Encoding a 128 bit integer in Python?

Inspired by the "coding scheme" of the answer to this question , I implemented my own coding algorithm in Python.

This is how it looks:

import random
from math import pow
from string import ascii_letters, digits

# RFC 2396 unreserved URI characters
unreserved = '-_.!~*\'()'
characters = ascii_letters + digits + unreserved
size = len(characters)
seq = range(0,size)

# Seed random generator with same randomly generated number

dictionary = dict(zip(seq, characters))
reverse_dictionary = dict((v,k) for k,v in dictionary.iteritems())

def encode(n):
    d = []
    n = n
    while n > 0:
        qr = divmod(n, size)
        n = qr[0]
    chars = ''
    for i in d:
        chars += dictionary[i]
    return chars

def decode(str):
    d = []
    for c in str:
    value = 0
    for i in range(0, len(d)):
        value += d[i] * pow(size, i)
    return value


The problem I'm running into is encoding and decoding very large integers. For example, this is how large numbers are currently being encoded and decoded:

s = encode(88291326719355847026813766449910520462)
# print s -> "3_r(AUqqMvPRkf~JXaWj8"
i = decode(s)
# print i -> "8.82913267194e+37"
# print long(i) -> "88291326719355843047833376688611262464"


The highest 16 places match up perfectly, but after them the number deviates from the original.

I guess this is a precision issue with very large integers when dividing in Python. Is there a way to get around this problem? Or is there another problem that I am not aware of?


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1 answer

The problem is this line:

value += d[i] * pow(size, i)


You seem to be using math.pow

the inline method here instead pow

. It returns a floating point number, so you lose precision for your large numbers. You must use the built-in operator, pow

or **

, better yet, store the current cardinality of the base in an integer variable:

def decode(s):
    d = [reverse_dictionary[c] for c in s]
    result, power = 0, 1
    for x in d:
        result += x * power
        power *= size
    return result


Now it gives me the following output:

print decode(encode(88291326719355847026813766449910520462))
# => 88291326719355847026813766449910520462




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