String Hashing

Template

As mentioned in the articles above, there is no need to calculate modular inverses.

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class HashedString {
  private:
	// change M and B if you want
	static const long long M = 1e9 + 9;
	static const long long B = 9973;

	// pow[i] contains B^i % M
	static vector<long long> pow;

	// p_hash[i] is the hash of the first i characters of the given string

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import java.util.*;

public class HashedString {
	// Change M and B if you want
	public static final long M = (long)1e9 + 9;
	public static final long B = 9973;

	// pow[i] contains B^i % M
	private static ArrayList<Long> pow = new ArrayList<>();

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class HashedString:
	# Change M and B if you want
	M = int(1e9) + 9
	B = 9973

	# pow[i] contains B^i % M
	_pow = [1]

	def __init__(self, s: str):
		while len(self._pow) <= len(s):

This implementation calculates

The hash of any particular substring $S[a : b]$ is then calculated as

using prefix sums. This is nice because the highest power of $B$ in that polynomial will always be $B^{b - a}$.

Since $10^9 + 9$ is prime, the probability of collision when using this hash is at most $\frac{N}{10^9 + 9} < 10^{-4}$, by the Schwarz-Zippel lemma. This means that if you select any two different strings of length at most $N=10^5$ and a random base modulo $10^9 + 9$ (e.g. $9973$ in the code), the probability that they hash to the same value is at most $10^{-4}$.

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mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
const ll B = uniform_int_distribution<ll>(0, M - 1)(rng);

Searching For Strings

Explanation - One Hash

We'll use a sliding window over $H$ to find the "matches" with $N$.

Since we don't care about relative order when comparing two substrings, we can store frequency tables of the characters in the current window and in $N$. When we slide the window, at most two values in that table change. To compare two substrings, we simply compare the 26 values in each table.

If we only needed to count the number of matches, then the above alone would suffice (in fact, IOI 2006 Writing is just that). However, we need to count the distinct permutations of $N$ in $H$, so we need to be a bit more clever.

One way to solve this is by storing the polynomial hashes of each match in a set, since we expect different permutations to have different polynomial hashes. The answer would simply be the size of that set at the end.

Using a relatively small modulus such as $M=10^9+9$ will not pass (see the note above regarding the birthday paradox). Instead, we use $M=2^{61}-1$.

Implementation

Time Complexity: $\mathcal O((|N| + |H|) \cdot \Sigma)$, where $\Sigma$ is the size of the alphabet.

Failure Probability: $\mathcal O\left(\frac{|N||H|^2}{M}\right)$

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#include <bits/stdc++.h>
using namespace std;

using ll = long long;

 Code Snippet: HashedString (Click to expand)

int freq_target[26], freq_curr[26];
string n, h;

Explanation - Two Hashes

An alternative solution without frequency tables would be to hash the substrings that we're trying to match. Since order doesn't matter, we need to modify our hash function slightly.

In particular, instead of computing the polynomial hash of the substrings, compute the product $(B + s_1)(B + s_2) \dots (B + s_k) \bmod M$ as the hash (again, using two modulos). This hash is nice because the relative order of the letters doesn't matter, as multiplication is commutative. Furthermore, as any two strings with different frequency tables map to different polynomials in $B$, they hash to the same value with probability at most $\frac{|N|}{M}$ over the choice of $B$.

Since this hash requires the modular inverse, there's an extra $\log M$ factor in the time complexity.

Implementation

Time Complexity: $\mathcal O((|N| + |H|) \log M)$

Failure Probability: $\mathcal O\left(\frac{|N||H|^2}{M}\right)$

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#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

 Code Snippet: HashedString (Click to expand)

const auto M = HashedString::M;
const auto B = HashedString::B;
const auto mul = HashedString::mul;
const auto mod_mul = HashedString::mod_mul;

Problems

XOR Hashing/Zobrist Hashing

Hashing can also be used to check if sets of elements are equal. To do this, we first randomly generate a hash value for each unique element. Typically, the hash value is an integer in the range $[0, 2^{63}-1]$ since $2^{63}-1$ is the maximum value of a 64-bit signed integer. The hash of a set $S$ is the XOR sum of the hash values of all the elements in $S$. Since $x \oplus x = 0$ for all $x$, we can delete an element $s$ from set $S$ by applying the hash value of $s$ again on the hash. The probability of a collision of $N$ sets is approximately $\frac{N^2}{M}$, where $M$ is the maximum possible hash value.

Explanation

For each distinct numerical value in the arrays, we generate a random positive 64-bit integer. With this map, we can build the prefix XOR hashes for $a$ and $b$.

An issue we have to deal with is duplicate elements, as XORing an element with itself will result in a value of $0$ and will be equivalent it never having existed in the first place. To fix this, we use a set to detect duplicate subsequent values and only XOR an element with the prefix hash if it's new.

Now, to answer a query, we check if the XOR hashes at the given indices are the same.

Implementation

Time Complexity: $\mathcal{O}(N\log N + Q)$

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#include <chrono>
#include <iostream>
#include <map>
#include <random>
#include <set>
#include <vector>

using std::cout;
using std::endl;
using std::vector;

Problems