### Monogon's blog

By Monogon, history, 4 weeks ago, Randomization is known to be a powerful algorithmic technique, and I want to start a discussion about it specifically in the context of competitive programming. What extra considerations should we make about how online judges work when analyzing our randomized algorithms, and approving such problems for an official contest? In particular, I want to discuss algorithms whose running times are uncertain.

Let's say that the running time of an algorithm is a random variable $X$. Typically, we might say the algorithm is acceptable if the expected running time is a fraction of the time limit: $E(X)=T/k$. With Markov's inequality, we can say $P(X>T)\le 1/k$. That is, the probability of failing a test case due to timeout is at most $1/k$.

## TLE Trick

Online judges usually run your program a few times before giving the Time Limit Exceeded verdict. This fact is known to help randomized algorithms, as mentioned here: https://codeforces.com/blog/entry/78817

Let's apply this in the context of random running times. If your program is run $n$ times, the actual probability of failing a test case is at most $1/k^n$. If there are $m$ test cases total, the probability of failing at least one of the test cases is $1-(1-1/k^n)^m$

## Solving for $k$

Let's suppose we call a program acceptable if its probability of failure is at most $p$. Then we are interested in what the constant $k$ should be in order to be confident of success.

$k\ge \left[ 1-(1-p)^{1/m}\right]^{-1/n}$

Let's plug in some values. Suppose there are $m=200$ test cases, your program gets $n=3$ chances per test case, and we can tolerate a failure probability of $p=10^{-3}$. Then we should require $k\approx 58.47$, which is quite a considerable constant factor that we should be careful about when making randomized solutions.

## Future Directions

If we make extra assumptions on the distribution of $X$ that apply widely to randomized algorithms, can we be confident with a much smaller value of $k$? What will the analysis look like on some specific cases like treaps? I'm interested to hear your thoughts.

By Monogon, history, 3 months ago, I hope everyone enjoyed the contest!

UPD: Added implementations for all problems.

1405A - Permutation Forgery

Author: antontrygubO_o

Tutorial

Implementation

1405B - Array Cancellation

Author: hugopm

Tutorial

Implementation

1405C - Balanced Bitstring

Author: Kuroni

Tutorial

Implementation

1405D - Tree Tag

Author: Maripium

Tutorial

Implementation

1405E - Fixed Point Removal

Author: hugopm

Tutorial

Implementation

1404D - Game of Pairs

Author: Ari

Tutorial

Implementation

1404E - Bricks

Author: Monogon

Tutorial

Implementation Tutorial of Codeforces Round #668 (Div. 1) Tutorial of Codeforces Round #668 (Div. 2)
By Monogon, history, 3 months ago, Warning: The following contains graphic depictions of geometry, precision errors, and degenerate cases. Viewer discretion is advised.

### Prerequisites

I assume the reader is familiar with:

• 2D Convex Hulls
• 3D Vector Operations (dot and cross products)

### Introduction

Recall that in the 2D convex hull problem, you are given a set of 2D points, and you must compute the smallest convex polygon containing all the given points. By convex, we mean that for any two points $A$ and $B$ inside the polygon, the entire line segment $AB$ is also inside the polygon.

The problem in 3D is completely analogous. You are given $n$ 3D points, and you must compute the smallest convex polyhedron containing all the given points. Similarly, a polyhedron is convex if for any two points $A$ and $B$ inside the polyhedron, the line segment $AB$ is also inside the polyhedron.

In the 2D case, it is more obvious what the output is. We can simply output a circular list of vertices on the boundary of the polygon. But how do we represent a polyhedron? Well, we can represent it as a list of triangular faces. A non-triangular face can always be divided into triangles, so this requirement isn't limiting.

In the 2D case, there are some degenerate cases that are easy to miss if you're not careful. For example, if there are 3 collinear points, the polygon might have two adjacent sides of the same slope, or you might combine them into one segment. Or if two points have the same $x$ coordinate, you need to handle this carefully when sorting the points by $x$ coordinate. Or if all the points are on the same line, then the convex hull isn't even a non-degenerate polygon!

Now, degenerate cases are bad enough with two dimensions. When you enter the third dimension, it only gets worse. What if four points lie on the same plane? Since we are requiring triangular faces, we could triangulate this in multiple ways. Or maybe we could choose to have one triangle of three points and the forth point lies inside this triangle, so we ignore it. What if all the points are on the same line? Or on the same plane even? Then the convex hull isn't a non-degenerate polyhedron. I will ignore such degenerate cases for now, and revisit them when applicable.

### Brute Force Algorithm in $O(n^4)$

Suppose that $n$ is very small, and we are guaranteed that no four points are coplanar. Then how can we compute the 3D convex hull in the dumbest way possible?

We could simply consider every triplet of three points $(\vec{a}, \vec{b}, \vec{c})$, and check if they create a triangular face on the convex hull. In order for it to be a face, the remaining points must "see" the same side of the triangle. In other words, if we consider the plane containing this triangle, the remaining points should lie on the same side of the plane. To compute which side of a plane a point $\vec{p}$ is on, we can first take a vector $\vec{q}=(\vec{b}-\vec{a})\times (\vec{c}-\vec{a})$ orthogonal to the plane. Then the sign of $(\vec{p}-\vec{a})\cdot \vec{q}$ tells us the side of the plane. In particular, a result of $0$ tells us that $\vec{p}$ lies on the plane.

For each triplet, we perform this check with all points, so the total time complexity is $O(n^4)$. Because of its simplicity, this should be the approach when the constraints allow it. And by examining the brute force case, we learned how to perform the most fundamental operation in any 3D convex hull algorithm: checking which side of a plane a point is on.

### Incremental Algorithm in $O(n^2)$

Now we want a more practical algorithm. The strategy of the incremental algorithm is to build the convex hull for the first $i$ points, as $i$ goes from $1$ to $n$. All we have to do is figure out how to update the convex hull in order to accommodate one new point.

Let's start by making an analogy with the 2D convex hull. Suppose we currently have the convex hull of the first $i-1$ points, and we wish to add the $i$-th point. Well, if the point is already inside the hull, we should do nothing. Otherwise, let's pretend we are looking at the polygon from the perspective of the $i$-th point. We should delete all line segments it can see, and add two new line segments: one from the new point to each of the extreme points. A similar thing happens in the 3D case. If a point is already inside the polyhedron, we simply do nothing. Otherwise, we delete all faces the new point can see. But what's less obvious is what we should add. Well, we've left the polyhedron with a connected set of faces removed. This exposes a cycle of edges. For each of these exposed edges, we should create a new face with the new point and that edge. Effectively, we create a cone of faces to repair the opening.

Now, let's analyze the time complexity. For each iteration, we process all faces of the hull in that iteration. And it can be proven that the number of faces is at most $O(n)$. For the proof, we use Euler's Formula. It states that for any polyhedron with $V$ vertices, $E$ edges, and $F$ faces, $V-E+F=2$. All faces are triangles, so we can substitute $E=3F/2$ since each face has $3$ edges, and we count each edge twice for the $2$ faces it touches. Then we have $V-F/2=2$ and $F=2V-4=O(n)$. Since we process $O(n)$ faces in $O(n)$ iterations, the overall time complexity is $O(n^2)$.

Implementation

### Clarkson-Shor Algorithm in $O(n\log n)$

There exist deterministic algorithms to compute the 3D convex hull in $O(n\log n)$ time. Most famously, there is a divide-and-conquer algorithm that solves it, but it is notoriously difficult to implement correctly. But don't lose hope! There is also a randomized algorithm based on the same incremental strategy, which takes $O(n\log n)$ time in expectation if the points are added in random order.

Unfortunately, we can't just shuffle the points, run the usual incremental algorithm, and call it a day. In order to achieve $O(n\log n)$, we should eliminate extra work of processing faces invisible to the new point. How can we do this? Well, before we checked for each point, which faces it can see. Instead, we will maintain for each point a list of faces it can see. Then when a face is created, we add it to the lists of all points that will be able to see it.

"Wait a minute," I hear you ask, "isn't this still $O(n^2)$ because you still process the same thing for each face-point pair, just in a different order?" The answer is yes, we're still in $O(n^2)$ territory. But, using some cleverness we can avoid checking with every future point every time. The key idea is that a new face $F$ is attached to a preexisting edge $e$, and a point $p$ can see $F$ only if it was able to see one of the two faces of $e$. Instead of checking with all future points, we will only check the points in the lists associated with these two faces. Unfortunately, this requires more work to implement. This time, we need to start caring about how the faces touch each other in order to find these two lists.

### Precision Errors and Degenerate Cases

The only potential source for precision error is our check of which side of a plane a point is on. If all coordinates are integers, then our computations will stay as integers. If it overflows on 64-bit integer types, we can convert to floating points for computations, and set $\epsilon=0.5$. Note that if the coordinates are at most $X,Y,Z$ in absolute value (in the $x$, $y$, and $z$ directions, respectively), our computations are on the order of $XYZ$. Use this to determine whether it will overflow on integers or find what you should set $\epsilon$ to. Or just try every possible $\epsilon$ until it works.

To handle degenerate cases, we should make sure that the first $4$ points are not coplanar. If all the $n$ points do not lie in the same plane, we can find $4$ such points and move them to the beginning. Once this is taken care of, the only real degenerate case remaining is when we need to decide if a face is visible to a point in the same plane. I made the decision to consider the face invisible in such a case. This ensures that the algorithm won't unnecessarily subdivide faces. However, it is still possible that the algorithm can have non-unique output depending on the order the points are processed. This is inevitable since there is not a unique way to triangulate faces in degenerate cases. If we want to, though, we can combine adjacent coplanar faces afterward.

To my knowledge, the Clarkson-Shor algorithm doesn't make any guarantees about $O(n\log n)$ performance when there are degenerate cases. I think by considering coplanar faces invisible, we are safe. But if anyone has a nasty test case against it, I'm interested to hear it.

Here is my implementation of the $O(n\log n)$ algorithm with full handling of degenerate cases (I hope). If anyone finds bugs or ways to improve the efficiency and/or readability of this implementation, please let me know!

Implementation

### Application: Delaunay Triangulation and Voronoi Diagrams

Suppose we have $n$ 2D points $p_1,\ldots,p_n$. We may want to answer queries of the form "Which point is closest to $q_i$?" for $m$ points $q_1,\ldots,q_m$.

The Voronoi diagram divides the plane into $n$ regions in a way that can answer these kinds of queries. It is defined so that the region corresponding to the point $i$ consists of all points closer to $p_i$ than any other of the $n$ points. That is, it's the region of points whose nearest neighbor is $p_i$. The Voronoi diagram can be represented by $O(n)$ line segments that border the regions (some line segments will extend to a point at infinity). We can answer our queries by sorting all line segments and queries, and performing a sweep line.

The dual of the Voronoi diagram is called the Delaunay Triangulation. It is the graph that connects two points if their Voronoi cells share a boundary edge. Conversely, the points of a Voronoi diagram are the circumcenters of Delaunay triangles, and an edge in the Voronoi diagram connects the circumcircles of two adjacent Delaunay triangles.

The Delaunay triangulation has several nice properties that come in handy. Most notably, the Euclidean minimum spanning tree consists of a subset of Delaunay edges. A nice consequence of implementing 3D convex hull is that we get Delaunay triangulation for free. We can simply map each point $(x,y)$ into a 3D point $(x,y,x^2+y^2)$. Then the downward-facing triangles of the 3D convex hull are precisely the Delaunay triangles. The proof is left as an exercise to the reader.

### Practice Problems

By Monogon, history, 4 months ago, I hope you enjoyed the contest! I will add implementations later.

1382A - Common Subsequence

Tutorial

Implementation

1382B - Sequential Nim

Tutorial

Implementation

1382C1 - Prefix Flip (Easy Version)

Tutorial

1382C2 - Prefix Flip (Hard Version)

Tutorial

1382D - Unmerge

Tutorial

Implementation

1382E - Mastermind

Tutorial

Implementation

1381D - The Majestic Brown Tree Snake

Tutorial

Implementation

1381E - Origami

Tutorial

Implementation Tutorial of Codeforces Round #658 (Div. 1) Tutorial of Codeforces Round #658 (Div. 2)
By Monogon, history, 4 months ago, Hello, Codeforces!

I'm very glad to invite you to Codeforces Round #658 (Div. 1) and Codeforces Round #658 (Div. 2). This contest will take place on Jul/21/2020 17:35 (Moscow time). In both divisions, you will have 2 hours to solve 5 problems (and one subtask). The score distribution will be announced closer to the start of the round.

Huge thanks to:

I've worked hard to ensure the pretests are short and the statements are strong. Remember to only read the problems that you can solve, and may you have the best of luck!

Because I know you all have so many questions, I have compiled an FAQueue

• Q: Is it rated?
• A: Yes

UPD Here is the score distribution:

Div. 2: 500 — 1250 — (1000 + 1000) — 2250 — 3000

Div. 1: (500 + 500) — 1500 — 2000 — 2500 — 3000

UPD Editorial

UPD I filled in the answer to the FAQ. Also, congrats to the winners!

Div. 2:

Div. 1: Announcement of Codeforces Round #658 (Div. 2)
By Monogon, history, 7 months ago, Hello, Codeforces!

I'm very glad to invite you to Codeforces Round #639 (Div. 1) and Codeforces Round #639 (Div. 2). This contest will take place on May/06/2020 17:35 (Moscow time). In both divisions, you will have 2 hours 15 minutes to solve 6 problems. The score distribution will be announced closer to the start of the contest.

Of course, this round would not be possible without the help of many people, who I'd like to thank:

I've done my best to write clear problem statements and strong pretests. Don't forget to read all problems, and I hope you enjoy the round. Good luck!

Because I know you all have so many questions, I have compiled an FAQ:

• Q: Is it rated?
• A: Yes.

UPD

Here is the score distribution.

Div. 1: 500 — 1000 — 1500 — 1750 — 2500 — 2500

Div. 2: 500 — 1000 — 1500 — 2000 — 2500 — 2750

UPD Unfortunately, this contest is delayed due to problems in the Codeforces system. It is temporarily scheduled, but I will update this announcement when a final decision on scheduling has been made. It is sad that my contest is delayed, but let's be grateful that the issue was detected early instead of arising in the middle of the round. On the bright side, this may be the earliest announced score distribution in Codeforces history.

UPD The final decision on scheduling is to keep it as is. The contest will take place on May/06/2020 17:35 (Moscow time).

UPD Unfortunately due to the long queue and other issues, the contest is unrated. There was also a bug that gave us trouble in answering your questions. I'm sorry if you asked a question that went unanswered for a long time. I promise we did our best to answer as many questions as we could! Despite all the problems, I hope you enjoyed the problemset!

UPD Editorial Announcement of Codeforces Round #639 (Div. 1) Announcement of Codeforces Round #639 (Div. 2)
By Monogon, history, 7 months ago, 1345A - Puzzle Pieces

Tutorial

1345B - Card Constructions

Tutorial

1345C - Hilbert's Hotel

Tutorial

1345D - Monopole Magnets

Tutorial

1345E - Quantifier Question

Tutorial

1345F - Résumé Review

Tutorial

1344E - Train Tracks

Tutorial

1344F - Piet's Palette

Tutorial Tutorial of Codeforces Round #639 (Div. 1) Tutorial of Codeforces Round #639 (Div. 2) Tutorial of div2 B-C #1