It seem like since 43 is larger than your alphabet, you can just reverse the math. I don't know how to prove there are no hash collisions, so this may have edge cases. For example:
letters = 'weiojpknasdjhsuert'
def hash(string_input):
h = 3
for i in range(0, len(string_input)):
h = h * 43 + letters.index(string_input[i])
return h
n = hash('wepostjapansand')
print(n)
# 9533132150649729383107184
def rev(n):
s = ''
while n:
l = n % 43 # going in reverse this is the index of the next letter
n -= l # now reverse the math...subtract that index
n //= 43 # and divide by 43 to get the next n
if n:
s = letters[l] + s
return s
print(rev(n))
# wepostjapansand
With a more reasonable alphabet, like lowercase ascii and a space, this still seem to be okay:
from string import ascii_lowercase
letters = ascii_lowercase + ' '
def hash(string_input):
h = 3
for i in range(0, len(string_input)):
h = h * 43 + letters.index(string_input[i])
return h
n = hash('this is some really long text to test how well this works')
print(n)
# 4415562436659212948343839746913950248717359454765313646276852138403823934037244869651587521298
def rev(n):
s = ''
# with more compact logic
while n > 3:
s = letters[n % 43] + s
n = (n - (n % 43)) // 43
return s
print(rev(n))
# this is some really long text to test how well this works
The basic idea is that after all the math, the last number is:
prev * 43 + letter_index
This means you can recover the final letter index by taking the prev modulus 43. Then subtract that and divide by 43 (which is just the reverse of the math) and do it again until your number is zero.
