An alternative sorting order for Mo's algorithm

Revision en3, by gepardo, 2018-08-14 14:00:05

Hello, Codeforces!

I think many of Codeforces users know about Mo's algorithm. For those, who don't know, please read this blog.

Here, I consider an alternative (and faster) approach of sorting queries in Mo's algorithm.

# Canonical version of the algorithm

The canonical implementation of this algorithm gives time complexity if insertions and deletions work in O(1).

Usually, the two comparators for query sorting are used. The slowest one:

struct Query {
int l, r, idx;

inline pair<int, int> toPair() const {
return make_pair(l / block, r);
}
};

inline bool operator<(const Query &a, const Query &b) {
return a.toPair() < b.toPair();
}


We know it can be optimized a little if for even blocks we use the reverse order for the right boundary:

struct Query {
int l, r, idx;

inline pair<int, int> toPair() const {
return make_pair(l / block, ((l / block) & 1) ? -r : +r);
}
};

inline bool operator<(const Query &a, const Query &b) {
return a.toPair() < b.toPair();
}


# Achieving a better time complexity

We can achieve complexity for Mo's algorithm (still assuming that insertions and deletions are O(1). Note that is always better than . We can prove it as follows: The last statement is always true, so the statement above is proved.

But how this complexity can be achieved?

## Relation to TSP

At first, notice that number of operations to change the segment (l1;r1) to (l2;r2) will take |l1 - l2| + |r1 - r2| insertions and deletions.

Denote the queries as (l1;r1), (l2;r2), ..., (lq;rq). We need to permute them to minimize the number of operations, i. e. to find a permutation p1, p2, ... pq such that total number of operations is minimal possible.

Now we can see the relationship between Mo's algorithm and TSP on Manhattan metrics. Boundaries of queries can be represented as points on 2D plane. And we need to find an optimal route to visit all these points (process all the queries).

But TSP is NP-complete, so it cannot help us to find the optimal query sorting order. Instead, we should use a fast heuristic approach. A good approach would be the use of recursive curves. Wikipedia says about using Sierpiński curve as a basis to find a good TSP solution.

But Sierpiński curve is not very convenient in implementation for integer point coordinates, so we will use another recursive curve: Hilbert curve.

## Hilbert curve

Let's build a Hilbert curve on a 2k × 2k matrix and pass all the cells on matrix according to this curve. Denote ord(i, j, k) as the number of cells visited before the cell (i;j) in order of Hilbert curve on the 2k × 2k matrix. The following picture shows ord(i, j) for each cell on 8 × 8 matrix: Now we can write a function to determine ord(i, j, k) It will do the following:

• Divide the matrix onto four parts (as Hilbert curve is built recursively, each part is a rotated Hilbert curve).
• Determine, in which of the four parts the cell (i, j) is located.
• Add all the squares in the previous parts 4n - 1·k, where k is the number of parts visited before the current one.
• Recursively go into the selected part (do not forget that it can be rotated).

Code looks in the following way:

inline int64_t hilbertOrder(int x, int y, int pow, int rotate) {
if (pow == 0) {
return 0;
}
int hpow = 1 << (pow-1);
int seg = (x < hpow) ? (
(y < hpow) ? 0 : 3
) : (
(y < hpow) ? 1 : 2
);
seg = (seg + rotate) & 3;
const int rotateDelta = {3, 0, 0, 1};
int nx = x & (x ^ hpow), ny = y & (y ^ hpow);
int nrot = (rotate + rotateDelta[seg]) & 3;
int64_t subSquareSize = int64_t(1) << (2*pow - 2);
int64_t ans = seg * subSquareSize;
int64_t add = hilbertOrder(nx, ny, pow-1, nrot);
ans += (seg == 1 || seg == 2) ? add : (subSquareSize - add - 1);
return ans;
}


Assume the matrix has size 2k × 2k, where 2k ≤ n (you can take k = 20 for most cases or find minimal such k that 2k ≤ n). Now denote oi as ord(li, ri, k) on this matrix. Then sort the queries according to their oi.

Here is a proof why this works in time. Suppose we have n = 2k and

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, where k and l are some integers. (If n and q are not powers of 2 and 4 respectively, we increase them, it don't have any effect on asymptotic). Now divide the matrix onto squares with size 2l × 2l. To travel between a pair of adjacent squares, we need time, so we can travel between all the squares in time. Now consider the groups of queries inside a 2l × 2l square. Here we can travel from one query to another in , so we process all such groups of queries in . So the total time to process all the queries is , which was to be proved.

# Benchmarks

Let's compare the canonical version of Mo's algorithm and the version with Hilbert order. To do this, we will use the problem 617E - XOR и любимое число with different constraints for n and q. The implementations are here:

• Standard implementation: code
• Mo with Hilbert curves: code

To reduce the amount of input and output, the generators are built into the code and the output is hashed.

For benchmarks I used Polygon. The results are here:

n q n/q Standard Mo time Mo+Hilbert time Time ratio
400000 400000 1 2730 ms 2698 ms 1.012
500000 250000 2 3369 ms 2730 ms 1.234
600000 200000 3 4134 ms 2901 ms 1.425
600000 150000 4 4851 ms 2496 ms 1.944
700000 140000 5 6255 ms 2854 ms 2.192
750000 100000 7.5 5116 ms 2667 ms 1.918
1000000 100000 10 7425 ms 3977 ms 1.866
1000000 40000 25 9671 ms 2355 ms 4.107
1000000 20000 50 9016 ms 1590 ms 5.670
1000000 10000 100 6879 ms 1185 ms 5.805
1000000 5000 200 5802 ms 857 ms 6.770
1000000 2500 400 4897 ms 639 ms 7.664

What is interesting, when I ran these codes locally even on the test with n = q, Hilbert version worked about three times faster.

# Applicability

As you can see, such sorting order doesn't make Mo's algorithm work slower, and when q is significantly less than n, it works much faster than the classical version. So it is ideal for problems with n = 106andq = 104. For smaller q solutions with naive query processing can pass. mo, sqrt,

#### History

Revisions Rev. Lang. By When Δ Comment
en14 gepardo 2020-11-19 20:39:38 1 Tiny change: 'q)\sqrt{n}$. We can ' -> 'q)\sqrt{n})$. We can '
en13 gepardo 2018-09-20 18:30:53 77 Fix proof
en12 gepardo 2018-08-29 06:22:05 22 Bugfix in codes (use [L-1; R] segements instead of [L; R], as required in the problem)
en11 gepardo 2018-08-14 17:49:49 63 Added [cut]
en10 gepardo 2018-08-14 15:52:52 4 Fixed bugs, thanks Vovuh
en9 gepardo 2018-08-14 14:51:38 98 Adding notice about comments
en8 gepardo 2018-08-14 14:46:41 25 Release to the world! (published)
en7 gepardo 2018-08-14 14:44:03 140 Many-many-fixes
en6 gepardo 2018-08-14 14:27:08 90 Fix the proof
en5 gepardo 2018-08-14 14:24:32 2 Tiny change: ' $n = 10^6 and q = 10^4$.' -> ' $n = 10^6$ and $q = 10^4$.'
en4 gepardo 2018-08-14 14:22:57 690 More benchmarks added
en3 gepardo 2018-08-14 14:00:05 457 Added table of contents
en2 gepardo 2018-08-14 13:49:27 5448 The second part of the text is written
en1 gepardo 2018-08-14 12:23:55 3242 Initial revision (saved to drafts)