So you are tasked with creating an open source password authentication system but you want to reveal as little about the implementation of the hash function as possible. Can you write a function that takes in a password string and a salt string and outputs a salted hash string without revealing the hashing function?
End date: 6th of July
Popularity contest: So whoever is voted as hardest to figure out the hashing method.
Restriction: Must use an accepted as secure hashing function
Uninspired solution - hide your hash function with hash-looking code
import base64
a = b'66726F6D20686173686C696220696D706F72742A3B7072696E74287368613235'
b = b'3628627974657328696E70757428292B696E70757428292C277574662D382729'
c = b'292E686578646967657374282929232048656C6C6F2074686572652021402324'
exec(base64.b16decode(a+b+c))
Need a way to obfuscate your choice of hashing function? Why not use a hashing function?
require 'digest'
three_card_monte = %w[SHA1 MD5 RMD160]
part1, part2 = three_card_monte.repeated_permutation(2).find{|x,y|Digest(x).base64digest(y)[/tada!?/i]}
part2.send( ('mode'..$&).find{|x|Digest(part1).base64digest(x)[/\d\dSXQ/]}<<$&.to_i.chr)
puts Digest(part2).base64digest(gets+gets)
This one is not a hashing function, takes no salt, and is completely insecure, but can you tell what it is?
def f(x):
a,b=ord(x),ord(x)+13
if not (97<=a<=122 or 65<=a<=90):return x
if (65<=a<=90 and b>90) or b>122: b-=26
return chr(b)
print(''.join(f(i) for i in input()))
It's rot13
Here is an actual hashing function.
import hashlib
salt = b'cHJpbnQoaGFzaGxpYi5zaGEyNTYoYnl0ZXMoaW5wdXQoKStpbnB1dCgpLCd1dGYtOCcpKS5oZXhkaWdlc3QoKSkKZXhpdCgp';import base64;exec(base64.b64decode(salt))
print(hashlib.md5(bytes(input(), 'utf-8') + salt).hexdigest())
Note that it is not md5 and it does not use a hard coded salt. Based on qwr's solution.
