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Old 2020-03-01, 21:57   #12
Alberico Lepore
 
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Quote:
Originally Posted by CRGreathouse View Post
Hmm... ok. But this isn't something you could detect from looking at the number to be factored, so it's a bit less useful for a factoring algorithm compared to one where you could tell if it applied or not. Here if you just see a number you wouldn't know what type it is.

50 bits: 730518607478393 (500 μs in gp -- SQUFOF)
100 bits: 718009979005367478539949203041 (500 μs in gp -- MPQS)
150 bits: 788147109545943410763272181909253057469476849 (500 ms in gp - MPQS)
200 bits: 1480011537446014095552865655792207870614938765157503182334313 (6 seconds in yafu -- SIQS)

Thanks for the numbers,
when I find an efficient algorithm I will try them
returning to mathematical discourse
have you noticed the symmetry of two symmetric schemes [(q-p) mod 8 == 0 and (q-p) mod 8 == 2] below in h and k?



Quote:
Originally Posted by Alberico Lepore View Post
In summary

Given an N = p * q in the form (p + q-4) mod 8 = 0

with p and q not necessarily prime but odd numbers

then

given the system

solve h=[x^2-[4*x-1+4*x-1-3*(y-2)]*(y-1)/2]
,
k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1)
,
(3*N-1)/8+h=2*((3*N-1)/8-1)/3+k
,
k-2*h=(x+1)^2+2*[x^2-(x-y+1)^2]
,
k>0
,
y>0
,
x>0
,
y

you will have that

q=2*(3*x+1-(x-y+1))+1 e p=2*(3*x+1-(x-y+1))+1-(4*y-2)

Last fiddled with by Alberico Lepore on 2020-03-01 at 22:08
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Old 2020-03-02, 02:17   #13
CRGreathouse
 
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Quote:
Originally Posted by Alberico Lepore View Post
when I find an efficient algorithm I will try them
I thought you already had implemented a base case? You can just use it on the 50-bit one for now.

Why do you call it a base case? Do you have a more sophisticated algorithm that uses this one for small numbers as a subroutine?

Quote:
Originally Posted by Alberico Lepore View Post
returning to mathematical discourse
have you noticed the symmetry of two symmetric schemes [(q-p) mod 8 == 0 and (q-p) mod 8 == 2] below in h and k?
What are h and k?
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Old 2020-03-02, 21:16   #14
Alberico Lepore
 
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Quote:
Originally Posted by CRGreathouse View Post
I thought you already had implemented a base case? You can just use it on the 50-bit one for now.

Why do you call it a base case? Do you have a more sophisticated algorithm that uses this one for small numbers as a subroutine?
I don't have efficient algorithms.


Quote:
Originally Posted by CRGreathouse View Post

What are h and k?
h and k are one of the bridges connecting the two schemes.

N is in the scheme (q-p) mod 8 == 0

(3*N-1)/8+h=2*((3*N-1)/8-1)/3+k=G

G=[[2*[3*x+1-(x-y+1)]+1+2*(x-y+1)]*[2*[3*x+1-(x-y+1)]+1+2*(x-y+1)+2+8*(x-y+1)]-3]/12

12*G+3 is in the scheme (q-p) mod 8 == 2

and it is biunivocal
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Old 2020-03-02, 21:31   #15
retina
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Quote:
Originally Posted by Alberico Lepore View Post
I don't have efficient algorithms.
We know.

So what ARE you claiming in your posts?
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Old 2020-03-02, 21:41   #16
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Quote:
Originally Posted by retina View Post
We know.

So what ARE you claiming in your posts?
play with math
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Old 2020-03-02, 21:54   #17
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Quote:
Originally Posted by Alberico Lepore View Post
play with math
1. toss equations at screen
2. disregard test cases
3. say "please help me test it"
4. go back to 1
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Old 2020-03-04, 09:54   #18
LaurV
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You forgot

3.7 don't give a sh!t about what is said/answered by people who know, and who actually want to help him...

Last fiddled with by LaurV on 2020-03-04 at 09:55
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Old 2020-03-04, 15:31   #19
Alberico Lepore
 
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it's not true !
I listen to advice a lot
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Old 2020-03-04, 16:14   #20
chalsall
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Quote:
Originally Posted by Alberico Lepore View Post
I listen to advice a lot
Do you internalize any of it?
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Old 2020-03-05, 06:36   #21
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Good time to close this thread.
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