- **Alberico Lepore**
(*https://www.mersenneforum.org/forumdisplay.php?f=166*)

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

Lepore factorization nr. 105 (Bruteforce)[url]https://www.academia.edu/49978974/Lepore_factorization_nr_105_Bruteforce[/url]
What do you think? |

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 |

[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] |

I tried to implement it without good results
[url]https://github.com/Piunosei/Lepore-factorization-nr.-105-Bruteforce-/blob/main/lepore_105.c[/url] |

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 |

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 |

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 |

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 |

@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 |

[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. |

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 |

All times are UTC. The time now is 20:37. |

Powered by vBulletin® Version 3.8.11

Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.