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Prime factorization. Mathew White March 14, 2022 (Mon) 22:00:47 No. 145

The brain dump of a managerie of often contracdictory and fantastic characters, from a mossad member turned professor, to an indian student turned part time street criminal, a homeless poet in philly, an ex-skin head, a former chinese investment banker, the owner of a multimillion dollar startup, two military veterans, and a cast of other folks.

The result of a a simple discovery that a rich set of algebraic identities underlay the product of large primes.

For example, a series of variables related to n unknown value, d4, easily derived from known variables. d4a, d4u, d4z, d4H, etc.
Or c/d4, neither known by themselves, but the ratio of which is easily found, and the *product* of which, cd4, is the ratio of our product's factors, b/a.

Be warned the code is thick, mostly written by an indian guy, and translated by the rest of us over time. It is 198 pages of dense work, defining so many variables the original authors resorted to greek letters and elements from the periodic table for naming.

Good luck.

https://pastebin.com/Ad46Awcp

Anonymous March 21, 2022 (Mon) 03:22:41 No. 154

>list of primes
>Miller test (deterministic Miller-Rabin)
>a bunch of tests and variables I have no reason to memorize, followed by a schizo string list including elements of the periodic table..
>...
I'd like to converse with you about your findings, but I have no paper, no lemmas and no theorems.

Some observations Anonymous March 24, 2022 (Thu) 23:39:42 No. 155

"I'd like to converse with you about your findings, but I have no paper, no lemmas and no theorems. "

It appears there is no lemma or theorem and OP is working at best in 8th grade math.

I'm just a lowly front end developer but I'll take a crack at it.

All the key variables are on two apparent slopes, d4, and 'gold'.

I could see getting estimates on unknown factors by simply taking known products with common variables (d4a, d4u, d4z, d4H, etc), and then finding the smallest of the set, on the basis that would tightly constrain the upper and lower bound of the factors of those variables, but that assumes none of the unknown factors like u, d4, c, z, or H, are less than 1.

I also wrote some functions for converting variables between the gold line, the _gold line, and the d4 line.

def _gold_to_d4(var):
return (var**2)*Hc

def d4_to__gold(var):
return (var/Hc).sqrt()

This followed from the observation that

d4u/Hc/G
Decimal('1.332259228726959456136695243518871469428174838645985237776420270')

(_goldc**2/G)
Decimal('1.332259228726959456136695243518871469428174838645985237776420270')

meaning

d4u/Hc == some _gold**2

Shows d4 and _gold series are interconnected.

Taking any _gold, multiplying by the power of 2, and then multiplying by Hc should give
a d4 variable.

While dividing a d4 variable by Hc, and then rooting it, should give a _gold series variable

if d4 is inverted (less than 1), then (((_goldc)**2)*Hc)/d4 > d4u
To test this I ran the script a few times with various factors.

a
Decimal('84780491')
b
Decimal('7022437637')

>>> (((_goldc)**2)*Hc)/d4 > d4u
True

>>> a
Decimal('247017779')
>>> b
Decimal('8552169383')

(((_goldc)**2)*Hc)/d4 > d4u
True

>>> a
Decimal('39420301')
>>> b
Decimal('8490912585839')
(((_goldc)**2)*Hc)/d4 > d4u
False

As you can see with the above example, both c and d4 are normal, but c is less than d4,
so _cd4 is less than 1, and the outcome is of course false in this case.
Good to know.

And what this means is that when _cd4 is less than 1, Post too long. Click here to view the full text.

Anonymous March 25, 2022 (Fri) 00:07:22 No. 156

Making sense of all of what looks like dead code, the elListI() function appears to list elements based on some mysterious "I" value.
The elements that have matching I values tend to share lots of identities. Thats all I could tell about it though.

Anonymous April 01, 2022 (Fri) 09:59:56 No. 158

i love primes bro

File: image.png ( 736.33 KB , 800x647 , 1643053076107.png )

Anonymous January 24, 2022 (Mon) 19:37:56 No. 113

>>115 >>114

What the heck is a Euclid distance? Why is it different from the Minsk distance and Cosine distance?
If i have a matrix as so, i'm expected to make all three distance matrices from it

X = \begin{bmatrix} 0 & 2 \\ 2 & 0 \\ 3 & 1 \\ 5 & 1 \end{bmatrix}

Also this probably has something to do with statistics idk

Anonymous January 24, 2022 (Mon) 20:22:17 No. 114 >>116

>>113
Cosine similarity is

\frac{\sum_{k=1}^n x_ky_k}{\sqrt{\sum_{k=1}^n x_k^2}\cdot\sqrt{\sum_{k=1}^n y_k^2}}

Minkowski distance is

\left(\sum_{k=1}^n|x_k - y_k|^p\right)^\frac{1}{p}

Euclidean distance is a speacial case of Minkowski distance when

p=2

i.e.

\sqrt{\sum_{k=1}^n (x_k - y_k)^2}

Your matrix is

X

means:

\begin{bmatrix} x_1 & y_1\\x_2 & y_2\\x_3 & y_3\\x_4 & y_4\\\end{bmatrix}

Anonymous January 24, 2022 (Mon) 21:45:55 No. 115

File: image.png ( 891.54 KB , 707x1000 , 1643060754913.png )

>>113
I'm a dumbass, it's actually minkowski distance

Anyways, for the euclidian distance, apparently you do:

\sqrt{ (x_1-y_1)^2+(x_2-y_2)^2}

or for those playing along in calc,

=SQRT((B$3-B3)^2+(C$3-C3)^2)

and apply to the appropriate cells.

\begin{bmatrix} 0.0 & 2.8&3.2&5.1\\ 2.8&0.0&1.4&3.2\\ 3.2&1.4&0.0&2.0\\ 5.1&3.2&2.0&0.0 \end{bmatrix}

Since i was working in calc, the table containing the data spans over the

B3:C6

range

Then, for the minkowski distance, there's something called the P value, and if it's equal to 1, then it becomes the manhattan distance, so in this example the equation would be

|x1-y1|\ +\ |x2-y2|

and the corresponding calc formula for a particular cell (i'm not saying which one hehe)

=(($B$3-$B3)^1+($C$3-$C3)^1)^(1/RIGHT($E$14,1))

\begin{bmatrix} 0&0&2&4\\0&0&2&4\\-2&-2&0&2\\-4&-4&-2&0 \end{bmatrix}

And for the cosine similarity or distance idk, just do a vector dot product

\frac{x \cdot y}{|x|\cdot|y|}

This time it was a bit more difficult in calc, and the formula for the left upper cell is

=SUMPRODUCT(B$3:C$3,B3:C3)/(SQRT(SUMSQ(B$3:C$3))*SQRT(SUMSQ(B3:C3)))

\begin{bmatrix} 1.00&0.00&0.32&0.20\\0.00&1.00&0.95&0.98\\0.32&0.95&1.00&0.99\\0.20&0.98&0.99&1.00\\ \end{bmatrix}

Cool.

Anonymous January 24, 2022 (Mon) 21:48:06 No. 116

>>114
Thanks for the info. I probably made some mistake here but it's probably good