Hello, Codeforces!

I think many of you 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.

# Table of contents

# Canonical version of the algorithm

The canonical version of this algorithm has 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:

*n*^{2}*q* < *n*(*n* + *q*)^{2},

*nq* < *n*^{2} + 2*nq* + *q*^{2},

*n*^{2} + *nq* + *q*^{2} > 0

But how this complexity can be achieved?

## Relation to TSP

At first, notice that changing the segment (*l*_{1};*r*_{1}) to (*l*_{2};*r*_{2}) will take |*l*_{1} - *l*_{2}| + |*r*_{1} - *r*_{2}| insertions and deletions.

Denote the queries as (*l*_{1};*r*_{1}), (*l*_{2};*r*_{2}), ..., (*l*_{q};*r*_{q}). We need to permute them to minimize the number of operations, i. e. to find a permutation *p*_{1}, *p*_{2}, ... *p*_{q} 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 2^{k} × 2^{k} matrix and visit all the cells on the 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 2^{k} × 2^{k} matrix. The following picture shows *ord*(*i*, *j*) for each cell on the 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 cells in the previous parts. There are 4
^{n - 1}·*k*or them, 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).

The 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[4] = {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 2^{k} × 2^{k}, where 2^{k} ≥ *n* (you can take k = 20 for most cases or find minimal such *k* that 2^{k} ≥ *n*). Now denote *o*_{i} as *ord*(*l*_{i}, *r*_{i}, *k*) on this matrix. Then sort the queries according to their *o*_{i}.

Here is the proof why this works in time. Suppose we have *n* = 2^{k} and *q* = 4^{l}, 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 2^{k - l} × 2^{k - l}. To travel between a pair of adjacent squares, we need *O*(2^{k - l}) time, so we can travel between all the squares in time. Now consider the groups of queries inside a 2^{k - l} × 2^{k - l} square. Here we can travel from one query to another in *O*(2^{k - l}), so we process all such groups of queries in . So the total time to process all the queries in , 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 and Favorite Number with different constraints for *n* and *q*. The implementations are here:

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 | Standard Mo time | Mo+Hilbert time | Time ratio | |
---|---|---|---|---|---|

400000 | 400000 | 1 | 2730 ms | 2698 ms | 1.012 |

1000000 | 1000000 | 1 | 13602 ms | 10841 ms | 1.255 |

500000 | 250000 | 2 | 3369 ms | 2730 ms | 1.234 |

1000000 | 500000 | 2 | 10077 ms | 7644 ms | 1.318 |

600000 | 200000 | 3 | 4134 ms | 2901 ms | 1.425 |

1000000 | 333333 | 3 | 8767 ms | 6240 ms | 1.405 |

600000 | 150000 | 4 | 4851 ms | 2496 ms | 1.944 |

1000000 | 250000 | 4 | 8672 ms | 5553 ms | 1.561 |

700000 | 140000 | 5 | 6255 ms | 2854 ms | 2.192 |

1000000 | 200000 | 5 | 8423 ms | 5100 ms | 1.652 |

750000 | 100000 | 7.5 | 5116 ms | 2667 ms | 1.918 |

1000000 | 333333 | 7.5 | 7924 ms | 4009 ms | 1.976 |

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* (e. g. *n* = *q* = 400000), 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* = 10^{6} and *q* = 10^{4}. For smaller *q* solutions with naive query processing can pass.

Thanks to everyone who read this article! I hope it will be useful for someone.

If something is unclear in the post or the post contains bugs, feel free to write in comments.

Without a doubt that technic is musthave if you wanna squeeze your MO into TL)

Do you have any idea about the worst case for your solution?

I actually wrote a post asking about this problem a while ago: http://codeforces.com/blog/entry/57681. In the comments, a way to make this algorithm run just as fast as Mo's algorithm for O(n) queries is described, by setting

l,rfor all queries to multiples of .In the phrase "find

minimumsuchkthat 2^{k}≤n" do you meanmaximumsuchk? Or even 2^{k}≥n?I meant "find

minimumsuchkthat 2^{k}≥n". This is fixed now, thanks for reporting.Auto comment: topic has been updated by gepardo (previous revision, new revision, compare).The canonical version of Mo's also works in

provided you choose the right block size.Why? Let's fix some (yet unknown) block size

Mand recall what we do inside each block after standard sort (first by block, then by right border). Assume thatq_{i}is the number of queries insidei-th block. So:nsteps to the right;Msteps per query bouncing between the two borders of our block.Here we have operations per block. In general that will be , which turns into as soon as we remember that the is simply

q.The final step is to average these two addends; this happens when .

Voila!

Interesting idea :)

Just tried it and compared with the other versions. The results of the benchmark are here:

(the test order is the same as in the benchmark in the post)

Code

Must definitely try Hilbert curves next time since the outperformance is quite significant in most cases! Thank you for sharing this :)

is also a lower bound: you can't hope to do better. In particular, assume the

qqueries were (aS,bS) for all pairs , and Then any two of them have distanceSfrom each other, so all ordering have total distanceWow this is cool. I put the

`gilbertOrder`

code in existing AC solution and ithalfedmy runtime.If you are intrigued by the Hilbert Curve, I recommend that you watch this video by 3Blue1Brown.

Hilbert curve is also used by Google S2 for spatial indexing that is used by Google Maps.

Wonder how other space-filling curves, such as Peano curve or Moore curve, would fare?

Also these curves fill square spaces, but queries have

L<Rso the area in concern is actually triangular. Would a triangle-filling curve work better?"Supposen= 2^{k}andq= 4^{l}. Now divide the matrix onto squares with size 2^{l}× 2^{l}."The sizes here are not 2

^{l}× 2^{l}, they're actually 2^{k - l}× 2^{k - l}(i.e., ), for the proof to work. Not surprisingly, the same proof applies for the regular Mo's sorting criterion, therefore its complexity (see above comment).This however seems like a more general approach, that doesn't have some "magic" constants, and might even work fast enough for 3-dimensional queries as well (as long as we can design a space-filling curve for 3D). (original Mo's complexity here is , if

n=q)A good thing to experiment with is trying different space-filling curves for different sets of range queries, and comparing the TSP candidates. This can be done without actually adding the extra overhead of implementing an algorithm that solves some problem after the sorting; the total number of expected operations is all we need.

Thanks,

~~I'll fix it soon~~fixed.3D Hilbert curves? (By the way, Hilbert curves are good enough to be generalized to more dimensions)

For 3D case, if we use a 3D Hilbert curve, the time complexity will be .

Generalizing this to more dimensions, we get , where

dis the number of dimensions.