Can a cryptographic hash algorithm be used as a PRNG? - hash

Can MD5/SHA256/SHA512, etc., be used as a PRNG? E.g., given an integer seed, is the pseudo-code:
random_number = truncate_to_desired_range(
sha512( seed.toString() + ',' + i.toString() )
…a decent PRNG? (i is an increasing integer, e.g., the outputs are:
convert(sha512("<seed>,0"))
convert(sha512("<seed>,1"))
convert(sha512("<seed>,2"))
convert(sha512("<seed>,3"))
…
"Decent", in the context of this question, refers only to the distribution of the output: is the output of cryptographic hash functions uniform, when used this way? (Though I suppose it would depend on the hash function, all cryptographic hashes should also have uniform output, right?)
Note: I will concede that this is going to be a slow PRNG, compared to say Mersenne-Twister, due to the use of a cryptographic hash. I'm not interested in speed, and I'm not interested in the result being secure — just that the distribution is correct.
In my particular use case, I'm looking for something similar to XKCD's geohashing, in that it is easily implemented by distributed parties, who will all arrive at the same answer. Mersenne-Twister can be substituted, but it less available in many target languages. (Some languages lack it entirely, some lack access to the raw U32 output of it, etc. SHA512 is either built in, or easily available.)

Assuming the cryptographic hash function meets its design goals, the output will (provably) follow a uniform distribution over its period, as every input to the hash function is unique by design.
One of the goals of a hash function is to approximate a random oracle, that is, for any two distinct inputs A and B, the outputs H(A) and H(B) should (for a true random oracle) be uncorrelated. Hash functions get pretty close to that, but of course weaknesses creep in with time and cryptanalysis.
That said, cryptographic primitives are essentially the best mathematical algorithms we have available when it comes to quality, therefore it is safe to say that if they cannot solve your problem, nothing will.

It can be made to work (with good sized inputs, padding, etc. as mentioned in other answers/comments) and will provide reasonably good results, but it's going to be slow, so don't do that if you are doing simulations or something that require heavy PRNG use...

Related

improve hashing using genetic programming/algorithm

I'm writing a program which can significantly lessen the number of collisions that occur while using hash functions like 'key mod table_size'. For this I would like to use Genetic Programming/Algorithm. But I don't know much about it. Even after reading many articles and examples I don't know that in my case (as in program definition) what would be the fitness function, target (target is usually the required result), what would pose as the population/individuals and parents, etc.
Please help me in identifying the above and with a few codes/pseudo-codes snippets if possible as this is my project.
Its not necessary to be using genetic programming/algorithm, it can be anything using evolutionary programming/algorithm.
thanks..
My advice would be: don't do this that way. The literature on hash functions is vast and we more or less understand what makes a good hash function. We know enough mathematics not to look for them blindly.
If you need a hash function to use, there is plenty to choose from.
However, if this is your uni project and you cannot possibly change the subject or steer it in a more manageable direction, then as you noticed there will be complex issues of getting fitness function and mutation operators right. As far as I can tell off the top of my head, there are no obvious candidates.
You may look up e.g. 'strict avalanche criterion' and try to see if you can reason about it in terms of fitness and mutations.
Another question is how do you want to represent your function? Just a boolean expression? Something built from word operations like AND, XOR, NOT, ROT ?
Depending on your constraints (or rather, assumptions) the question of fitness and mutation will be different.
Broadly fitness is clearly minimize the number of collisions in your 'hash modulo table-size' model.
The obvious part is to take a suitably large and (very important) representative distribution of keys and chuck them through your 'candidate' function.
Then you might pass them through 'hash modulo table-size' for one or more values of table-size and evaluate some measure of 'niceness' of the arising distribution(s).
So what that boils down to is what table-sizes to try and what niceness measure to apply.
Niceness is context dependent.
You might measure 'fullest bucket' as a measure of 'worst case' insert/find time.
You might measure sum of squares of bucket sizes as a measure of 'average' insert/find time based on uniform distribution of amongst the keys look-up.
Finally you would need to decide what table-size (or sizes) to test at.
Conventional wisdom often uses primes because hash modulo prime tends to be nicely volatile to all the bits in hash where as something like hash modulo 2^n only involves the lower n-1 bits.
To keep computation down you might consider the series of next prime larger than each power of two. 5(>2^2) 11 (>2^3), 17 (>2^4) , etc. up to and including the first power of 2 greater than your 'sample' size.
There are other ways of considering fitness but without a practical application the question is (of course) ill-defined.
If your 'space' of potential hash functions don't all have the same execution time you should also factor in 'cost'.
It's fairly easy to define very good hash functions but execution time can be a significant factor.

How do I truncate a 64-bit hash into a 32-bit hash? [duplicate]

We're trying to settle an internal debate on our dev team:
We're looking for a 64-bit PHP hash function. We found a PHP implementation of MurmurHash3, but MurmurHash3 is either 32-bit or 128-bit, not 64-bit.
Co-worker #1 believes that to produce a 64-bit hash from MurmurHash3, we can simply slice the first (or last, or any) 64 bits of the 128-bit hash and that it will be as collision-proof as a native 64-bit hash function.
Co-worker #2 believes that we must find a native 64-bit hash function to reduce collisions and that 64-bit slices of a 128-bit hash will not be as collision proof as a native 64-bit hash.
Who's correct?
Does the answer change if we take the first (or last, or any) 64-bits of a cryptographic hash like SHA1 instead of Murmur3?
If you had real random, uniformly distributed values, then "slicing" would yield exactly the same results as if you had started with the smaller value right from the start. To see why, consider this very simple example: Let's say your random generator outputs 3 random bits, but you only need one random bit to work with. Let's assume the output is
b1 b2 b3
The possible values are
000, 001, 010, 011, 100, 101, 110, 111
and all are to occur with equal probability of 1/8. Now whatever bit you slice from those three for your purpose - the first, second or third - the probability of having a '1' is always going to be 1/2, regardless of the position - and the same is true for a '0'.
You can easily scale this experiment to the 64 out of 128 bit case: regardless of which bits you slice, the probability of ending up with a one or a zero in a certain position is going to be one half. What this means is that if you had a sample taken from a uniformly distributed random variable, then slicing wouldn't make the probability for collisions more or less likely.
Now a good question is whether random functions are really the best we can do to prevent collisions. But as it turns out, it can be shown that the probability of finding collisions increases whenever a function deviates from random.
Cryptographic hash functions: co-worker #1 wins
The problem in real life is that hash functions are not random at all, on the contrary, they are boringly deterministic. But a design goal of cryptographic hash functions is as follows: if we didn't know their initial state, then their output would be computationally indistinguishable from a real random function, that is there's no computationally efficient way to tell the difference between the hash output and real random values. This is why you'd consider a hash already as kind of broken if you can find a "distinguisher", a method to tell the hash from real random values with a probability higher than one half. Unfortunately, we can't really prove these properties for existing cryptographic hashes, but unless somebody breaks them, we may assume these properties hold with some confidence. Here is an example of a paper about a distinguisher for one of the SHA-3 submissions that illustrates the process.
To summarize, unless a distinguisher is found for a given cryptographic hash, slicing is perfectly fine and does not increase the probability of a collision.
Non-cryptographic hash functions: co-worker #2 might win
Non-cryptographic hashes do not have to satisfy the same set of requirements as cryptographic hashes do. They are usually defined to be very fast and satisfy certain properties "under sane/benevolent conditions", but they might easily fall short if somebody tries to maliciously manipulate them. A good example for what this means in practice is the computational complexity attack on hash table implementations (hashDoS) presented earlier this year. Under normal conditions, non-crypto hashes work perfectly fine, but their collision resistance may be severely undermined by some clever inputs. This can't happen with cryptographic hash functions, because their very definition requires them to be immune to all sorts of clever inputs.
Because it is possible, sometimes even quite easy, to find a distinguisher like above for the output of non-cryptographic hashes, we can immediately say that they do not qualify as cryptographic hash functions. Being able to tell the difference means that somewhere there is a pattern or bias in the output.
And this fact alone implies that they deviate more or less from a random function, and thus (after what we said above) collisions are probably more likely than they would be for random functions. Finally, since collisions occur with higher probability for the full 128 bits already, this will not get better with shorter ouptputs, collisions will probably be even more likely in that case.
tl;dr You're safe with a cryptographic hash function when truncating it. But you're better off with a "native" 64 bit cryptographic hash function compared to truncating a non-cryptographic hash with a larger output to 64 bits.
Due to the avalanche effect, a strong hash is one where a single bit of change in the source results in half the bits of the hash flipping on average. For a good hash, then, the "hashness" is evenly distributed, and so each section or slice is affected by an equal and evenly distributed amount of source bits, and therefore is just as strong as any other slice of the same bit length could be.
I would agree with co-worker 1 as long as the hash has good properties and even distribution.
This question seems incomplete without this being mentioned:
Some hashes are provably perfect hashes for a specific class of inputs (eg., for input of length n for some reasonable value of n). If you truncate that hash then you are likely to destroy that property, in which case you are, by definition, increasing the rate of collisions from zero to non-zero and you have weakened the hash in that use case.
It's not the general case, but it's an example of a legitimate concern when truncating hashes.

Hashing Similarity

Normally, the goal of hashing is to turn a continuous function into a discrete one: a small change in the input should cause a large change in the output. However, is there any hashing algorithm that will, (very) roughly speaking, return similar but (still different) hashes for similar inputs?
(An example of the use of this would be to check whether two files are "similar" by checking their hashes for similarity. Of course, some failure is always acceptable.)
Look at Locality Sensitive Hashing (LSH). That is a probabilistic way of quickly finding a bunch of points near a given one, for example.
Given a distance function that tells you how similar or different are your objects, you can also employ distance permutations:
http://www.computer.org/portal/web/csdl/doi/10.1109/TPAMI.2007.70815
or sketches:
http://portal.acm.org/citation.cfm?id=1638180
For an implementation of the latter approach:
http://obsearch.net
You really don't want to see similar hashes. Hashing is to insure Integrity, therefore the slightest change in your file/app/program will produce an entirely different hash. If two different strings show the same hash, this is called a collision, and the hashing algorithm is now compromised. MD5 has some collisions but is still used today.

How are hash functions like MD5 unique?

I'm aware that MD5 has had some collisions but this is more of a high-level question about hashing functions.
If MD5 hashes any arbitrary string into a 32-digit hex value, then according to the Pigeonhole Principle surely this can not be unique, as there are more unique arbitrary strings than there are unique 32-digit hex values.
You're correct that it cannot guarantee uniqueness, however there are approximately 3.402823669209387e+38 different values in a 32 digit hex value (16^32). That means that, assuming the math behind the algorithm gives a good distribution, your odds are phenomenally small that there will be a duplicate. You do have to keep in mind that it IS possible to duplicate when you're thinking about how it will be used. MD5 is generally used to determine if something has been changed (I.e. it's a checksum). It would be ridiculously unlikely that something could be modified and result in the same MD5 checksum.
Edit: (given recent news re: SHA1 hashes)
The answer above, still holds, but you shouldn't expect an MD5 hash to serve as any kind of security check against manipulation. SHA-1 Hashes as 2^32 (over 4 billion) times less likely to collide, and it has been demonstrated that it is possible to contrive an input to produce the same value. (This was demonstrated against MD5 quite some time ago). If you're looking to ensure nobody has maliciously modified something to produce the same hash value, these days, you need at SHA-2 to have a solid guarantee.
On the other hand, if it's not in a security check context, MD5 still has it's usefulness.
The argument could be made that an SHA-2 hash is cheap enough to compute, that you should just use it anyway.
You are absolutely correct. But hashes are not about "unique", they are about "unique enough".
As others have pointed out, the goal of a hash function like MD5 is to provide a way of easily checking whether two objects are equivalent, without knowing what they originally were (passwords) or comparing them in their entirety (big files).
Say you have an object O and its hash hO. You obtain another object P and wish to check whether it is equal to O. This could be a password, or a file you downloaded (in which case you won't have O but rather the hash of it hO that came with P, most likely). First, you hash P to get hP.
There are now 2 possibilities:
hO and hP are different. This must mean that O and P are different, because using the same hash on 2 values/objects must yield the same value. Hashes are deterministic. There are no false negatives.
hO and hP are equal. As you stated, because of the Pigeonhole Principle this could mean that different objects hashed to the same value, and further action may need to be taken.
a. Because the number of possibilities is so high, if you have faith in your hash function it may be enough to say "Well there was a 1 in 2128 chance of collision (ideal case), so we can assume O = P. This may work for passwords if you restrict the length and complexity of characters, for example. It is why you see hashes of passwords stored in databases rather than the passwords themselves.
b. You may decide that just because the hash came out equal doesn't mean the objects are equal, and do a direct comparison of O and P. You may have a false positive.
So while you may have false positive matches, you won't have false negatives. Depending on your application, and whether you expect the objects to always be equal or always be different, hashing may be a superfluous step.
Cryptographic one-way hash functions are, by nature of definition, not Injective.
In terms of hash functions, "unique" is pretty meaningless. These functions are measured by other attributes, which affects their strength by making it hard to create a pre-image of a given hash. For example, we may care about how many image bits are affected by changing a single bit in the pre-image. We may care about how hard it is to conduct a brute force attack (finding a prie-image for a given hash image). We may care about how hard it is to find a collision: finding two pre-images that have the same hash image, to be used in a birthday attack.
While it is likely that you get collisions if the values to be hashed are much longer than the resulting hash, the number of collisions is still sufficiently low for most purposes (there are 2128 possible hashes total so the chance of two random strings producing the same hash is theoretically close to 1 in 1038).
MD5 was primarily created to do integrity checks, so it is very sensitive to minimal changes. A minor modification in the input will result in a drastically different output. This is why it is hard to guess a password based on the hash value alone.
While the hash itself is not reversible, it is still possible to find a possible input value by pure brute force. This is why you should always make sure to add a salt if you are using MD5 to store password hashes: if you include a salt in the input string, a matching input string has to include exactly the same salt in order to result in the same output string because otherwise the raw input string that matches the output will fail to match after the automated salting (i.e. you can't just "reverse" the MD5 and use it to log in because the reversed MD5 hash will most likely not be the salted string that originally resulted in the creation of the hash).
So hashes are not unique, but the authentication mechanism can be made to make it sufficiently unique (which is one somewhat plausible argument for password restrictions in lieu of salting: the set of strings that results in the same hash will probably contain many strings that do not obey the password restrictions, so it's more difficult to reverse the hash by brute force -- obviously salts are still a good idea nevertheless).
Bigger hashes mean a larger set of possible hashes for the same input set, so a lower chance of overlap, but until processing power advances sufficiently to make brute-forcing MD5 trivial, it's still a decent choice for most purposes.
(It seems to be Hash Function Sunday.)
Cryptographic hash functions are designed to have very, very, very, low duplication rates. For the obvious reason you state, the rate can never be zero.
The Wikipedia page is informative.
As Mike (and basically every one else) said, its not perfect, but it does the job, and collision performance really depends on the algo (which is actually pretty good).
What is of real interest is automatic manipulation of files or data to keep the same hash with different data, see this Demo
As others have answered, hash functions are by definition not guaranteed to return unique values, since there are a fixed number of hashes for an infinite number of inputs. Their key quality is that their collisions are unpredictable.
In other words, they're not easily reversible -- so while there may be many distinct inputs that will produce the same hash result (a "collision"), finding any two of them is computationally infeasible.

What are the important points about cryptographic hash functions?

I was reading this question on MD5 hash values and the accepted answer confuses me. One of the main properties, as I understand it, of a cryptopgraphic hash function is that it is infeasible to find two different messages (inputs) with the same hash value.
Yet the consensus answer to the question Why aren't MD5 hash values reversible? is Because an infinite number of input strings will generate the same output. This seems completely contradictory to me.
Also, what perplexes me somewhat is the fact that the algorithms are public, yet the hash values are still irreversible. Is this because there is always data loss in a hash function so there's no way to tell which data was thrown away?
What happens when the input data size is smaller than the fixed output data size (e.g., hashing a password "abc")?
EDIT:
OK, let me see if I have this straight:
It is really, really hard to infer the input from the hash because there are an infinite amount of input strings that will generate the same output (irreversible property).
However, finding even a single instance of multiple input strings that generate the same output is also really, really hard (collision resistant property).
Warning: Long answer
I think all of these answers are missing a very important property of cryptographic hash functions: Not only is it impossible to compute the original message that was hashed to get a given hash, it's impossible to compute any message that would hash to a given hash value. This is called preimage resistance.
(By "impossible" - I mean that no one knows how to do it in less time than it takes to guess every possible message until you guess the one that was hashed into your hash.)
(Despite popular belief in the insecurity of MD5, MD5 is still preimage resistant. Anyone who doesn't believe me is free to give me anything that hashes to 2aaddf751bff2121cc51dc709e866f19. What MD5 doesn't have is collision resistance, which is something else entirely.)
Now, if the only reason you can't "work backwards" in a cryptographic hash function was because the hash function discards data to create the hash, then it would not guarantee preimage resistance: You can still "work backwards", and just insert random data wherever the hash function discards data, and while you wouldn't come up with the original message, you'd still come up with a message that hashes to the desired hash value. But you can't.
So the question becomes: Why not? (Or, in other words, how do you make a function preimage resistant?)
The answer is that cryptographic hash functions simulate chaotic systems. They take your message, break it into blocks, mix those blocks around, have some of the blocks interact with each other, mix those blocks around, and repeat that a lot of times (well, one cryptographic hash function does that; others have their own methods). Since the blocks interact with each other, block C not only has to interact with block D to produce block A, but it has to interact with block E to produce block B. Now, sure, you can find values of blocks C, D, E that would produce the blocks A and B in your hash value, but as you go further back, suddenly you need a block F that interacts with C to make D, and with E to make B, and no such block can do both at the same time! You must have guessed wrong values for C, D, and E.
While not all cryptographic hash functions are exactly as described above with block interaction, they have the same idea: That if you try to "work backwards", you're going to end up with a whole lot of dead ends, and the time it takes for you to try enough values to generate a preimage is on the order of hundreds to millions of years (depending on the hash function), not much better than the time it would take just to try messages until you find one that works.
1: The primary purpose of a hash is to map a very, very large space to a smaller but still very large space (e.g., MD5, which will take 'anything' and convert it into a space of size 2^128 -- big, but not nearly as big as aleph-0.)
In addition to other features, good hashes fill the destination space homogeneously. Bad hashes fill the space in a clumpy fashion, coming up with the same hash for many common inputs.
Imagine the idiotic hash function sum(), which just adds all the digits of the input number: it succeeds in mapping down, but there are a bunch of collisions (inputs with the same output, like 3 and 12 and 21) at the low end of the output space and the upper end of the space is nearly empty. As a result it makes very poor use of the space, is easy to crack, etc.
So a good hash that makes even use of the destination space will make it difficult to find two inputs with the same output, just by the odds: if MD5 were perfect, the odds that two inputs would have the same output would be 2^-128. That's pretty decent odds: the best you can do without resorting to a larger output space. (In truth MD5 isn't perfect, which is one of the things that makes it vulnerable.)
But it will still be true that a huge number of inputs will map to any given hash, because the input space is 'infinite', and dividing infinity by 2^128 still gives you infinity.
2: Yes, hashes always cause data loss, except in the case where your output space is the same as, or larger than, your input space -- and in that case you probably didn't need to hash!
3: For smaller inputs, best practice is to salt the input. Actually, that's good practice for any cryptographic hashing, because otherwise an attacker can feed you specific inputs and try to figure out which hash you are using. 'Salt' is just a set of additional information that you append (or prepend) to your input; you then hash the result.
edit: In cryptography, it is also important that the hash function is resistant to preimage attacks, intuitively, that is hard to guess the input for a given output even knowing many other input/output pairs. The "sum" function could probably be guessed rather easily (but since it destroys data still might not be easy to reverse).
You may be confused, because the answer to the question you cite is confusing.
One of the requirements for a cryptographic hash function is that it should be preimage resistant. That is, if you know MD5(x) but not the message x, then it is difficult to find any x' (either equal x or different from x) such that MD5(x') = MD5(x).
Being preimage resistant is a different property than being reversible. A function is reversible if given y = f(x) there is exactly one x which fits (whether this is easy or not). For example define f(x) = x mod 10.
Then f is not reversible. From f(x) = 7 you can't determine whether x was 17, 27 or something else. But f is not preimage resistant, since values x' such that f(x) = 7 are easy to find. x' = 17, 27, 12341237 etc all work.
When doing crypto you usually need functions that are preimage resistant (and other properties such as collision resistance), not just something that is not reversible.
These are the properties of hash functions in general.
A word of caution though, MD5 shouldn't be used anymore because of vulnerabilities that have been found in it. Check the 'Vulnerabilities' section and external links detailing these attacks. http://en.wikipedia.org/wiki/Md5 You can make an MD5 collision by changing only 128 bits in a message.
SHA-1 is safe for simple hashing although there are some attacks that would make it weaker against well-funded entities (Governments, large corporations)
SHA-256 is a safe starting point against technology for the next couple decades.
Yet the consensus answer to the question "why aren't MD5 hash values reversible?" is because "an infinite number of input strings will generate the same output."
This is true for any hash function, but it is not the essence of a cryptographic hash function.
For short input strings such as passwords it is theoretically possible to reverse a cryptographic hash function, but it ought to be computationally infeasible. I.e. your computation would run too long to be useful.
The reason for this infeasibility is that the input is so thoroughly "mixed together" in the hash value that it becomes impossible to disentangle it with any less effort than the brute force attack of computing the hash value for all inputs
"why aren't MD5 hash values reversible?" is because "an infinite number of input strings >will generate the same output"
this is the reason that it isn't possible to reverse the hash function (get the same input).
cryptographic hash functions are collision resistant, that means that it's also hard to find another input value that maps to the same output (if your hash function was mod 2 : 134 mod 2 = 0; now you can't get the 134 back from the result, but we can stil find number 2 with the same output value (134 and 2 collide)).
When the input is smaller than the block size, padding is used to fit it to the block size.