5 ms·

I can't wait for the detailed HN post explaining how this scheme was defeated.

by oneJob 11y ago

diabeetusman 11y ago
A website with rules like "passwords must be 8 characters or less, contain no dictionary words inside them, and cannot have double letters" would put a stop to that pretty quick
- johnpowell 11y ago
  I simply don't use those websites. I have never had a password stolen and I can retrieve them if I forget my phone and have to borrow someones to visit my password page. This happens a lot since I find my phone to be a anchor and my life is better when I leave it at home. A few weeks ago I had to return something to Target I ordered online. I didn't have my phone with me but my sister had hers and they wanted a order number that was only emailed to me. So with a pencil and some paper I was able to visit the page with my email password and convert it and login to my email on her phone. It saved a few hours of driving back to my apartment and back to the store.
  - thomnottom 11y ago
    Life is better when you leave your phone at home, as long as you can rely on other people to have their phone with them so you can borrow it?
dsfyu404ed 11y ago
# this should be a start, it still needs output all permutations of the cipher applied over all possible substrings in the ciphertext because not all parts may have the same cipher applied and it also needs to brute force a salt up to a reasonable length to precede/follow the output text since salting passwords with a memorized constant is common. def ceasar_permutations(ciphertext): alpabets = [ascii_lowercase, ascii_uppercase,'0123456789',' ~!@#$%^&*()_+``-=[]{}:";\\\<>?,./]'] for perm in permutations(alpabets): alphabet = "" for p in perm: alphabet+=p for i in range(0, alphabet.__len__()): shifted_alphabet = alphabet[i:] + alphabet[:i] table = str.maketrans(alphabet, shifted_alphabet) print( ciphertext.translate(table))
- johnpowell 11y ago
  It is almost like I left out important parts of how I retrieve my passwords.

I can't wait for the detailed HN post explaining how this scheme was defeated.

by oneJob 11y ago

I can't wait for the detailed HN post explaining how this scheme was defeated.

A website with rules like "passwords must be 8 characters or less, contain no dictionary words inside them, and cannot have double letters" would put a stop to that pretty quick

johnpowell 11y ago
I simply don't use those websites. I have never had a password stolen and I can retrieve them if I forget my phone and have to borrow someones to visit my password page. This happens a lot since I find my phone to be a anchor and my life is better when I leave it at home. A few weeks ago I had to return something to Target I ordered online. I didn't have my phone with me but my sister had hers and they wanted a order number that was only emailed to me. So with a pencil and some paper I was able to visit the page with my email password and convert it and login to my email on her phone. It saved a few hours of driving back to my apartment and back to the store.
- thomnottom 11y ago
  Life is better when you leave your phone at home, as long as you can rely on other people to have their phone with them so you can borrow it?

dsfyu404ed 11y ago

# this should be a start, it still needs output all permutations of the cipher applied over all possible substrings in the ciphertext because not all parts may have the same cipher applied and it also needs to brute force a salt up to a reasonable length to precede/follow the output text since salting passwords with a memorized constant is common. def ceasar_permutations(ciphertext): alpabets = [ascii_lowercase, ascii_uppercase,'0123456789',' ~!@#$%^&*()_+``-=[]{}:";\\\<>?,./]'] for perm in permutations(alpabets): alphabet = "" for p in perm: alphabet+=p for i in range(0, alphabet.__len__()): shifted_alphabet = alphabet[i:] + alphabet[:i] table = str.maketrans(alphabet, shifted_alphabet) print( ciphertext.translate(table))

johnpowell 11y ago
It is almost like I left out important parts of how I retrieve my passwords.