fix some formatting in articles

jakobkogler · jakobkogler · commit 72f7b48d49e7 · 2022-01-08T19:52:20.000+01:00
diff --git a/src/algebra/phi-function.md b/src/algebra/phi-function.md
@@ -8,6 +8,7 @@ hide:
 Euler's totient function, also known as **phi-function** $\phi (n)$, counts the number of integers between 1 and $n$ inclusive, which are coprime to $n$. Two numbers are coprime if their greatest common divisor equals $1$ ($1$ is considered to be coprime to any number).
 
 Here are values of $\phi(n)$ for the first few positive integers:
+
 $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
 \hline
 n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 \\\\ \hline
diff --git a/src/algebra/polynomial.md b/src/algebra/polynomial.md
@@ -65,33 +65,47 @@ $$A = a_0 + a_1 x + \dots + a_n x^n$$
 
 $$B = b_0 + b_1 x + \dots + b_m x^m$$
 
-You have to compute polynomial $C = A \cdot B$, which is defined as $$\boxed{C = \sum\limits_{i=0}^n \sum\limits_{j=0}^m a_i b_j x^{i+j}}  = c_0 + c_1 x + \dots + c_{n+m} x^{n+m}.$$
-It can be computed in $O(n \log n)$ via the [Fast Fourier transform](./algebra/fft.html) and almost all methods here will use it as subroutine.
+You have to compute polynomial $C = A \cdot B$, which is defined as
+
+$$\boxed{C = \sum\limits_{i=0}^n \sum\limits_{j=0}^m a_i b_j x^{i+j}}  = c_0 + c_1 x + \dots + c_{n+m} x^{n+m}.$$
+
+It can be computed in $O(n \log n)$ via the [Fast Fourier transform](fft.md) and almost all methods here will use it as subroutine.
 
 ### Inverse series
 
 If $A(0) \neq 0$ there always exists an infinite formal power series $A^{-1}(x) = q_0+q_1 x + q_2 x^2 + \dots$ such that $A^{-1} A = 1$. It often proves useful to compute first $k$ coefficients of $A^{-1}$ (that is, to compute it modulo $x^k$). There are two major ways to calculate it.
 
 #### Divide and conquer
 
-This algorithm was mentioned in [Schönhage's article](http://algo.inria.fr/seminars/sem00-01/schoenhage.pdf) and is inspired by [Graeffe's method](https://en.wikipedia.org/wiki/Graeffe's_method). It is known that for $B(x)=A(x)A(-x)$ it holds that $B(x)=B(-x)$, that is, $B(x)$ is an even polynomial. It means that it only has non-zero coefficients with even numbers and can be represented as $B(x)=T(x^2)$. Thus, we can do the following transition: $$A^{-1}(x) \equiv \frac{1}{A(x)} \equiv \frac{A(-x)}{A(x)A(-x)} \equiv \frac{A(-x)}{T(x^2)} \pmod{x^k}$$
+This algorithm was mentioned in [Schönhage's article](http://algo.inria.fr/seminars/sem00-01/schoenhage.pdf) and is inspired by [Graeffe's method](https://en.wikipedia.org/wiki/Graeffe's_method). It is known that for $B(x)=A(x)A(-x)$ it holds that $B(x)=B(-x)$, that is, $B(x)$ is an even polynomial. It means that it only has non-zero coefficients with even numbers and can be represented as $B(x)=T(x^2)$. Thus, we can do the following transition:
+
+$$A^{-1}(x) \equiv \frac{1}{A(x)} \equiv \frac{A(-x)}{A(x)A(-x)} \equiv \frac{A(-x)}{T(x^2)} \pmod{x^k}$$
 
 Note that $T(x)$ can be computed with a single multiplication, after which we're only interested in the first half of coefficients of its inverse series. This effectively reduces the initial problem of computing $A^{-1} \pmod{x^k}$ to computing $T^{-1} \pmod{x^{\lfloor k / 2 \rfloor}}$.
 
-The complexity of this method can be estimated as $$T(n) = T(n/2) + O(n \log n) = O(n \log n).$$
+The complexity of this method can be estimated as
+
+$$T(n) = T(n/2) + O(n \log n) = O(n \log n).$$
 
 #### Sieveking–Kung algorithm
 
 The generic process described here is known as Hensel lifting, as it follows from Hensel's lemma. We'll cover it in more detail further below, but for now let's focus on ad hoc solution. "Lifting" part here means that we start with the approximation $B_0=q_0=a_0^{-1}$, which is $A^{-1} \pmod x$ and then iteratively lift from $\bmod x^a$ to $\bmod x^{2a}$.
 
-Let $B_k \equiv A^{-1} \pmod{x^a}$. The next approximation needs to follow the equation $A B_{k+1} \equiv 1 \pmod{x^{2a}}$ and may be represented as $B_{k+1} = B_k + x^a C$. From this follows the equation $$A(B_k + x^{a}C) \equiv 1 \pmod{x^{2a}}.$$
+Let $B_k \equiv A^{-1} \pmod{x^a}$. The next approximation needs to follow the equation $A B_{k+1} \equiv 1 \pmod{x^{2a}}$ and may be represented as $B_{k+1} = B_k + x^a C$. From this follows the equation
+
+$$A(B_k + x^{a}C) \equiv 1 \pmod{x^{2a}}.$$
+
+Let $A B_k \equiv 1 + x^a D \pmod{x^{2a}}$, then the equation above implies
 
-Let $A B_k \equiv 1 + x^a D \pmod{x^{2a}}$, then the equation above implies $$x^a(D+AC) \equiv 0 \pmod{x^{2a}} \implies D \equiv -AC \pmod{x^a} \implies C \equiv -B_k D \pmod{x^a}.$$
+$$x^a(D+AC) \equiv 0 \pmod{x^{2a}} \implies D \equiv -AC \pmod{x^a} \implies C \equiv -B_k D \pmod{x^a}.$$
 
 From this, one can obtain the final formula, which is
+
 $$x^a C \equiv -B_k x^a D  \equiv B_k(1-AB_k) \pmod{x^{2a}} \implies \boxed{B_{k+1} \equiv B_k(2-AB_k) \pmod{x^{2a}}}$$
 
-Thus starting with $B_0 \equiv a_0^{-1} \pmod x$ we will compute the sequence $B_k$ such that $AB_k \equiv 1 \pmod{x^{2^k}}$ with the complexity $$T(n) = T(n/2) + O(n \log n) = O(n \log n).$$
+Thus starting with $B_0 \equiv a_0^{-1} \pmod x$ we will compute the sequence $B_k$ such that $AB_k \equiv 1 \pmod{x^{2^k}}$ with the complexity
+
+$$T(n) = T(n/2) + O(n \log n) = O(n \log n).$$
 
 The algorithm here might seem a bit more complicated than the first one, but it has a very solid and practical reasoning behind it, as well as a great generalization potential if looked from a different perspective, which would be explained further below.
 
@@ -152,10 +166,17 @@ and $G(x, y)$ is some formal power series of $x$ and $y$. With this result we ca
 
 Let $F(Q_k) \equiv 0 \pmod{x^{a}}$. We need to find $Q_{k+1} \equiv Q_k + x^a C \pmod{x^{2a}}$ such that $F(Q_{k+1}) \equiv 0 \pmod{x^{2a}}$.
 
-Substituting $x = Q_{k+1}$ and $y=Q_k$ in the formula above, we get $$F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) + (Q_{k+1} - Q_k)^2 G(x, y) \pmod x^{2a}.$$
+Substituting $x = Q_{k+1}$ and $y=Q_k$ in the formula above, we get
+
+$$F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) + (Q_{k+1} - Q_k)^2 G(x, y) \pmod x^{2a}.$$
+
+Since $Q_{k+1} - Q_k \equiv 0 \pmod{x^a}$, it also holds that $(Q_{k+1} - Q_k)^2 \equiv 0 \pmod{x^{2a}}$, thus
+
+$$0 \equiv F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) \pmod{x^{2a}}.$$
+
+The last formula gives us the value of $Q_{k+1}$:
 
-Since $Q_{k+1} - Q_k \equiv 0 \pmod{x^a}$, it also holds that $(Q_{k+1} - Q_k)^2 \equiv 0 \pmod{x^{2a}}$, thus $$0 \equiv F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) \pmod{x^{2a}}.$$
-The last formula gives us the value of $Q_{k+1}$: $$\boxed{Q_{k+1} = Q_k - \dfrac{F(Q_k)}{F'(Q_k)} \pmod{x^{2a}}}$$
+$$\boxed{Q_{k+1} = Q_k - \dfrac{F(Q_k)}{F'(Q_k)} \pmod{x^{2a}}}$$
 
 Thus, knowing how to invert polynomials and how to compute $F(Q_k)$, we can find $n$ coefficients of $P$ with the complexity $$T(n) = T(n/2) + f(n),$$ where $f(n)$ is the time needed to compute $F(Q_k)$ and $F'(Q_k)^{-1}$ which is usually $O(n \log n)$.
 
@@ -273,7 +294,7 @@ You want to know if $A(x)$ and $B(x)$ have any roots in common. There are two in
 
 ### Euclidean algorithm
 
-Well, we already have an [article](./algebra/euclid-algorithm.html) about it. For an arbitrary  domain you can write the Euclidean algorithm as easy as:
+Well, we already have an [article](euclid-algorithm.md) about it. For an arbitrary  domain you can write the Euclidean algorithm as easy as:
 
 ```cpp
 template<typename T>
diff --git a/src/geometry/convex-hull.md b/src/geometry/convex-hull.md
@@ -118,6 +118,7 @@ with the line connecting the last point in the lower convex hull and the current
 the previous point once added to the hull.
 
 The final convex hull is obtained from the union of the upper and lower convex hull, forming a clockwise hull, and the implementation is as follows.
+
 If you need collinear points, you just need to check for them in the clockwise/counterclockwise routines.
 However, this allows for a degenerate case where all the input points are collinear in a single line, and the algorithm would output repeated points.
 To solve this, we check whether the upper hull contains all the points, and if it does, we just return the points in reverse, as that
