cp-algorithms
diff --git a/‎src/algebra/fft.md
Lines changed: 4 additions & 4 deletions b/‎src/algebra/fft.md
Lines changed: 4 additions & 4 deletions
diff --git a/‎src/algebra/phi-function.md
Lines changed: 0 additions & 1 deletion b/‎src/algebra/phi-function.md
Lines changed: 0 additions & 1 deletion
diff --git a/‎src/algebra/polynomial.md
Lines changed: 78 additions & 72 deletions b/‎src/algebra/polynomial.md
Lines changed: 78 additions & 72 deletions
diff --git a/‎src/combinatorics/burnside.md
Lines changed: 1 addition & 1 deletion b/‎src/combinatorics/burnside.md
Lines changed: 1 addition & 1 deletion
@@ -299,26 +299,26 @@ Therefore if we reverse the bits of the position of each coefficient, and sort t
 
 For example the desired order for $n = 8$ has the form:
 
-$$a = \left\{ \left[ (a_0, a_4), (a_2, a_6) \right], \left[ (a_1, a_5), (a_3, a_7) \right] \right\}$$
+$$a = \bigg\{ \Big[ (a_0, a_4), (a_2, a_6) \Big], \Big[ (a_1, a_5), (a_3, a_7) \Big] \bigg\}$$
 
 Indeed in the first recursion level (surrounded by curly braces), the vector gets divided into two parts $[a_0, a_2, a_4, a_6]$ and $[a_1, a_3, a_5, a_7]$.
 As we see, in the bit-reversal permutation this corresponds to simply dividing the vector into two halves: the first $\frac{n}{2}$ elements and the last $\frac{n}{2}$ elements.
 Then there is a recursive call for each halve.
 Let the resulting DFT for each of them be returned in place of the elements themselves (i.e. the first half and the second half of the vector $a$ respectively.
 
-$$a = \left\{[y_0^0, y_1^0, y_2^0, y_3^0], [y_0^1, y_1^1, y_2^1, y_3^1]\right\}$$
+$$a = \bigg\{ \Big[y_0^0, y_1^0, y_2^0, y_3^0\Big], \Big[y_0^1, y_1^1, y_2^1, y_3^1 \Big] \bigg\}$$
 
 Now we want to combine the two DFTs into one for the complete vector.
 The order of the elements is ideal, and we can also perform the union directly in this vector.
 We can take the elements $y_0^0$ and $y_0^1$ and perform the butterfly transform.
 The place of the resulting two values is the same as the place of the two initial values, so we get:
 
-$$a = \left\{[y_0^0 + w_n^0 y_0^1, y_1^0, y_2^0, y_3^0], [y_0^0 - w_n^0 y_0^1, y_1^1, y_2^1, y_3^1]\right\}$$
+$$a = \bigg\{ \Big[y_0^0 + w_n^0 y_0^1, y_1^0, y_2^0, y_3^0\Big], \Big[y_0^0 - w_n^0 y_0^1, y_1^1, y_2^1, y_3^1\Big] \bigg\}$$
 
 Similarly we can compute the butterfly transform of $y_1^0$ and $y_1^1$ and put the results in their place, and so on.
 As a result we get:
 
-$$a = \left\{[y_0^0 + w_n^0 y_0^1, y_1^0 + w_n^1 y_1^1, y_2^0 + w_n^2 y_2^1, y_3^0 + w_n^3 y_3^1], [y_0^0 - w_n^0 y_0^1, y_1^0 - w_n^1 y_1^1, y_2^0 - w_n^2 y_2^1, y_3^0 - w_n^3 y_3^1]\right\}$$
+$$a = \bigg\{ \Big[y_0^0 + w_n^0 y_0^1, y_1^0 + w_n^1 y_1^1, y_2^0 + w_n^2 y_2^1, y_3^0 + w_n^3 y_3^1\Big], \Big[y_0^0 - w_n^0 y_0^1, y_1^0 - w_n^1 y_1^1, y_2^0 - w_n^2 y_2^1, y_3^0 - w_n^3 y_3^1\Big] \bigg\}$$
 
 Thus we computed the required DFT from the vector $a$.
 
 
@@ -33,7 +33,6 @@ $$\phi(p^k) = p^k - p^{k-1}.$$
 
     This relation is not trivial to see. It follows from the [Chinese remainder theorem](chinese-remainder-theorem.md). The Chinese remainder theorem guarantees, that for each $0 \le x < a$ and each $0 \le y < b$, there exists a unique $0 \le z < a b$ with $z \equiv x \pmod{a}$ and $z \equiv y \pmod{b}$. It's not hard to show that $z$ is coprime to $a b$ if and only if $x$ is coprime to $a$ and $y$ is coprime to $b$. Therefore the amount of integers coprime to $a b$ is equal to product of the amounts of $a$ and $b$.
 
-
   - In general, for not coprime $a$ and $b$, the equation
 
     \[\phi(ab) = \phi(a) \cdot \phi(b) \cdot \dfrac{d}{\phi(d)}\]
 
@@ -10,40 +10,30 @@ In this article we will cover common operations that you will probably have to d
 
 ## Basic Notion and Facts
 
-Consider a polynomial $A(x) = a_0 + a_1 x + \dots + a_n x^n$ such that $a_n \neq 0$.
+Let $A(x) = a_0 + a_1 x + \dots + a_n x^n$ be a polynomial over some field $\mathbb F$. It For simplicity we will write $A$ instead of $A(x)$ wherever possible, which will be understandable from the context. It is assumed that either $a_n \neq 0$ or $A(x)=0$.
 
-  - For simplicity we will write $A$ instead of $A(x)$ wherever possible, which will be understandable from the context.
-  - We will define the degree of polynomial $A$ as $\deg A = n$. It is convenient to say that $\deg A = -\infty$ for $A(x) = 0$.
-  - For arbitrary polynomials $A$ and $B$ it holds that $\deg AB = \deg A + \deg B$.
-  - Polynomials form an euclidean ring which means that for any polynomials $A$ and $B \neq 0$ we can uniquely represent $A$ as:
+The degree of polynomial $A$ with $a_n \neq 0$ is defined as $\deg A = n$. For consistency, degree of $A(x) = 0$ is defined as $\deg A = -\infty$. In this notion, $\deg AB = \deg A + \deg B$ for arbitrary polynomials $A$ and $B$.
 
-    \[ A = D \cdot B + R,~ \deg R < \deg B. \]
+Polynomials form an Euclidean ring which means that for any polynomials $A$ and $B \neq 0$ we can uniquely represent $A$ as $$A = D \cdot B + R,~ \deg R < \deg B.$$ Here $R$ is the remainder of $A$ modulo $B$ and $D$ is called the quotient. If $A$ and $B$ have the same remainder modulo $C$, they're said to be equivalent modulo $C$, which is denoted as $A \equiv B \pmod{C}$. Several important properties of polynomial Euclidean division:
 
-    Here $R$ is called remainder of $A$ modulo $B$ and $D$ is called quotient.
+- $A$ is a multiple of $B$ if and only if $A \equiv 0 \pmod B$.
 
-  - If $A$ and $B$ have the same remainder modulo $C$, they're said to be equivalent modulo $C$, which is denoted as:
+- It implies that $A \equiv B \pmod C$ if and only if $A-B$ is a multiple of $C$.
 
-    \[A \equiv B \pmod{C}\]
+- In particular, $A \equiv B \pmod{C \cdot D}$ implies $A \equiv B \pmod{C}$.
 
-  - For any linear polynomial $x-r$ it holds that:
+- For any linear polynomial $x-r$ it holds that $A(x) \equiv A(r) \pmod{x-r}$.
 
-    \[A(x) \equiv A(r) \pmod{x-r}\]
+- It implies that $A$ is a multiple of $x-r$ if and only if $A(r)=0$.
 
-  - In particular:
-    
-    \[A(r) = 0 \iff A(x) \equiv 0 \pmod {x-r}\]
-
-    Which means that $A$ is divisible by $x-r$ $\iff$ $A(r)=0$.
-
-  - If $A \equiv B \pmod{C \cdot D}$ then $A \equiv B \pmod{C}$
-  - $A \equiv a_0 + a_1 x + \dots + a_{k-1} x^{k-1} \pmod{x^k}$
+- For modulo being $x^k$, it holds that $A \equiv a_0 + a_1 x + \dots + a_{k-1} x^{k-1} \pmod{x^k}$.
 
 ## Basic implementation
 [Here](https://github.com/e-maxx-eng/e-maxx-eng-aux/blob/master/src/polynomial.cpp) you can find the basic implementation of polynomial algebra.
 
-It supports all trivial operations and some other useful methods. The main class is `poly<T>` for polynomials with coefficients of class `T`.
+It supports all trivial operations and some other useful methods. The main class is `poly<T>` for polynomials with coefficients of type `T`.
 
-All arithmetic operation `+`, `-`, `*`, `%` and `/` are supported, `%` and `/` standing for remainder and quotient in integer division.
+All arithmetic operation `+`, `-`, `*`, `%` and `/` are supported, `%` and `/` standing for remainder and quotient in Euclidean division.
 
 There is also the class `modular<m>` for performing arithmetic operations on remainders modulo a prime number `m`.
 
@@ -69,97 +59,113 @@ Other useful functions:
 
 ### Multiplication
 
-The very core operation is the multiplication of two polynomials, that is, given polynomial $A$ and $B$:
+The very core operation is the multiplication of two polynomials. That is, given the polynomials $A$ and $B$:
 
 $$A = a_0 + a_1 x + \dots + a_n x^n$$
 
 $$B = b_0 + b_1 x + \dots + b_m x^m$$
 
-You have to compute polynomial $C = A \cdot B$:
+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.
 
-$$\boxed{C = \sum\limits_{i=0}^n \sum\limits_{j=0}^m a_i b_j x^{i+j}}  = c_0 + c_1 x + \dots + c_{n+m} x^{n+m}$$
+### Inverse series
 
-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.
+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.
 
-### Inverse series
+#### 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: $$A^{-1}(x) \equiv \frac{1}{A(x)} \equiv \frac{A(-x)}{A(x)A(-x)} \equiv \frac{A(-x)}{T(x^2)} \pmod{x^k}$$
+
+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 $$T(n) = T(n/2) + O(n \log n) = O(n \log n).$$
+
+#### 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 $$A(B_k + x^{a}C) \equiv 1 \pmod{x^{2a}}.$$
 
-If $A(0) \neq 0$ there always exists an infinite series $A^{-1}(x) = \sum\limits_{i=0}^\infty a_i'x^i$ such that $A^{-1} A = 1$.
+Let $A B_k \equiv 1 + x^a D \pmod{x^{2a}}$, then the equation above implies $$x^a(D+AC) \equiv 0 \pmod{x^{2a}} \implies D \equiv -AC \pmod{x^a} \implies C \equiv -B_k D \pmod{x^a}.$$
 
-It may be reasonable for us to calculate first $k$ coefficients of $A^{-1}$:
+From this, one can obtain the final formula, which is
+$$x^a C \equiv -B_k x^a D  \equiv B_k(1-AB_k) \pmod{x^{2a}} \implies \boxed{B_{k+1} \equiv B_k(2-AB_k) \pmod{x^{2a}}}$$
 
- 1. Let's say that $A^{-1} \equiv B_k \pmod{x^{a}}$. That means that $A B_k \equiv 1 \pmod {x^{a}}$.
- 2. We want to find $B_{k+1} \equiv B_k + x^{a}C \pmod{x^{2a}}$ such that $A B_{k+1} \equiv 1 \pmod{x^{2a}}$:
- 
-    \[A(B_k + x^{a}C) \equiv 1 \pmod{x^{2a}}\]
-   
- 3. Note that since $A B_k \equiv 1 \pmod{x^{a}}$ it also holds that $A B_k \equiv 1 + x^a D \pmod{x^{2a}}$. Thus:
- 
-    \[x^a(D+AC) \equiv 0 \pmod{x^{2a}} \implies D \equiv -AC \pmod{x^a} \implies C \equiv -B_k D \pmod{x^a}\]
- 
- 4. From this we obtain that:
- 
-    \[x^a C \equiv -B_k x^a D  \equiv B_k(1-AB_k) \pmod{x^{2a}} \implies \boxed{B_{k+1} \equiv B_k(2-AB_k) \pmod{x^{2a}}}\]
+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.
 
-### Division with remainder
+### Euclidean division
 
-Consider two polynomials $A(x)$ and $B(x)$ of degrees $n$ and $m$. As it was said earlier you can rewrite $A(x)$ as:
+Consider two polynomials $A(x)$ and $B(x)$ of degrees $n$ and $m$. As it was said earlier you can rewrite $A(x)$ as
 
-$$A(x) = B(x) D(x) + R(x), \deg R < \deg B$$
+$$A(x) = B(x) D(x) + R(x), \deg R < \deg B.$$
 
-Let $n \geq m$, then you can immediately find out that $\deg D = n - m$ and that leading $n-m+1$ coefficients of $A$ don't influence $R$.
+Let $n \geq m$, it would imply that $\deg D = n - m$ and the leading $n-m+1$ coefficients of $A$ don't influence $R$. It means that you can recover $D(x)$ from the largest $n-m+1$ coefficients of $A(x)$ and $B(x)$ if you consider it as a system of equations.
 
-That means that you can recover $D(x)$ from the largest $n-m+1$ coefficients of $A(x)$ and $B(x)$ if you consider it as a system of equations.
+The system of linear equations we're talking about can be written in the following form:
 
-The formal way to do it is to consider the reversed polynomials:
+$$\begin{bmatrix} a_n \\ \vdots \\ a_{m+1} \\ a_{m} \end{bmatrix} = \begin{bmatrix}
+b_m & \dots & 0 & 0 \\
+\vdots & \ddots & \vdots & \vdots \\
+\dots & \dots & b_m & 0 \\
+\dots & \dots & b_{m-1} & b_m
+\end{bmatrix} \begin{bmatrix}d_{n-m} \\ \vdots \\ d_1 \\ d_0\end{bmatrix}$$
+
+From the looks of it, we can conclude that with the introduction of reversed polynomials
 
 $$A^R(x) = x^nA(x^{-1})= a_n + a_{n-1} x + \dots + a_0 x^n$$
 
 $$B^R(x) = x^m B(x^{-1}) = b_m + b_{m-1} x + \dots + b_0 x^m$$
 
 $$D^R(x) = x^{n-m}D(x^{-1}) = d_{n-m} + d_{n-m-1} x + \dots + d_0 x^{n-m}$$
 
-Using these terms you can rewrite that statement about the largest $n-m+1$ coefficients as:
+the system may be rewritten as
 
-$$A^R(x) \equiv B^R(x) D^R(x) \pmod{x^{n-m+1}}$$
+$$A^R(x) \equiv B^R(x) D^R(x) \pmod{x^{n-m+1}}.$$
 
-From which you can unambiguously recover all coefficients of $D(x)$:
+From this you can unambiguously recover all coefficients of $D(x)$:
 
 $$\boxed{D^R(x) \equiv A^R(x) (B^R(x))^{-1} \pmod{x^{n-m+1}}}$$
 
-And from this in turn you can easily recover $R(x)$ as $R(x) = A(x) - B(x)D(x)$.
+And from this, in turn, you can recover $R(x)$ as $R(x) = A(x) - B(x)D(x)$.
+
+Note that the matrix above is a so-called triangular [Toeplitz matrix](https://en.wikipedia.org/wiki/Toeplitz_matrix) and, as we see here, solving system of linear equations with arbitrary Toeplitz matrix is, in fact, equivalent to polynomial inversion. Moreover, inverse matrix of it would also be triangular Toeplitz matrix and its entries, in terms used above, are the coefficients of $(B^R(x))^{-1} \pmod{x^{n-m+1}}$.
 
 ## Calculating functions of polynomial
 
 ### Newton's method
-Let's generalize the inverse series approach.
-You want to find a polynomial $P(x)$ satisfying $F(P) = 0$ where $F(x)$ is some function represented as:
 
-$$F(x) = \sum\limits_{i=0}^\infty \alpha_i (x-\beta)^k$$
+Let's generalize the Sieveking–Kung algorithm. Consider equation $F(P) = 0$ where $P(x)$ should be a polynomial and $F(x)$ is some polynomial-valued function defined as
+
+$$F(x) = \sum\limits_{i=0}^\infty \alpha_i (x-\beta)^k,$$
+
+where $\beta$ is some constant. It can be proven that if we introduce a new formal variable $y$, we can express $F(x)$ as
+
+$$F(x) = F(y) + (x-y)F'(y) + (x-y)^2 G(x,y),$$
+
+where $F'(x)$ is the derivative formal power series defined as
+
+$$F'(x) = \sum\limits_{i=0}^\infty (k+1)\alpha_{i+1}(x-\beta)^k,$$
+
+and $G(x, y)$ is some formal power series of $x$ and $y$. With this result we can find the solution iteratively.
+
+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 $$F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) + (Q_{k+1} - Q_k)^2 G(x, y) \pmod x^{2a}.$$
 
-Where $\beta$ is some constant. It can be proven that if we introduce a new formal variable $y$, we can express $F(x)$ as:
+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 $Q_{k+1}$: $$\boxed{Q_{k+1} = Q_k - \dfrac{F(Q_k)}{F'(Q_k)} \pmod{x^{2a}}}$$
 
-$$F(x) = F(y) + (x-y)F'(y) + (x-y)^2 G(x,y)$$
+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 $O(n \log n)$.
 
-Where $F'(x)$ is the derivative formal power series defined as $F'(x) = \sum\limits_{i=0}^\infty (k+1)\alpha_{i+1}(x-\beta)^k$ and $G(x, y)$ is some formal power series of $x$ and $y$.
+The iterative rule above is known in numerical analysis as [Newton's method](https://en.wikipedia.org/wiki/Newton%27s_method).
 
-Given this we can find the coefficients of the solution iteratively:
+#### Hensel's lemma
 
- 1. Assume that $F(Q_k) \equiv 0 \pmod{x^{a}}$, we want to find $Q_{k+1} \equiv Q_k + x^a C \pmod{x^{2a}}$ such that $F(Q_{k+1}) \equiv 0 \pmod{x^{2a}}$.
- 2. Pasting $x = Q_{k+1}$ and $y=Q_k$ in the formula above we get:
-    
-    \[F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) + (Q_{k+1} - Q_k)^2 G(x, y) \pmod x^{2a}\]
-    
- 3. Since $Q_{k+1} - Q_k \equiv 0 \pmod{x^a}$ we can say that $(Q_{k+1} - Q_k)^2 \equiv 0 \pmod{x^{2a}}$, thus: 
-    
-    \[0 \equiv F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) \pmod{x^{2a}}\]
-    
- 4. From the last formula we derive the value of $Q_{k+1}$:
- 
-    \[\boxed{Q_{k+1} = Q_k - \dfrac{F(Q_k)}{F'(Q_k)} \pmod{x^{2a}}}\]
+As was mentioned earlier, formally and generically this result is known as [Hensel's lemma](https://en.wikipedia.org/wiki/Hensel%27s_lemma) and it may in fact used in even broader sense when we work with a series of nested rings. In this particular case we worked with a sequence of polynomial remainders modulo $x$, $x^2$, $x^3$ and so on.
 
-Thus knowing how to invert arbitrary polynomial and how to compute $F(Q_k)$ quickly, we can find $n$ coefficients of $P$ with the complexity: \[T(n) = T(n/2) + f(n)\] Where $f(n)$ is the maximum of $O(n \log n)$ needed to invert series and time needed to compute $F(Q_k)$ which is usually also $O(n \log n)$.
+Another example where Hensel's lifting might be helpful are so-called [p-adic numbers](https://en.wikipedia.org/wiki/P-adic_number) where we, in fact, work with the sequence of integer remainders modulo $p$, $p^2$, $p^3$ and so on. For example, Newton's method can be used to find all possible [automorphic numbers](https://en.wikipedia.org/wiki/Automorphic_number) (numbers that end on itself when squared) with a given number base. The problem is left as an exercise to the reader. You might consider [this](https://acm.timus.ru/problem.aspx?space=1&num=1698) problem to check if your solution works for $10$-based numbers.
 
 
 ### Logarithm
@@ -267,7 +273,7 @@ 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](euclid-algorithm.md) about it. For an arbitrary euclidean domain you can write the Euclidean algorithm as easy as:
+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:
 
 ```cpp
 template<typename T>
 
@@ -119,7 +119,7 @@ $$|\text{Classes}| \cdot |G| = \sum_{f} J(f)$$
 ## Pólya enumeration theorem
 
 The Pólya enumeration theorem is a generalization of Burnside's lemma, and it also provides a more convenient tool for finding the number of equivalence classes.
-It should noted that this theorem was already discovered before Pólya by Redfield in 1927, but his publication went unnoticed by mathematicians.
+It should be noted that this theorem was already discovered before Pólya by Redfield in 1927, but his publication went unnoticed by mathematicians.
 Pólya independently came to the same results in 1937, and his publication was more successful.
 
 Here we discuss only a special case of the Pólya enumeration theorem, which will turn out very useful in practice.