I have a hash from my school work that I need to decrypt.
The hash is: 68728d8fa7977d2567c6363381eda037.
It looks like it uses either MD4 or MD5 hashing algorithm
How to decrypt it?
You "decrypt" a hash by making a lookup table. In your case, if you know that the password is exactly 9 digits long, you'd create a hash for every possible 9-digit password and compare each one to the hash you're trying to decrypt.
By definitions hash is one way encryption so you can't decrypt it. You can use kind of tables with all possible combinations. Which is CPU consuming process. Here is one simple code on meta language:
for i from 0 to 999999999
h=md5(i)
write(file,h)
endoffor
line=search(file,inputhash)
print(line)
Related
I have data I need to hash.
I have list of numbers to compare for hashed numbers.
as far as someone could tell me, the data was hashed with SHA_256
I have only one example for input and output and I need to find out the hashing logic in SQL Server.
This is the original number: 02229747
And this is the hash number: 4ad54f5b376038f49a44d411e6d551ae4c8dd147c8605a7eec32ba850080b326
I have tried using the following but I can't manage to get the same result.
declare #number bigint = 022529747
DECLARE #HashId varbinary(50) = HashBytes('SHA2_256', cast(#number as varbinary(50)))
select #HashId
Hashbytes can take a number of different algorithms as input. Try each one:
select
a = hashbytes('MD2', '022529747'),
b = hashbytes('MD4', '022529747'),
c = hashbytes('MD5', '022529747'),
d = hashbytes('SHA', '022529747'),
e = hashbytes('SHA1', '022529747'),
f = hashbytes('SHA2_256', '022529747'),
g = hashbytes('SHA2_512', '022529747')
Column f returns the value you are looking for, so the algorithm used was SHA2_256. Note that I am putting the data in as a string (varchar), not an integer (bigint).. The bytes which represent 022529747 as a varchar are very different from the bytes which represent 022529747 as a bigint.
Background:
Hashing and encryption are different.
SHA stands for "secure hashing algorithm". It takes some input, and produces an output hash. If the input changes, the hash changes (with the limit of the birthday problem. But you can't go backwards. You can't take the output hash, and turn it back into the input data. The best you can do is try every different possible input, and see if that input generates the hash. See this 3Blue1Brown video for an illustrative explanation.
SHA is a family of cryptographic hash functions, but don't let the name fool you. "Cryptographic" doesn't mean the same thing as "encryption". It just means that it's "hard to guess" what the input data might be based on the output, because the output appears random. See This thread for the difference between a hash function and a cryptographic hash function
AES stands for "advanced encryption standard". This is a symmetric key encryption. Data encrypted with AES can be decrypted back to the original input. The "symmetric" part means one key is used to both encrypt, and decrypt (compared to, e.g., PGP encryption, which uses different keys to encrypt and decrypt).
The SQL hashbytes function can use a number of different algorithms, but none of the are reversible. They are all hashing algorithms, not encryption algorithms.
If you need encryption and decryption in code, the correct SQL functionality to use is EncryptByKey and DecryptByKey
I am faced with the need of deriving a single ID from N IDs and at first a i had a complex table in my database with FirstID, SecondID, and a varbinary(MAX) with remaining IDs, and while this technically works its painful, slow, and centralized so i came up with this:
simple version in C#:
Guid idA = Guid.NewGuid();
Guid idB = Guid.NewGuid();
byte[] data = new byte[32];
idA.ToByteArray().CopyTo(data, 0);
idB.ToByteArray().CopyTo(data, 16);
byte[] hash = MD5.Create().ComputeHash(data);
Guid newID = new Guid(hash);
now a proper version will sort the IDs and support more than two, and probably reuse the MD5 object, but this should be faster to understand.
Now security is not a factor in this, none of the IDs are secret, just saying this 'cause everyone i talk to react badly when you say MD5, and MD5 is particularly useful for this as it outputs 128 bits and thus can be converted directly to a new Guid.
now it seems to me that this should be just dandy, while i may increase the odds of a collision of Guids it still seems like i could do this till the sun burns out and be no where near running into a practical issue.
However i have no clue how MD5 is actually implemented and may have overlooked something significant, so my question is this: is there any reason this should cause problems? (assume sub trillion records and ideally the output IDs should be just as global/universal as the other IDs)
My first thought is that you would not be generating a true UUID. You would end up with an arbitrary set of 128-bits. But a UUID is not an arbitrary set of bits. See the 'M' and 'N' callouts in the Wikipedia page. I don't know if this is a concern in practice or not. Perhaps you could manipulate a few bits (the 13th and 17th hex digits) inside your MD5 output to transform the hash outbut to a true UUID, as mentioned in this description of Version 4 UUIDs.
Another issue… MD5 does not do a great job of distributing generated values across the range of possible outputs. In other words, some possible values are more likely to be generated more often than other values. Or as the Wikipedia article puts it, MD5 is not collision resistant.
Nevertheless, as you pointed out, probably the chance of a collision is unrealistic.
I might be tempted to try to increase the entropy by repeating your combined value to create a much longer input to the MD5 function. In your example code, take that 32 octet value and use it repeatedly to create a value 10 or 1,000 times longer (320 octects, 32,000 or whatever).
In other words, if working with hex strings for my own convenience here instead of the octets of your example, given these two UUIDs:
78BC2A6B-4F03-48D0-BB74-051A6A75CCA1
FCF1B8E4-5548-4C43-995A-8DA2555459C8
…instead of feeding this to the MD5 function:
78BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C8
…feed this:
78BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C878BC2A6B-4F03-48D0-BB74-051A6A75CCA1FCF1B8E4-5548-4C43-995A-8DA2555459C8
…or something repeated even longer.
In our app we're going to be handed png images along with a ~200 character byte array. I want to save the image with a filename corresponding to that bytearray, but not the bytearray itself, as i don't want 200 character filenames. So, what i thought was that i would save the bytearray into the database, and then MD5 it to get a short filename. When it comes time to display a particular image, i look up its bytearray, MD5 it, then look for that file.
So far so good. The problem is that potentially two different bytearrays could hash down to the same MD5. Then, one file would effectively overwrite another. Or could they? I guess my questions are
Could two ~200 char bytearrays MD5-hash down to the same string?
If they could, is it a once-per-10-ages-of-the-universe sort of deal or something that could conceivably happen in my app?
Is there a hashing algorithm that will produce a (say) 32 char string that's guaranteed to be unique?
It's logically impossible to get a 32 byte code from a 200 byte source which is unique among all possible 200 byte sources, since you can store more information in 200 bytes than in 32 bytes.
They only exception would be that the information stored in these 200 bytes would also fit into 32 bytes, in which case your source date format would be extremely inefficient and space-wasting.
When hashing (as opposed to encrypting), you're reducing the information space of the data being hashed, so there's always a chance of a collision.
The best you can hope for in a hash function is that all hashes are evenly distributed in the hash space and your hash output is large enough to provide your "once-per-10-ages-of-the-universe sort of deal" as you put it!
So whether a hash is "good enough" for you depends on the consequences of a collision. You could always add a unique id to a checksum/hash to get the best of both worlds.
Why don't you use a unique ID from your database?
The probability of two hashes will likely to collide depends on the hash size. MD5 produces 128-bit hash. So for 2128+1 number of hashes there will be at least one collision.
This number is 2160+1 for SHA1 and 2512+1 for SHA512.
Here this rule applies. The more the output bits the more uniqueness and more computation. So there is a trade off. What you have to do is to choose an optimal one.
Could two ~200 char bytearrays MD5-hash down to the same string?
Considering that there are more 200 byte strings than 32 byte strings (MD5 digests), that is guaranteed to be the case.
All hash functions have that problem, but some are more robust than MD5. Try SHA-1. git is using it for the same purpose.
It may happen that two MD5 hashes collides (are the same). In 1996, a flaw was found in MD5 algorithm, and cryptanalysts advised to switch to SHA-1 hashing algorithm.
So, I will advise you to switch to SHA-1 (40 characters). But do not worry: I doubt that your two pictures will get the same hash. I think you can assume this risk in your application.
As other said before. Hash doesnt give you what you need unless you are fine with risk of collision.
Database is helpful here.
You get unique index for each 200 long string. No collisions here, and you need to set your 200 long names to be indexed, in that way it will use extra memory but it will sort it for you making search very very fast. You get unique id which can be easily used for filenames.
I have'nt worked much on hashing algorithms but as per my understanding there is always a chance of collison in hashing algorithm i.e. two differnce object may be hashed to same hash value but it is guaranteed that every time a object will be hashed to same hash value. There are other techniques that may be used for this , like linear probing.
With a knowledge of how md5 works, would it be possible to use a population based algorithm such as genetic programming to break simple passwords?
For examples, given a md5 hash for a string that is between 5 to 10 characters, we are to try to get the string back.
If yes, what could be
A good representation for an individual of the population
Selction criteria
Recombination methods
This is to understand the application of genetic algorithms and to know if anyone has done anything of this sort.
Not really.
With just 5 characters, you could brute force it in not too unreasonable amounts of time, but presumably you're asking more about GAs than you are about breaking MD5. The problem is that there's no exploitable structure in an MD5 hash. Strings that are "close together" do not generate hashes that are "close together" under any useful distance relationship. The fitness function will basically be random.
I think the answer is "no". Because you are not able to get any crossover function. And the fitness function will be boolean. GA with only mutation operator and such fitness function is a bruteforce.
No, it is highly unlikely.
The genetic algorithm is used eg. for finding local/global maximum/minimum of some function. In case of md5 hash, if you change the value for which you calculate md5 hash, the md5 hash changes completely, thus narrowing the input value range is completely unuseful. MD5 algorithm was designed to hash the generated value if input data changes in any way. The only possibility to find the correct value is when you apply mutation, but it results in checking random input values for whether they generate the given hash (which - as oxilumin said - is just a brute force attack).
You can read more about finding value that generated specific md5 hash here (rainbow tables).
Although the answer is probably "no", there is one caveat to consider: The published collisions are strings that only differ by a few key bytes: https://en.wikipedia.org/wiki/MD5#Collision_vulnerabilities
Guessing the plaintext with a genetic algorithm isn't guaranteed, but it may be more efficient to discover a collision that way.
Or if it's in PHP and compares the md5 hash with the == operator... https://eval.in/108854
At the moment I am trying to crack the TEA block cipher in C. It is an assignment and the tea cipher has been weakend so that the key is 2 16-bit numbers.
We have been given the code to encode plaintext using the key and to decode the cipher text with the key also.
I have the some plaintext examples:
plaintext(1234,5678) encoded (3e08,fbab)
plaintext(6789,dabc) encoded (6617,72b5)
Update
The encode method takes in plaintext and a key, encode(plaintext,key1). This occurs again with another key to create the encoded message, encode(ciphertext1,key), which then creates the encoded (3e08,fbab) or encoded (6617,72b5).
How would I go about cracking this cipher?
At the moment, I encode the known plaintext with every possible key; the key size being hex value ffffffff. I write this to file.
But now I am stuck and in need of direction.
How could I use the TEA's weakness of equivalent keys to lower the amount of time it would take to crack the cipher? Also, I am going to use a man in the middle attack.
As when I encode with known plaintext and all key 1s it will create all the encrypted text with associated key and store it in a table.
Then I will decrypt with the known ciphertext that is in my assignment with all the possible values of key2. This will leave me with a table of decrypts that has only been decrypted once.
I can then compare the 2 tables together to see if any of encrpts with key1 match the decrypts with key2.
I would like to use the equilenvent weakness as well, if someone could help me with implmenting this in code that would be great. Any ideas?
This is eerily similar to the Double Crypt problem from the IOI '2001 programming contest. The general solution is shown here, it won't give you the code but might point you in the right direction.
Don't write your results to a file -- just compare each ciphertext you produce to the known ciphertext, encoding the known plain text with every possible key until one of them produces the right ciphertext. At that point, you've used the correct key. Verify that by encrypting the second known plaintext with the same key to check that it produces the correct output as well.
Edit: the encoding occurring twice is of little consequence. You still get something like this:
for (test_key=0; test_key<max; test_key++)
if (encrypt(plaintext, test_key) == ciphertext)
std::cout << "Key = " << test_key << "\n";
The encryption occurring twice means your encrypt would look something like:
return TEA_encrypt(TEA_encrypt(plaintext, key), key);
Edit2: okay, based on the edited question, you apparently have to do the weakened TEA twice, each with its own 16-bit key. You could do that with a single loop like above, and split up the test_key into two independent 16-bit keys, or you could do a nested loop, something like:
for (test_key1=0; test_key1<0xffff; test_key1++)
for (test_key2=0; test_key2<0xffff; test_key2++)
if (encrypt(encrypt(plaintext, test_key1), test_key2) == ciphertext)
// we found the keys.
I am not sure if this property holds for 16-bit keys, but 128-bit keys have the property that four keys are equivalent, reducing your search space by four-fold. I do not off the top of my head remember how to find equivalent keys, only that the key space is not as large as it appears. This means that it's susceptible to a related-key attack.
You tagged this as homework, so I am not sure if there are other requirements here, like not using brute force, which it appears that you are attempting to do. If you were to go for a brute force attack, you would probably need to know what the plaintext should look like (like knowing it English, for example).
The equivalent keys are easy enough to understand and cut key space by a factor of four. The key is split into four parts. Each cycle of TEA has two rounds. The first uses the first two parts of the key while the second uses the 3rd and 4th parts. Here is a diagram of a single cycle (two rounds) of TEA:
(unregistered users are not allowed to include images so here's a link)
https://en.wikipedia.org/wiki/File:TEA_InfoBox_Diagram.png
Note: green boxes are addition red circles are XOR
TEA operates on blocks which it splits into two halves. During each round, one half of the block is shifted by 4,0 or -5 bits to the left, has a part of the key or the round constant added to it and then the XOR of the resulting values is added to the other half of the block. Flipping the most significant bit of either key segment flips the same bit in the sums it is used for and by extension the XOR result but has no other effect. Flipping the most significant bit of both key segments used in a round flips the same bit in the XOR product twice leaving it unchanged. Flipping those two bits together doesn't change the block cipher result making the flipped key equivalent to the original. This can be done for both the (first/second) and (third/fourth) key segments reducing the effective number of keys by a factor of four.
Given the (modest) size of your encryption key, you can afford to create a pre-calculated table (use the same code given above, and store data in large chuncks of memory - if you don have enough RAM, dump the chuncks to disk and keep an addressing scheme so you can lookup them in a proper order).
Doing this will let you cover the whole domain and finding a solution will then be done in real-time (one single table lookup).
The same trick (key truncation) was used (not a long time ago) in leading Office software. They now use non-random data to generate the encryption keys -which (at best) leads to the same result. In practice, the ability to know encryption keys before they are generated (because the so-called random generator is predictable) is even more desirable than key-truncation (it leads to the same result -but without the hurdle of having to build and store rainbow tables).
This is called the march of progress...