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 Index > Heap > what do you think is the lengt shortest pair of md5 collide
Author
l4m2

Joined: 15 Jan 2015
Posts: 648
l4m2
if md5(a)=md5(b)then len(a)+len(b)is
12 Feb 2015, 08:40
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17259
revolution
len(a)=len(b)=0
12 Feb 2015, 08:51
l4m2

Joined: 15 Jan 2015
Posts: 648
l4m2
revolution wrote:
len(a)=len(b)=0

not a collide
12 Feb 2015, 10:05
HaHaAnonymous

Joined: 02 Dec 2012
Posts: 1180
Location: Unknown
HaHaAnonymous
[ Post removed by author. ]

Last edited by HaHaAnonymous on 28 Feb 2015, 17:58; edited 1 time in total
12 Feb 2015, 13:39
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17259
revolution
It is unknown if all possible output vectors are attainable.

It is unknown if all attainable output vectors have collisions.
12 Feb 2015, 13:49
HaHaAnonymous

Joined: 02 Dec 2012
Posts: 1180
Location: Unknown
HaHaAnonymous
[ Post removed by author. ]

Last edited by HaHaAnonymous on 28 Feb 2015, 17:58; edited 1 time in total
12 Feb 2015, 13:58
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17259
revolution
Yes. All hashes have this property. There can be a large number of collisions for many (perhaps most) of the attainable outputs (although this is not infinite because hashes have a limited maximal input length). This is basically the definition of a hash. A short fixed length output for any length of allowable input will guarantee many collisions.
12 Feb 2015, 14:08
HaHaAnonymous

Joined: 02 Dec 2012
Posts: 1180
Location: Unknown
HaHaAnonymous
[ Post removed by author. ]

Last edited by HaHaAnonymous on 28 Feb 2015, 17:58; edited 1 time in total
12 Feb 2015, 14:12
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17259
revolution
The trick with designing a "good" hash is to make constructable collisions very difficult to obtain. MD5 has famously been "broken" with a number of constructed collisions, thus showing its weakness.
12 Feb 2015, 14:18
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