Can two different strings generate the same MD5 ha

问题:

For each of our binary assets we generate a MD5 hash. This is used to check whether a certain binary asset is already in our application. But is it possible that two different binary assets generate the same MD5 hash. So is it possible that two different strings generate the same MD5 hash?

回答1:

For a set of even billions of assets, the chances of random collisions are negligibly small -- nothing that you should worry about. Considering the birthday paradox, given a set of 2^64 (or 18,446,744,073,709,551,616) assets, the probability of a single MD5 collision within this set is 50%. At this scale, you'd probably beat Google in terms of storage capacity.

However, because the MD5 hash function has been broken (it's vulnerable to a collision attack), any determined attacker can produce 2 colliding assets in a matter of seconds worth of CPU power. So if you want to use MD5, make sure that such an attacker would not compromise the security of your application!

Also, consider the ramifications if an attacker could forge a collision to an existing asset in your database. While there are no such known attacks (preimage attacks) against MD5 (as of 2011), it could become possible by extending the current research on collision attacks.

If these turn out to be a problem, I suggest looking at the SHA-2 series of hash functions (SHA-256, SHA-384 and SHA-512). The downside is that it's slightly slower and has longer hash output.

回答2:

MD5 is a hash function – so yes, two different strings can absolutely generate colliding MD5 codes.

In particular, note that MD5 codes have a fixed length so the possible number of MD5 codes is limited. The number of strings (of any length), however, is definitely unlimited so it logically follows that there must be collisions.

回答3:

Yes, it is possible. This is in fact a Birthday problem. However the probability of two randomly chosen strings having the same MD5 hash is very low.

See this and this questions for examples.

回答4:

Yes, of course: MD5 hashes have a finite length, but there are an infinite number of possible character strings that can be MD5-hashed.

回答5:

Yes, it is possible that two different strings can generate the same MD5 hash code.

Here is a simple test using very similar binary message in hex string:

$ echo '4dc968ff0ee35c209572d4777b721587d36fa7b21bdc56b74a3dc0783e7b9518afbfa200a8284bf36e8e4b55b35f427593d849676da0d1555d8360fb5f07fea2' | xxd -r -p | tee >/dev/null >(md5) >(sha1sum)
c6b384c4968b28812b676b49d40c09f8af4ed4cc  -
008ee33a9d58b51cfeb425b0959121c9

$ echo '4dc968ff0ee35c209572d4777b721587d36fa7b21bdc56b74a3dc0783e7b9518afbfa202a8284bf36e8e4b55b35f427593d849676da0d1d55d8360fb5f07fea2' | xxd -r -p | tee >/dev/null >(md5) >(sha1sum)
c728d8d93091e9c7b87b43d9e33829379231d7ca  -
008ee33a9d58b51cfeb425b0959121c9

They generate different SHA-1 sum, but the same MD5 hash value. Secondly the strings are very similar, so it's difficult to find the difference between them.

The difference can be found by the following command:

$ diff -u <(echo 4dc968ff0ee35c209572d4777b721587d36fa7b21bdc56b74a3dc0783e7b9518afbfa200a8284bf36e8e4b55b35f427593d849676da0d1555d8360fb5f07fea2 | fold -w2) <(echo 4dc968ff0ee35c209572d4777b721587d36fa7b21bdc56b74a3dc0783e7b9518afbfa202a8284bf36e8e4b55b35f427593d849676da0d1d55d8360fb5f07fea2 | fold -w2)
--- /dev/fd/63  2016-02-05 12:55:04.000000000 +0000
+++ /dev/fd/62  2016-02-05 12:55:04.000000000 +0000
@@ -33,7 +33,7 @@
 af
 bf
 a2
-00
+02
 a8
 28
 4b
@@ -53,7 +53,7 @@
 6d
 a0
 d1
-55
+d5
 5d
 83
 60

Above collision example is taken from Marc Stevens: Single-block collision for MD5, 2012; he explains his method, with source code (alternate link to the paper).

---

Another test:

$ echo '0e306561559aa787d00bc6f70bbdfe3404cf03659e704f8534c00ffb659c4c8740cc942feb2da115a3f4155cbb8607497386656d7d1f34a42059d78f5a8dd1ef' | xxd -r -p | tee >/dev/null >(md5) >(sha1sum)
756f3044edf52611a51a8fa7ec8f95e273f21f82  -
cee9a457e790cf20d4bdaa6d69f01e41

$ echo '0e306561559aa787d00bc6f70bbdfe3404cf03659e744f8534c00ffb659c4c8740cc942feb2da115a3f415dcbb8607497386656d7d1f34a42059d78f5a8dd1ef' | xxd -r -p | tee >/dev/null >(md5) >(sha1sum)
6d5294e385f50c12745a4d901285ddbffd3842cb  -
cee9a457e790cf20d4bdaa6d69f01e41

Different SHA-1 sum, the same MD5 hash.

Difference is in one byte:

$ diff -u <(echo 0e306561559aa787d00bc6f70bbdfe3404cf03659e704f8534c00ffb659c4c8740cc942feb2da115a3f4155cbb8607497386656d7d1f34a42059d78f5a8dd1ef | fold -w2) <(echo 0e306561559aa787d00bc6f70bbdfe3404cf03659e744f8534c00ffb659c4c8740cc942feb2da115a3f415dcbb8607497386656d7d1f34a42059d78f5a8dd1ef | fold -w2)
--- /dev/fd/63  2016-02-05 12:56:43.000000000 +0000
+++ /dev/fd/62  2016-02-05 12:56:43.000000000 +0000
@@ -19,7 +19,7 @@
 03
 65
 9e
-70
+74
 4f
 85
 34
@@ -41,7 +41,7 @@
 a3
 f4
 15
-5c
+dc
 bb
 86
 07

Above example is adapted from Tao Xie and Dengguo Feng: Construct MD5 Collisions Using Just A Single Block Of Message, 2010.

---

Related:

• Are there two known strings which have the same MD5 hash value? at Crypto.SE

回答6:

Yes, it is possible. It is called a Hash collision.

Having said that, algorithms such as MD5 are designed to minimize the probability of a collision.

The Wikipedia entry on MD5 explains some vulnerabilities in MD5, which you should be aware of.

回答7:

Just to be more informative. From a math point of view, Hash functions are not injective.
It means that there is not a 1 to 1 (but one way) relationship between the starting set and the resulting one.

Bijection on wikipedia

EDIT: to be complete injective hash functions exist: it's called Perfect hashing.

回答8:

Yes, it is! Collision will be a possibility (although, the risk is very small). If not, you would have a pretty effective compression method!

EDIT: As Konrad Rudolph says: A potentially unlimited set of input converted to a finite set of output (32 hex chars) will results in an endless number of collisions.

回答9:

As other people have said, yes, there can be collisions between two different inputs. However, in your use case, I don't see that being a problem. I highly doubt that you will run into collisions - I've used MD5 for fingerprinting hundreds of thousands of image files of a number of image (JPG, bitmap, PNG, raw) formats at a previous job and I didn't have a collision.

However, if you are trying to fingerprint some kind of data, perhaps you could use two hash algorithms - the odds of one input resulting in the same output of two different algorithms is near impossible.

回答10:

I think we need to be careful choosing the hashing algorithm as per our requirement, as hash collisions are not as rare as I expected. I recently found a very simple case of hash collision in my project. I am using Python wrapper of xxhash for hashing. Link: https://github.com/ewencp/pyhashxx

s1 = 'mdsAnalysisResult105588'
s2 = 'mdsAlertCompleteResult360224'
pyhashxx.hashxx(s1) # Out: 2535747266
pyhashxx.hashxx(s2) # Out: 2535747266

It caused a very tricky caching issue in the system, then I finally found that it's a hash collision.

回答11:

I realize this is old, but thought I would contribute my solution. There are a 2^128 possible hash combinations. And thus a 2^64 probability of a birthday paradox. While the solution below won't eliminate possibility of collisions, it surely will reduce the risk by a very substantial amount.

2^64 = 18,446,744,073,709,500,000 possible combinations

What I have done is I put a few hashes together based on the input string to get a much longer resulting string that you consider your hash...

So my pseudo-code for this is:

Result = Hash(string) & Hash(Reverse(string)) & Hash(Length(string))

That is to practical improbability of a collision. But if you want to be super paranoid and can't have it happen, and storage space is not an issue (nor is computing cycles)...

Result = Hash(string) & Hash(Reverse(string)) & Hash(Length(string)) 
         & Hash(Reverse(SpellOutLengthWithWords(Length(string)))) 
         & Hash(Rotate13(string)) Hash(Hash(string)) & Hash(Reverse(Hash(string)))

Okay, not the cleanest solution, but this now gets you a lot more play with how infrequently you will run into a collision. To the point I might assume impossibility in all realistic senses of the term.

For my sake, I think the possibility of a collision is infrequent enough that I will consider this not "surefire" but so unlikely to happen that it suits the need.

Now the possible combinations goes up significantly. While you could spend a long time on how many combinations this could get you, I will say in theory it lands you SIGNIFICANTLY more than the quoted number above of

2^64 (or 18,446,744,073,709,551,616) 

Likely by a hundred more digits or so. The theoretical max this could give you would be

Possible number of resulting strings:

528294531135665246352339784916516606518847326036121522127960709026673902556724859474417255887657187894674394993257128678882347559502685537250538978462939576908386683999005084168731517676426441053024232908211188404148028292751561738838396898767036476489538580897737998336