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when I hash a file with Md5 what is hashed?
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Closed 7 years ago.
I know a file's MD5 Hash is like a digital fingerprint used to confirm integrity and authenticity. There are many utilities to get the MD5 Hash of a file but what does that hash base on? File size? File low level binaries? Code?
MD5 is a so-called cryptographic hash function.
This basically means that you can give in any bitstring as input for the function, and you will get out a fixed-size bitstring (128-bit in the case of MD5) as output. The output is usually called "digest".
The digest depends solely on the input and nothing else. Thus in itself it can be used as an integrity proof, but not as authenticity, if the underlying hash function has the necessary properties (in this case collision-resistance). This means that for two different outputs the digest itself should be also different. The problem is that the digest's size is fixed, which in turn means that with sufficient number of messages it will always be possible to find a collision (i.e., two different inputs yielding the same output).
One should also note that there is nowadays no justification to use MD5, as weaknesses have been discovered (namely post-fix collision attacks). Also using SHA-256/512 on modern hardware is usually faster then MD5.
Shortly: the output of the cryptographic hash functions (and so MD5's) depends on the input bitstring.
Update: based on your comment for the other answer, you are looking for this: https://en.wikipedia.org/wiki/MD5#Algorithm
You can read about it here:
https://en.wikipedia.org/wiki/Md5sum
In general, the algorithm runs over the file and it's output is checksum, that means that if someone changes a bit in the file the checksum will be changed. so it's a way to validate that the file you looking at is the file you think you looking at and lowering the probability that someone put a melicius code in it
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I have got a repository where I store all my image files. I know that there are much images which are duplicated and I want to delete each one of duplicated ones.
I thought if I generate checksum for each image file and rename the file to its checksum, I can easily find out if there are duplicated ones by examining the filename. But the problem is that, I cannot be sure about selecting the checksum algorithm to use. For example, if I generate the checksums using MD5, can I exactly trust if the checksums are the same that means the files are exactly the same?
Judging from the response to a similar question in security forum (https://security.stackexchange.com/a/3145), the collision rate is about 1 collision per 2^64 messages. If your files are differenet and your collection is not huge (i.e. close to this number), md5 can be used safely.
Also, see response to a very similar question here: How many random elements before MD5 produces collisions?
The chances of getting the same checksum for 2 different files are extremely slim, but can never be absolutely guaranteed (Pigeonhole principle). An indication of how slim may be that GIT uses the SHA-1 checksum for software development source code including Linux and has never caused any known problems so I would say that you are safe. I would use SHA-1 instead of MD5 because it is slightly better if you are really paranoid.
To make really sure you best follow a two-step-procedure: first calculate a checksum for every file. If the checksums differ you're sure the files are not identical. If you happen to find some files with equal checksums there's no way around doing a bit-by-bit-comparison to make 100% sure if they are really identical. This holds regardless of the hashing-algorithm used.
What you'll get is a massive time-saving as doing bit-by-bit comparison for every possible pair of files will take forever and a day while comparing a hand full of possible candidates is fairly easy.
I received a MD5 hash and a Regular Expression which have the same plaintext..
How do I use the Regular Expression to crack the MD5 hash and find the text behind the MD5?
b89e49cab317f2681be60fb3d1c0f8f8
[(a|c|d)n-t\|]{8}
The idea would be to use the regex as a template and generate inputs that satisfy it.
You can search for a regex visualizer to see this, but what that one says is any of the characters ()acd| or any character between n and t (inclusive) in any order, repeated eight times. I tested this in hashcat, and the regex is correct despite it looking like it means something else. A shorter way to write that would be [acd|()n-t]{8}.
So you start generating 8 character strings with those values and taking the md5 of them. You can do this in almost any programming language but Python is a good choice. Look up the hashlib library, it has a function md5. You'll call the function hexdigest on that and compare it to the provided hash.
>>> import hashlib
>>> hashlib.md5(b'cybering').hexdigest()
'61e4feebe66ad22349e292d1462afd3a'
Additionally, if you want to use cracking software, look up JohnTheRipper or hashcat. You should be able to provide them a dictionary and have it attempt to break the hash. I was able to solve this with hashcat on my 980ti in ~5 seconds. This tutorial helped me set up the custom charset and mask to perform the attack.
Have fun!
One approach would be to generate all possible eight-character combinations (with repetition) of the 19 characters allowed by the regex. Test each combination by computing the md5 hash and comparing it to the one you were given.
That would be 13^8 = 815,730,721 possible combinations to check. The answer will likely be found before checking all of them.
I was able to whip out a little Node.js program on my laptop that found the solution in about 4 minutes (I split the problem up using workers to take advantage of multiple CPU cores).
Edit: I thought the regex had n-z instead of n-t so the search space was actually much smaller.
You cant crack the md5 hash value it has used one way hashing algorithm.
I'm writing a file system deduper. The first pass generates md5 checksums, and the second pass compares the files with identical checksums.
Is there a collection of strings which differ but generate identical md5 checksums I can incorporate into my test case collection?
Update: mjv's answer points to these two files, perfect for my test case.
http://www.win.tue.nl/~bdeweger/CollidingCertificates/MD5Collision.certificate1.cer
http://www.win.tue.nl/~bdeweger/CollidingCertificates/MD5Collision.certificate2.cer
You can find a couple of different X.509 certificate files with the same MD5 hash at this url.
I do not know of MD5 duplicate files repositories, but you can probably create your own, using the executables and/or the techniques described on Vlastimil Klima's page on MD5 Collision
Indeed MD5 has been know for its weakness with regards to collision resistance, however I wouldn't disqualify it for a project such as your file system de-duper; you may just want to add a couple of additional criteria (which can be very cheap, computationally speaking) to further decrease the possibility of duplicates.
Alternatively, for test purposes, you may simply modify your MD5 compare logic so that it deems some MD5 values identical even though they are not (say if the least significant byte of the MD5 matches, or systematically, every 20 comparisons, or at random ...). This may be less painful than having to manufacture effective MD5 "twins".
http://www.nsrl.nist.gov/ might be what you want.
I am basically creating an API in php, and one of the parameters that it will accept is an md5 encrypted value. I don't have much knowledge of different programming languages and also about the MD5. So my basic question is, if I am accepting md5 encrypted values, will the value remain same, generated from any programing language like .NET, Java, Perl, Ruby... etc.
Or there would be some limitation or validations for it.
Yes, correct implementation of md5 will produce the same result, otherwise md5 would not be useful as a checksum. The difference may come up with encoding and byte order. You must be sure that text is encoded to exactly the same sequence of bytes.
It will, but there's a but.
It will because it's spec'd to reliably produce the same result given a repeated series of bytes - the point being that we can then compare that results to check the bytes haven't changed, or perhaps only digitally sign the MD5 result rather than signing the entire source.
The but is that a common source of bugs is making assumptions about how strings are encoded. MD5 works on bytes, not characters, so if we're hashing a string, we're really hashing a particular encoding of that string. Some languages (and more so, some runtimes) favour particular encodings, and some programmers are used to making assumptions about that encoding. Worse yet, some spec's can make assumptions about encodings. This can be a cause of bugs where two different implementations will produce different MD5 hashes for the same string. This is especially so in cases where characters are outside of the range U+0020 to U+007F (and since U+007F is a control, that one has its own issues).
All this applies to other cryptographic hashes, such as the SHA- family of hashes.
Yes. MD5 isn't an encryption function, it's a hash function that uses a specific algorithm.
Yes, md5 hashes will always be the same regardless of their origin - as long as the underlying algorithm is correctly implemented.
A vital point of secure hash functions, such as MD5, is that they always produce the same value for the same input.
However, it does require you to encode the input data into a sequence of bytes (or bits) the same way. For instances, there are many ways to encode a string.
Are there algorithms for putting a digest into the file being digested?
In otherwords, are there algorithms or libraries, or is it even possible to have a hash/digest of a file contained in the file being hashed/digested. This would be handy for obvious reasons, such as built in digests of ISOs. I've tried googling things like "MD5 injection" and "digest in a file of a file." No luck (probably for good reason.)
Not sure if it is even mathematically possible. Seems you'd be able to roll through the file but then you'd have to brute the last bit (assuming the digest was the last thing in the file or object.)
Thanks,
Chenz
It is possible in a limited sense:
Non-cryptographically-secure hashes
You can do this with insecure hashes like the CRC family of checksums.
Maclean's gzip quine
Caspian Maclean created a gzip quine, which decompresses to itself. Since the Gzip format includes a CRC-32 checksum (see the spec here) of the uncompressed data, and the uncompressed data equals the file itself, this file contains its own hash. So it's possible, but Maclean doesn't specify the algorithm he used to generate it:
It's quite simple in theory, but the helper programs I used were on a hard disk that failed, and I haven't set up a new working linux system to run them on yet. Solving the checksum by hand in particular would be very tedious.
Cox's gzip, tar.gz, and ZIP quines
Russ Cox created 3 more quines in Gzip, tar.gz, and ZIP formats, and wrote up in detail how he created them in an excellent article. The article covers how he embedded the checksum: brute force—
The second obstacle is that zip archives (and gzip files) record a CRC32 checksum of the uncompressed data. Since the uncompressed data is the zip archive, the data being checksummed includes the checksum itself. So we need to find a value x such that writing x into the checksum field causes the file to checksum to x. Recursion strikes back.
The CRC32 checksum computation interprets the entire file as a big number and computes the remainder when you divide that number by a specific constant using a specific kind of division. We could go through the effort of setting up the appropriate equations and solving for x. But frankly, we've already solved one nasty recursive puzzle today, and enough is enough. There are only four billion possibilities for x: we can write a program to try each in turn, until it finds one that works.
He also provides the code that generated the files.
(See also Zip-file that contains nothing but itself?)
Cryptographically-secure digests
With a cryptographically-secure hash function, this shouldn't be possible without either breaking the hash function (particularly, a secure digest should make it "infeasible to generate a message that has a given hash"), or applying brute force.
But these hashes are much longer than 32 bits, precisely in order to deter that sort of attack. So you can write a brute-force algorithm to do this, but unless you're extremely lucky you shouldn't expect it to finish before the universe ends.
MD5 is broken, so it might be easier
The MD5 algorithm is seriously broken, and a chosen-prefix collision attack is already practical (as used in the Flame malware's forged certificate; see http://www.cwi.nl/news/2012/cwi-cryptanalist-discovers-new-cryptographic-attack-variant-in-flame-spy-malware, http://arstechnica.com/security/2012/06/flame-crypto-breakthrough/). I don't know of what you want having actually been done, but there's a good chance it's possible. It's probably an open research question.
For example, this could be done using a chosen-prefix preimage attack, choosing the prefix equal to the desired hash, so that the hash would be embedded in the file. A
preimage attack is more difficult than collision attacks, but there has been some progress towards it. See Does any published research indicate that preimage attacks on MD5 are imminent?.
It might also be possible to find a fixed point for MD5; inserting a digest is essentially the same problem. For discussion, see md5sum a file that contain the sum itself?.
Related questions:
Is there any x for which SHA1(x) equals x?
Is a hash result ever the same as the source value?
The only way to do this is if you define your file format so the hash only applies to the part of the file that doesn't contain the hash.
However, including the hash inside a file (like built into an ISO) defeats the whole security benefit of the hash. You need to get the hash from a different channel and compare it with your file.
No, because that would mean that the hash would have to be a hash of itself, which is not possible.