 mersenneforum.org Lepore factorization nr. 105 (Bruteforce)
 Register FAQ Search Today's Posts Mark Forums Read  2021-07-30, 13:19 #12 Alberico Lepore   May 2017 ITALY 23·32·7 Posts range ok h >=1 I tried to write the pseudo code Code: i=0 while(i<10) { solve this system and memorize y(i) and r(i) sqrt(N/((10+i)/10))=a , ((10+i)/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 } j=0 while (!(N mod p ==0 && p!=1 && p!=N)){ i=0 while(i<10) { solve this system with unique integer solution of h 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 2021-07-30, 16:11 #13 Alberico Lepore   May 2017 ITALY 23·32·7 Posts (*) here I have to improve, multiplying the range by 10 ^ (tot), but I still don't know tot Last fiddled with by Uncwilly on 2021-07-30 at 21:37 Reason: Removed unneeded self quote of immediately preceding post.   2021-07-30, 17:55 #14 Alberico Lepore   May 2017 ITALY 23·32·7 Posts Code: i=0 while(i<10) { solve this system and memorize y(i) and r(i) sqrt(N/((10+i)/10))=a , ((10+i)/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]/2=r } j=0 while (!(N mod p ==0 && p!=1 && p!=N)){ i=0 while(i<10) { solve this system with unique integer solution of h 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[range of x]>=1 if exist (*) if exist { choose the only possible integer solution of x x-(sqrt(32*x+1)+1)/2=h 2*(x)*(x+1)-y*(y-1)/2=(N-3)/8 calculate p p=4*x+1-2*(y-1) } i++ } j++ } Example N=390644893234047643 , sqrt(N/(15/10))=a , (15/10*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]/2=r r=71437,..... N=390644893234047643 , 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= 1 & <2   2021-07-31, 10:20 #15 Alberico Lepore   May 2017 ITALY 23×32×7 Posts If we establish how far x can be from the capofila at mos (order size)t, here r=71501 , r=(2*h+sqrt(32*h+81)+9)/2-(2*h-sqrt(32*h+49)+7)/2 , (2*x-sqrt(32*x+1)-1)/2 2021-08-01, 13:48 #16 Alberico Lepore   May 2017 ITALY 23×32×7 Posts Code: check=0 i=0 while(i<10) { solve this system sqrt(N/((10+i)/10))=a , ((10+i)/10*a+a-4)/8=x , 2*x*(x+1)-b*(b-1)/2=(N-3)/8 } memorize b(i) C=0 while (check==0){ i=0 while(i<10 && check==0) { h=x-b(i)/2-C // b/2 must be integer , [2*(h-1)*(h-1+1)] < N-3)/8-b(i)/2*(4*x+1-2*(y-1)) <= [2*h*(h+1)] , 2*(x)*(x+1)-y*(y-1)/2=(N-3)/8 while(min_h <= max_h && check==0){ x=h+b/2+C 2*(x)*(x+1)-y*(y-1)/2=(N-3)/8 calculate p p=4*x+1-2*(y-1) if(N mod p ==0 && p!=1 && p!=N){ check=1 break } min_h++ } i++ } C++ }` Example N=390644893234047643 , sqrt(N/(15/10))=a , (15/10*a+a-4)/8=x , 2*x*(x+1)-b*(b-1)/2=(N-3)/8 b = 63790420 h=x-(63790420)/2-C , [2*(h-1)*(h-1+1)] < (390644893234047643-3)/8-(63790420)/2*(4*x+1-2*(y-1)) <= [2*h*(h+1)] , 2*(x)*(x+1)-y*(y-1)/2=(390644893234047643-3)/8 per C=-7454 127855236<=h<127855255 -> size range = 19 per h=127855241 , h=x-(63790420)/2-7454 -> x=159757905 the problem is that size range of h is decreasing for C=0 127580838<=h<127584034 -> size range = 3196 for C=3727 127775161<=h<127775188 -> size range =27 an exponential decrease would seem to our advantage ********************************************************************* UPDATE: I tried to solve in x and I noticed that the first valid value is our x = 159757905 if it would always happen the cost of factoring 390644893234047643 would be 7454 * 10 = 74540 Tomorrow morning I will continue with other tests https://www.wolframalpha.com/input/?...%2F8+%2Ch+%2Cy Last fiddled with by Alberico Lepore on 2021-08-01 at 19:00 Reason: UPDATE   2021-08-02, 08:41   #17
Alberico Lepore

May 2017
ITALY

23·32·7 Posts Quote:
 Originally Posted by Alberico Lepore ********************************************************************* UPDATE: I tried to solve in x and I noticed that the first valid value is our x = 159757905 if it would always happen the cost of factoring 390644893234047643 would be 7454 * 10 = 74540 Tomorrow morning I will continue with other tests https://www.wolframalpha.com/input/?...%2F8+%2Ch+%2Cy
unfortunately this is not true.

But x is very close to min_range_x

I tested on a number of 30 digits with p and q of 15 digits and the result is 37

N=188723059539473758658629052963

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

b=64759908643727

h=x-(64759908643726)/2-88973930
,
[2*(h-1)*(h-1+1)]
<
(188723059539473758658629052963-3)/8-(64759908643726)/2*(4*x+1-2*(y-1))
<=
[2*h*(h+1)]
,
2*(x)*(x+1)-y*(y-1)/2=(188723059539473758658629052963-3)/8

range x 113364197263548<=x<=113364197263741

size_range 193

distance x 37

x=113364197263585

I need to be able to quantify distance x or size_range_x

total cost [10*my_quantify *88973930] about N ^ (1/3)

It's still a very heavy bruteforce, but I'm happy

Last fiddled with by Alberico Lepore on 2021-08-02 at 08:50 Reason: [10*my_quantify *88973930]   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Alberico Lepore Alberico Lepore 3 2021-07-13 14:43 Alberico Lepore Alberico Lepore 2 2019-12-02 08:37 Alberico Lepore Alberico Lepore 43 2018-01-17 15:55 Alberico Lepore Alberico Lepore 48 2017-12-30 09:43 Alberico Lepore Alberico Lepore 61 2017-09-23 21:52

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