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

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l4m2



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


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



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

not a collide
Post 12 Feb 2015, 10:05
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HaHaAnonymous



Joined: 02 Dec 2012
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HaHaAnonymous
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Last edited by HaHaAnonymous on 28 Feb 2015, 17:58; edited 1 time in total
Post 12 Feb 2015, 13:39
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revolution
When all else fails, read the source


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revolution
It is unknown if all possible output vectors are attainable.

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



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


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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.
Post 12 Feb 2015, 14:08
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HaHaAnonymous



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


Joined: 24 Aug 2004
Posts: 17259
Location: In your JS exploiting you and your system
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.
Post 12 Feb 2015, 14:18
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