Re: SHA Collision Resistance

stan <stan@theman.com> writes:

>On 26 Dec 2005 15:30:45 GMT, Unruh <unruh-spam@physics.ubc.ca> wrote:

>>stan <stan@theman.com> writes:
>>
>>>I Googled and also looked at FIPS 180-2 and have been unable to figure
>>>out the following questions:
>>
>>>For SHA-1 at 160 bits (which appears to be default)
>>
>>>And if I (hypothetically) had processing power to do this
>>
>>>If I take all values of integers from 0 to (2^160 - 1) or in hex
>>
>>>0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
>>
>>>to
>>
>>>FFFF FFFF FFFF FFFF FFFF FFFF FFFF FFFF FFFF FFFF
>>
>>>and run them through SHA-1, will I get 2^160 distinct 160bit values?
>>
>>No.
>>
>>
>>>IOW no collisions in that range?
>>
>>There almost certainly are many many collisions in that range.
>>SHA is NOT an encryption.

>So the cryptographic hash is designed to make collisions unlikely but
>they are still possible. But based on your comment encryption will not
>collide?. Is this what is meant by 1 to 1?

Yes. Yes.

>So if I (hypothetically) had processing power to do this

>And if I encrypted 128 bits of zeros using a 128 bit passphrase on
>AES-128 and the 128 bit passphrase used every integer value between
>0 and (2^128 - 1) would there be no collisions?

There would be collisions. This is not an encryption.
If you encrypted every integer from 0 to 2^128-1 with the same key then
there would be no collsions.
Received on Tue Jan 3 03:41:27 2006