* Stephan V Bechtolsheim wrote on Fri, Jun 05, 2009 at 18:20 -0700:
> > This is hardly anything remotely resembling a formal proof,
> > of course. But it should give you the basic idea -- it's a
> > difficult problem because the numbers are big.
> Your argument only applies to "your algorithm". The
On Sun, Jun 07, 2009 at 07:18:19AM -0500, Michael S. Zick wrote:
> On Sun June 7 2009, Victor Duchovni wrote:
> >
> > No proof is known that better algorithms won't come along, but for now
> > state of the-art number theory gives us GNFS.
> >
>
> Mathematics is an open-ended field on any subjec
On Sun June 7 2009, Victor Duchovni wrote:
>
> No proof is known that better algorithms won't come along, but for now
> state of the-art number theory gives us GNFS.
>
Mathematics is an open-ended field on any subject for which
a proof does not exist.
Some 'great mind' may come along at any tim
On Sun, Jun 07, 2009, jaze lee wrote:
> That is , n = q*p , we can choose the prime has given bits, but we
> can not know all that prime in that range.if
> we want to know the range , we should test it every odd number in that
> range, or should find a function that can do the job efficiently .
>
2009/6/7 Victor Duchovni :
> On Sun, Jun 07, 2009 at 09:51:14AM +0800, jaze lee wrote:
>
>> The problem is we can not find the function yet ? or some other ways
>> to judge a big integer whether it's a prime. Is it so-called
>> mathematics problem that many cipher based on it ?
>
> No answer to you
On Sun, Jun 07, 2009 at 09:51:14AM +0800, jaze lee wrote:
> The problem is we can not find the function yet ? or some other ways
> to judge a big integer whether it's a prime. Is it so-called
> mathematics problem that many cipher based on it ?
No answer to your questions beyond "it's magic, trus
2009/6/6 Michael S. Zick :
> On Sat June 6 2009, jaze lee wrote:
>>
>> i still not understand the problem. although i don''t get the result.
>>
>
> Q1: Why is this problem "hard" - as in: "computationally hard" ?
>
> A1: There are two many number trials (computations) required in a
> "cryptographi
On Sat, Jun 06, 2009 at 04:58:24AM -0500, Michael S. Zick wrote:
> DAQ1: How many integer numbers are there? (an uncountable value)
Not "uncountable", countably infinite.
> DAQ2: How many after we throw away all the even ones? (an uncountable value)
Ditto.
> Obviously, this isn't leading to
On Sat June 6 2009, jaze lee wrote:
>
> i still not understand the problem. although i don''t get the result.
>
Q1: Why is this problem "hard" - as in: "computationally hard" ?
A1: There are two many number trials (computations) required in a
"cryptographically hard" number.
True, for any br
On Sat, Jun 06, 2009 at 03:39:21PM +0800, jaze lee wrote:
> may be you are wright, i try , but i can not get the result.
> if a integer with m bits and another integer with n bits, if the
> multiple , there product has m+n bits or m+n-1 bits.
> 248911498900030209107 is a 21 bits number,
No it i
Jaze lee what exactly you can't understand ?
--
Best Regards Rustam !!!
On Sat June 6 2009, jaze lee wrote:
> 2009/6/6 Rustam Rakhimov :
> > So if you are so brave try the example given before.
> > Than you will feel reality.
> may be you are wright, i try , but i can not get the result.
> if a integer with m bits and another integer with n bits, if the
> multiple , t
2009/6/6 Rustam Rakhimov :
> So if you are so brave try the example given before.
> Than you will feel reality.
may be you are wright, i try , but i can not get the result.
if a integer with m bits and another integer with n bits, if the
multiple , there product has m+n bits or m+n-1 bits.
2489114
> This is hardly anything remotely resembling a formal proof, of course. But
> it should give you the basic idea -- it's a difficult problem because the
> numbers are big.
Your argument only applies to "your algorithm". The question is whether there
exists something
else besides a trial / brute fo
So if you are so brave try the example given before.
Than you will feel reality.
-
Best Regards Rustam !!!
2009/6/6 David Schwartz :
>
>> hello,
>> when i read some books about cryptography, it always go that the
>> cryptography is based on the difficult math problem, for example big
>> integer decomposition,
>> i don't understand it, for if we know that n = p*q , p, q are prime ,
>> why it's diff
> hello,
> when i read some books about cryptography, it always go that the
> cryptography is based on the difficult math problem, for example big
> integer decomposition,
> i don't understand it, for if we know that n = p*q , p, q are prime ,
> why it's difficult to get p and q ? i think ,i
.
StvB
From: Victor Duchovni
To: openssl-users@openssl.org
Sent: Friday, June 5, 2009 8:32:29 AM
Subject: Re: about the integer decomposition
On Fri, Jun 05, 2009 at 03:52:07PM +0800, jaze lee wrote:
> hello,
> when i read some books about cryptograp
On Fri, Jun 05, 2009 at 03:52:07PM +0800, jaze lee wrote:
> hello,
> when i read some books about cryptography, it always go that the
> cryptography is based on the difficult math problem, for example big
> integer decomposition,
> i don't understand it, for if we know that n = p*q , p, q ar
hello,
when i read some books about cryptography, it always go that the
cryptography is based on the difficult math problem, for example big
integer decomposition,
i don't understand it, for if we know that n = p*q , p, q are prime ,
why it's difficult to get p and q ? i think ,if we know the
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