- Hi

The absolute limit to which the fermat factorization would have

to go to find a solution at the worst would be when the number, N, is

prime and no tests have been carried out: the maximum value would be

given by (N + 1 - (2s))/2, where s= flr(sqrt(N)).

if some trial division has been carried out to say D, the maximum

limit can be reduced to:

(((s - D)^2 + (N - (s*s))/D)/2

The number of operations can be reduced by looking at (N + 1 - (2s))

mod 4, if the result is 0 then only even values need be looked at, and

if the result is 2 then only odd numbers need be looked at.

example N = 13199 s = 114

(13199 + 1 - (2*114)) = 12972

12972 mod 4 = 0

the numbers that need to be looked at are (114 + 2aa) aa =

1, 2, 3, ... upto (12972/2)/2.

example N = 1803601 s = 1342

(1803601 + 1 - (2*1342) = 1800918

1800918 mod 4 = 2

the numbers that need to be looked at are (1342 + (2aa -1)) aa =

1, 2, 3 .... upto (1800918/2)/2

For large numbers the amount of processing would still be

considerable and then one would need to look at a sieve for the values

of a.

Ron

>----Original Message----

complexity of fermat factorization method?

>"kad"

>

>wrote:

>> 1. Fermat factorization

>> Posted by: "kad"yourskadhir@... yourskadhir

>> Date: Sun Nov 11, 2012 12:48 am ((PST))

>>

>> Hi,

>> Can any one tell me Whats the exact logic lie behind the worst time

>>

factorization we have to search 'a' > sqrt(N) so that a^2 - N = b^2.

>>

>> In order to factorize N = pq any semiprime in order to do fermat

Then factorization of N = (a - b)(a + b). My question is when worst

running time complexity will occur and whats the reason for worst

running time.

>

/2

>Suppose N = 3 y, where y is prime.

>

>Then if a^2 - N = b^2,

>

>then a = (y + 3)/2 and b = (y-3)/2.

>

>Thus in this worst case, you will need to advance "a" to (N/3 + 3)

>before finding the Fermat factors.

>

>

>Kermit Rose - Dear all

He has also sent a message directly to me.

He clearly does not understand some very basic concepts:

1) If you think that you have found an error, look for the error

in your own work and if you cannot spot the error put the work away

for sometime (3 months is good) then go back to it.

2) If a message has been posted to a group and no one has found an

error, the likelihood is that the error is in your own work; usually

true 999 times out of a thousand.

3) if the formula work on small numbers, try the same formula on

medium size numbers and than on large numbers; and not just one of

each, it would not be that first time that spurious results can result

from special factors relating to the number(s).

4) Do not use results, ie the difference, which can only be found

after the complete factorization has been completed.

Ron

>----Original Message----

large

>From: yourskadhir@...

>Date: 23/11/2012 13:24

>To: "ronhallam@..."<ronhallam@...>

>Subj: Re: [PrimeNumbers] Re: Fermat factorization

>

>The logic that factors with least difference will factored easily and

>difference will factored hardly is wrong. Please follow the link which

i

>found about Fermat factorization.

N, is

>http://kadinumberprops.blogspot.in/2012/11/fermats-factorization-running-time.html

>

>On Mon, Nov 12, 2012 at 5:08 PM, ronhallam@... <

>ronhallam@...> wrote:

>

>> **

>>

>>

>> Hi

>> The absolute limit to which the fermat factorization would have

>> to go to find a solution at the worst would be when the number,

>> prime and no tests have been carried out: the maximum value would

be

>> given by (N + 1 - (2s))/2, where s= flr(sqrt(N)).

(2s))

>>

>> if some trial division has been carried out to say D, the maximum

>> limit can be reduced to:

>>

>> (((s - D)^2 + (N - (s*s))/D)/2

>>

>> The number of operations can be reduced by looking at (N + 1 -

>> mod 4, if the result is 0 then only even values need be looked at,

and

>> if the result is 2 then only odd numbers need be looked at.

values

>>

>> example N = 13199 s = 114

>>

>> (13199 + 1 - (2*114)) = 12972

>>

>> 12972 mod 4 = 0

>>

>> the numbers that need to be looked at are (114 + 2aa) aa =

>> 1, 2, 3, ... upto (12972/2)/2.

>>

>> example N = 1803601 s = 1342

>>

>> (1803601 + 1 - (2*1342) = 1800918

>>

>> 1800918 mod 4 = 2

>>

>> the numbers that need to be looked at are (1342 + (2aa -1)) aa =

>> 1, 2, 3 .... upto (1800918/2)/2

>>

>> For large numbers the amount of processing would still be

>> considerable and then one would need to look at a sieve for the

>> of a.

time

>>

>> Ron

>>

>>

>> >----Original Message----

>>

>> >"kad"

>>

>> >

>>

>> >wrote:

>>

>> >> 1. Fermat factorization

>>

>> >> Posted by: "kad"yourskadhir@... yourskadhir

>>

>> >> Date: Sun Nov 11, 2012 12:48 am ((PST))

>>

>> >>

>>

>> >> Hi,

>>

>> >> Can any one tell me Whats the exact logic lie behind the worst

>> complexity of fermat factorization method?

b^2.

>>

>> >>

>>

>> >>

>>

>> >> In order to factorize N = pq any semiprime in order to do fermat

>> factorization we have to search 'a' > sqrt(N) so that a^2 - N =

>> Then factorization of N = (a - b)(a + b). My question is when worst

>> running time complexity will occur and whats the reason for worst

>> running time.

>>

>> >

>>

>> >Suppose N = 3 y, where y is prime.

>>

>> >

>>

>> >Then if a^2 - N = b^2,

>>

>> >

>>

>> >then a = (y + 3)/2 and b = (y-3)/2.

>>

>> >

>>

>> >Thus in this worst case, you will need to advance "a" to (N/3 + 3)

>> /2

>>

>> >before finding the Fermat factors.

>>

>> >

>>

>> >

>>

>> >Kermit Rose

>>

>>

>>

>