# Dictionary help needed

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
LVL 1
###### Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

consultant Commented:
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.
0
Commented:
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))
0

Experts Exchange Solution brought to you by

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Author Commented:
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?
0
Commented:
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.
0
###### It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Python

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.