Using bitset to Count Dominating Pairs in High Dimension

Правка en2, от robinyqc, 2024-07-10 03:36:10

Hi Codeforces! Here I want to share a method to count dominating pairs in high dimension, while the number of point($$$n$$$) is small but dimensionality($$$k$$$) is large (like $$$n = 32768, k = 8$$$).

The problem of counting dominating pairs in K-Dimension is as follows:

For two $$$k$$$-tuples $$$a = (a_0, a_1, \dots, a_{k - 1})$$$ and $$$b = (b_0, b_1, \dots, b_{k - 1})$$$, define $$$dom(a, b) = \prod\limits_{i = 0}^{k - 1}[a_i \geq b_i]$$$, which outputs $$$1$$$ if each $$$a_i$$$ is greater than or equal to $$$b_i$$$ ($$$a$$$ dominates $$$b$$$), otherwise $$$0$$$.

Given $$$n$$$ $$$k$$$-tuples $$$t_0, t_1, \dots, t_{n - 1}$$$, for each $$$i \in [0, n)$$$, compute $$$g(i) = \sum\limits_{j = 0}^{n - 1} dom(t_i, t_j)$$$.

When $$$k = 2$$$, the problem simplifies to counting inversions, solvable directly using a Fenwick Tree.

For $$$k = 3$$$, one can consider Divide and Conquer (D&C) approach, achieving a complexity of $$$O(n \log^2 n)$$$, which is excellent. Alternatively, a K-D Tree approach achieves $$$O(n \sqrt{n})$$$ complexity, also a good option.

For $$$k = 4$$$, D&C embedded with D&C or just K-D Tree solutions are less optimal, with reduced complexity benefits.

At $$$k = 5$$$, the Divide and conquer approach is notably less efficient than brute force, and the K-D Tree offers no significant advantage over brute force.

An optimized brute force approach involves using bitsets to enhance efficiency.

Brute Force Approach

An obvious brute force approach is to enumerate $$$i, j$$$ and check whether $$$p_i$$$ dominates $$$p_j$$$ by each dimension in $$$O(k)$$$ time. The overall complexity is $$$O(n^2 k)$$$.

A key property is that $$$dom[(a_{0..i}), (b_{0..i})] = 1$$$ implies $$$dom[(a_{0..i + 1}), (b_{0..i + 1})] = 1$$$. Therefore, the enumeration order can be swapped. First, enumerating dimensions (tuple indices) $$$d$$$, maintaining the dominant relation up to dimension $$$d$$$, and computing the relation up to next dimension based on the current order:

Let $$$b_{d, i, j} = f(p_{i, 0..d - 1}, p_{j, 0..d - 1})$$$. Then:

$$$ b_{d + 1, i, j} \gets b_{d, i, j} \cdot [p_{i, d} \geq p_{j, d}] $$$

In practice, dimension $$$d$$$ can be omitted because each $$$(i, j)$$$ pair is independent, simplifying to:

$$$ b_{i, j} \gets b_{i, j} \cdot [p_{i, d} \geq p_{j, d}] $$$

Total complexity is $$$O(n^2 k)$$$.

Bitset Optimization

It's observed that this problem can be optimized using bitset. Utilize a 2D array bitset<N> b[N]. Considering the transition $$$[p_{i, d} \geq p_{j, d}]$$$, sort points by $$$p_{i, d}$$$ to obtain $$$q$$$, where $$$q_i$$$ is located at $$$p$$$ index $$$l_i$$$.

Enumerating from $$$0$$$ to $$$n - 1$$$, deduce a relationship array $$$r$$$, where $$$r_j$$$ indicates $$$p_{l_j, d} \leq q_{i, d}$$$. Then update $$$b_{l_i}$$$ using $$$b_{l_i} \gets \left(b_{l_i} \text{ bitand } r\right)$$$.

Time complexity is $$$O(\frac{n^2 k}{w})$$$, space complexity is $$$O(\frac{n^2}{z})$$$, where $$$w = 64, z = 8$$$, with excellent constant factors.

Space Optimization

If space constraints are strict, use a grouping method: process $$$S (S \geq w)$$$ points $$$p_{l..l + S - 1}$$$ together. Complexity includes sorting $$$O(nk \log n)$$$, deriving $$$r$$$ for each group $$$O(nk)$$$, and updating $$$b$$$ for each group $$$O(\frac{Snk}{w})$$$, achieving $$$O(\frac{n^2 k}{w})$$$ time and $$$O(\frac{Sk}{z})$$$ space complexity, where $$$S \geq w$$$.

Divide $$$S$$$ (where $$$S \geq w$$$) points $$$p_{l..l + S - 1}$$$ into groups, and each time focus only on the values of $$$b_{l} \dots b_{l + S - 1}$$$. Analyzing the complexity:

  • Sorting takes $$$O(nk\log n)$$$ time;
  • Calculating $$$r$$$ for each group is $$$O(nk)$$$, thus the total complexity is $$$O(\frac{n^2k}{S})$$$. Thus, without space optimization, calculating $$$r$$$ has a complexity of $$$O(nk)$$$;
  • Calculating the array $$$b$$$ for each group takes $$$O(\frac{Snk}{w})$$$, resulting in a total complexity of $$$O(\frac{n^2k}{w})$$$.

Since $$$S \geq w$$$, the time complexity is $$$O(\frac{n^2k}{w})$$$, and the space complexity is $$$O(\frac{Sn}{z})$$$.

Implementation (C++)

Related problem: CF1826E.

There is also a zh-cn version of this post. If you have any interest, try this link.

Теги bitmask, bitset, efficiency, multidimensional

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en8 Английский robinyqc 2024-10-22 14:27:03 35 Fixed a typo. (And solved the mistake of revisions).
en7 Английский robinyqc 2024-10-22 14:25:48 35 Reverted to en4
en6 Английский robinyqc 2024-10-22 14:24:21 0 Fixed a typo.
en5 Английский robinyqc 2024-10-22 14:22:08 35
en4 Английский robinyqc 2024-07-10 04:26:47 4 Tiny change: ', i, j} = f(p_{i, 0..' -> ', i, j} = dom(p_{i, 0..'
en3 Английский robinyqc 2024-07-10 04:14:18 421 publish (published)
en2 Английский robinyqc 2024-07-10 03:36:10 325
en1 Английский robinyqc 2024-07-09 17:19:30 5604 Initial revision (saved to drafts)