# PRBHA-10: A hashing algorithm in Python

So I've been working on a simple little hashing algorithm for Python that I like to call PRBHA-10 for short, or Pseudo-Random-Based Hash Algorithm. This is a hash algorithm in Python that is meant to be simplistic and secure. As you could see from the title, the algorithm implements a system with seeding the pseudo-random number generator to be able to produce the correct integer values when provided the correct password, therefore enabling the decryption of an encrypted byte sequence or string.

Here's the advantages of my method:

• The functions are tiny and simple, unlike some hashing methods out there.

• Despite the simplicity of the algorithm, it is secure due to the generation of large random numbers that scale with the password size.

• There is no chance of revealing even part of the encrypted string without the full password due to the huge numbers created by the seeded PRNG and multiplication that completely messes up the string if the wrong password is entered.

• It is fast. I have not done timeits yet, but the functions run in a trivial amount of time to generate their results.

The disadvantages:

• A somewhat large password is required to ensure security - usually anything over seven characters is completely cryptographically secure.

• It is vulnerable to the same brute-force attacks that any cryptographic encryption method is vulnerable to.

• With larger passwords comes greater security, but also a larger hash. The hash size slightly increases with each character added onto the password.

Here's the whole module program, prhba.py:

"""Pseudo-Random Based Hash Algorithm (PRHBA-10) is a simple
Python cipher that uses the integer representation of a
string, along with taking advantage of the built-in
pseudo-random number generator to be able to securely encrypt
strings and bytearrays.

The algorithm uses the integer representation of byte sequences,
seeding the random number generator using the password to
create a predictable random number for later use, and a special
encoded form to turn any byte sequence into an ASCII string
using only the letters 'aceilnorst'. This is secure due to the
extremely large random number generated ensuring the security
of the result.

The result can be later decrypted using a special algorithm to
decipher the text into a large integer number. Then, the
password is used as a seed to once again get the random number
used earlier in the process. That integer is used to modify
the large integer to be able to get the integer form of the
original string, which can then be converted to a string.

This cipher currently only works in Python due to the specific
random number generation implementation in the random module.
"""

import math
import random

__all__ = ["encrypt", "decrypt"]

LETTERS = "aceilnorst"

def to_num(s):
return int.from_bytes(s.encode(), 'little')

def from_num(n):
return n.to_bytes(math.ceil(n.bit_length() / 8), 'little')

def encrypt(data, password):
assert len(password) > 1, "Password length cannot be less than two"
random.seed(to_num(password))
unique_id = to_num(data) * random.getrandbits(len(password))
chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)

def decrypt(encrypted_data, password):
random.seed(to_num(password))
partnum_digits = []
for char in encrypted_data:
partnum_digits.append(str(LETTERS.index(char)))
partnum = int("".join(partnum_digits))
return from_num(partnum // random.getrandbits(len(password)))


Here's the questions I have:

• Is it PEP-8 compliant?

• Are the variable names confusing?

• Is there any parts that could be improved or further reduced?

• Do you have any other suggestions?

Thanks in advance!

• If you want to understand how this stuff works, try reading about how real hash functions work. Most hashes use what's called the Merkle-Damgard construction, and understanding how that works and how the blocks that make up real hashes will probably help you understand the problem better. – AJMansfield Jun 8 '19 at 2:18
• Also this isn't a hash function at all, it's an encryption algorithm. And if you don't understand the difference you definitely shouldn't be writing either. – AJMansfield Jun 8 '19 at 2:21

## 1 Answer

Don’t roll your own encryption. It takes a team of experts to develop new secure encryption methods, and even they can get it wrong from time to time.

Huge hole in your DIY encryption:

If I use a 2-character password, I might naïvely expect I’d have $$\62^2\$$ possible passwords that I can encrypt the data with. I’d be really shocked when it turns out there are only 4.

random.getrandbits(len(password))


generates 2 random bits, for $$\2^2\$$ possible values to multiply to_num(data) by. Only 4 possibilities is a lot easier to attack than $$\62^2\$$ different possibilities!

And one of those possibilities ... all bits zero ... destroys all the data you want to encode. So we’re down to actually only 3 possible numbers to test to reverse the encryption.

Any encryption mechanism worth its salt uses a salt (initial random data) to prevent the same message with the same password from being encrypted as the same text.

Code improvements: use generator expressions. Eg)

chunk = []
for digit in str(unique_id):
chunk.append(LETTERS[int(digit)])
return "".join(chunk)


would be more efficient rewritten as:

return "".join(LETTERS[int(digit)] for digit in str(unique_id))


The chunk list is never created in memory; instead, the join method pulls items one at a time from the generator expression.

Finally, don’t roll your own encryption.

• Thanks for the suggestion. I agree, it is lacking in the department of small passwords. That's why I said it needs sufficiently large passwords in the post. This was just something I was playing with - I don't actually expect it to compete with any standard cryptographic methods. – miike3459 Jun 1 '19 at 21:05
• You missed the point. A two character pass phrase creates a two bit encryption key; an 10 character pass phrase creates a 10 bit encryption key!!! – AJNeufeld Jun 1 '19 at 22:28
• No, yes, I understand that. It's a limitation I've been trying to get past. – miike3459 Jun 1 '19 at 22:29