Reversible scramble for integers of any number of digits

What you want to do

I wanted to reversibly convert a numerical value with an arbitrary number of digits, and when I searched variously, I found the following articles of great ancestors. Reversible scrambling of integers --C Sharpens you up Reversible scrambling of integers in Python

Changed so that the digit part can be received as an argument. It seems that the bit-reversal part cannot be realized well unless it is a multiplier of 2. (Maybe I didn't find out enough) The reverse part was a little cut off.

scramble.py

# -*- coding: utf-8 -*-
"""
Digit-free reversible integer creation
"""

class Scramble:
    def __init__(self, bit_digit):
        self.__bit_digit = bit_digit
        self.__mask = (1 << bit_digit) - 1

    def scramble(self, number, salt, inverse_salt):
        return self.__trim(self.__reverse(self.__trim(number * salt)) * inverse_salt)

    def __reverse(self, number):
        bit = '0' * self.__bit_digit + bin(number)[2:]
        bit = bit[-self.__bit_digit:]
        bit = ''.join(reversed(list(bit)))
        return long(bit, 2)

    def __trim(self, number):
        return number & self.__mask

def scramble(number, bit_digit, salt, inverse_salt):
    return Scramble(bit_digit).scramble(number, salt, inverse_salt)

When you actually use it

from scramble import scramble

print scramble(1, 48, 0x717b9f2dded3, 0xb784b8b6295b)
>> 186674888497786L

print scramble(186674888497786L, 48, 0x717b9f2dded3, 0xb784b8b6295b
>> 1

It was converted reversibly like this.

important point

If number is 48bit, salt and inverse_salt should also be 48bit.