Recently I happened to encounter the problem of having to convert big hexadecimal numbers(base-16) to base 93 for storing inside a datastore(since datastores don’t compress data by themselves and by changing the number base to a higher one significantly decreases the text length and thus the bytes being used) however a problem raised as I was trying to figure out the solution to the problem. When trying to change a number base most of the time the process is the following:
- Convert the number from the start base(base-16) to base 10
- Convert the base 10 result to the wanted base(base-93)
The algorithms used for those operations are the following:
--For topic simplicity, let's assume that this module encodes and decodes b93 and that's the alphabet
--A similar process is used for encoding and decoding hex as well
local b93a = " !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz|~"
--Encodes a base10 number to a base93 string
function Encode(n: number): string
local result = ""
repeat
result ..= string.sub(b93a, n%93+1, n%93+1)
n = math.floor(n/93)
until n == 0
return result
end
--Decodes a base93 string to a base10 number
function Decode(text: string): number
local num = 0
for i = 1, #text do
local char = string.sub(text, i, i)
local byte = string.find(b93a, char, nil, true)
num += (byte-1)*93^(i-1)
end
return num
end
So what’s the issue? The issue is large numbers, you see when performing operations such as decoding a base16 that is huge to a base10 lua string and math functions won’t function properly due to not having enough bytes to store the number properly and thus perform operations on it. So I’m asking for a way of converting to convert base16 to base93 while bypassing the above issues(with math and string libraries acting weird) either through a weird way to bypass the base10 step or by using some sort of custom library for the operations(that store numbers as a string instead and have custom simple equations in them for handling them).
