Optimal Integer Factorization and Quantum Theory of Numbers

by **Guest** » Fri Jun 19, 2020 6:43 pm

Updated Link Title:

Guest wrote:Relevant Reference Link:

'Randomness can be a useful tool for solving problems.',

https://www.math10.com/forum/viewtopic.php?f=1&t=8855&start=80

by **Guest** » Fri Jun 19, 2020 6:38 pm

Relevant Reference Link:

'Optimal Integer Factorization and Quantum Theory of Numbers',

https://www.math10.com/forum/viewtopic.php?f=1&t=8855&start=80

by **Guest** » Wed Jul 18, 2018 7:46 am

Guest wrote:C=1
X=(N-1)/2
If X is even c=c+1, X/2
If X is odd X=N-X
When X=1 , N mod C = (p+q)-1, after just use fermat

N=33. X=(33-1)/2 = 16, C =1.
X(2) = 16/2 =8, C = 2.
X(3) = 8/2 = 4, C = 3.
X(4) = 4/2 = 2, C = 4.
X(5) = 2/2 = 1, C = 5.

33 mod 5 = 3, while p+q-1 = 11+3-1 = 13.

The method is wrong.

by **Guest** » Wed Apr 25, 2018 2:58 pm

Relevant Reference Link:

https://www.researchgate.net/post/What_are_the_algebraic_equations_of_integer_multiplication_in_base_10

by **Guest** » Mon Apr 10, 2017 5:13 am

Guest wrote:C=1
X=(N-1)/2
If X is even c=c+1, X/2
If X is odd X=N-X
When X=1 , N mod C = (p+q)-1, after just use fermat

Nope, that's not even close to being fast. It's exponential as hell.

by **Guest** » Wed Apr 05, 2017 7:06 pm

C=1
X=(N-1)/2
If X is even c=c+1, X/2
If X is odd X=N-X
When X=1 , N mod C = (p+q)-1, after just use fermat

by **Guest** » Fri Nov 13, 2015 11:42 am

[tex]r_{10 }[/tex] = [tex]a_{0 }*b_{0}[/tex] (mod 10); [tex]c_{11}[/tex] = ([tex]a_{0}b_{0} - r_{10}[/tex])/10;
[tex]r_{11 }[/tex] = [tex]c_{11} + a_{1 }*b_{0}[/tex] (mod 10); [tex]c_{12}[/tex] = ([tex]c_{11} + a_{1 }*b_{0} - r_{11}[/tex])/10;
[tex]r_{12 }[/tex] = [tex]c_{12} + a_{2 }*b_{0}[/tex] (mod 10); [tex]c_{1}[/tex] = ([tex]c_{12} + a_{2 }*b_{0} - r_{12}[/tex])/10;

So, [tex]a_{2 }a_{1 }a_{0 }[/tex]

X [tex]b_{0 }[/tex]
__________________________________________
= [tex]c_{1}r_{12 }r_{11 }r_{10 }[/tex] (Stage 1)

[tex]a_{2 }a_{1 }a_{0 }[/tex]

X [tex]b_{1 }[/tex]
__________________________________________
= [tex]c_{2}r_{22 }r_{21 }r_{20 }[/tex] 0 (Stage 2)

[tex]a_{2 }a_{1 }a_{0 }[/tex]

X [tex]b_{2 }[/tex]
__________________________________________
= [tex]c_{3}r_{32 }r_{31 }r_{30 }[/tex] 0 0 (Stage 3)

However, the 'devil' is not knowing the details... Please complete the multiplication process (modulus-10) for stages, two and three. Next, combine (add) all stages to complete the multiplication process (modulus-10). And correct any mistakes too.

by **Guest** » Fri Nov 13, 2015 11:13 am

Guest wrote:
Guest wrote:
Guest wrote:Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = [tex]\sum_{i=0}^{4 }r_{i}*10^i[/tex],

then [tex]r_{0}[/tex] = 9, [tex]r_{1}[/tex] = 8, [tex]r_{2}[/tex] = 9, [tex]r_{3}[/tex] = 1, and [tex]r_{4}[/tex] = 4.

We assume, p = [tex]\sum_{i=0}^{2}a_{i}*10^i[/tex] and q = [tex]\sum_{i=0}^{2 }b_{i}*10^i[/tex] since the square root of n is roughly a three-digit number.

So we have:
[tex]a_{2 }a_{1 }a_{0 }[/tex]

x [tex]b_{2 }b_{1 }b_{0 }[/tex]
______________________________

We must generate the right system of equations to solve the problem... good luck

by **Guest** » Tue Nov 10, 2015 3:53 pm

Guest wrote:
Guest wrote:Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = [tex]\sum_{i=0}^{4 }r_{i}*10^i[/tex],

then [tex]r_{0}[/tex] = 9, [tex]r_{1}[/tex] = 8, [tex]r_{2}[/tex] = 9, [tex]r_{3}[/tex] = 1, and [tex]r_{4}[/tex] = 4.

We assume, p = [tex]\sum_{i=0}^{2}a_{i}*10^i[/tex] and q = [tex]\sum_{i=0}^{2 }b_{i}*10^i[/tex] since the square root of n is roughly a three-digit number.

So we have:
[tex]a_{2 }a_{1 }a_{0 }[/tex]

x [tex]b_{2 }b_{1 }b_{0 }[/tex]
______________________________

We must generate the right system of equations to solve the problem... good luck

by **Guest** » Tue Nov 10, 2015 3:43 pm

Guest wrote:Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = [tex]\sum_{i=0}^{4 }r_{i}*10^i[/tex]

, then

[tex]r_{0}[/tex] = 9, [tex]r_{1}[/tex] = 8, [tex]r_{2}[/tex] = 9, [tex]r_{3}[/tex] = 1, and [tex]r_{4}[/tex] = 4.

We assume, p = [tex]\sum_{i=0}^{2}a_{i}*10^i[/tex] and q = [tex]\sum_{i=0}^{2 }b_{i}*10^i[/tex] since the square root of n is roughly a three-digit number.

by **Guest** » Tue Nov 10, 2015 3:34 pm

Howdy,

If n =41989, and we know it to be a product two distinct primes, p and q, what are they?

Furthermore, n = [tex]\sum_{i=0}^{4 }*r_{i}*10^i[/tex]

by **primework123** » Sat Nov 07, 2015 11:10 am

Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = [tex]\sum_{i=0}^{j}a_{i}*10^{i}[/tex], and q = [tex]\sum_{i=0}^{k}b_{i}*10^{i}[/tex] where

[tex]a_{i}, b_{i}\in Z_{10 }[/tex]={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume [tex]500\le j \le k[/tex], and [tex]p \ne q[/tex].[/quote][/quote]

So, n = [tex]\sum_{i=0}^{l}r_{i}*10^{i}[/tex] where [tex]r_{0} \in[/tex] {1, 3, 7, 9}.[/quote]

primework123,

So, given n as a product two large, unknown, and distinct prime numbers, p and q, we need to solve a system of equations to find

[tex]a_{i}[/tex] and [tex]b_{i}[/tex] for all i, [tex]0 \le i \le j[/tex] and [tex]0 \le i \le k[/tex], respectively, in order to construct the unknown prime factors, p and q.

But, how do optimization and handshaking helps us to solve the system of equations efficiently (in polynomial time or less)?
_________________________________________
Hi,

Yes, you're on the right path to a solution which was discovered back in the fall of 2010. Keep up the good work, and you'll sure to find the proper solution. I have given enough hints...

Best wishes,

David Cole
(aka primework123)
Keep the faith (effort and hope) and keep an open mind.

http://biblia.com/verseoftheday/image/Ro8.28

by **Guest** » Fri Nov 06, 2015 3:25 pm

Guest wrote:
Guest wrote:
Guest wrote:Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = [tex]\sum_{i=0}^{j}a_{i}*10^{i}[/tex], and q = [tex]\sum_{i=0}^{k}b_{i}*10^{i}[/tex] where

[tex]a_{i}, b_{i}\in Z_{10 }[/tex]={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume [tex]500\le j \le k[/tex], and [tex]p \ne q[/tex].

So, n = [tex]\sum_{i=0}^{l}r_{i}*10^{i}[/tex] where [tex]r_{0} \in[/tex] {1, 3, 7, 9}.

primework123,

So, given n as a product two large, unknown, and distinct prime numbers, p and q, we need to solve a system of equations to find

[tex]a_{i}[/tex] and [tex]b_{i}[/tex] for all i, [tex]0 \le i \le j[/tex] and [tex]0 \le i \le k[/tex], respectively, in order to construct the unknown prime factors, p and q.

But, how do optimization and handshaking helps us to solve the system of equations efficiently (in polynomial time or less)?

by **Guest** » Fri Nov 06, 2015 2:11 pm

Guest wrote:
Guest wrote:Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = [tex]\sum_{i=0}^{j}a_{i}*10^{i}[/tex], and q = [tex]\sum_{i=0}^{k}b_{i}*10^{i}[/tex] where

[tex]a_{i}, b_{i}\in Z_{10 }[/tex]={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume [tex]500\le j \le k[/tex], and [tex]p \ne q[/tex].

So, n = [tex]\sum_{i=0}^{l}r_{i}*10^{i}[/tex] where [tex]r_{0} \in {1, 3, 7, 9}.[/tex]

by **Guest** » Fri Nov 06, 2015 2:07 pm

Guest wrote:Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = [tex]\sum_{i=0}^{j}a_{i}*10^{i}[/tex], and q = [tex]\sum_{i=0}^{k}b_{i}*10^{i}[/tex] where

[tex]a_{i}, b_{i}\in Z_{10 }[/tex]={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume [tex]500\le j \le k[/tex], and [tex]p \ne q[/tex].

by **Guest** » Fri Nov 06, 2015 12:26 pm

Hey primework123, I like your previous post.

Suppose, we are given a positive integer, n, which is the product of two large primes, p and q, how do you factor n?

In base-10, we can let p = [tex]\sum_{i=0}^{j}a_{i}*10^{i}[/tex], and q = [tex]\sum_{i=0}^{k}b_{i}*10^{i}[/tex] where

[tex]a_{i}, b_{i}\in Z_{10 }[/tex]={0,1,2,3,4,5,6,7,8,9}. Furthermore, we assume [tex]500\le j \le k[/tex], and [tex]p \ne q[/tex].

by **primework123** » Wed Nov 04, 2015 4:58 pm

"I_C

We begin far apart with simple ideas and with little knowledge of the path before us,

But we shall meet at the middle and handshake our grand triumph." --David Cole (aka primework123)

Stage 0:

Keywords: Prime decomposition of n = p * q, primes are the atoms of the integers, base-10 and the finite field, [tex]Z_{2 }[/tex], are sub-atomic levels of the positive integers, Complexity and Change,...

Relevant Keywords: Carry-bit/number input, Residue bit/number, Carry-bit/number output, Optimization and Synchronization (or Handshaking), ...

If we let s, t[tex]\in[/tex] [tex]Z_{10 }[/tex] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, the product, s* t, can be easily computed as follows:

Carry-bit input or [tex]c_{i0 }[/tex] = 0 (always);

Residue bit or [tex]r_{0 }[/tex] = s * t (mod 10);

Carry-bit output or [tex]c_{o0 }[/tex] = (s * t - [tex]r_{0 }[/tex])/10 ;

Simple Example: Let s = 4 and s = t, then we have the following results:

[tex]c_{i0 }[/tex] = 0 (always);

[tex]r_{0 }[/tex] = s * t (mod 10) = 12 (mod 10) = 2;

[tex]c_{o0 }[/tex] = (s * t - [tex]r_{0 }[/tex])/10 = ( 12 - 2 ) /10 = 10/10 =1;

So, s * t = 12.

... End of Stage 0.

David Cole
(aka primework123)
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!

by **primework123** » Wed Nov 04, 2015 12:31 pm

"I_C

We begin far apart with simple ideas and with little knowledge of the path before us,

But we shall meet at the middle and handshake our grand triumph." --David Cole (aka primework123)

Stage 0:

Keywords: Decomposition of n = p * q, primes are the atoms of the integers, base-10 and the finite field, [tex]Z_{2 }[/tex], are sub-atomic levels of the integers,...

Relevant Keywords: Carry-bit/number input, Residue bit/number, Carry-bit/number output, Optimization and Synchronization (or Handshaking), ...

If we let s, t[tex]\in[/tex] [tex]Z_{10 }[/tex] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, the product, s* t, can be easily computed as follows:

Carry-bit input or [tex]c_{i0 }[/tex] = 0 (always);

Residue bit or [tex]r_{0 }[/tex] = s * t (mod 10);

Carry-bit output or [tex]c_{o0 }[/tex] = (s * t - [tex]r_{0 }[/tex])/10 ;

Simple Example: Let s = 4 and s = t, then we have the following results:

[tex]c_{i0 }[/tex] = 0 (always);

[tex]r_{0 }[/tex] = s * t (mod 10) = 12 (mod 10) = 2;

[tex]c_{o0 }[/tex] = (s * t - [tex]r_{0 }[/tex])/10 = ( 12 - 2 ) /10 = 10/10 =1;

So, s * t = 12.

... End of Stage 0.

David Cole
(aka primework123)
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!