I'm trying to find 2 different plain text words that create very similar hashes.
I'm using the hashing method 'whirlpool', but I don't really need my question to be answered in the case or whirlpool, if you can using md5 or something easier that's ok.
The similarities i'm looking for is that they contain the same number of letters (doesnt matter how much they're jangled up)
i.e
plaintext 'test'
hash 1: abbb5 has 1 a , 3 b's , one 5
plaintext 'blahblah'
hash 2: b5bab must have the same, but doesnt matter what order.
I'm sure I can read up on how they're created and break it down and reverse it, but I am just wondering if what I'm talking about occurs.
I'm wondering because I haven't found a match of what I'm explaining (I created a PoC to run threw random words / letters till it recreated a similar match), but then again It would take forever doing it the way i was dong it. and was wondering if anyone with real knowledge of hashes / encryption would help me out.
So you can do it like this:
create an empty sorted map \
create a 64 bit counter (you don't need more than 2^63 inputs, in all probability, since you would be dead before they would be calculated - unless quantum crypto really takes off)
use the counter as input, probably easiest to encode it in 8 bytes;
use this as input for your hash function;
encode output of hash in hex (use ASCII bytes, for speed);
sort hex on number / alphabetically (same thing really)
check if sorted hex result is a key in the map
if it is, show hex result, the old counter from the map & the current counter (and stop)
if it isn't, put the sorted hex result in the map, with the counter as value
increase counter, goto 3
That's all folks. Results for SHA-1:
011122344667788899999aaaabbbcccddeeeefff for both 320324 and 429678
I don't know why you want to do this for hex, the hashes will be so large that they won't look too much alike. If your alphabet is smaller, your code will run (even) quicker. If you use whole output bytes (i.e. 00 to FF instead of 0 to F) instead of hex, it will take much more time - a quick (non-optimized) test on my machine shows it doesn't finish in minutes and then runs out of memory.
Related
is there any tool or method to figure out what is this hash/cipher function?
i have only a 500 item list of input and output plus i know all of the inputs are numeric, and output is always 2 Byte long hexadecimal representation.
here's some samples:
794352:6657
983447:efbf
479537:0796
793670:dee4
1063060:623c
1063059:bc1b
1063058:b8bc
1063057:b534
1063056:b0cc
1063055:181f
1063054:9f95
1063053:f73c
1063052:a365
1063051:1738
1063050:7489
i looked around and couldn't find any hash this short, is this a hash folded on itself? (with xor maybe?) or maybe a simple trivial cipher?
is there any tool or method for finding the output of other numbers?
(i want to figure this out; my next option would be training a Neural Network or Regression, so i thought i ask before taking any drastic action )
Edit: The Numbers are directory names, and for accessing them, the Hex parts are required.
Actually, Wikipedia's page on hashes lists three CRCs and three checksum methods that it could be. It could also be only half the output from some more complex hashing mechanism. Cross your fingers and hope that it's of the former. Hashes are specifically meant to be difficult (if not impossible) to reverse engineer.
What it's being used for should be a very strong hint about whether or not it's more likely to be a checksum/CRC or a hash.
I want code to render n bits with n + x bits, non-sequentially. I'd Google it but my Google-fu isn't working because I don't know the term for it.
For example, the input value in the first column (2 bits) might be encoded as any of the output values in the comma-delimited second column (4 bits) below:
0 1,2,7,9
1 3,8,12,13
2 0,4,6,11
3 5,10,14,15
My goal is to take a list of integer IDs, and transform them in a way they can still be used for persistent URLs, but that can't be iterated/enumerated sequentially, and where a client cannot determine programmatically if a URL in a search result set has been visited previously without visiting it again.
I would term this process "encoding". You'll see something similar done to permit the use of communications channels that have special symbols that are not permitted in data. Examples: uuencoding and base64 encoding.
That said, you still need to (and appear at first blush to have) ensure that there is only one correct de-code; and accept the increase in size of the output (in the case above, the output will be double the size, bit-for-bit as the input).
I think you'd be better off encrypting the number with a cheap cypher + a constant secret key stored on your server(s), adding a random character or four at the end, and a cheap checksum, and simply reject any responses that don't have a valid checksum.
<encrypt(secret)>
<integer>+<random nonsense>
</encrypt>
+
<checksum()>
<integer>+<random nonsense>
</checksum>
Then decrypt the first part (remember, cheap == fast), validate the ciphertext using the checksum, throw off the random nonsense, and use the integer you stored.
There are probably some cryptographic no-no's here, but let's face it, the cost of this algorithm being broken is a touch on the low side.
As the Title says, i am trying out this last year's problem that wants me to write a program that works the same as scanf().
Ubuntu:
Here is my code:
#include<unistd.h>
#include<stdio.h>
int main()
{
int fd=0;
char buf[20];
read(0,buf,20);
printf("%s",buf);
}
Now my program does not work exactly the same.
How do i do that both the integer and character values can be stored since my given code just takes the character strings.
Also how do i make my input to take in any number of data, (only 20 characters in this case).
Doing this job thoroughly is a non-trivial exercise.
What you show does not emulate sscanf("%s", buffer); very well. There are at least two problems:
You limit the input to 20 characters.
You do not stop reading at the first white space character, leaving it and other characters behind to be read next time.
Note that the system calls cannot provide an 'unget' functionality; that has to be provided by the FILE * type. With file streams, you are guaranteed one character of pushback. I recently did some empirical research on the limits, finding values that the number of pushed back characters ranged from 1 (AIX, HP-UX) to 4 (Solaris) to 'big', meaning up to 4 KiB, possibly more, on Linux and MacOS X (BSD). Fortunately, scanf() only requires one character of pushback. (Well, that's the usual claim; I'm not sure whether that's feasible when distinguishing between "1.23e+f" and "1.23e+1"; the first needs three characters of lookahead, it seems to me, before it can tell that the e+f is not part of the number.)
If you are writing a plug-in replacement for scanf(), you are going to need to use the <stdarg.h> mechanism. Fortunately, all the arguments to scanf() after the format string are data pointers, and all data pointers are the same size. This simplifies some aspects of the code. However, you will be parsing the scan format string (a non-trivial exercise in its own right; see the recent discussion of print format string parsing) and then arranging to make the appropriate conversions and assignments.
Unless you have unusually stringent conditions imposed upon you, assume that you will use the character-level Standard I/O library functions such as getchar(), getc() and ungetc(). If you can't even use them, then write your own variants of them. Be aware that full integration with the rest of the I/O functions is tricky - things like fseek() complicate matters, and ensuring that pushed-back characters are properly consumed is also not entirely trivial.
This is a Vigenere cipher-text
EORLL TQFDI HOEZF CHBQN IFGGQ MBVXM SIMGK NCCSV
WSXYD VTLQS BVBMJ YRTXO JCNXH THWOD FTDCC RMHEH
SNXVY FLSXT ICNXM GUMET HMTUR PENSU TZHMV LODGN
MINKA DTLOG HEVNI DXQUG AZGRM YDEXR TUYRM LYXNZ
ZGJ
The index of coincidence gave a shift of six (6): I know this is right (I used an online Java applet to decrypt the whole thing using the key 'QUARTZ').
However, in this question we are only told the first and last two letters of the Key - 'Q' and 'TZ.'
So far I have split the ciphertext into slices using this awesome applet. So the first slice is 0, k, 2k, 3k, 4k; the second is 1, k + 1, 2k + 1, 3k + 1; et cetera.
KeyPos=0: EQEQQSCXQJJHDEYIUTSVMTVUMTYJ
KeyPos=1: OFZNMICYSYCWCHFCMUULILNGYUX
KeyPos=2: RDFIBMSDBRNOCSLNERTONOIADYN
KeyPos=3: LICFVGVVVTXDRNSXTPZDKGDZERZ
KeyPos=4: LHHGXKWTBXHFMXXMHEHGAHXGXMZ
KeyPos=5: TOBGMNSLMOTTHVTGMNMNDEQRRLG
My idea was to calculate the highest-frequency letter in each block, hoping that the most frequent letter would give me some clue as to how to find 'U,' 'A' and 'R.' However, the most frequent letters in these blocks are:
KeyPos=0: Q,4 T,3 E,3, J,3
KeyPos=1: C,4 U,3 Y,3
KeyPos=2: N,4 O,3 R,3 D,3 B,2
KeyPos=3: V,4 D,3 Z,3
KeyPos=4: H,6 X,6 M,3 G,3
KeyPos=5: M,4 T,4 N,3 G,3
Which yields QCNVHM, or QUNVHM (being generous), neither of which are that close to QUARTZ. There are online applets that can crack this no problem, so it mustn't be too short a text to yield decent frequency counts from the blocks.
I guess I must be approaching this the wrong way. I just hoped one of you might be able to offer some clue as to where I am going wrong.
p.s. This is for a digital crypto class.
Interesting question...
I don't have a programmatic solution for cracking the original ciphertext, but I was able to solve it with a little mind power and some helpful JavaScript.
I started by using this page and the information you supplied. Provide the ciphertext, a key length of 6 and hit initialize. What's nice about the approach here is that unknowns in either the plaintext or key are left as hyphens.
Update the key, adding only what you know Q---TZ and click 'update plaintext'. At this point we know:
o---sua---opo---oca---nha---enc---rom---dth---ama---int---ept---our---mun---tio---ewi---eus---the---ond---loc---onf---now---hed---off---ere---nsw---esd---tmi---ght
Here's where I applied a bit of brain power. You start recognizing bits of the plaintext. the, now and off make an appearance. At the end, there's ght - this made me think the prior letter is likely a vowel. For example light or thought. I replaced the corresponding hyphen with u and clicked update keyword to find what letter would have produced that combination. The matching letter turns out to be F. I think updated the plaintext to see the results. They didn't look promising. So I tried i instead which resulted in:
o--usua--ropo--loca--onha--eenc--prom--edth--eama--eint--cept--gour--mmun--atio--wewi--beus--gthe--cond--yloc--ionf--mnow--thed--poff--mere--insw--nesd--atmi--ight
Now we're getting somewhere. At the start I see something that might be usual, and further in I see int--cept and near the end w--nesd-- at mi--ight. Voila. Filling in the letters for wednesday and updating the keyword yielded QUARTZ.
... So, how to port this approach to code? Not sure about the best way to do that just yet. The idea of using the known characters in the key, partially decrypting the ciphertext and brute forcing the rest is appealing. But without a dictionary handy, I'm not sure what the best brute-forcing method would be...
To be continued (maybe)...
An algorithm wouldn't just consider the most frequent letters but the frequency pattern of the whole alphabet. Technically you compute the index of coincidence for each possible shift and consider the maximal ones.
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.