[Tutor] All possible 16 character alphanumeric strings?

Dave Angel d at davea.name
Sun Sep 16 05:27:53 CEST 2012


On 09/15/2012 10:03 PM, Scurvy Scott wrote:
>>
>> That list would fill all the PC's on the planet a few billions times.
>> The number of items in the list has 25 digits in it.  print 32**16
>>
>> I actually should've specified that the list I'm trying to create would
> not start at say "0000000000000001".

Of course not.  Your defined set of characters didn't have a zero in it.

> I'm attempting to generate all possible .onion addressess which look like "
> kpvz7ki2v5agwt35.onion" for instance.
> 

"Look like" is pretty vague.  To me,  2222222222222222.onion certainly
looks like the one you said.  And if it's missing, then you don't have
them all.

> 
> The TOR network generates these numbers with this method:
> "If you decide to run a hidden service Tor generates an
> RSA-1024<http://en.wikipedia.org/wiki/RSA>keypair. The .onion name is
> computed as follows: first the
>  SHA1 <http://en.wikipedia.org/wiki/SHA_hash_functions> hash of the
> DER<http://en.wikipedia.org/wiki/Distinguished_Encoding_Rules>-encoded
>  ASN.1 <http://en.wikipedia.org/wiki/Abstract_Syntax_Notation_One> public
> key is calculated. Afterwards the first half of the hash is encoded to
> Base32 <http://en.wikipedia.org/wiki/Base32> and the suffix ".onion" is
> added. Therefore .onion names can only contain the digits 2-7 and the
> letters a-z and are exactly 16 characters long."
> -from the Tor Hidden Services DOC page.
> 
> I'm not sure if that changes anything as far as the impossible size of my
> dataset.
> 

I can't see any reason why it changes anything.  The world doesn't have
enough disk space to store *every* address.  If you need to calculate
all of them, you don't have enough time.

You need to rethink whatever your real problem is.  For example, if you
were really trying to crack a safe with 32 numbers on the dial and 16
settings to open it, perhaps you should forget all of it and get some
nitro.  Or a stethoscope.  Or bribe somebody who knows the combination.
 If you have to try all of the combinations systematically, you'll never
get there.

> Again, any input is useful.
> 
> Scott
> 


-- 

DaveA


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