|A, O| is now +- (A, O)(x)

 

Step 3: What cancelled and what is the bank now?

The order in which this segment was performed:

|A, O| + |A, A| + |A, O| - |O, A| + |O, A| - |A, O| = |A, O| + |A, A|

 

+- A, O
+- A, A = +-AOAA or:

 

+A, O
-A, O
-A, A
+A, A

 

If we chose to take this segment instead:

|A, O| + |A, A| + |A, O| - |A, O| + |O, A| - |O, A|

 

We would be left with just |A, O| or +- A, O or:

+A, O
-A, O

 

Lets try a segment without choosing |A, A| as the second term: |A, O|

 

|A, O| - |A, O| + |A, O| + |A, A| + |O, A| - |O, A| = 

|A, O| + |A, A| + |A, O| - |A, O| + |O, A| - |O, A| = |A, O| + |A, A|

 

We found that these two are similar but not the same, because the method
on which they were used created a different pattern, but same outcome.

Lets try another segment without choosing |A, A| or |A, O| as the second
term: |O, A|

 

|A, O| + |O, A| - |A, O| + |A, A| - |O, A| + |A, O| =

|A, O| + |A, A| -- the same result as the first one, but a different
pattern.

 

Lets try one last segment to complete the series we will try |O, A| once
more but with a different pattern to see the outcome: |O, A|

 

|A, O| + |O, A| - |A, O| + |A, O| - |O, A| + |A, A| =

|A, O| + |A, A| =

 

1. |A, O| + |A, A| + |A, O| - |O, A| + |O, A| - |A, O|
2. |A, O| - |A, O| + |A, O| + |A, A| + |O, A| - |O, A|
3. |A, O| + |O, A| - |A, O| + |A, O| - |O, A| + |A, A|
4. |A, O| + |O, A| - |A, O| + |A, O| - |A, A| - |O, A| 

 

-- I just switched the last two terms around, for the fourth one that is
the same outcome but different pattern.

 

1. |A, O| + |A, A| + |A, O| - |A, O| + |O, A| - |O, A|

 

We are left with just |A, O|.

 

Then if we just do a quick cancel out for both O, A and A, O we are left with:

 

1. |A, O| + |O, A| - |A, O| + |A, O| - |O, A| + |A, A|

 

We are left with just |A, A|.

 

Therefore we only have three terms that will be a direct outcome. This is good for categorizing all the different types of patterns.
 

 

 

(s)tatic

0000//// +-(O, A +-A, O) +-(A, O)

0000\\\\ +-(A, O) +-(A, A) +-(O, A).

1(S1). |O, A| + |A, O| - |A, O| + |A, A| + |A, O| - |O, A| = |A, A| + |A, O|
2(S1). |O, A| + |A, A| + |A, O| - |O, A| - |A, O| + |A, O| = |A, A| + |A, O|
3(S1). |O, A| - |O, A| + |A, O| + |A, A| - |A, O| + |A, O| = |A, A| + |A, O|
4(S1). |O, A| + |A, A| + |A, O| - |A, O| + |A, O| - |O, A| = |A, A| + |A, O|

 

5(X1). |O, A| - |O, A| + |A, O| - |A, O| + |A, O| + |A, A| = |A, O| + |A, A|

1(S2). |A, O| - |A, O| + |O, A| + |A, A| + |A, O| - |O, A| = |A, A| + |A, O|
2(S2). |A, O| + |A, A| + |O, A| - |A, O| + |A, O| - |O, A| = |A, A| + |A, O|
3(S2). |A, O| + |O, A| - |O, A| + |A, A| - |A, O| + |A, O| = |A, A| + |A, O|
4(S2). |A, O| + |A, A| + |O, A| - |O, A| - |A, O| + |A, O| = |A, A| + |A, O|

 

5(X2). |A, O| - |A, O| + |O, A| - |O, A| + |A, O| + |A, A| = |A, O| + |A, A|

 

 

Notice that the (X1) and (X2) are opposite in each term regarding the order. Also notice that there is an "X" label for this term, in which regards to the outcome being only two terms. This means that there will be an equal representation in either format with the same outcomes. This is why this is labeled direct.

 

 

 

If you work out the static part you only get |A, A| or |A, O|. Never pairs, this is static. Either XOR, AND, OR, never true or false. We aren't comparing, we are selecting. And it doesn't matter the pattern, its only for decisive measures. Put it this way, if I had apple and oranges, and I select one, does it matter the pattern in which I selected them if I am just doing a comparison?

 

4 ** 2 for four options, and each can have two comparsions. 16 ** 2 for the (S2) and their comparisons with the added pattern sequence in (S1 + S2). 256; 256 ** 2 for 5(X1) only. 65,536; total plus another 65,536 for 5(X2). We didn't square because they aren't combined. 131,072 total original patterns.

 

131,072 ** 2 for how each pattern got there meaning that the potential energy and kinetic action have two separate divisions, because the way the idea works is that it predicts, therefore if I predict one side, and I have the kinetic side it would be squared because every item has a prediction counter. 17,179,869,184 what am I leaving out?

 

Another way to describe this is looking below:

 

|A, A| + |A, O|

|A, A| + |A, O|

|A, A| + |A, O|

|A, A| + |A, O|

 

|A, O| + |A, A|

 

|A, A| + |A, O|

|A, A| + |A, O|

|A, A| + |A, O|

|A, A| + |A, O|

 

|A, O| + |A, A|

 

I can stack them ontop of each other and have two different patterns and still have the same outcome but without using the second term, just the first. |A, A|, now we add 1 to each of those outcomes. It's already applied to the X1 and X2, therefore we just square. It doesn't matter that we don't use the second term because it is associated with the first term. Technically if we acually counted a cancel, which I haven't applied it and probably never will with the difficulty of expressing a pattern reserve for the specific term. I'm already dedicating to math and GPU, if I were to cancel one out I would have to pick from the GPU or the math and only if the pattern library was uploaded into a programming language could we leave out math; 17,179,869,184 what am I leaving out?

 

Oh yes, 131,072 ** 2 + 131,072 ** 2 because of a secondary pattern annexing the first one. Because we have four options, and we already accounted for eight of the styles. 34,359,738,368 patterns possible in sequence. And that's not using static to declare secondary sequence after first pattern. 34,359,738,368 * 2 not squared because we already declared the pattern system. 68,719,476,736

 

 

 

 

 

 

 

 

 

 

    When applying expressions, one has to consider the remainder that happens when external conflicts happen. In this following example, described by the notorious backslashes, (y - 1) + (x - z)...   to make an expression the knowledge of "X" has to be described other than "unknown". This example shows that the above expression is simplified into A^2 + AX + X^2a + X^3

 

Below, Xo and 2AA are prepared as the same expression as above and below. Here's why: We are given that "X" is unknown and can be considered into a point {0, 0} of origin and also can be a slope. If it were a slope, because of *2(A * A) we would have a dynamic state of solution, and would have to refer to basics to HOW the problem became a solution through a boolean.

 

(A - X^2)*(A + X) \\ (y - 1) + (x - z)...

(X - O)    *2(A * A)

 

A^2 + AX + X^2A + X^3

AAAX XXAX XX(+1)

-AOAA

-AOAO = [2AOA, -AO] (y)

 

+AOAA

+AOAO = [2AOA, +AO] (z)

 

 

Another reason why they are the same resolution is that Xo, 2AA has a restricted compound. This compound is a strict form, where the point is hit twice, and the next coordinate creates a mathematical error or dictates the slope. As you see AAAX XXAX XX[X](+1) is not a system, yet describes itself into a deep probability that using the formula from origin, the sequence would stem into a molecular format, and dictate this expression automatically. Making this solution a compound solution. If someone were to hit the same coordinate twice in one point, there is no clear representation that there is a line. Therefore, these expressions cannot be used as graphing coordinates yet. All restricted compound solutions are the same, since they cannot be perceived as a real solution and cannot be organized through the formula, and are imaginary.

 

Imaginary expressions can still be used in programming, manipulation and overall affecting the outcome of an ordinary solution by transitioning an addition or subtraction from the stylization of bank (supply) or memory suchas: (y - 1) + (x - z)= "+1". A number can represent real equations outside the formula, and resolve issues that develop within the numerical errors or representation of the form.