# dict_crack.py
# Cracks a password has using a dictionary of common passwords.
# Author: Euan McCall
# November 2015
# Last tested: November 28th 2015
# Functional? NO


import hashlib

passwd_hash = '4297f44b13955235245b2497399d7a93'

# user input type
# passwd_hash = raw_input();

# set up list of common passwords
dic = ['123','1234','12345','123456','1234567','12345678','password','qwerty','abc','abcd','abc123','111111','monkey','arsenal','letmein','trustno1','dragon','baseball','superman','iloveyou','starwars','montypython','cheese','123123','football','password','batman']

# Our dictionary attack subroutine, which does the checks and conversions.
def dict_attack(password_hash):
    """ Checks pw hash, against a dictionary of common pw hashes. """
    # Calculate md5 from list of pws and compare with provided value passwd_hash
    passwd_found = False;
    print 'Finding matching hash for:', passwd_hash



    # Iteration and conversion to md5
    for x in dic:
        #print "Checking Password: " +x;
        md5hash = hashlib.md5(x);
        #print "Hash: %s"% (md5hash.hexdigest())

        # Conditional check for md5 (Doesn't work)
        if md5hash == passwd_hash:
            passwd_found = True;
            print "Hash found: %s"% (passwd_found) ;
            print passwd_hash;
            print x;
        else:
            passwd_found = False;
            print "No Match: %s"% (passwd_found);


        # Print respective message to user
      #  if passwd_found == True:
       #     print '[+] Pasword found: %s'% (passwd)
        #else:
         #   print '[-] Password not found'


def main():
    print '[dict_crack] Tests'
    dict_attack(passwd_hash)



if __name__=='__main__':
    main()