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Dictionary help needed

Posted on 2007-03-19
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Last Modified: 2010-04-16
I'm trying to implement a forward search attack (i.e. a brute force generating and testing all possible combinations) on 2 words encrypted in RSA whose contents may include A-Z and [space] (which might be used for padding), that is 65-90 and 32 ASCII. I have some (very limited) knowledge of Python but can't get the dictionary contents right. I've included the code below, can someone please give me some pointers.

scrambled1 = "3648141"
scrambled2 = "5608387"
dictionary1 = {}
code=0
e=5
n=21508387

for a in range(65,91):
    for b in range(65,91):
        for c in range(65,91):
            plain=(a<<16)+(b<<8)+c
            code = pow(plain,e,n)
          dictionary1[code]=plain

for d in range(65,91):
      for e in range(65,91):
            plain=(f<<16)+(g<<8)+32
              code = pow(plain,e,n)
            
            dictionary1[code]=plain

dictionary1[code]=scrambled2
dictionary1[code]=scrambled2
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Question by:mark_667
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4 Comments
 
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Expert Comment

by:ramrom
Comment Utility
I don't know enough about RSA encryption to even begin to guess what you mean by "can't get the dictionary contents right". Perhaps if you explain what the contents should be and what you are getting we could help.
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Accepted Solution

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mish33 earned 125 total points
Comment Utility
I don't understand why you need dictionary at all:

scrambled1 = "3648141"
scrambled2 = "5608387"
e=5
n=21508387
AZ = range(65,91)
AZspace = list(AZ)+[32]

for a in AZ:
    for b in AZ:
        for c in AZspace:
            plain=(a<<16)+(b<<8)+c
            code = pow(plain,e,n)
            if code==scrambled1:
               print 'found 1', ''.join(chr(x) for x in (a, b, c))
            if code==scrambled2:
               print 'found 2', ''.join(chr(x) for x in (a, b, c))
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Author Comment

by:mark_667
Comment Utility
The dictionary holds all plaintext possibilities for the specified range in one column and stores the encrypted versions of these in the other giving a lookup table.

This should allow me to match the scrambled words to one of the encrypted entries and look at the corresponding plaintext entry to get a result. As you can see if you run it, this does not work. I tried the code mish33 posted above but when I printed the contents of code to a file to check  what was being produced the first scrambled word was present but not the second.

This could only happen if I had not covered all possibilities when generating values for code. Also, I had no way of looking up the corresponding plaintext when a match was found. Here's a modified version I've tried:

scrambled1 = "3648141"
scrambled2 = "5608387"
e=5
n=21508387
AZ = range(65,91)
AZspace = list(AZ)+[32]
outfd=open('outfile.txt','wb')

for a in AZspace:
    for b in AZspace:
        for c in AZspace:
            plain=(a<<16)+(b<<8)+c
            code = pow(plain,e,n)

          outfd.write('%s \n' % str(plain))
          outfd.write('%s \n' % str(code))
              outfd.write('%s \n')

            if code==scrambled1:
               print 'found 1', ''.join(chr(x) for x in (a, b, c))
            if code==scrambled2:
               print 'found 2', ''.join(chr(x) for x in (a, b, c))

outfd.close()

I think the problem is the space. Instead of encrypting it with the characters in the range A-Z it seems to just be adding it on to the end of what is encrypted, thus not generating all of the necessary possibilities. Any further thoughts?
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Expert Comment

by:mish33
Comment Utility
Check that scrambled2 really is 5608387, because that number is encoded by combination
a,b,c = 209, 106, 126 (string 'Ñj~') which far outside AZspace.
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