Re: SHA Collision Resistance

On 27 Dec 2005 00:03:29 GMT, Unruh <unruh-spam@xxxxxxxxxxxxxx> wrote:

>stan <stan@xxxxxxxxxx> writes:
>
>>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?
>
>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.

OK. Yes. I get it. Thank you.

.

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