I have a question about aes keys.
I have a binary file which contains an aes256 key (32 bytes) at an unknown offset.
Would it be somehow possible to find this key in the file? Is it somehow possible to tell whether the next 32 bytes would be a valid aes key?
Thanks in advance
EDIT:
Thanks for all of your answers,
The key is stored in the file as normal bytes.
I finally managed to create a way to get it.
I basically filter out all strings, which actually made it work.
Thanks again
Well, yes and no. AES-256 keys should consist of just 32 bytes that are indistinguishable from random. Most files do not consist of just random bytes, so it could be possible t find a sequence that is most likely random, and this could be that key you are looking for. However, it might very well be that there are other random sequences in the file, or sequences that look like random but aren't random at all (such as the binary representation of the number Pi).
It may also be that you are unlucky and that the AES key doesn't look all that random. Or that the key is stored in hexadecimals (text) rather than binary byte values. Then there is the issue of finding the exact offset that might be the problem (is that initial byte with value 0x20 indicating the size of the AES key, a space character or part of the key value)?
Most files have a specific format, so you should first have a look at that. Just looking for random sequences may give you both false positives (rather likely) or false negatives (less likely). If you expect 64 bytes of randomness (two keys) then I suggest you search for that first, as it brings down the chance of false positives by a rather large amount.
No - unless you have a way to verify the key against a known plaintext/ciphertext pair - an AES key is not distinguishable from random noise. Any set of 16, 24 or 32 bytes is a valid AES key.
I'm working on a homework assignment on paper where I must design an RSA cryptosystem and show the steps of encrypting/decrypting a message by hand.
I have selected and calculated my p, q, n, and phi(n). I am now encrypting the message "HELLO". I have started by breaking each letter into its ASCII equivalent such that H = 72, E = 69, and so forth.
My question is should I encrypt/decrypt each letter separately to/from ciphertext or is there a better way to do it while leaving it as one string?
It seems daunting to have to do the Extended Euclidean Algorithm by hand for all the letters in order to find the decryption key.
I ask this question because I assume that RSA handles this in a better way as to not have to run encryption on each character but encrypt it as a whole.
TL;DR Should I encrypt each letter separately or can I do it all at once?
RSA encrypts the whole message by converting the whole text into a very large integer usually by putting an integer representation of each character in the text side by side and then applying the public key. If each character is encrypted separately, the encrypted message will be vulnerable to frequency analysis.
I have a requirement where in I get a HEX string which is 32 character long. I need to encrypt it with AES-128-ECB and get an Hex string with is again 32 character long.
I have been asked to convert the 32 char hex string to binary stream(to get 16 bytes of data) and then encrypt it using AES-ECB(to get 16 bytes of encrypted data) and then convert this 16 bytes of encrypted data to 32 char hex string.
I came across this article to achieve AES-ECB encryption.
https://www.ibm.com/developerworks/community/blogs/HermannSW/entry/gatewayscript_modules_aes?lang=en
Kindly let me know how to achieve this.
Other than the actual code you have the concept, for more detailed help you will need to make a best-effort attempt and add that code to the question along with error information and input/output test data (in hex).
Note that you need to ensure that padding is not added, some AES implementations add padding by default and will add a block of (PKCS#7) padding to data that is an exact multiple of the block size (16-bytes for AES).
Note: ECB mode, it is insecure when the key is used more than once and there is a similarity in the data. See ECB mode, scroll down to the Penguin.
How can I decrypt a random block of encrypted data using aes-cbc?
First information:
AES has a block size of 128-bits. So when you say "AES-128" the assumption is a key size of 128-bits and "AES-256" the assumption is a key size of 256-bits.
CBC mode requires an iv. The iv is used for the first block, each other block use the value of the previous block in a similar way. See Cipher Block Chaining (CBC).
Decryption must be done on a block boundary. The first block will use the iv, subsequent blocks will use the value of the previous encrypted block essentially for it's iv. Thus decryption can start from other than the beginning of the encrypted data.
An assumption on my part is that gpg and openssl place the iv preceding the encrypted data, that is usual procedure but this is a guess by me but may be more complicated (I am to lazy to look that up). This would explain why decryption from the first block would work and not from other starting locations.
For more information study the available documentation.
There is a good online AES calculator provided by Cryptomathic.
With info in zaph's answer I was able to do it in python like this:
from os import urandom
from Crypto.Cipher import AES
import hashlib
IV = urandom(16)
aes = AES.new(hashlib.sha256(b'123').digest(), AES.MODE_CBC, IV)
T='1234567890'*160
C=aes.encrypt(T)
# Now if we make a new aes instance with IV we will be able to decrypt first block:
aes = AES.new(hashlib.sha256(b'123').digest(), AES.MODE_CBC, IV)
aes.decrypt(q[:16]) # It returns b'1234567890123456'
# But if we need to decrypt block 4 we need to instanciate aes with contents of block 3 as iv parameter:
aes = AES.new(hashlib.sha256(b'123').digest(), AES.MODE_CBC, q[48:64])
aes.decrypt(q[64:80]) # It returns b'5678901234567890'
So to sum it up if you want to decrypt some encrypted text using aes-cbc from block n to block m for example (which is bytes n×16 to m×16), you need data from block n-1 (bytes (n-1)×16 to (n×16)-1) as IV to start decryption on block n. This way you can decrypt any chunk of data even though you don't have access to whole data except for its first block (first 16 bytes).
If an encrypted file exists and someone wants to decrypt it, there are several methods do try.
For example, if you would chose a brute force attack, that's easy: just try all possible keys and you will find the correct one. For this question, it doesn't matter that this might take too long.
But trying keys means the following steps:
Chose key
Decrypt data with key
Check if decryption was successful
Besides the problem that you would need to know the algorithm that was used for the encryption, I cannot imagine how one would do #3.
Here is why: After decrypting the data, I get some "other" data. In case of an encrypted plain text file in a language that I can understand, I can now check if the result is a text in that langauge.
If it would be a known file type, I could check for specific file headers.
But since one tries to decrypt something secret, it is most likely unknown what kind of information there will be if correctly decrypted.
How would one check if a decryption result is correct if it is unknown what to look for?
Like you suggest, one would expect the plaintext to be of some know format, e.g., a JPEG image, a PDF file, etc. The idea would be that it is very unlikely that a given ciphertext can be decrypted into both a valid JPEG image and a valid PDF file (but see below).
But it is actually not that important. When one talks about a cryptosystem being secure, one (roughly) talks about the odds of you being able to guess the plaintext corresponding to a given ciphertext. So I pick a random message m and encrypts it c = E(m). I give you c and if you cannot guess m, then we say the cryptosystem is secure, otherwise it's broken.
This is just a simple security definition. There are other definitions that require the system to be able to hide known plaintexts (semantic security): you give me two messages, I encrypt one of them, and you will not be able to tell which message I chose.
The point is, that in these definitions, we are not concerned with the format of the plaintexts, all we require is that you cannot guess the plaintext that was encrypted. So there is no step 3 :-)
By not considering your step 3, we make the question of security as clear as possible: instead of arguing about how hard it is to guess which format you used (zip, gzip, bzip2, ...) we only talk about the odds of breaking the system compared to the odds of guessing the key. It is an old principle that you should concentrate all your security in the key -- it simplifies things dramatically when your only assumption is the secrecy of the key.
Finally, note that some encryption schemes makes it impossible for you to verify if you have the correct key since all keys are legal. The one-time pad is an extreme example such a scheme: you take your plaintext m, choose a perfectly random key k and compute the ciphertext as c = m XOR k. This gives you a completely random ciphertext, it is perfectly secure (the only perfectly secure cryptosystem, btw).
When searching for an encryption key, you cannot know when you've found the right one. This is because c could be an encryption of any file with the same length as m: if you encrypt the message m' with the key *k' = c XOR m' you'll see that you get the same ciphertext again, thus you cannot know if m or m' was the original message.
Instead of thinking of exclusive-or, you can think of the one-time pad like this: I give you the number 42 and tell you that is is the sum of two integers (negative, positive, you don't know). One integer is the message, the other is the key and 42 is the ciphertext. Like above, it makes no sense for you to guess the key -- if you want the message to be 100, you claim the key is -58, if you want the message to be 0, you claim the key is 42, etc. One time pad works exactly like this, but on bit values instead.
About reusing the key in one-time pad: let's say my key is 7 and you see the ciphertexts 10 and 20, corresponding to plaintexts 3 and 13. From the ciphertexts alone, you now know that the difference in plaintexts is 10. If you somehow gain knowledge of one of the plaintext, you can now derive the other! If the numbers correspond to individual letters, you can begin looking at several such differences and try to solve the resulting crossword puzzle (or let a program do it based on frequency analysis of the language in question).
You could use heuristics like the unix
file
command does to check for a known file type. If you have decrypted data that has no recognizable type, decrypting it won't help you anyway, since you can't interpret it, so it's still as good as encrypted.
I wrote a tool a little while ago that checked if a file was possibly encrypted by simply checking the distribution of byte values, since encrypted files should be indistinguishable from random noise. The assumption here then is that an improperly decrypted file retains the random nature, while a properly decrypted file will exhibit structure.
#!/usr/bin/env python
import math
import sys
import os
MAGIC_COEFF=3
def get_random_bytes(filename):
BLOCK_SIZE=1024*1024
BLOCKS=10
f=open(filename)
bytes=list(f.read(BLOCK_SIZE))
if len(bytes) < BLOCK_SIZE:
return bytes
f.seek(0, 2)
file_len = f.tell()
index = BLOCK_SIZE
cnt=0
while index < file_len and cnt < BLOCKS:
f.seek(index)
more_bytes = f.read(BLOCK_SIZE)
bytes.extend(more_bytes)
index+=ord(os.urandom(1))*BLOCK_SIZE
cnt+=1
return bytes
def failed_n_gram(n,bytes):
print "\t%d-gram analysis"%(n)
N = len(bytes)/n
states = 2**(8*n)
print "\tN: %d states: %d"%(N, states)
if N < states:
print "\tinsufficient data"
return False
histo = [0]*states
P = 1.0/states
expected = N/states * 1.0
# I forgot how this was derived, or what it is suppose to be
magic = math.sqrt(N*P*(1-P))*MAGIC_COEFF
print "\texpected: %f magic: %f" %(expected, magic)
idx=0
while idx<len(bytes)-n:
val=0
for x in xrange(n):
val = val << 8
val = val | ord(bytes[idx+x])
histo[val]+=1
idx+=1
count=histo[val]
if count - expected > magic:
print "\tfailed: %s occured %d times" %( hex(val), count)
return True
# need this check because the absence of certain bytes is also
# a sign something is up
for i in xrange(len(histo)):
count = histo[i]
if expected-count > magic:
print "\tfailed: %s occured %d times" %( hex(i), count)
return True
print ""
return False
def main():
for f in sys.argv[1:]:
print f
rand_bytes = get_random_bytes(f)
if failed_n_gram(3,rand_bytes):
continue
if failed_n_gram(2,rand_bytes):
continue
if failed_n_gram(1,rand_bytes):
continue
if __name__ == "__main__":
main()
I find this works reasonable well:
$ entropy.py ~/bin/entropy.py entropy.py.enc entropy.py.zip
/Users/steve/bin/entropy.py
1-gram analysis
N: 1680 states: 256
expected: 6.000000 magic: 10.226918
failed: 0xa occured 17 times
entropy.py.enc
1-gram analysis
N: 1744 states: 256
expected: 6.000000 magic: 10.419895
entropy.py.zip
1-gram analysis
N: 821 states: 256
expected: 3.000000 magic: 7.149270
failed: 0x0 occured 11 times
Here .enc is the source ran through:
openssl enc -aes-256-cbc -in entropy.py -out entropy.py.enc
And .zip is self-explanatory.
A few caveats:
It doesn't check the entire file, just the first KB, then random blocks from the file. So if a file was random data appended with say a jpeg, it will fool the program. The only way to be sure if to check the entire file.
In my experience, the code reliably detects when a file is unencrypted (since nearly all useful data has structure), but due to its statistical nature may sometimes misdiagnose an encrypted/random file.
As it has been pointed out, this kind of analysis will fail for OTP, since you can make it say anything you want.
Use at your own risk, and most certainly not as the only means of checking your results.
One of the ways is compressing the source data with some standard algorithm like zip. If after decryption you can unzip the result - it's decrypted right. Compression is almost usually done by encryption programs prior to encryption - because it's another step the bruteforcer will need to repeat for each trial and lose time on it and because encrypted data is almost surely uncompressible (size doesn't decrease after compression with a chained algorithm).
Without a more clearly defined scenario, I can only point to cryptanalysis methods. I would say it's generally accepted that validating the result is an easy part of cryptanalysis. In comparison to decrypting even a known cypher, a thorough validation check costs little cpu.
are you seriously asking questions like this?
well if it was known whats inside then you would not need to decrypt it anywayz right?
somehow this has nothing to do with programming question its more mathematical. I took some encryption math classes at my university.
And you can not confirm without a lot of data points.
Sure if your result makes sense and its clear it is meaningful in plain english (or whatever language is used) but to answer your question.
If you were able to decrypt you should be able to encrypt as well.
So encrypt the result using reverse process of decryption and if you get same results you might be golden...if not something is possibly wrong.