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String hashing using Polynomial rolling hash function

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  • Difficulty Level : Medium
  • Last Updated : 04 Jun, 2022

Hash Function

A Hash function is a function that maps any kind of data of arbitrary size to fixed-size values. The values returned by the function are called Hash Values or digests. There are many popular Hash Functions such as DJBX33A, MD5, and SHA-256. This post will discuss the key features, implementation, advantages and drawbacks of the Polynomial Rolling Hash Function.

Note that if two strings are equal, their hash values should also be equal. But the inverse need not be true.

The polynomial rolling hash function

Polynomial rolling hash function is a hash function that uses only multiplications and additions. The following is the function:

\text{hash(s)} = \text{s}[0] + \text{s}[1]\cdot p + \text{s}[2]\cdot p^2 + \dots + \text{s}[n - 1]\times p^{n - 1}\quad \text{mod}\ m

or simply,

\text{hash(s)} = \displaystyle\sum_{i = 0}^{n - 1} s[i]\cdot p^i\quad \text{mod}\ m

Where

  • The input to the function is a string s      of length n     .
  • p      and m      are some positive integers.
  • The choice of p      and m      affects the performance and the security of the hash function.
  • If the string s      consists of only lower-case letters, then p = 31      is a good choice.
    • Competitive Programmers prefer using a larger value for p     . Examples include 29791     11111     111111     .
  • m      shall necessarily be a large prime since the probability of two keys colliding (producing the same hash) is nearly \cfrac{1}{m}     10^9 + 7      and 10^9 + 9      are widely used values for m     .
  • The output of the function is the hash value of the string s      which ranges between 0      and (m - 1)      inclusive.

Below is the implementation of the Polynomial Rolling Hash Function

C




#include <stdio.h>
#include <string.h>
 
int get_hash(const char* s, const int n) {
    const int p = 31, m = 1e9 + 7;
    int hash = 0;
    long p_pow = 1;
    for(int i = 0; i < n; i++) {
        hash = (hash + (s[i] - 'a' + 1) * p_pow) % m;
        p_pow = (p_pow * p) % m;
    }
    return hash;
}
 
int main() {
    char s[] = "geeksforgeeks";
    int n = strlen(s);
    printf("Hash of %s is %d\n", s, get_hash(s, n));
    return 0;
}

C++




#include <bits/stdc++.h>
using namespace std;
 
struct Hash {
    const int p = 31, m = 1e9 + 7;
    int hash_value;
    Hash(const string& s) {
        int hash_so_far = 0;
        long p_pow = 1;
        const int n = s.length();
        for (int i = 0; i < n; ++i) {
            hash_so_far = (hash_so_far + (s[i] - 'a' + 1) * p_pow) % m;
            p_pow = (p_pow * p) % m;
        }
        hash_value = hash_so_far;
    }
    bool operator==(const Hash& other) {
        return (hash_value == other.hash_value);
    }
};
 
int main() {
    const string s = "geeksforgeeks";
    Hash h(s);
    cout << "Hash of " << s << " is: " << h.hash_value << '\n';
    return 0;
}

Java




class Hash {
    final int p = 31, m = 1000000007;
    int hash_value;
    Hash(String S) {
        int hash_so_far = 0;
        final char[] s = S.toCharArray();
        long p_pow = 1;
        final int n = s.length;
        for (int i = 0; i < n; i++) {
            hash_so_far = (int)((hash_so_far + (s[i] - 'a' + 1) * p_pow) % m);
            p_pow = (p_pow * p) % m;
        }
        hash_value = hash_so_far;
    }
}
 
class Main {
    public static void main(String[] args) {
        String s = "geeksforgeeks";
        Hash h = new Hash(s);
        System.out.println("Hash of " + s + " is " + h.hash_value);
    }
}

Python3




class Hash:
    def __init__(self, s: str):
        self.hash_value = 0
        self.p, self.m = 31, 10**9 + 7
        self.length = len(s)
        hash_so_far = 0
        p_pow = 1
        for i in range(self.length):
            hash_so_far = (hash_so_far + (1 + ord(s[i]) - ord('a')) * p_pow) % self.m
            p_pow = (p_pow * self.p) % self.m
        self.hash_value = hash_so_far
     
    def __eq__(self, other):
        return self.hash_value == other.hash_value
 
 
if __name__ == '__main__':
    s = "geeksforgeeks"
    h = Hash(s)
    print("Hash value of {} is {}".format(s, h.hash_value))

Output

Hash of geeksforgeeks is 609871790

Collisions in Polynomial Rolling Hash

Since the output of the Hash function is an integer in the range [0, m)     , there are high chances for two strings producing the same hash value.

For instance, the strings \text{``countermand''}      and \text{``furnace''}      produce the same hash value for p = 31      and m = 10^9 + 7     .

Also, the strings \text{``answers''}      and  \text{``stead''}      produce the same hash value for p = 37      and m = 10^9 + 9     .

We can guarantee a collision within a very small domain. Consider a set of strings, S     , consisting of only lower-case letters, such that the length of any string in S      doesn’t exceed 7     .

We have |S| = (26 + 26^2 + 26^3 + 26^4 + 26^5 + 26^6 + 26^7) = 8353082582\gt 10^9 + 7     . Since the range of the Hash Function is [0, m)     , one-one mapping is impossible. Hence, we can guarantee a collision by arbitrarily generating two strings whose length doesn’t exceed 7     .

Collision Resolution

We can note that the value of m      affects the chances of collision. We have seen that the probability of collision is \cfrac{1}{m}     . We can increase the value of m      to reduce the probability of collision. But that affects the speed of the algorithm. Larger the value of m     , the slower the algorithm. Also, some languages (C, C++, Java) have a limit on the size of the integer. Hence, we can’t increase the value of m      to a very large value.

Then how can we minimise the chances of a collision?

Note that the hash of a string depends on two parameters: p      and m     .

We have seen that the strings \text{``countermand''}      and \text{``furnace''}      produce the same hash value for p = 31      and m = 10^9 + 7     . But for p = 37      and m = 10^9 + 9     , they produce different hashes.

Observation:

If two strings produce the same hash values for a pair (p1, m1)     , they will produce different hashes for a different pair, (p2, m2)     .

Strategy:

We cannot, however, nullify the chances of collision because there are infinitely many strings. But, surely, we can reduce the probability of two strings colliding.

We can reduce the probability of collision by generating a pair of hashes for a given string. The first hash is generated using p = 31      and m = 10^9 + 7     , while the second hash is generated using p = 37      and m = 10^9 + 9     .

Why will this work?

We are generating two hashes using two different modulo values, m1      and m2     . The probability of a collision is now \cfrac{1}{m1} \times \cfrac{1}{m2}     . Since both m1      and m2      are greater than 10^9     , the probability that a collision occurs is now less than \displaystyle10^{-18}      which is so much better than the original probability of collision, 10^{-9}     .

Below is the implementation for the same

C




#include <stdio.h>
#include <string.h>
 
int get_hash1(const char* s, int length) {
    const int p = 31, m = 1e9 + 7;
    int hash_value = 0;
    long p_pow = 1;
    for(int i = 0; i < length; i++) {
        hash_value = (hash_value + (s[i] - 'a' + 1) * p_pow) % m;
        p_pow = (p_pow * p) % m;
    }
    return hash_value;
}
 
int get_hash2(const char* s, int length) {
    const int p = 37, m = 1e9 + 9;
    int hash_value = 0;
    long p_pow = 1;
    for(int i = 0; i < length; i++) {
        hash_value = (hash_value + (s[i] - 'a' + 1) * p_pow) % m;
        p_pow = (p_pow * p) % m;
    }
    return hash_value;
}
 
int main() {
    char s[] = "geeksforgeeks";
    int length = strlen(s);
    int hash1 = get_hash1(s, length);
    int hash2 = get_hash2(s, length);
    printf("Hash values of %s are: (%d, %d)\n", s, hash1, hash2);
    return 0;
}

C++




#include <bits/stdc++.h>
using namespace std;
 
struct Hash {
    const int p1 = 31, m1 = 1e9 + 7;
    const int p2 = 37, m2 = 1e9 + 9;
    int hash1 = 0, hash2 = 0;
    Hash(const string& s) {
        compute_hash1(s);
        compute_hash2(s);
    }
 
    void compute_hash1(const string& s) {
        long p_pow = 1;
        for(char ch: s) {
            hash1 = (hash1 + (ch + 1 - 'a') * p_pow) % m1;
            p_pow = (p_pow * p1) % m1;
        }
    }
 
    void compute_hash2(const string& s) {
        long p_pow = 1;
        for(char ch: s) {
            hash2 = (hash2 + (ch + 1 - 'a') * p_pow) % m2;
            p_pow = (p_pow * p2) % m2;
        }
    }
 
    // For two strings to be equal
    // they must have same hash1 and hash2
    bool operator==(const Hash& other) {
        return (hash1 == other.hash1 && hash2 == other.hash2);
    }
};
 
int main() {
    const string s = "geeksforgeeks";
    Hash h(s);
    cout << "Hash values of " << s << " are: ";
    cout << "(" << h.hash1 << ", " << h.hash2 << ")" << '\n';
    return 0;
}

Java




class Hash {
    final int p1 = 31, m1 = 1000000007;
    final int p2 = 37, m2 = 1000000009;
    int hash_value1, hash_value2;
    Hash(String s) {
        compute_hash1(s);
        compute_hash2(s);
    }
    void compute_hash1(String s) {
        int hash_so_far = 0;
        final char[] s_array = s.toCharArray();
        long p_pow = 1;
        final int n = s_array.length;
        for (int i = 0; i < n; i++) {
            hash_so_far = (int)((hash_so_far + (s_array[i] - 'a' + 1) * p_pow) % m1);
            p_pow = (p_pow * p1) % m1;
        }
        hash_value1 = hash_so_far;
    }
    void compute_hash2(String s) {
        int hash_so_far = 0;
        final char[] s_array = s.toCharArray();
        long p_pow = 1;
        final int n = s_array.length;
        for (int i = 0; i < n; i++) {
            hash_so_far = (int)((hash_so_far + (s_array[i] - 'a' + 1) * p_pow) % m2);
            p_pow = (p_pow * p2) % m2;
        }
        hash_value2 = hash_so_far;
    }
}
 
class Main {
    public static void main(String[] args) {
        String s = "geeksforgeeks";
        Hash h = new Hash(s);
        System.out.println("Hash values of " + s + " are: " + h.hash_value1 + ", " + h.hash_value2);
    }
}

Python3




class Hash:
    def __init__(self, s: str):
        self.p1, self.m1 = 31, 10**9 + 7
        self.p2, self.m2 = 37, 10**9 + 9
        self.hash1, self.hash2 = 0, 0
        self.compute_hashes(s)
 
    def compute_hashes(self, s: str):
        pow1, pow2 = 1, 1
        hash1, hash2 = 0, 0
        for ch in s:
            seed = 1 + ord(ch) - ord('a')
            hash1 = (hash1 + seed * pow1) % self.m1
            hash2 = (hash2 + seed * pow2) % self.m2
            pow1 = (pow1 * self.p1) % self.m1
            pow2 = (pow2 * self.p2) % self.m2
        self.hash1, self.hash2 = hash1, hash2
 
    def __eq__(self, other):
        return self.hash1 == other.hash1 and self.hash2 == other.hash2
 
    def __str__(self):
        return f'({self.hash1}, {self.hash2})'
 
 
if __name__ == '__main__':
    s = "geeksforgeeks"
    hash = Hash(s)
    print("Hash of " + s + " is " + str(hash))

Output

Hash values of geeksforgeeks are: (609871790, 642799661)

Features of Polynomial rolling hash function

Calculation of Hashes of any substring of a given string in O(1)

Note that computing the hash of the string S will also compute the hashes of all of the prefixes. We just have to store the hash values of the prefixes while computing. Say \text{hash[i]} denotes the hash of the prefix \text{S[0…i]}, we have

\text{hash[i...j]}\cdot p^i = \text{hash[0...j]} - \text{hash[0...(i - 1)]}

This allows us to quickly compute the hash of the substring \text{S[i...j]}      in O(1)      provided we have powers of p      ready.

The behaviour of the hash when a character is changed

Recall that the hash of a string s      is given by

\text{hash(s)} = \displaystyle\sum_{i = 0}^{n - 1} s[i]\cdot p^i\quad \text{mod}\ m

Say, we change a character ch1      at some index i      to some other character ch2     . How will the hash change?

If \text{hash\_old}      denotes the hash value before changing and \text{hash\_new}      is the hash value after changing, then the relation between them is given by

\text{hash\_new} = \text{hash\_old} - p^i\cdot(ch1) + p^i\cdot(ch2)

Therefore, queries can be performed very quickly instead of recalculating the hash from beginning, provided we have the powers of p      ready.

A more elegant implementation is provided below. 

C++




#include <bits/stdc++.h>
using namespace std;
 
long long power(long long x, long long y, long long p) {
    long long result = 1;
    for(; y; y >>= 1, x = x * x % p) {
        if(y & 1) {
            result = result * x % p;
        }
    }
    return result;
}
 
long long inverse(long long x, long long p) {
    return power(x, p - 2, p);
}
 
class Hash {
private:
    int length;
    const int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
    const int p1 = 31, p2 = 37;
    vector<int> hash1, hash2;
    pair<int, int> hash_pair;
 
public:
    inline static vector<int> inv_pow1, inv_pow2;
    inline static int inv_size = 1;
     
    Hash() {}
 
    Hash(const string& s) {
        length = s.size();
        hash1.resize(length);
        hash2.resize(length);
 
        int h1 = 0, h2 = 0;
        long long p_pow1 = 1, p_pow2 = 1;
        for(int i = 0; i < length; i++) {
            h1 = (h1 + (s[i] - 'a' + 1) * p_pow1) % mod1;
            h2 = (h2 + (s[i] - 'a' + 1) * p_pow2) % mod2;
            p_pow1 = (p_pow1 * p1) % mod1;
            p_pow2 = (p_pow2 * p2) % mod2;
            hash1[i] = h1;
            hash2[i] = h2;
        }
        hash_pair = make_pair(h1, h2);
 
        if(inv_size < length) {
            for(; inv_size < length; inv_size <<= 1);
             
            inv_pow1.resize(inv_size, -1);
            inv_pow2.resize(inv_size, -1);
 
            inv_pow1[inv_size - 1] = inverse(power(p1, inv_size - 1, mod1), mod1);
            inv_pow2[inv_size - 1] = inverse(power(p2, inv_size - 1, mod2), mod2);
             
            for(int i = inv_size - 2; i >= 0 && inv_pow1[i] == -1; i--) {
                inv_pow1[i] = (1LL * inv_pow1[i + 1] * p1) % mod1;
                inv_pow2[i] = (1LL * inv_pow2[i + 1] * p2) % mod2;
            }
        }
    }
 
    int size() {
        return length;
    }
 
    pair<int, int> prefix(const int index) {
        return {hash1[index], hash2[index]};
    }
 
    pair<int, int> substr(const int l, const int r) {
        if(l == 0) {
            return {hash1[r], hash2[r]};
        }
        int temp1 = hash1[r] - hash1[l - 1];
        int temp2 = hash2[r] - hash2[l - 1];
        temp1 += (temp1 < 0 ? mod1 : 0);
        temp2 += (temp2 < 0 ? mod2 : 0);
        temp1 = (temp1 * 1LL * inv_pow1[l]) % mod1;
        temp2 = (temp2 * 1LL * inv_pow2[l]) % mod2;
        return {temp1, temp2};
    }
 
    bool operator==(const Hash& other) {
        return (hash_pair == other.hash_pair);
    }
};
 
 
int main() {
    string my_str = "geeksforgeeks";
    const int n = my_str.length();
    auto hash = Hash(my_str);
    auto hash_pair = hash.substr(0, n - 1);
    cout << "Hashes of the string " << my_str << " are:\n";
    cout << hash_pair.first << ' ' << hash_pair.second << '\n';
    return 0;
}

In the above implementation, we are computing the inverses of powers of $p$ in linear time.

Applications:

Consider this problem: Given a sequence S of N strings and Q queries. In each query, you are given two indices, i and j, your task is to find the length of the longest common prefix of the strings S[i] and S[j].

Before getting into the approach to solve this problem, note that the constraints are:

1\le N \le 10^5\\ 1\le Q \le 10^5\\ 1\le |S| \le 10^5\\ \text{The Sum of |S| over all test cases doesn't exceed } 10^6

Using Hashing, the problem can be solved in O(N + Q/log|S|_{max}). The approach is to compute hashes for all the strings in O(N) time, Then for each query, we can binary search the length of the longest common prefix using hashing. The implementation for this approach is provided below.

C++14




#include <bits/stdc++.h>
using namespace std;
 
long long power(long long x, long long y, long long p) {
    long long result = 1;
    for(; y; y >>= 1, x = x * x % p) {
        if(y & 1) {
            result = result * x % p;
        }
    }
    return result;
}
 
long long inverse(long long x, long long p) {
    return power(x, p - 2, p);
}
 
class Hash {
private:
    int length;
    const int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
    const int p1 = 31, p2 = 37;
    vector<int> hash1, hash2;
    pair<int, int> hash_pair;
 
public:
    inline static vector<int> inv_pow1, inv_pow2;
    inline static int inv_size = 1;
     
    Hash() {}
 
    Hash(const string& s) {
        length = s.size();
        hash1.resize(length);
        hash2.resize(length);
 
        int h1 = 0, h2 = 0;
        long long p_pow1 = 1, p_pow2 = 1;
        for(int i = 0; i < length; i++) {
            h1 = (h1 + (s[i] - 'a' + 1) * p_pow1) % mod1;
            h2 = (h2 + (s[i] - 'a' + 1) * p_pow2) % mod2;
            p_pow1 = (p_pow1 * p1) % mod1;
            p_pow2 = (p_pow2 * p2) % mod2;
            hash1[i] = h1;
            hash2[i] = h2;
        }
        hash_pair = make_pair(h1, h2);
 
        if(inv_size < length) {
            for(; inv_size < length; inv_size <<= 1);
             
            inv_pow1.resize(inv_size, -1);
            inv_pow2.resize(inv_size, -1);
 
            inv_pow1[inv_size - 1] = inverse(power(p1, inv_size - 1, mod1), mod1);
            inv_pow2[inv_size - 1] = inverse(power(p2, inv_size - 1, mod2), mod2);
             
            for(int i = inv_size - 2; i >= 0 && inv_pow1[i] == -1; i--) {
                inv_pow1[i] = (1LL * inv_pow1[i + 1] * p1) % mod1;
                inv_pow2[i] = (1LL * inv_pow2[i + 1] * p2) % mod2;
            }
        }
    }
 
    int size() {
        return length;
    }
 
    pair<int, int> prefix(const int index) {
        return {hash1[index], hash2[index]};
    }
 
    pair<int, int> substr(const int l, const int r) {
        if(l == 0) {
            return {hash1[r], hash2[r]};
        }
        int temp1 = hash1[r] - hash1[l - 1];
        int temp2 = hash2[r] - hash2[l - 1];
        temp1 += (temp1 < 0 ? mod1 : 0);
        temp2 += (temp2 < 0 ? mod2 : 0);
        temp1 = (temp1 * 1LL * inv_pow1[l]) % mod1;
        temp2 = (temp2 * 1LL * inv_pow2[l]) % mod2;
        return {temp1, temp2};
    }
 
    bool operator==(const Hash& other) {
        return (hash_pair == other.hash_pair);
    }
};
 
void query(vector<Hash>& hashes, const int N) {
    int i = 0, j = 0;
    cin >> i >> j;
    i--, j--;
    int lb = 0, ub = min(hashes[i].size(), hashes[j].size());
    int max_length = 0;
    while(lb <= ub) {
        int mid = (lb + ub) >> 1;
        if(hashes[i].prefix(mid) == hashes[j].prefix(mid)) {
            if(mid + 1 > max_length) {
                max_length = mid + 1;
            }
            lb = mid + 1;
        }
        else {
            ub = mid - 1;
        }
    }
    cout << max_length << '\n';
}
 
int main() {
    int N = 0, Q = 0;
    cin >> N >> Q;
    vector<Hash> hashes;
    for(int i = 0; i < N; i++) {
        string s;
        cin >> s;
        hashes.push_back(Hash(s));
    }
    for(; Q > 0; Q--) {
        query(hashes, N);
    }
    return 0;
}

Input:
5 4
geeksforgeeks geeks hell geeksforpeaks hello
1 2
1 3
3 5
1 4
Expected Output:
5
0
4
8

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