02-möbius inversion formula of "combinatorial mathematics"

There are many kinds of counting problems, we need to pay attention to some common methods and conclusions in order to avoid getting into the aimless burning brain movement. Mathematical abstraction and universality is what we have been advocating, from a number of special problems found in the general method, but also always exciting and sigh with regret. Most of the general teaching materials are arranged by the beginning of the combination, using a number of techniques very strong elementary methods to discuss the combination of counting, I think you can directly grasp some sharp tools, to see those problems, there will be the pleasure of the Gordian knot.

1. Associative algebra 1.1 An example

In order to have an intuitive understanding of the inversion formula, we start with a simple question and examine the summation formula of the series (1). The left-hand means that when you know each item of the sequence (a_n\), you can get the first \ (n\) and \ (s_n\), the right type means know and \ (s_n\) can also be pushed back to the item \ (a_n\). To sort out the elements in this statement, there are two functions defined on the set of natural numbers \ (f (n), g (n) \), \ (f (n) \) can be represented by \ (g (1), \cdots,g (n) \) linear, \ (g (n) \) can also be by \ (f (1), \cdots,f (n) \) The linear table out, the conclusion is that these two linear table out is equivalent, by one can be pushed to export another.

\[s_n=a_1+a_2+\cdots+a_n\;\leftrightarrow\;a_n=s_n-s_{n-1}\tag{1}\]

Such reciprocal derivation is called inversion , and is now in deep search for patterns. With the knowledge of linear algebra, the above derivation is actually a simple linear equation group (equation (2)). There are two points to emphasize, the point is that because of the recursive characteristics, both sides of the matrix are the lower triangular invertible matrix. The other point is that the expression is n\ to any of the arbitrary \ (s), where the vectors and squares can be regarded as infinite dimensions. 2nd let us draw attention to the Square (a,b\), because no matter how much (n\) is how many, both sides of the square in the \ (a_{ij},b_{ij}\) The value is the same!

\[\begin{bmatrix}s_1\\s_2\\\vdots\\s_n\end{bmatrix}=\begin{bmatrix}1&&&\\1&1&&\\\vdots &\vdots&\ddots&\\1&\cdots&1&1\end{bmatrix}\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{ bmatrix}\;\leftrightarrow\;\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}=\begin{bmatrix}1&&&\\ -1&1&&\\&\ddots&\ddots&\\&&-1&1\end{bmatrix}\begin{bmatrix}s_1\\s_2\\\ Vdots\\s_n\end{bmatrix}\tag{2}\]

1.2 Partial-order set

It seems that we have introduced the problem to the infinite dimension of the triangular square, and the object of discussion is its inverse element. In order to get such an extension, we use another method to describe the lower triangular phalanx. First it defines the two-tuple relationship of the positive integer set ((x, y) \), and second only when \ (X\geqslant y\) is meaningful. We know that the set of positive integers is a fully ordered set, and any two numbers are "comparable", and the lower triangular matrix is a function of the "comparable" relationship. Since we want to extend the definition, we may as well define this function in a slightly weaker "order set", only the local is "comparable", the inversion in the local should also be meaningful.

Specifically, we review the partial-order set in set theory, which defines the two-element relationship of set \ (x\) (\geqslant\), which satisfies the reflexive rate, anti-symmetry, transitivity, and is called partial order . A set that defines a partial order is also known as a partial-order set (partially ordered set), where two elements of a two-tuple relationship are called comparable, and all elements between elements are called truncation and are recorded as \ ([x,y]\). The partial-order set can be visually represented by the Hasse graph, where only directly comparable elements (upper and lower) are connected, and the whole looks like a mesh structure. In addition to the integer set, the common partial-order set includes the inclusion relationship between collections, the division of positive integers, and so on.

The partial-order set can be infinite, but for ease of discussion, we assume that the local is finite, i.e. \ ([x,y]\) is a finite set. If the partial order relationship is reversed, the obtained is obviously a partial order set, and the original set is isomorphic, there is no essential difference, they are called dual . In addition, for partial-order sets \ (x_1,x_2,\cdots,x_n\), it is easy to know that the direct product \ (X_1\times\cdots\times x_n\) is also a partial-ordered set under the formula (3) definition. If all subsets of the record \ (n\) tuple are in a partial-order set of \ (B (n) \) under the containing relationship, it is easy to know that it is isomorphic to the n\ \ (l_2^n\) of \ (2\) meta chain \ (l_2\). In the second memory \ (n\), the integer in the division relationship under the partial order set is \ (D (n) \), set \ (N=p_1^{e_1}\cdots p_k^{e_k}\), it is easy to know that it is isomorphic to \ (L_{e_1+1}\times\cdots\times l_{e_k+1}\ )。

\[(x_1,x_2,\cdots,x_n) \geqslant (y_1,y_2,\cdots,y_n) \;\leftrightarrow\;x_1\geqslant Y_1\wedge\cdots\wedge x_n\ Geqslant Y_n\tag{3}\]

1.3 Associative algebra

The function \ (\alpha (x, y) \in x\times x\to\bbb{r}\) is now defined on the partial-order relationship \ (x\), when \ (X\not\geqslant y\) (\alpha (x, y) =0\). To discuss their algebraic structure, remember that the set of all functions is \ (I (X) \), which defines addition and multiplication as trivial. As a reference matrix multiplication, the multiplication of functions (\alpha\beta\) can be defined by the formula (4), which proves that the multiplication satisfies the binding law. In abstract algebra we know that such a structure is called associative algebra, where we call it a partial-order set \ (x\) of associative algebra (incidence algebra), also known as \ (I (X) \).

\[(\alpha\beta) (x, y) =\sum_{x\geqslant z\geqslant y}\alpha (x,z) \beta (z,y) \tag{4}\]

It is easy to verify that the following function \ (\delta\) is the unit element of the associative algebra, and the inverse of the associative algebra is discussed below. Suppose \ (\beta\alpha=\delta\), the fixed \ (x\) value, is defined by the multiplication of the available formula (6). If \ (\alpha (x,x) \ne 0\), it is obvious that you can recursively get any one \ (\beta (x, y) \), which gives the necessary and sufficient conditions for the existence of the inverse of \ (\alpha\): \ (\alpha (x,x) \ne 0\). You can also determine that \ (\beta (x, y) \) depends entirely on \ (z\in[x,y]\) \ (\alpha (z,y) \) values. It is also worth noting that inverse \ (\beta\) is not only related to \ (\alpha\), but to the structure of the partial-order set.

\[\delta (x, y) =\left\{\begin{matrix}1,& (x=y) \\0,& (X\ne y) \end{matrix}\right.\tag{5}\]

\[\beta (x,x) \alpha (x,x) =1,\;\sum_{x\geqslant z\geqslant Y}\beta (x,z) \alpha (z,y) =0\tag{6}\]

In particular, the commonly used function \ (\zeta\) in the definition formula (7) is the zeta function , and its inverse \ (\mu\) is obviously present, and is called the Möbius function . The \mu\ (6) can be obtained by the recursive (8), which only contains \ (0,\pm 1\), the specific value and the structure of the partial-order set. Set the partial order set \ (x_1,\cdots,x_k\) on the function \ (\delta_i,\zeta_i,\mu_i\), for their direct product \ (x\), first of all obviously have the formula (9), the derivation of the formula (10) and the uniqueness of the inverse element also illustrates the third form.

\[\zeta (x, y) =\left\{\begin{matrix}1,& (x\geqslant y) \\0,& (x\not\geqslant y) \end{matrix}\right.\tag{7}\]

\[\mu (x,x) =1,\;\mu (x, y) =-\sum_{x\geqslant z>y}\mu (x,z) \tag{8}\]

\[\delta (x, y) =\prod_{i=1}^k\delta_i (x_i,y_i) \;\zeta (x, y) =\prod_{i=1}^k\zeta_i (x_i,y_i); \;\mu (x, y) =\prod_{i=1} ^k\mu_i (x_i,y_i) \tag{9}\]

\[\delta (x, y) =\prod_{i=1}^k\sum_{x_i\geqslant z_i\geqslant y_i}\zeta_i (x_i,z_i) \mu_i (z_i,y_i) =\sum_{x\geqslant z\ Geqslant y}\prod_{i=1}^k\zeta_i (x_i,z_i) \mu_i (z_i,y_i) \tag{10}\]

2. Möbius inversion Formula 2.1 inversion formula

With the above preparatory work, the problems in the example are now extended. In order to avoid the infinite discussion, first limit to arbitrary \ (x\), less than its element is limited, such a partial order set is called under finite . The same can be defined as limited , with the conclusion that there are also corresponding results in the upper limit. Set \ (f (x), g (x) \) is the function defined on the partial-order set \ (x\), and \ (\alpha,\beta\) is the reciprocal element in \ (I (x) \), then the certificate (11) is required.

\[f (×) =\sum_{y\leqslant X}\alpha (x, y) g (Y) \;\leftrightarrow\;g (x) =\sum_{y\leqslant X}\beta (x, y) f (y) \tag{11}\]

For argument convenience, you can replace \ (\sum\) subscript with any value \ (y\) (because when \ (Y\not\leqslant x\) \ (\alpha (x, y) =0\), prove \ (\rightarrow\), just put \ (f (Y) =\sum\ Limits_z\alpha (Y,z) g (z) \) is brought into the right-hand (\sum\limits_y\beta (x, y) f (y) \). Procedure withheld, the same method can be verified \ (\leftarrow\). In particular, the Möbius inversion formula (12) for the zeta function was established, while formula (11) was the form of its generalization, which was published by American mathematician Rota in 1964.

\[f (x) =\sum_{y\leqslant x}g (y) \;\leftrightarrow\;g (x) =\sum_{y\leqslant x}\mu (y) F \tag{12}\]

When the partial-order set is a finite set, the relational algebra can actually be expressed as a phalanx, when the inverse formula is actually the solution of the linear equation Group. In order to explicitly represent the recursive relationship of the formula (11), we need to arrange the rows (columns) of the square matrix according to the order of the partial ordinal set elements from small to large, which requires the extension of the partial order relationship. This is not difficult to do, first of \ (n=1\) The partial order set is already a full-order set, when \ (n>1\) First put a minimum value as \ (x_1\), the other \ (n-1\) elements still constitute a partial sequence set. It is known from the inductive method that they can extend the x_2\leqslant\cdots\leqslant x_n\, when form set \ (X_1\leqslant\cdots\leqslant x_n\) is the extension that satisfies the condition. The whole sequence of associative algebra is the lower triangular matrix, and the necessary and sufficient conditions for the invertible elements are: the diagonal element of the square is nonzero.

2.2 Application of inversion formula 2.2.1 recursive sequence

Next, we discuss several common partial-order sets, starting with the simplest single-strand \ (l_n\), then the inverse formula (13). The series \ (f (n), g (n), respectively) as the line vector \ (f,g\), and then the recurrence of their relationship as the reciprocal of the lower triangular matrix \ (a,b\), inversion formula is actually ordinary \ (F=ag\leftrightarrow g=bf\). In particular, \ (l_n\) \ (\zeta,\mu\) can be represented as a matrix in the formula (2) respectively. The above conclusion is also the same as n=\infty\, \ (f,g\) becomes the infinite Koriyuki vector, \ (a,b\) is the infinite dimension triangle square.

\[f (n) =\sum_{k=1}^na_{nk}g (k) \;\leftrightarrow\;g (n) =\sum_{k=1}^nb_{nk}f (k) \tag{13}\]

Sometimes we need to get an inverse function (square) other than \ (\zeta,\mu\), and we can reverse the reciprocal function by using two simple sequences of relationships. Here is a polynomial example (the following assumes that the reader has a high school permutation knowledge), assuming there are two polynomial sequences \ (\{p_n (x) \},\{q_n (x) \}\), where \ (P_k (x), q_k (x) \) Order are \ (k\), it is easy to prove that they can be a unique linear representation of each other. If an association function can find the appropriate polynomial sequence, its inverse is easy to obtain. For example, (P_n (x) =x^n\) and \ (Q_n (x) = (x-1) ^n\), it is obvious that the following form is established, so the inverse of the two-item matrix \ (a_{ij}=\binom{i}{j}\) is \ (b_{ij}= ( -1) ^{i-j}\binom{i} {j}\), which is called the inverse formula of the two-item equation.

\[p_n (x) =\sum_{k=0}^n\binom{n}{k}q_k (x); \;\;q_n (x) =\sum_{k=0}^n ( -1) ^{n-k}\binom{n}{k}p_k (x) \tag{14}\]

The two-item equation is a very common sequence, and using this inversion formula can solve some difficult counting problems. For example, consider the \ (n\) Letter to disassemble the reload, remember \ (k\) Letter loaded wrong loading method has \ (d_k\), this number is called dislocation permutation number, will be discussed later. Because \ (n\) Letter of random arrangement has \ (n!\) species, and this can be divided into a \ (0,1,\cdots,n\) Encapsulation of the case, so that there is a formula (15) left-hand, by the two-item inversion formulas to get the equation (15) right-type.

\[n!=\sum_{k=0}^k\binom{n}{k}d_k\;\leftrightarrow\;D _n=\sum_{k=0}^n ( -1) ^{n-k}\binom{n}{k}k!\tag{15}\]

In particular, when \ (Q_n (x) =p_n (×) \), their associated functions are reciprocal with themselves. The following notes \ (((x) _n=x (x-1) \cdots (x-n+1) \), Investigate \ (P_n (x) = (-X) _n\) and \ (Q_n (x) = (x) _n\), because the arbitrary integer \ (m\) has the formula (16) established. Because the linear representation of the polynomial is unique, and the \ (m\) is replaced by the \ (x\) equation, this gets the formula (17), and the corresponding inverse equation is called the lah inversion formula .

\[( -1) ^n\dfrac{(-m) _n}{n!} =\binom{m+n-1}{n}=\sum_{k=0}^n\binom{n-1}{k-1}\dfrac{(m) _k}{k!} \tag{16}\]

\[(-X) _n=\sum_{k=0}^nl_{nk} (x) _n\;\leftrightarrow\;(x) _n=\sum_{k=0}^nl_{nk} (-X) _n,\;l_{nk}= ( -1) ^n\dfrac{n!} {k!} \binom{n-1}{k-1}\tag{17}\]

2.2.2 Principle of tolerance and repulsion

This section discusses the subset of collections \ (a\) \ (B (A) \) The partial-order set under the containing relationship, where only the \ (\zeta,\mu\) function is discussed. As previously known, \ (B (A) \) is isomorphic to \ (l_2^n\), and \ (l_2\) \ (\mu\) function is \ (\begin{bmatrix}1&0\\-1&1\end{bmatrix}\). If subset \ (S\supset t\), use formula (9) easy to know \ (\mu (s,t) = (-1) ^{| s|-| T|} \), this conclusion is to be used immediately.

Now we are going to discuss \ (B (A) \) functions \ (f (s), g (s) \), and they have a relationship \ (f (s) =\sum\limits_{t\subseteq s}g (T) \). To make the problem intuitive and meaningful, you can create a model in which each element of the set has a subset of attributes in the attribute set \ (x=\{p_1,p_2,\cdots,p_n\}\), and we care about two collections: The number of all attribute elements in the property subsets \ (t\) \ (n_{=} (t) \), and at least \ (t\) The number of elements of all attributes \ (N_{\geqslant} (t) \).

This model also has one we are more familiar with the description method, remember that all containing properties \ (p_k\) element is \ (a_k\), then \ (a_1,\cdots,a_n\) can be regarded as a complete \ (a\) \ (n\) subset. The N\ property is précis-writers ([n]\), the subset of attributes \ (t\) is a subset of \ ([n]\), and the formula (18) illustrates the equivalence relationship between the two descriptions. For another familiar set \ (A_1\cup\cdots\cup a_k\), you can first discuss its inverse \ (\bar{a}_1\cap\cdots\cap \bar{a}_k\), reduce the model to \ (K\) properties, it is actually \ (n_{=} (\ varnothing) \).

\[n_{\geqslant} (T) =\left|\bigcap_{i\in t}a_i\right|;\; n_{=} (T) =\left| (\bigcap_{i\in t}a_i) \cap (\bigcap_{j\not\in t}a_j) \right|\tag{18}\]

In a real-world problem, \ (N_{\geqslant} (T) \) is easier to obtain because it simply focuses on having a property set \ (t\) element. \ (n_{=} (T) \) is more difficult to calculate, but it is easy to have a formula (19) left side relationship. They satisfy the inverse formula on the dual partial order set of the attribute subset, it is easy to have the conclusion of the right side of the formula (19). In particular, the formula (20), which has an intuitive description, does not contain any of the nature of the elements can be counted: first in the Complete \ (a\) to remove the number of \ (a_i\), and then the repeated removal of the \ (A_i\cap a_j\), plus the more removed parts .... This process is repeatedly removed and then added, so the formula (20) is also called the repulsion principle .

\[n_{\geqslant} (T) =\sum_{s\supseteq t}n_{=} (S) \;\leftrightarrow\; n_{=} (t) =\sum_{s\supseteq T} (-1) ^{| s|-| T|} N_{\geqslant} (S) \tag{19}\]

\[| A|-\left| A_1\cup\cdots\cup a_n\right|=n_{=} (\varnothing) =\sum_{k=0}^n ( -1) ^k\sum_{| S|=k}n_{\geqslant} (S) \tag{20}\]

The repulsion principle is a very old conclusion, and the proof of the inversion formula is clearer than any elementary proof. In fact, we have already used this conclusion, and here are a few more examples. First of all, to review the problem of dislocation arrangement, the first (i\) bit is not wrong as the nature of (p_i\). There are at least \ (k\) bits in the correct number of \ (\binom{n}{k} (n-k)!\), the use of the principle of tolerance and collation formula (21) left, it is the same as the formula (15) is actually the same. Interestingly, there is also the formula (21) Right, which shows that the probability of the envelope being completely wrong is tending to \ (1/e\).

\[d_n=n!\sum_{k=0}^n\dfrac{( -1) ^k}{k!};\;\;\ lim_{n\to\infty}\dfrac{d_n}{n!} =\dfrac{1}{e}\tag{21}\]

Then look at the Euler function \ (\varphi (n) \), which represents the number of \ ([n]\) and \ (n\) in the count. Set \ (n\) of all factorization is \ (p_1,\cdots,p_k\), to be divided by \ (p_i\) is the nature of (p_i\), \ ([n]\) is at least by the \ (m\) factorization the number of divisible by \ (\sum \dfrac{n}{p_{i_1}\cdots p _{i_m}}\). Using the principle of tolerance and collation (22), it also has obvious probability theory meaning: not being divisible by \ (p_i\) and not divisible by \ (p_j\) is an independent event.

\[\varphi (n) =n (1-\dfrac{1}{p_1}) (1-\dfrac{1}{p_2}) \cdots (1-\dfrac{1}{p_k) \tag{22}\]

　　• sum of the number of n]\ (n\) in the calculation of \ (hint: the sum of the numbers that can be divisible by \ (p_1\cdots p_i\), Answer \ (\dfrac{1}{2}n\varphi (n) \))

　　• the letter \ (a_1,\cdots,a_n\) each has two, uses it to form the \ (2n\) meta-word, the same letter is not adjacent the word how many?

2.2.3 Classical Inversion formula

Finally, we discuss the partial sequence set of the positive integer set in the division of integers, which is the original prototype of the Möbius inversion formula. When \ (d|n\), first calculate \ (\mu (n,d) \), consider the partial-order set \ (d (n) \), where \ (N=p_1^{e_1}\cdots p_k^{e_k}\). Previously known that it is isomorphic to \ (L_{e_1+1}\times\cdots\times l_{e_k+1}\), \ (\mu_i (n_i,d_i) \) (\ (n_i,d_i\) is \ (n,d\) the power of \ (the p_i\)) when \ (n_i=d_i \) equals \ (1\) when \ (n_i=d_ip_i\) is \ ( -1\), and the other is \ (0\).

using the formula (9) can prove that \ (\mu (n,d) \) only with \ (m=\dfrac{n}{d}\) , which can be directly recorded as \ (\mu (m) \). It is the classical Möbius function , it is easy to know that it has an expression (23), under which the classical Möbius inversion formula is type (24), the previous inversion formula is the extension of the conclusion.

\[\mu (M) =\left\{\begin{matrix} ( -1) ^r,&\text{all}e^i=1\\0,&\text{others}\end{matrix}\right.,\;\;(m=p_1^ {e_1}\cdots p_r^{e_r},\;\text{all}e^i>0) \tag{23}\]

\[f (n) =\sum_{d|n}g (d) \;\leftrightarrow\;g (n) =\sum_{d|n}\mu (\dfrac{n}{d}) F (d) \tag{24}\]

02-möbius inversion formula of "combinatorial mathematics"