Re: SHA Collision Resistance

On 26 Dec 2005 15:30:45 GMT, Unruh <unruh-spam@xxxxxxxxxxxxxx> wrote:

>stan <stan@xxxxxxxxxx> 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?

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?

Again, thank you.

>
>
>
>>Any help is appreciated and thank you.
>
>
>>stan

.

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