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-   -   Lepore factorization nr. 105 (Bruteforce) (https://www.mersenneforum.org/showthread.php?t=27005)

Alberico Lepore 2021-07-16 13:03

Lepore factorization nr. 105 (Bruteforce)
 
[url]https://www.academia.edu/49978974/Lepore_factorization_nr_105_Bruteforce[/url]

What do you think?

Alberico Lepore 2021-07-17 07:20

ERRATA CORRIGE

[2*(x-1)*((x-1)+1)] < [2*x*(x+1)-y*(y-1)/2] <= [2*(x)*(x+1)]

1 <= y < (sqrt(32*x+1)+1)/2

Alberico Lepore 2021-07-17 19:29

[CODE]
/*
This algorithm is generic and does not exploit that q / p < 2

Plus it uses a single A and not many A
*/


A=9+16*a;//choose A with many small factors

if(M % 4 ==1){
M=3*M;
}
if((M % 8 == 3){
N=M;
}else{
N=5*M;
}
while(1){
if([1/4*(sqrt(N+1)+2)] != (int)[1/4*(sqrt(N+1)+2)]){
x=(int)[1/4*(sqrt(N+1)+2)];
}else{
x=[1/4*(sqrt(N+1)+2)]-1;
}


P=4*x+2-sqrt[4*(2*x+1)^2-N];
Q=N/P;
if (P is integer && (P % M) !=0 && (Q % M) !=0){
breack;
}
N=N*A
}
p=GCD(P,M);


[/CODE]

Alberico Lepore 2021-07-20 12:15

I tried to implement it without good results

[url]https://github.com/Piunosei/Lepore-factorization-nr.-105-Bruteforce-/blob/main/lepore_105.c[/url]

Alberico Lepore 2021-07-22 11:25

the new theory applied to Lepore Factorization nr. 105
If the theory is correct to factor RSA with q / p <2 it would seem O ((log_2 (n)) ^ 2) but I am studying how to implement it in O (2 * (log_2 (n))),after lunch I will study this third hypothesis and the implementation tonight

Alberico Lepore 2021-07-24 21:10

Hello

Unfortunately
O((log_2(n))^2) does not work
O(2*(log_2(n))) does not work
It would seem that O(K*[sqrt(8*sqrt(n)-31)-1]/16) work
K depends on the number n

I still don't know the order of size of K
for example for n=390644893234047643 -> K=4

Alberico Lepore 2021-07-25 10:14

with a parallel distribution on many computers, finding p and q takes as long as possible
Example on 10 computers for n = 390644893234047643 would take 16150 cycles or 1615 for each computer

Alberico Lepore 2021-07-28 20:02

use m not 2*m here [url]https://www.academia.edu/50318218/Crypto_factorization_example[/url]

15 digit 3194383
12 digit 63245
9 digit 3195


p= 9 digit

[2*(h-1)*((h-1)+1)]
<
[2500000062500000-1/2 (h - x) (-1 + h - x + 2 y)+2 (h - x) (1 + h + x)]
<=
[2*(h)*(h+1)]
,
2*(x)*(x+1)-y*(y-1)/2=2500000062500000
,
h=62500001

37500000<=x<37503195 -> 3195

p=12 digit

[2*(h-1)*((h-1)+1)]
<
[2500000000062500000000-1/2 (h - x) (-1 + h - x + 2 y)+2 (h - x) (1 + h + x)]
<=
[2*(h)*(h+1)]
,
2*(x)*(x+1)-y*(y-1)/2=2500000000062500000000
,
h=12500000000

37500000000<=<<37500063245 -> 63245

P=15 digit

[2*(h-1)*((h-1)+1)]
<
[2500000000000062500000000000-1/2 (h - x) (-1 + h - x + 2 y)+2 (h - x) (1 + h + x)]
<=
[2*(h)*(h+1)]
,
2*(x)*(x+1)-y*(y-1)/2=2500000000000062500000000000
,
h=62500000000001

37500000000000<=x<37500003194383 -> 3194383

Alberico Lepore 2021-07-29 19:25

@CRGreathouse I know you don't talk to me anymore but I have achieved an extraordinary result with your number

N=390644893234047643

sqrt(390644893234047643/(15/10))=a
,
(15/10*a+a-4)/8=x
,
2*x*(x+1)-y*(y-1)/2=(390644893234047643-3)/8

->

y=63790420,........




[2*(h)*(h-1)]
<
[(390644893234047643-3)/8+k*(k-1)/2]
<=
[2*(h)*(h+1)]
,
2*(x)*(x+1)-y*(y-1)/2=(390644893234047643-3)/8
,
x-(sqrt(32*x+1)+1)/2 <h<x+(sqrt(32*x+1)+1)/2
,
k=63790420+j*100000

for j=25

see please range h here

[url]https://www.wolframalpha.com/input/?i=%5B2*%28h%29*%28h-1%29%5D++%3C+++%5B%28390644893234047643-3%29%2F8%2Bk*%28k-1%29%2F2%5D+++%3C%3D+++%5B2*%28h%29*%28h%2B1%29%5D++%2C++2*%28x%29*%28x%2B1%29-y*%28y-1%29%2F2%3D%28390644893234047643-3%29%2F8++%2C+x-%28sqrt%2832*x%2B1%29%2B1%29%2F2+%3Ch%3Cx%2B%28sqrt%2832*x%2B1%29%2B1%29%2F2%2Ck%3D63790420%2Bj*100000%2Cj%3D25[/url]

total cost for factorize N=390644893234047643 is 25*10=250 step

Alberico Lepore 2021-07-30 09:10

[QUOTE=VBCurtis;584367]I don't think you count steps very well.

Consider trial factoring. Each prime tried is a single step, and a bunch of them don't work. If you find one that does, the number of steps to find that factor is the number of TOTAL trials, including all the things that didn't work.

Your "method" appears to be such a thing- try a bunch of parameter selections until something works, and then claim "it only took 250 steps!". How many parameters did you try against this number that took thousands of steps, or didn't work at all? That's the actual "number of steps" you took to factor it.[/QUOTE]





yes, I'll try to explain myself better

knowing that the ratio q / p <2 then I will test:


q / p
>

1


1.1


1.2


1.3


1.4


1.5


1.6


1.7


1.8


1.9


and I will execute them at the same time

so it will be the actual time for 10

we consider a 30-digit number with p and q also not prime numbers

188723059539473758658629052963=323456789054341*583456789054343

q/p=1,8038167965499768547404880957269



N=188723059539473758658629052963
,
sqrt(N/(18/10))=a
,
(18/10*a+a-4)/8=x
,
2*x*(x+1)-y*(y-1)/2=(N-3)/8
->
y=64759908643727,........


N=188723059539473758658629052963
,
2*(h)*(h-1)<(N-3)/8+k*(k-1)/2<=2*(h)*(h+1)
,
2*(x)*(x+1)-y*(y-1)/2=(N-3)/8
,
x-(sqrt(32*x+1)+1)/2<h<x+(sqrt(32*x+1)+1)/2
,
k=64759908643727+2399*[COLOR="Red"]100000000[/COLOR]

I still can't establish the exact size order of the number in red, it would appear from the first tests to be
10 ^ [((digit p) +1) / 2]
, but I'm still studying this number.

In theory, if the above were confirmed,
since k <= y <p with y being the order size of p-1 and the first digit of y is given by the 10 ratios we will have our solution in
10 * {[[ digit p] -1] -1 - [((digit p) +1) / 2]}

I repeat still do not know well the number in red.

Alberico Lepore 2021-07-30 11:01

maybe i quantified the r number in red

red value = r

N=188723059539473758658629052963
,
sqrt(N/(18/10))=a
,
(18/10*a+a-4)/8=x
,
2*x*(x+1)-y*(y-1)/2=(N-3)/8
,
(sqrt(32*x+1)+1)/2=b
,
b*(b-1)/2-(sqrt(32*(x-b)+1)+1)/2*[(sqrt(32*(x-b)+1)+1)/2-1]/2=r

r=120441770,......


N=188723059539473758658629052963
,
2*(h)*(h-1)<(N-3)/8+k*(k-1)/2<=2*(h)*(h+1)
,
2*(x)*(x+1)-y*(y-1)/2=(N-3)/8
,
x-(sqrt(32*x+1)+1)/2<h<x+(sqrt(32*x+1)+1)/2
,
k=64759908643727+1992*120441770


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