mersenneforum.org Lepore factorization nr. 105 (Bruteforce)
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 2021-07-16, 13:03 #1 Alberico Lepore     May 2017 ITALY 52210 Posts Lepore factorization nr. 105 (Bruteforce) https://www.academia.edu/49978974/Le...105_Bruteforce What do you think?
 2021-07-17, 07:20 #2 Alberico Lepore     May 2017 ITALY 20A16 Posts 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
 2021-07-17, 19:29 #3 Alberico Lepore     May 2017 ITALY 2·32·29 Posts 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); Last fiddled with by Alberico Lepore on 2021-07-17 at 19:37 Reason: }
 2021-07-20, 12:15 #4 Alberico Lepore     May 2017 ITALY 20A16 Posts I tried to implement it without good results https://github.com/Piunosei/Lepore-f...n/lepore_105.c
 2021-07-22, 11:25 #5 Alberico Lepore     May 2017 ITALY 2×32×29 Posts 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
 2021-07-24, 21:10 #6 Alberico Lepore     May 2017 ITALY 2×32×29 Posts 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
 2021-07-25, 10:14 #7 Alberico Lepore     May 2017 ITALY 2·32·29 Posts 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 Last fiddled with by Uncwilly on 2021-07-30 at 21:36 Reason: Removed unneeded self quote of immediately preceding post.
 2021-07-28, 20:02 #8 Alberico Lepore     May 2017 ITALY 52210 Posts use m not 2*m here https://www.academia.edu/50318218/Cr...zation_example 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
 2021-07-29, 19:25 #9 Alberico Lepore     May 2017 ITALY 52210 Posts @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
2021-07-30, 09:10   #10
Alberico Lepore

May 2017
ITALY

20A16 Posts

Quote:
 Originally Posted by VBCurtis 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.

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*100000000

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.

 2021-07-30, 11:01 #11 Alberico Lepore     May 2017 ITALY 2·32·29 Posts 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

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