know wich characters are in a hash sha256 sha512 with vortex maths
Hello everyone, I think this is a more appropriate topic to share this, I emphasize that these are applications of my studies in this mathematics, studies that I have not completed but have made great progress and since they serve me for nothing more than the pleasure of finding I share something new with all of you.

I will continue with these studies, so far I have some time in them, but I hope to finish them.

I'm doing this from my cell phone, so excuse the will.

I will be as brief and clear as possible. And for this I will not teach or teach vortex and vector mathematics, I myself am studying and applying it, however I should read a little of them to understand the concepts.

Ok, a sha256 hash


Which contains letters ranging from A to F and a variety of number combinations. I create groups of 4 digits, forming 16 groups in total, from which we take the index number and create 4 groups with each of them.

(2923) (3843) (1637) (6464)
    A.      B.      C.        D
He gave each group a name in order to recognize it later in the geometry.
Following the vortex rules, each number corresponds to a group of letters creating an enneagram.

1 (ab c t u)
2 (d e v w)
3 (f g x y z)
4 ( h i )
6 (lm)
7 (not)
8 (qp)
9 (rs)

It is worth statistics that it is clear that this group contains 98% of the letters of the alphabet, which makes it impossible to decipher in a very short period of time, however the hash only contains 6, to know what those 5 are, I continue with fractal geometry and vector mathematics, (did not change formulas continuum with vortex mathematics.)

I don't know how to upload a photo or image so I'll try to explain the figure you should use.
If we understand that each group belongs to an enneagram in a circle, if we divide it into 40° angles, it will give us 9 points, the starting point being the north of the circle that will include the number 9, and turning according to the clock the next will be 1, 2 etc up to 8.

As I use 4 groups, they are 4 circles, and since the 4 are part of the whole, the circles are connected at the point that corresponds to the number 9, in this way the 9 of each circle are in the center of the figure.

I suggest that each circle is divided into quadrants so that it is understood since that is how I put it, however any other simpler method will surely work.

Well we already have the figure, I make 2 copies, because doing everything in one is a bit messy at sight. Since you have to compare fractal geometries according to polarity to know which numbers are true and which are not, this is where he eliminated numbers that do not contain the letters of the word in the hash.

I create the fractal geometry in each circle. In the first figure of 4 circles, I start from group A and continue with group B, which I place in the next circle to the right in an anticlockwise direction, and so on until I end up in group D. In the other copy of circles, I did the same but in opposite direction, this means that starting from group A I continue to group B but this group is on the left and I continue clockwise until group D.

With this comparison you will see that there are 3 groups that intercede and a 3 never does, (A B C) leaving D out. In this case I do not take into account group D to obtain the information I am looking for. (I do not rule it out, it is important to understand that each object contains information, the one I have obtained so far is the one expressed here, however I know that there are other relationships with the groups that are not analyzed).

Well, now only with groups A B and C, he made a comparison of the geometric figures, and here the elementary concepts of vortex mathematics are applied according to polarity to rule out numbers.

I start from group A, I only do this this way because I take 9 as the starting point. Analyzing the fractal geometries, it can be seen with the naked eye that group B is repeated in 2 different intersection sets, the first (according to the polarity I chose starting from group A), intersection is (BC) and the second (BA).

Already here we have a result and it is that the number 8 of group B, is part of the word of the hash, and is repeated 2 times, (I have not continued studying this but with patience I know that it is simpler, but up to this point it tells me or that the word comprises the letters Q and P, or QQ or PP, thus reducing the probabilities.

This geometric comparison and the index numbers of the fractal figures can already be obtained the following implied numbers that contain the letters of the word. 1, 2, 9 and 6. Thus reducing the 8 numbers found in the groups to 4, and knowing that 8 is repeated and the word contains 6 letters, for an obvious and simple reason it is 1 letter of each of the remaining numbers. You can also know what they are, you have to repeat the process of the figure for the 4 initial groups, but in this case the groups will be the numbers 1, 2, 9 and 6, since each letter has a numerical value, we return to the fractal geometry, the polarity and index geometry number of each figure.

Thus we obtain that the letters are u, r, l, and e the 4 letters are missing three.

The word is purple, but to add alphanumeric complexity I added 124 to the end being "purple124". I made it simple for it to be understood, 124 has 4 as its index number, in the vortex 124 is in the eleventh position, and this number can be obtained with the sum of the total sum of the numbers of the initial groups. But in this case I use the group that I did not work with initially, the D, (6464), we apply the geometry again, it is a repetition of 8 and 3, we add it as a result of 11, which is the vortex level where the number 124. I have some hypotheses on how to get it but so far at level 11, there are only 9 number possibilities, including 124. This is how I progress with this so far, I get all the letters and numbers that can be involved. Thus, according to what I was very effectively instructed in this forum, performing a brute force comparison of combinations is reduced to 9 possible numbers, and 4 letters and the possibility of the permutation of QP. What would take about 6 days without data with a tesla gpu, only takes 3hs with a gtx430 gpu.

I go back and repeat, I am not a scholar in this, it is my studies and discoveries, I know and I realize that more can be done, (alphanumeric words with disordered characters and adding signs too), it is a little more work but it can be done I invite you to practice it.
It had been weeks since I came to this forum, I thought I would come back when I had this clear and everything was done, but I left everything until here since other responsibilities afflict me. However glad to share.
I don't know if email can be shared in this post, but I'll check the forum more often in the future.

What I am leaving in the inkwell in hypotyposis;

1°A simpler model to do all this, (an equation is what I think would be adequate since there are few variables and the others are constant.)

2° Adding one of the ascii signs to this makes it more tedious, because there are 3 figures, (being letters and numbers using 2 I see enough, but adding one more variable is cleaner adding a third figure for fractals).

3°I have noticed certain similarities between the index numbers regarding the hash, with what I think the order could be obtained, but it is a theoretical thought, I have not put anything on paper regarding this, I should analyze more exercises to reach a conclusion.

4°In the initial post where I asked how hashes work, I was informed that it is not only used for a word, password. But it can be a complete file, a folder of files, image or videos. (I guess audio too?), well all this is shared by others regarding what can be done with a hash system, so moving on to that other type of analysis would be entertaining, once I finish what I do Now, I confess that making a "view" (I don't know what to call it without putting an inappropriate concept to what can be done with VMs), of an alphanumeric word can be a long job, but once the concept is understood it is like any mathematical exercise, it is to apply the rule and everything comes out easy.
Hello all there. 
First let me remember you guys, english isnt my mother language. I speak spanish. 
I share my studies here beause I know many importante people use hashcat and i know this metodh will be a usefull tool to get more info about hash that you need to "hashcat",
Is this method easy to do? I spend almost 3 years working on this maths, wasnt easy, and still work to ger more info, allready do more news things, for ejample, for an alphanumeric word from 10 to 20 caracteres i need almost 24hs to do all calc maths to ger all caracteres, I dont get the specific position of the charcters but im working on that step. Also with a hash that have sing (I mean sacii sings) need to add another step that make the work a litle more longer but I can find it to.
Why I say all this? Because many ppl mail me asking me that if I can get the charcters from a hash, yes guys I can do it. But I dont know if you guys want to do bad things with it thats Why I dont answer mails. I left personal email if serius ppl want to chat about my studies