It's highly improbable, because you're starting from the bottom and not the top, in a sense. What you are doing is calculating the next sequence in pragma that is inverted, when you shape it like a tree hierarchy, or pyramid of vertices. Because it branches downwards, and you're calculating the first sequence, e.g. a=b{1}, or any pragma. You would depend on the result of that function to say e.g. c == a=b{2}, which is not the label but the result of the function, in other words you are guessing that d == c{3}, which depends on the result of a=b{1}. In other words with this approach you're just labeling the next function, which is highly improbable to guess at higher levels of math and abstraction, because some formula live in another stratum in the functions of math. Therefor you need to start somewhere in the middle of the pyramid, and not at the top. Because if the top is {1}, you are going to a next iteration in a hierarchy {n+n1} as a result, but you're still essentially calculating {n-n1}, which is the same as building the math, or essentially calculating the results of the function A{1} to get to the content or output of function B{2}. This will not work because the difficulty will be linear. What you want to do, is to find an algorithm that adds those functions, and tosses with the results of the function of B{2}. (Read: The numbers in parenthesis are just the logical order of the functions.) In other words: {n}=A{1} < {n+1}=B{2}. So in essence we could say: {n+1} == n(n1) or {n+1} == B{n1} When we are taking this approach we see that n becomes a label, or an increment in the function, but the problem is you want the result of the function, and not just the incremental logic of the function, which in this case is already laid out in the structure of the model to describe it. Possible Solution: ----- A probable theory could be that you use each function only as label for the results, and use these results as heavier weights. In theory, then the function becomes the incremental logic in the function, the increment (f). So, when we have the function as an increment as a label for the function which is *(f) .. that would be the same as the function of f(n), or the prototype for the function (f)n. Why it will not solve, unless you are calculating the results, which are really the weights of the function. E.g. n=1, n=2, n=3, etc. So, if: n=1{1} == f(n)=1 or f(n+1)=B{2} == ( f(B)-n(1) == *(A) )... Where A and B are the label for the next iteration in the hierarchy, n = the increment, and f() is a function. If you use results of functions, somewhere in the middle of the "visible math", you are really saying I have given: 300n w=300 ; n=300 ; where w is weight (or RES of DIV/1). Why you have the same mirror in the function, a sort of razor. If you want the sequence take the 300 as value, and not the n as value. Because that way you write the first function as: *c == a=b. That's when all you have to do is apply the weight in the formula. Because if: c==300n or A{1} n So, essentially n is 300. The goal is a sort of Rainbow tables, where you cross out linear algorithms, and calculate any carrots. You fit the functions into the result. Why the only thing you need to do is to decide on the number of parameters, and fit those in the result. So, you could e.g. start with a number like 2 or 3, and see how a=b fits in 2 or 3.
Sum less than the parts. Kid, you work for a man with a big ego. Your #1 job is to boost him up, even when you challenge him. If that feels undignified, find a new job.
That all or nothing fallacy, is what corrupts all. It is a very brutish ideology you are presenting and one cannot tell if you are being serious or what. The number one job is to solve the physics issues.
I'll help for a little longer but if I'm continually ignored, I'll just start my own campaign. I am already way ahead of what they were doing but thought I they would enjoy contributors. I seem to only be getting the cold shoulder. I am willing to give them the benefit of the doubt.
![Wolfram Physics Project: Working Session Tuesday, July 28, 2020 [Metamathematics | Part 3]](http://i.ytimg.com/vi/EszjCn8HJe8/mqdefault.jpg)
![Wolfram Physics Project: Working Session Tuesday, July 28, 2020 [Metamathematics | Part 3]](/img/tr.png)
![Wolfram Physics Project: Working Session Thursday, July 23, 2020 [Metamathematics | Part 1]](http://i.ytimg.com/vi/x5v3KFFWv2o/mqdefault.jpg)
Find the notebooks for this session here: www.wolframcloud.com/obj/wolframphysics/WorkingMaterial/2020/Metamathematics-05.nb & www.wolframcloud.com/obj/wolframphysics/WorkingMaterial/2020/Metamathematics-06.nb & www.wolframcloud.com/obj/wolframphysics/WorkingMaterial/2020/Metamathematics-07.nb & www.wolframcloud.com/obj/wolframphysics/WorkingMaterial/2020/MultiwaySystemGrowth-04.nb & www.wolframcloud.com/obj/wolframphysics/WorkingMaterial/2020/ShorAlgorithm-01.nb & www.wolframcloud.com/obj/wolframphysics/WorkingMaterial/2020/ShorAlgorithm-02.nb & www.wolframcloud.com/obj/wolframphysics/WorkingMaterial/2020/ShorAlgorithm-03.nb
It's highly improbable, because you're starting from the bottom and not the top, in a sense.
What you are doing is calculating the next sequence in pragma that is inverted, when you shape it like a tree hierarchy, or pyramid of vertices.
Because it branches downwards, and you're calculating the first sequence, e.g. a=b{1}, or any pragma. You would depend on the result of that function to say e.g. c == a=b{2}, which is not the label but the result of the function, in other words you are guessing that d == c{3}, which depends on the result of a=b{1}.
In other words with this approach you're just labeling the next function, which is highly improbable to guess at higher levels of math and abstraction, because some formula live in another stratum in the functions of math.
Therefor you need to start somewhere in the middle of the pyramid, and not at the top. Because if the top is {1}, you are going to a next iteration in a hierarchy {n+n1} as a result, but you're still essentially calculating {n-n1}, which is the same as building the math, or essentially calculating the results of the function A{1} to get to the content or output of function B{2}.
This will not work because the difficulty will be linear.
What you want to do, is to find an algorithm that adds those functions, and tosses with the results of the function of B{2}.
(Read: The numbers in parenthesis are just the logical order of the functions.)
In other words:
{n}=A{1} < {n+1}=B{2}.
So in essence we could say:
{n+1} == n(n1) or {n+1} == B{n1}
When we are taking this approach we see that n becomes a label, or an increment in the function, but the problem is you want the result of the function, and not just the incremental logic of the function, which in this case is already laid out in the structure of the model to describe it.
Possible Solution:
-----
A probable theory could be that you use each function only as label for the results, and use these results as heavier weights.
In theory, then the function becomes the incremental logic in the function, the increment (f).
So, when we have the function as an increment as a label for the function which is *(f) .. that would be the same as the function of f(n), or the prototype for the function (f)n.
Why it will not solve, unless you are calculating the results, which are really the weights of the function.
E.g. n=1, n=2, n=3, etc.
So, if:
n=1{1} == f(n)=1 or f(n+1)=B{2} == ( f(B)-n(1) == *(A) )...
Where A and B are the label for the next iteration in the hierarchy, n = the increment, and f() is a function.
If you use results of functions, somewhere in the middle of the "visible math", you are really saying I have given:
300n w=300 ; n=300 ; where w is weight (or RES of DIV/1).
Why you have the same mirror in the function, a sort of razor.
If you want the sequence take the 300 as value, and not the n as value.
Because that way you write the first function as:
*c == a=b.
That's when all you have to do is apply the weight in the formula.
Because if:
c==300n or A{1} n
So, essentially n is 300.
The goal is a sort of Rainbow tables, where you cross out linear algorithms, and calculate any carrots.
You fit the functions into the result. Why the only thing you need to do is to decide on the number of parameters, and fit those in the result.
So, you could e.g. start with a number like 2 or 3, and see how a=b fits in 2 or 3.
I fall asleep with youtube on and end up here. idk why
I can't wait for next fall to see Doctor Strange and the Domain of Discourse
A bunch of interesting comments starting ~02:54:00
Can't seem to get above 480p
Wolfram search engine was trash free of spam
State transitions, another thing in front of my face, take time. 🤔
Was a factor in the self improving program AI in yesterday's internet surf.
Limits on simultaneous actions?
...black hole surface area. 🤔
Lol. Chocolate covered face.
Stable subsets of the ABABABBA strings ...observer pegs?
Sum less than the parts. Kid, you work for a man with a big ego. Your #1 job is to boost him up, even when you challenge him. If that feels undignified, find a new job.
Last time I checked, my #1 job was to do interesting science. Egos have nothing to do with it ;)
@@jontology3173 oy
That all or nothing fallacy, is what corrupts all. It is a very brutish ideology you are presenting and one cannot tell if you are being serious or what. The number one job is to solve the physics issues.
An ego booster is probably cheaper.
I'll help for a little longer but if I'm continually ignored, I'll just start my own campaign. I am already way ahead of what they were doing but thought I they would enjoy contributors. I seem to only be getting the cold shoulder. I am willing to give them the benefit of the doubt.