I have a hash value and I want to convert it into string formate but I do not know how to do that.
Here is the hash value
7616db6c232292d2e56a2de9da49ea810d5bb80d53c10e7b07d9521dc88b3177
Hashing technique is a one way algorithm. So that is not possible, sorry.
You can't. A hash is a one-way function. Plus, two strings could theoretically generate the same hash.
BUT. If you had a large enough, pre-computed list of as many hashes as you could generate for each unique string, you could potentially convert your hash back into a string. Although it would require thousands of GB in entries and a lot of time to compute each and every hash. So it practically isn't possible.
Related
I am writing a Huffman Coding/Decoding algorithm and I am running into the problem that the storing the Huffman tree is taking up way to much room. Currently, I am converting the tree into a hashMap as such -> hashMap<Character(s),Huffman Code> and then storing that hash map. The issue is that, while the string is compressed great, adding the Huffman Tree data stored in the hash map is adding so much overhead that it's actually ending up bigger than the original. Currently I am just naively writing [data, value] pairs to the file, but I imagine there must be some sort of trickier way to do that. Any ideas?
You do not need the tree in order to encode. All you need is the bit lengths for each symbol and a way to order the symbols. See Canonical Huffman Code.
In fact, all you need is the symbols that are coded ordered by bit length, and within bit length sorted by symbol, and then the number of codes of each length. With just those two things you can encode.
This is hash type MD5 '9931BF135E464FE91E444DF4E046006A' but I can't change it to string is there any website that can do that.
There are several reasons why what you're asking is not possible.
First, the process by which the MD5 hash code is created loses information. MD5 is a 128-bit hash code. So if you take the hash code of anything larger than 128 bits (that's 16 bytes), by definition you're going to lose information.
Related: there is an infinite number of possible strings. By the pigeonhole principle, there is more than one string that will hash to any particular MD5 value.
I'm currently using NSCoding to serialize a tree of objects, but 1 of them contains as data member a native C float array with 1,000,000 entries, so in order to serialize it using encodeFloat:forKey: for each array entry, I need to apply 1,000,000 useless keys , that might be very slow. what the prefered way to handle this?
for each array entry, I need to apply 1,000,000 useless keys
No, you definitely do not need separate keys for each element. A C array is a contiguous block of memory, so you can simply create a NSData object from that block and store that as Hot Licks suggested. Or, since a million floats will require a fair bit of storage, you might compress the data before storing it. And in fact, you don't really even need NSData -- you can encode a range of bytes directly with -encodeBytes:length:forKey:.
Just wondering if knowing the original string length means that you can better unlash a SHA1 encryption.
No, not in the general case: a hash function is not an encryption function and it is not designed to be reversible.
It is usually impossible to recover the original hash for certain. This is because the domain size of a hash function is larger than the range of the function. For SHA-1 the domain is unbounded but the range is 160bits.
That means that, by the Pigeonhole principle, multiple values in the domain map to the same value in the range. When such two values map to the same hash, it is called a hash collision.
However, for a specific limited set of inputs (where the domain of the inputs is much smaller than the range of the hash function), then if a hash collision is found, such as through an brute force search, it may be "acceptable" to assume that the input causing the hash was the original value. The above process is effectively a preimage attack. Note that this approach very quickly becomes infeasible, as demonstrated at the bottom. (There are likely some nice math formulas that can define "acceptable" in terms of chance of collision for a given domain size, but I am not this savvy.)
The only way to know that this was the only input that mapped to the hash, however, would be to perform an exhaustive search over all the values in the range -- such as all strings with the given length -- and ensure that it was the only such input that resulted in the given hash value.
Do note, however, that in no case is the hash process "reversed". Even without the Pigeon hole principle in effect, SHA-1 and other cryptographic hash functions are especially designed to be infeasible to reverse -- that is, they are "one way" hash functions. There are some advanced techniques which can be used to reduce the range of various hashes; these are best left to Ph.D's or people who specialize in cryptography analysis :-)
Happy coding.
For fun, try creating a brute-force preimage attack on a string of 3 characters. Assuming only English letters (A-Z, a-z) and numbers (0-9) are allowed, there are "only" 623 (238,328) combinations in this case. Then try on a string of 4 characters (624 = 14,776,336 combinations) ... 5 characters (625 = 916,132,832 combinations) ... 6 characters (626 = 56,800,235,584 combinations) ...
Note how much larger the domain is for each additional character: this approach quickly becomes impractical (or "infeasible") and the hash function wins :-)
One way password crackers speed up preimage attacks is to use rainbow tables (which may only cover a small set of all values in the domain they are designed to attack), which is why passwords that use hashing (SHA-1 or otherwise) should always have a large random salt as well.
Hash functions are one-way function. For a given size there are many strings that may have produced that hash.
Now, if you know that the input size is fixed an small enough, let's say 10 bytes, and you know that each byte can have only certain values (for example ASCII's A-Za-z0-9), then you can use that information to precompute all the possible hashes and find which plain text produces the hash you have. This technique is the basis for Rainbow tables.
If this was possible , SHA1 would not be that secure now. Is it ? So no you cannot unless you have considerable computing power [2^80 operations]. In which case you don't need to know the length either.
One of the basic property of a good Cryptographic hash function of which SHA1 happens to be one is
it is infeasible to generate a message that has a given hash
Theoretically, let's say the string was also known to be solely of ASCII characters, and it's of size n.
There are 95 characters in ASCII not including controls. We'll assume controls weren't used.
There are 95ⁿ possible such strings.
There are 1.461501×10⁴⁸ possible SHA-1 values (give or take) and a just n=25, there are 2.7739×10⁴⁹ possible ASCII-only strings without controls in them, which would mean guaranteed collisions (some such strings have the same SHA-1).
So, we only need to get to n=25 when this becomes impossible even with infinite resources and time.
And remember, up until now I've been making it deliberately easy with my ASCII-only rule. Real-world modern text doesn't follow that.
Of course, only a subset of such strings would be anything likely to be real (if one says "hello my name is Jon" and the other says "fsdfw09r12esaf" then it was probably the first). Stil though, up until now I was assuming infinite time and computing power. If we want to work it out sometime before the universe ends, we can't assume that.
Of course, the nature of the attack is also important. In some cases I want to find the original text, while in others I'll be happy with gibberish with the same hash (if I can input it into a system expecting a password).
Really though, the answer is no.
I posted this as an answer to another question, but I think it is applicable here:
SHA1 is a hashing algorithm. Hashing is one-way, which means that you can't recover the input from the output.
This picture demonstrates what hashing is, somewhat:
As you can see, both John Smith and Sandra Dee are mapped to 02. This means that you can't recover which name was hashed given only 02.
Hashing is used basically due to this principle:
If hash(A) == hash(B), then there's a really good chance that A == B. Hashing maps large data sets (like a whole database) to a tiny output, like a 10-character string. If you move the database and the hash of both the input and the output are the same, then you can be pretty sure that the database is intact. It's much faster than comparing both databases byte-by-byte.
That can be seen in the image. The long names are mapped to 2-digit numbers.
To adapt to your question, if you use bruteforce search, for a string of a given length (say length l) you will have to hash through (dictionary size)^l different hashes.
If the dictionary consists of only alphanumeric case-sensitive characters, then you have (10 + 26 + 26)^l = 62^l hashes to hash. I'm not sure how many FLOPS are required to produce one hash (as it is dependent on the hash's length). Let's be super-unrealistic and say it takes 10 FLOP to perform one hash.
For a 12-character password, that's 62^12 ~ 10^21. That's 10,000 seconds of computations on the fastest supercomputer to date.
Multiply that by a few thousand and you'll see that it is unfeasible if I increase my dictionary size a little bit or make my password longer.
I have a user model on my app, and my password field uses sha1. What i want is to, when i get the sha1 from the DB, to make it a string again. How do i do that?
You can't - SHA1 is a one-way hash. Given the output of SHA1(X), is not possible to retrieve X (at least, not without a brute force search or dictionary/rainbow table scan)
A very simple way of thinking about this is to imagine I give you a set of three-digit numbers to add up, and you tell me the final two digits of that sum. It's not possible from those two digits for me to work out exactly which numbers you started out with.
See also
Is it possible to reverse a sha1?
Decode sha1 string to normal string
Thought relating MD5, these other questions may also enlighten you:
Reversing an MD5 Hash
How can it be impossible to “decrypt” an MD5 hash?
You can't -- that's the point of SHA1, MDB5, etc. Most of those are one-way hashes for security. If it could be reversed, then anyone who gained access to your database could get all of the passwords. That would be bad.
Instead of dehashing your database, instead hash the password attempt and compare that to the hashed value in the database.
If you're talking about this from a practical viewpoint, just give up now and consider it impossible. Finding the original string is impossible (except by accident). Most of the point of a cryptographically secure hash is to ensure you can't find any other string that produces the same hash either.
If you're interested in research into secure hash algorithms: finding a string that will produce a given hash is called a "preimage". If you can manage to do so (with reasonable computational complexity) for SHA-1 you'll probably become reasonably famous among cryptanalysis researchers. The best "break" against SHA-1 that's currently known is a way to find two input strings that produce the same hash, but 1) it's computationally quite expensive (think in terms of a number of machines running 24/7 for months at a time to find one such pair), and does not work for an arbitrary hash value -- it finds one of a special class of input strings for which a matching pair is (relatively) easy to find.
SHA is a hashing algorithm. You can compare the hash of a user-supplied input with the stored hash, but you can't easily reverse the process (rebuild the original string from the stored hash).
Unless you choose to brute-force or use rainbow tables (both extremely slow when provided with a sufficiently long input).
You can't do that with SHA-1. But, given what you need to do, you can try using AES instead. AES allows encryption and decryption.