Re: Algorithm For Ensuring p & q Sufficiently Large For n in RSA
- From: Le Chaud Lapin <jaibuduvin@xxxxxxxxx>
- Date: Wed, 4 Jun 2008 13:30:34 -0700 (PDT)
On Jun 4, 2:39 pm, David Eather <eat...@xxxxxxxxxx> wrote:
Le Chaud Lapin wrote:
Hi All,
What are the prevaling recommendations for choosing p, q, such that
the product is indeed, say, 512 bits, for a 512-bit modulus.
There are things that can be done to the bit-pattern of p and q during
randomization phase obviously, and I recall a while back in Coutinho's
there was a recommendation, but I wanted to explore more.
TIA,
-Le Chaud Lapin-
It's a game of Logs (base 2)
Add P and Q.
If the result is longer than the length of N/2 the product is too big
(just look for the most significant "1"). If less than N/2 -1 the
product will be too small. No division, no multiplication (you already
know the length of N so N/2 is not hard to figure out first)
Somebody else probably got here first.
Hmm...
Let [x] denoting bit-width of x.
I choose [N] to be 8 bits.
I choose P = 11. Then [P] = 4.
I choose Q = 13. Then [Q] = 4.
N = 11*13 = 143. [N] = 8, as expected.
P+Q = 11+13 = 24. [P+Q] = 5.
[N]/2 = 8/2 = 4.
5 is greater than 4.
Since 5 is greater than 4, according to your suggestion, I should
reject P = 11, Q = 13 as candidates for generating an 8-bit N,
because N would be greater than 8 bits, an erroneous conclusion,
unless I misunderstand what you wrote.
-Le Chaud Lapin-
.
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