1
$\begingroup$

I was learning the Polynomial Hash function in python, the one used in Rabin Karp Algorithm

This is the implementation I was taught:

def GivenHash(S,p=113,x=10):
  hash = 0
  for i in range(len(S)-1,-1,-1):
    hash = (hash*x + S[i]) % p
  return hash

Here's the (naive) implementation I came up with:

def MyHash(S,p=113,x=10):
  hash = 0
  for i in range(len(S)):
    hash += S[i]*(x**i)
  return hash % p

These equations operate on a list of integers S. I realize that they are equivalent (for eg. they give the same ans 27 on the list S = [1,2,3,4]).

My question is essentially that for 2 integer lists, how do i show that:

(S[0] + (S[1] mod p)*x) mod p = (S[0] + S[1]*x) mod p

I'm new to the concept of modular arithmetic, so I'm sure there's something I'm missing there, but I can't find a way of manipulating the expression using identities to make the left hand side equal to the right.

Similarly, for the rolling hash function, how do i show that:

if h(S[i..i+|P|-1]) = [S[i]*x^0 + S[i+1]*x^1 + .. S[i+|P|-1]*x^(|P|-1)] mod p

then h(S[i+1..i+|P|]) = [ (h(S[i..i+|P|-1]) - S[i]*x^0)/x + S[i+|P|]*x^|P| ] mod p

Why does taking mod p at the end work instead of having to use it everywhere due to the modulo distributive properties?

(Here h(S[i..i+n]) is the rolling hash of S[i..i+n] and |P| is the length of subarray to be hashed)

New contributor
reverseambition is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

1 Answer 1

2
$\begingroup$

It's a Math question, you need to learn modular arithmetic. I will use '%' to denote mod operation, with the same priority as '/':

1. (a+b) % p = (a%p + b%p) % p
2. a%p % p = a%p
3. (a%p * x) % p = (a * x) % p

These rules should be enough to prove your first equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.