Re: brute-force estimation
- From: tomstdenis@xxxxxxxxx
- Date: 22 Feb 2006 09:37:24 -0800
Richard Revis wrote:
"Elenhil" <elenhil@xxxxxxxxx> wrote in message
news:1140599172.853058.63330@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Say, I have a 20 character long password consisting of latin
characters (capital and lower-case), numbers and basic punctuation.
That is roughly 80 symbols to play with.
How many combinations one must look through from 1 to 20 (the exact
length is unknown) with unoptimized brute force?
I'm ashamed to admit that I haven't done any mathematics more complicated
than adding up my shopping bill since the second year of University, but I
don't think the problem is as complicated as you are making it.
Assuming 80 valid symbols in the password field and a length of exactly 20
with no limitation on the number of repeats of symbols the number of
combinations is 20^80.
If the password can be of any length up to 20 characters then the number of
combinations is 20^80+19^80+...+1^80.
you got the base backwards. It's 80^20. But otherwise sounds right
[right idea that is] so far.
Tom
.
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