Блог пользователя gautam94

How to hash a substring efficiently?
Автор gautam94, 11 лет назад, По-английски

I was reading about hashing from here and I am unable to understand the part about calculation of hash of a substring. I am calculating the hash of the entire input string in this way : h (S) = S [0] + S [1] * P + S [2] * P ^ 2 + S [3] * P ^ 3 + ... + S [N] * P ^ N

Suppose P = 31 and a = 1, b = 2, c = 3 and so on. Then for the input string abcdab, h[0] = 1, h[1] = 32, h[2] = 2915, h[3] = 122079, h[4] = 1045600, h[5] = 58303902. Now from these values, how can I calculate h[0..1] or h[3..5]?

  • Проголосовать: нравится
  • +6
  • Проголосовать: не нравится

»
11 лет назад, # |
← Rev. 3   Проголосовать: нравится +6 Проголосовать: не нравится

Let's store suffix hashes instead.

We can now calculate hashes like that: h[i] = h[i + 1] * p + s[i]

So, we want to find h[l..r]. Let len = r - l + 1

Let's write h[l] and h[r + 1]:

h[l] = s[l] * p^0 + s[l + 1] * p^1 + s[l + 2] * p^2 + ... + s[r] * p^(len - 1) + s[r + 1] * p^len + ...
h[r + 1] = s[r + 1] * p^0 + s[r + 2] * p^1 + ...
h[r + 1] * p^len = s[r + l] * p^len + s[r + 2] * p^(len+1) + ...
h[l] - h[r + 1] * p^len = s[l] * p^0 + s[l + 1] * p^1 + s[l + 2] * p^2 + ... + s[r] * p^(len - 1)

That's exactly what we need.

→ Ответить
  • »
    »
    11 лет назад, # ^ |
      Проголосовать: нравится 0 Проголосовать: не нравится

    Thanks for replying. Can you also tell me how can I use this for longer strings, say len = 10000. P^len will not fit in 64-bit int and I read this which suggests that we shouldn't do calculations mod 2^64. So, what mod should I use and how can I calculate hash of a substring when doing calculations mod some number?

    → Ответить
    • »
      »
      »
      11 лет назад, # ^ |
        Проголосовать: нравится +6 Проголосовать: не нравится

      You should always use a prime modulo, for example, 10^9 + 7 or 10^9 + 9. You can do all the operations by that modulo. Just calculate powers of p by that modulo, all calculations will be fine.

      BTW, when you substract 2 numbers by modulo, you can get a negative number in result. If you use just a % b, in many programming languages it would be negative. For example you can use cout << -3 % 2;, it'll output -1, but you would expect 1. So, the best way to prevent negative numbers is using (a % b + b) % b. For example a = -3, b = 2: (-3 % 2 + 2) % 2 = (-1 + 2) % 2 = 1 % 2 = 1

      → Ответить
      • »
        »
        »
        »
        10 лет назад, # ^ |
          Проголосовать: нравится 0 Проголосовать: не нравится

        can you explain where i have to use (a % b + b) % b in this code http://codeforces.net/contest/182/submission/6734689 i think my error is because of this.

        → Ответить
        • »
          »
          »
          »
          »
          10 лет назад, # ^ |
          ← Rev. 2   Проголосовать: нравится 0 Проголосовать: не нравится

          So, we want to know, what is remainder if we divide a by b. As you know, mathematical remainder always belongs to the range of [0, b)

          So, if we write a % b, we expect to get the remainder. If a >= 0 we'll indeed get the remainder. But if a < 0 we'll not get the remainder (in most languages: C++, Java, but not Python). But actually, if we add b to this, we'll get the remainder (it will be in range (-b, 0]).

          For example:

          a % b = [what we get in C++] (what the remainder actually is)
-4 % 5 = -4 (1)
-153 % 42 = -27 (15)
-89 % 2 = -1 (1)
-88 % 2 = 0 (0)

          So, if we write a % b + b, we'll always get a positive integer. In case of negative a, it will be the remainder. In case of a non-negative a, it will be r + b. To get a true remainer, we'll write (a % b + b) % b. This will work for all positive values of a.

          → Ответить
    • »
      »
      »
      10 лет назад, # ^ |
        Проголосовать: нравится 0 Проголосовать: не нравится

      You can also use a pair of hashes, both of which are computed modulo a different prime. That way you get much better collision resistance and you don't need to multiply 64-bit numbers.

      → Ответить
  • »
    »
    11 лет назад, # ^ |
      Проголосовать: нравится 0 Проголосовать: не нравится

    suppose we have this:

    h[k] means hash of substring [k, N]

    can we get hash(l, r) in O(1)?

    → Ответить
    • »
      »
      »
      11 лет назад, # ^ |
        Проголосовать: нравится +5 Проголосовать: не нравится

      Of course! o.0

      Just store powers of p! Now the result will be h[l] - h[r + 1] * ppow[r - l + 1]

      → Ответить
    • »
      »
      »
      11 лет назад, # ^ |
        Проголосовать: нравится +3 Проголосовать: не нравится

      Here's the implementation in CPP.

      → Ответить
      • »
        »
        »
        »
        11 лет назад, # ^ |
          Проголосовать: нравится +3 Проголосовать: не нравится

        same

        Anti-hash test is coming for you, so don't forget to make all the calculations by some modulo!

        Here's my implementation:

        struct SingleHash {
    vector<int> suf, b;
    int mod;

    SingleHash(string s, int base = 153, int _mod = 1000000009) {
        int n = s.length();
        suf.assign(n + 1, 0); // suf[n] = 0
        b.assign(n + 1, 0);
        b[0] = 1;
        b[1] = base;
        mod = _mod;
        for (int i = n - 1; i >= 0; --i) {
            suf[i] = (s[i] + (ll)suf[i + 1] * b[1]) % mod;
        }
        for (int i = 2; i <= n; ++i) {
            b[i] = (ll)b[i - 1] * b[1] % mod;
        }
    }

    int substr(int l, int r) const { // [l, r]
        ll v = suf[l] - (ll)suf[r + 1] * b[r - l + 1];
        v %= mod;
        v += mod;
        v %= mod;
        return v;
    }
};
        → Ответить
        • »
          »
          »
          »
          »
          11 лет назад, # ^ |
            Проголосовать: нравится 0 Проголосовать: не нравится

          I think this one's correct too. Anyways, thanks a lot for our help. Can you also suggest a few problems that can be solved by hashing?

          → Ответить
          • »
            »
            »
            »
            »
            »
            11 лет назад, # ^ |
              Проголосовать: нравится 0 Проголосовать: не нравится

            the second one (with #define BASE 163) is the correct way to do it.
            having said that, ur first code (despite not having MOD) was "right" because . ur second code wouldn't work without MOD as .

            → Ответить
  • »
    »
    10 лет назад, # ^ |
      Проголосовать: нравится 0 Проголосовать: не нравится

    what if h[l] lesser than h[r+1] * p^len?? i will obviously get negative value.

    → Ответить
  • »
    »
    6 лет назад, # ^ |
      Проголосовать: нравится 0 Проголосовать: не нравится

    Great method !

    → Ответить