"Combinatorial mathematics" 05-Classic Counting method

1. Basic count of the parent function

Now, using the parent function to solve the basic counting problem, the parent function can not only complete the automatic counting, but also can represent the count itself, such as Stirling number can only be represented by the parent function. Automatic counting is useful for counting problems that can be stepped, and the target value is the sum of each step value, which is related to the operation nature of the polynomial.

1.1 Number of combinations and number of strokes

The most intuitive feature of this is Model 2, which is selected from \ (n\) distinguishable objects \ (M \). Limit the number of times a \ (K\) object is taken in the set \ (M_k\), where the parent function of the selected condition is \ (\sum\limits_{i\in m_k}x^i\), all elements are selected can be automatically counted with the help of the parent function (1), the number of total selected \ (m\) elements is \ (x ^m\) coefficients. In particular, the non-repeating number of the parent function is \ ((1+x) ^n\), the number of repeatable combinations of the parent function is \ (\dfrac{1}{(1-x) ^n}\) (note \ (1+x+x^2+\cdots=\dfrac{1}{1-x}\)), the full-shot parent function is \ (\ dfrac{x^n}{(1-x) ^n}\).

\[\sum_{k=0}^{\infty}c_kx^k=\prod_{i=1}^n\left (\sum_{j\in m_i}x^j\right) \tag{1}\]

For the split number of Model 4, considering that the \ (1^{\lambda_1}2^{\lambda_2}\cdots m^{\lambda_m}\) split is to satisfy the previous (14), it can also be automatically counted with the parent function. For divisions of length \ (K\), the parent function of the division with a length of \ (k\) is \ (1+x^k+x^{2k}+\cdots=\dfrac{1}{1-x^k}\). So the parent function of \ (p (m) \) is the formula (2), because \ (P (m,k) \) equals the maximum number of partitions for \ (k\), so its parent function is the formula (3).

\[\sum_{n=0}^{\infty}p (n) x^n=\prod_{i=1}^{\infty} (1-x^i) ^{-1}\tag{2}\]

\[\sum_{n=0}^{\infty}p (n,k) x^n=x^k\prod_{i=1}^k (1-x^i) ^{-1}\tag{3}\]

It is worth mentioning that Euler initially discovered the parent function method in the study of the division number. Even if there is a formula (2), the nature of \ (p (m) \) is unclear, we notice that (2) inverse \ (q (x) \) is relatively simple and worth discussing. \ (q (x) \) has an obvious combination of meaning, each coefficient of absolute value is partial division of the number of divisions, the meaning of the symbol is the number of parts of odd and even time division of the difference. It is not difficult to obtain the (q (m) \) satisfying formula (4) by using this combinatorial meaning to assist the discussion (the process is shown in the textbook). The recursive relation (5) of the \ (p (m) \) is then obtained, which is called the Euler formula .

\[q (M) =\left\{\begin{matrix} ( -1) ^k,&\text{if}\;m=\frac{1}{2} (3k^2\pm k) \\0,&\text{if}\;m\ne\frac{1}{2} ( 3K^2\PM k) \end{matrix}\right.\tag{4}\]

\[p (M) =\sum_{k=1}^{\infty} ( -1) ^{k-1}\left (P (m-\frac{3k^2-k}{2}) +p (m-\frac{3k^2+k}{2}) \right) \tag{5}\]

　　• Use the parent function to prove that the division is an odd number of divisions equal to the number of divisions that are divided.

1.2 Exponential type female function

For the number of permutations of model 1, the simple addition is no longer satisfied, and the last article (2) is explored, which inspires us to use \ (\dfrac{x^k}{k!} \) instead of \ (x^k\) (Type (6)). For the series \ (c_n\), the series (6) is called the exponential type of series, which is very suitable for permutation problems. With the above analysis, \ (n\) The cross-m\ of the "(") arrangement of the parent function is the formula (7). In particular, the parent function of the non-repeating permutation number is \ ((1+x) ^n\), and the parent function of the repeatable permutation number is \ (e^{nx}\) (Note \ (1+x+\dfrac{x^2}{2!} +\cdots=e^x\), the parent function of the full-scale array is \ ((e^x-1) ^n\).

\[\sum_{k=0}^{\infty}c_k\dfrac{x^k}{k!} =c_0+c_1\dfrac{x}{1!} +c_2\dfrac{x^2}{2!} +c_3\dfrac{x^3}{3!} \cdots\tag{6}\]

\[\sum_{k=0}^{\infty}c_k\dfrac{x^k}{k!} =\prod_{i=1}^n\left (\sum_{j\in m_i}\dfrac{x^j}{j!} \right) \tag{7}\]

Finally, take a look at the number of Stirling in Model 3 because \ (k! S (m,k) \) is the full shot in Model 1, using the full-beam array of the parent function is easy to know \ (S (m,k) \) The exponential female function (formula (8)), the use of exponential female function can also be obtained in the previous (26). In order to get the exponential female function \ (s (m,k) \) \ (g_k (x) \), from the previous (27) right to get inspired, first calculate \ (g_k (x) \) The parent function get \ ((1+x) ^y\) (process slightly, the second layer of the parent function using \ (y\)), expand \ (y\) The power series will get \ (s (m,k) \) Exponential-type female function (9). With the exponential-type female function, the properties of the Stirling number can be studied by the parent function, for example, the derivation of the parent function can get the recursive relation (10) (11).

\[\sum_{m=0}^{\infty}s (m,k) \dfrac{x^m}{m!} =\dfrac{1}{k!} (e^x-1) ^k\tag{8}\]

\[\sum_{m=0}^{\infty}s (m,k) \dfrac{x^m}{m!} =\dfrac{1}{k!} (\ln (1-x)) ^k\tag{9}\]

\[s (m,k) =\sum_{i=k-1}^{m-1} \binom{m-1}{i}s (i,k-1) \tag{10}\]

\[s (M,k) =\sum_{i=k-1}^{m-1} ( -1) ^{m-i-1} (m-1) _{m-i-1}s (i,k-1) \tag{11}\]

　　• take each element to an even number of permutations.

2. Pólya counting theorem 2.1 introduction and ring character

The basic counting problem involves only two of the most extreme topologies, but in the process of elaboration, we have emphasized the idea of isomorphic classes. This idea will be promoted and used in more topological structures. Isomorphism is the permutation of elements in a topological structure, the relationship between elements and the permutation, and its algebraic meaning is the familiar ((A, B) = (f (a), G (a)) \). We're not going to say here. Elements can be distinguished and indistinguishable, and each element is indistinguishable from its own, and its differences are determined entirely by its position in the topological structure.

than the ring topology, intuitively, each element is "indistinguishable", but they can not be arbitrarily replaced. When the \ (a\) is replaced with \ (a ' \), in order to maintain \ (a,b\) relationship, \ (b\) can only be substituted to \ (b ' \), so the isomorphism in this topology can only be the entire graph rotation. Then another loop topology, in addition to the rotation, the rollover of the dashed line is also a homogeneous permutation. The final cube, which includes more homogeneous permutations, corresponds to our intuition that the changes are still the same as before. For the sake of simplicity, we will only discuss the original image as a complex topology, rather like a distinguishable set of issues. Its equivalent model is to dye the elements of the topological structure, and the isomorphism of course also requires that the elements be replaced with elements of the same color.

If the image is on the left (m\) Order of the Ring, with \ (n\) color for its dyeing (or write \ (n\) kind of letters), the result is called \ (m\) Yuan ring Word , the number of homogeneous ring words (c_n (m) \). First, the ring is fixed and dyed, the ring character is rotated, the different dyeing schemes are the number of the dyeing isomorphism. If the minimum period for a color is \ (d\), it is common to get \ (d\) a different stain, then there is \ (N^M=\SUM\LIMITS_{D|M}DM (d) \), where \ (m (d) \) is the minimum period of \ (d\) staining number (isomorphic meaning). The inverse formula can be used to obtain the \ (MM (m) =\sum\limits_{d|m}\mu (\dfrac{m}{d}) n^d\), which has a counting formula (12) of the ring word.

\[c_n (m) =\sum_{d|m}\sum_{d ' |d}\mu (\dfrac{d}{d '}) \dfrac{n^{d '}}{d}\tag{12}\]

2.2 Pólya counting theorem

Now for a general discussion of the problem, the first hypothesis is that the original image is distinguishable and dyed on it, and the staining scheme set \ (x\) is obtained. There is also a transform set \ (g\) for homogeneous permutation of the original topology (color-free), and for any \ (g\in g\), it is a transform \ (g:x\to x\). When it comes to \ (X_1,x_2\in x\), when the presence \ (g\in g\) makes \ (g (x_1) =x_2\), the name \ (x_1,x_2\) is equivalent, and our problem is to find the number of equivalence classes.

It is easy to prove that for each topological structure, all of its isomorphic permutations are closed under the compound operation, thus constituting a group. Because it is a subset of permutation group, it is also called symmetry group . \ (g\) to \ (x\) The transformation satisfies the condition (13), the transformation is \ (g\) on the \ (x\) The "function", you need to review "the role of the group" related knowledge. From the two dimensions discussed \ (g (x) =x\), you can get the number of equivalence class \ (N (g) \) The Burnside theorem of the satisfying formula (14), where \ (f_g\) represents the number of \ (x\) to meet \ (g (x) =x\). If each \ (x\in x\) weighted \ (w (x) \) and the element weights in the same equivalence class are the same, there is a weighted version of the Burnside Theorem (equation (15)), where \ (x_k\) is the representative of the equivalent class of the nth (k\).

\[(G_1g_2) (x) =g_1 (g_2 (x)); \;\;e (x) =x\tag{13}\]

\[n (G) =\dfrac{1}{| G|} \sum_{g\in g}| F_g|\tag{14}\]

\[\sum_{k=1}^{n (G)}w (x_k) =\dfrac{1}{| G|} \sum_{g\in g} (\sum_{x\in f_g}w (x)) \tag{15}\]

In the dyeing problem, we are concerned about a color combination used by a staining scheme, in order to embody this information, assign weights to each color \ (y_i\). The weight of a staining scheme is \ (Y_1^{k_1}y_2^{k_2}\cdots y_n^{k_n}\), where \ (k_i\) is the number of times the color \ (i\) appears and has \ (k_1+k_2+\cdots+k_n=m\). Formula (15) contains weights for all staining schemes (homogeneous meanings), in which we can get the number of staining schemes per color combination \ (k_1,k_2,\cdots,k_n\). The calculation of the formula (15) is obtained by the right side, and the right side focuses on the nature of each permutation \ (g\), making the problem only related to the topological structure.

We know that each permutation \ (g\) is actually a combination of some rotations, which is obviously (and crucially), and the necessary and sufficient condition for (g (x) =x\) is the same color in the same rotation. Here the idea of the parent function is borrowed to calculate the weight of \ (f_g\), and the sum of all possible weights for the rotation of length \ (k\) is \ (y_1^k+y_2^k+\cdots+y_n^k\). Thus the weights of the \ (1^{\lambda_1}2^{\lambda_2}\cdots m^{\lambda_m}\) (g\), \ (f_g\) are as follows (16). For ease of expression, Bashi (17) is known as the rotation index (cycle index) of \ (g\), obviously when \ (\delta_k=y_1^k+y_2^k+\cdots+y_n^k\) Gets the weight of the complete expression, which is also known as Model List . In particular, take \ (y_1=y_2=\cdots=y_n=1\) (i.e. \ (\delta_1=\delta_2=\cdots=\delta_m=n\)), you get the number of all styles (number of staining) \ (p_g (n,\cdots,n) \), These conclusions are the Polya (Pálya) counting theorem .

\[w (F_g) =\prod_{k=1}^n (y_1^k+y_2^k+\cdots+y_n^k) ^{\lambda_k}\tag{16}\]

\[p_g (\sigma_1,\cdots,\sigma_m) =\dfrac{1}{| G|} \sum_{g\in g}\sigma_1^{\lambda_1 (g)}\cdots\sigma_m^{\lambda_m (g)}\tag{17}\]

2.3 Typical examples

Polya theorem adopts the idea of similar parent function, can complete automatic counting accurately, but the conclusion does not give the final counting value, so it is more in its theoretical value. The following is an example of the three topological structures in the figure, which illustrates the application of the Polya theorem, and the key to the problem is the calculation of formula (17).

First, there is a forward loop, and it is clear that its symmetric subgroup is a \ (m\) Order Loop group \ (c_m=\{g,g^2,\cdots,g^m=e\}\), where \ (g\) is a single-step rotation. Easy to know, \ (g^k\) is \ ((k,m) \) A (\dfrac{m}{(k,m)}\) Order rotation of the product, where \ ((k,m) \) is the greatest common divisor of the mark. In turn, for arbitrary \ (d|m\), there is a \ (\varphi (\dfrac{m}{d}) \) (g^k\) rotation indicator is \ (\sigma_{m/d}^d\). So its rotation indicator is the formula (18), and then get the number of the ring Word (19), it is equivalent to the formula (12) (\ (\varphi (m) =\sum\limits_{d|m}\mu (d) \frac{m}{d}\)).

\[p_{c_m}=\dfrac{1}{m}\sum_{d|m}\varphi (\dfrac{m}{d}) \sigma_{m/d}^d=\dfrac{1}{m}\sum_{d|m}\varphi (d) \sigma_d^{ M/d}\tag{18}\]

\[c_n (M) =\dfrac{1}{m}\sum_{d|m}\varphi (d) n^{\frac{m}{d}}\tag{19}\]

To look at the non-direction ring, a typical question is how many different necklaces can be strung with pearls of different colors. Unlike a forward ring, it can be flipped, and its symmetric subgroup \ (d_m=\{g,g^2,\cdots,g^m=e,ag,ag^2,\cdots,ag^m=a\}\) is called a two-facet group . where the latter \ (m\) is a rollover transformation, when \ (m\) is an odd time is \ (\dfrac{n-1}{2}\) a swap, when \ (n\) is an even number, half is \ (\dfrac{n}{2}\) A swap, half is \ (\dfrac{n-2}{2}\) a swap, The rotation indicator of the flip is the formula (20).

\[\SUM_{K=1}^MW (ag^k) =\left\{\begin{matrix}\dfrac{1}{2}x_1x_2^{\frac{m-1}{2}},\;\;\;\;\;\;\;\;\;\;\;&\text {if M is odd}\\\\\dfrac{1}{4} (x_2^{\frac{m}{2}}+x_1^2x_2^{\frac{m-2}{2}}), &\text{if m is even}\end{matrix}\right . \tag{20}\]

Finally, the cube, the symmetry group is called the cube rotation group , in addition to the unit element, there are three types of rotation: (1) 3*3=9 to the opposite center for the rotation of the axis, (2) 2*4=8 to the vertex line as the axis of rotation; (4) 1* 6 rotates with an over-aligned midpoint line as an axis. These transformations can be used to work on vertices, edges and polygons respectively, and their corresponding rotation indicators are shown in the following three.

\[P_0=\FRAC{1}{24} (x_1^8+9x_2^4+6x_4^2+8x_1^2x_3^2) \;\;\;\;\;\;\tag{21}\]

\[P_1=\FRAC{1}{24} (x_1^{12}+3x_2^6+6x_4^2+8x_3^4+6x_1^2x_2^5) \;\tag{22}\]

\[P_2=\FRAC{1}{24} (x_1^6+3x_1^2x_2^2+6x_1^4x_4+6x_2^3+8x_3^2) \tag{23}\]

　　• discussion of the number of necklace types (n\), including the number of colors specified;

　　• discussion of the number of color-colored cube points, edges, and polygons, including the number of colors specified in the n\.

"Combinatorial mathematics" 05-Classic Counting method