While the solution below won't eliminate possibility of collisions, it surely will reduce the risk by a very substantial amount. 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 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 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. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Collectives on Stack Overflow. Learn more.
Can two different strings generate the same MD5 hash code? Ask Question. Asked 11 years, 11 months ago. Active 4 years, 11 months ago. Viewed 82k times. Improve this question. Dency G B 7, 9 9 gold badges 44 44 silver badges 74 74 bronze badges. Lieven Cardoen Lieven Cardoen MD5 is not used for such things, it is usually used as a checksum to determine if something has been changed it's just a hash.
It would be very unlikely that the same data could be changed and produce the same MD5 checksum again. Add a comment. Active Oldest Votes. Improve this answer. True, I updated my post.
The random collision attack is very fast indeed. The MD5 prefix collision attack can take days -- but is generally much more useful to an attacker — intgr. Here's an example in PHP: ideone. Show 3 more comments. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown.
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The above files were generated by exploiting two facts: the block structure of the MD5 function, and the fact that Wang and Yu's technique works for an arbitrary initialization vector. To understand what this means, it is useful to have a general idea of how the MD5 function processes its input.
This is done by an iteration method known as the Merkle-Damgard method. A given input file is first padded so that its length will be a multiple of 64 bytes.
It is then divided into individual byte blocks M 0 , M 1 , The MD5 hash is computed by computing a sequence of byte states s 0 , Here, the initial state s 0 is fixed, and is called the initialization vector. The final state s n is the computed MD5 hash. It is important that this works for any initialization vector s , and not just for the standard initialization vector s 0. Combining these observations, it is possible to find pairs of files of arbitrary length, which are identical except for bytes somewhere in the middle of the file, and which have identical MD5 hash.
Indeed, let us write the two files as sequences of byte blocks: M 0 , M 1 , The blocks at the beginning of the files, M 0 , Suppose that the internal state of the MD5 hash function after processing these blocks is s i. The probability of just two hashes accidentally colliding is approximately: 4. As you can see, the slower and longer the hash is, the more reliable it is. But, as you can imagine, the probability of collision of hashes even for MD5 is terribly low.
That probability is lower than the number of water drops contained in all the oceans of the earth together. So the common sense tells you that the possibility of collision should not be considered as a factor because it looks like a very remote possibility. In the case of MD5 that is also true… at least for casual collisions. As you probably know, MD5 has been compromised almost 20 years ago. So, nowadays it is actually possible to artificially produce MD5 collisions.
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