常规语言的字数

根据维基百科，对于任何正则语言存在常数和多项式，使得对于每数长度的字的在满足方程$L$$\lambda_1,\ldots,\lambda_k$$p_1(x),\ldots,p_k(x)$$n$$s_L(n)$$n$$L$

$\qquad \displaystyle s_L(n)=p_1(n)\lambda_1^n+\dots+p_k(n)\lambda_k^n$。

语言是常规语言（与之匹配）。如果n为偶数，则，否则为。$L =\{ 0^{2n} \mid n \in\mathbb{N} \}$$(00)^*$$s_L(n) = 1$$s_L(n) = 0$

但是，我找不到和（上面已经存在）。由于必须是可微的并且不是常数，所以它必须以某种方式像波浪一样，而且我看不到如何用多项式和指数函数来做到这一点，而不必以无穷大的求和数结束，例如泰勒展开式 谁能启发我？$\lambda_i$$p_i$$s_L(n)$

对于你的语言，则可以采取，λ 0 = 1，p 1（X ）= 1 / 2，λ 1 = - 1，和p 我（X ）= λ 我 = 0为我> 1？维基百科上的文章没有说系数是正数还是整数。我选择的总和是$p_0(x) = 1/2$$\lambda_0 = 1$$p_1(x) = 1/2$$\lambda_1 = -1$$p_i(x) = \lambda_i = 0$$i > 1$

$\qquad \displaystyle 1/2 + 1/2(-1)^n = 1/2 (1 + (-1)^n)$

对于偶数似乎为1 ，对于奇数n为0 。确实，归纳证明似乎很简单。$n$$n$

@ Patrick87为您的特定情况提供了一个很好的答案，我想我会给您一个提示，在可以用不可约的DFA表示的任何语言L的更一般情况下，如何找到从任何状态进入任何状态）。请注意，您的语言是这种类型的。$s_L(n)$$L$

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不可约DFA的定理证明

令为m态 DFA 的转移矩阵，因为它是不可约的，所以该矩阵是正常的并且具有完整的本征基| λ 1 ⟩ 。。。| λ 米 ⟩。让| 一个⟩是接受载体：即⟨ 我| 甲⟩是1，如果我是接受状态，和否则为0。WLOG假定 | 1 ⟩是初始状态，因为我们有一个完整的eigenbasis，我们知道| 1 ⟩ = c ^ 1 $D$$m$$|\lambda_1\rangle ... |\lambda_m\rangle$$|A\rangle$$\langle i | A \rangle$$i$$|1\rangle$对于一些系数 Ç 1。。。Ç 米（注意， c ^ 我 = ⟨ λ 我|我⟩）。$|1\rangle = c_1|\lambda_1\rangle + ... + c_m|\lambda_m\rangle$$c_1 ... c_m$$c_i = \langle \lambda_i | i \rangle$

现在我们可以证明问题中定理的一个受限情况（仅限于不可约的DFA；作为练习，将此证明推广到整个定理）。由于是转移矩阵D | 1 ⟩为状态的向量读取任何一个字符之后到达，d 2 | 1 ⟩是两个字符是相同的，等给定一个向量| X ⟩，⟨ 一个| X ⟩根本的组成部分的总和| X ⟩是接受状态。从而：$D$$D|1\rangle$$D^2|1\rangle$$|x\rangle$$\langle A|x\rangle$$|x\rangle$

$$\begin{align} s_L(n) & = \langle A |D^n| 1 \rangle \\ & =\langle A | D^n (c_1 |\lambda_1\rangle ... c_m |\lambda_m\rangle) \\ & = c_1 \lambda_1^n \langle A | \lambda_1 \rangle + ... + c_m \lambda_m^n \langle A | \lambda_m \rangle \\ & = \langle A | \lambda_1 \rangle\langle\lambda_1|1\rangle \lambda_1^n + ... + \langle A | \lambda_m \rangle\langle \lambda_m | 1 \rangle \lambda_m^n \\ & = p_1\lambda_1^n + ... + p_m \lambda_m^m \end{align}$$

Now we know that for an irreducible m-state DFA, $p_1 ... p_m$ will be zero order polynomials (i.e. constants) that depends on the DFA and $\lambda_1 ... \lambda_m$ will be eigenvalues of the transition matrix.

Generality note

If you want to prove this theorem for arbitrary DFA, then you will need to look at the Schur decomposition of $D$ and then polynomials of non-zero degree will pop up because of the nilpotent terms. It is still enlightening to do this, since it will let you bound the max degree of the polynomials. You will also find a relationship between how complicated the polynomials are and how many $\lambda$s you will have.

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Application to specific question

For your language $L$ we can select the DFA with transition matrix:

$$D = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

and accept vector:

$$A = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

Find the eigenvectors and their eigenvalues $\lambda_1 = 1$ with $| \lambda_1 \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\lambda_2 = -1$ with $| \lambda_2 \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. We can use this to find $p_1 = 1/2$ and $p_2 = 1/2$. To give us:

$$s_L(n) = \frac{1}{2} + \frac{1}{2}(-1)^n$$

Continuing Artem's answer, here is a proof of the general representation. As Artem shows, there is an integer matrix $A$ and two vectors $x,y$ such that

$$s_L(n) = x^T A^n y.$$ (The vector $x$ is the characteristic vector of the start state, the vector $y$ is the characteristic vector of all accepting state, and $A_{ij}$ is equal to the number of transitions from state $i$ to state $j$ in a DFA for the language.)

Jordan's theorem states that over the complex numbers, $A$ is similar to a matrix with blocks of one of the forms

$$\begin{pmatrix} \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{pmatrix}, \ldots$$ If $\lambda \neq 0$, then the $n$th powers of these blocks are $$\begin{pmatrix} \lambda^n \end{pmatrix}, \begin{pmatrix} \lambda^n & n\lambda^{n-1} \\ 0 & \lambda^n \end{pmatrix}, \begin{pmatrix} \lambda^n & n\lambda^{n-1} & \binom{n}{2} \lambda^{n-2} \\ 0 & \lambda^n & n\lambda^{n-1} \\ 0 & 0 & \lambda^n \end{pmatrix}, \begin{pmatrix} \lambda^n & n\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \binom{n}{3}\lambda^{n-3} \\ 0 & \lambda^n & n\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} \\ 0 & 0 & \lambda^n & n\lambda^{n-1} \\ 0 & 0 & 0 & \lambda^n \end{pmatrix}, \ldots$$ Here's how we got to these formulas: write the block as $B = \lambda + N$. Successive powers of $N$ are successive secondary diagonals of the matrix. Using the binomial theorem (using the fact that $\lambda$ commutes with $N$), $$B^n = (\lambda + n)^N = \lambda^n + n \lambda^{n-1} N + \binom{n}{2} \lambda^{n-2} N^2 + \cdots.$$ When $\lambda = 0$, the block is nilpotent, and we get the following matrices (the notation $[n = k]$ is $1$ if $n=k$ and $0$ otherwise): $$\begin{pmatrix} [n=0] \end{pmatrix}, \begin{pmatrix} [n=0] & [n=1] \\ 0 & [n=0] \end{pmatrix}, \begin{pmatrix} [n=0] & [n=1] & [n=2] \\ 0 & [n=0] & [n=1] \\ 0 & 0 & [n=0] \end{pmatrix}, \begin{pmatrix} [n=0] & [n=1] & [n=2] & [n=3] \\ 0 & [n=0] & [n=1] & [n=2] \\ 0 & 0 & [n=0] & [n=1] \\ 0 & 0 & 0 & [n=0] \end{pmatrix}$$

Summarizing, every entry in $A^n$ is either of the form $\binom{n}{k} \lambda^{n-k}$ or of the form $[n=k]$, and we deduce that

$$s_L(n) = \sum_i p_i(n) \lambda_i(n) + \sum_j c_j [n=j],$$ for some complex $\lambda_i,c_j$ and complex polynomials $p_i$. In particular, for large enough $n$, $$s_L(n) = \sum_i p_i(n) \lambda_i(n).$$