নিয়মিত ভাষায় শব্দের সংখ্যা

17

উইকিপিডিয়া অনুসারে , যে কোনও নিয়মিত ভাষার জন্য নিয়মিত $L$ থাকে $\lambda_1,\ldots,\lambda_k$ এবং বহুপদী $p_1(x),\ldots,p_k(x)$ যেমন প্রতিটি জন্য দৈর্ঘ্যের শব্দের $n$ সংখ্যা থাকে মধ্যে সন্তুষ্ট সমীকরণ $s_L(n)$ $n$ $L$

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

ভাষা $L =\{ 0^{2n} \mid n \in\mathbb{N} \}$ নিয়মিত (হয় $(00)^*$ মিলবে)। $s_L(n) = 1$ iff n সমান, এবং $s_L(n) = 0$ অন্যথায়।

তবে আমি $\lambda_i$ এবং $p_i$ (যা উপরের দ্বারা বিদ্যমান থাকতে হবে) খুঁজে পাচ্ছি না । হিসাবে $s_L(n)$ differentiable হতে হয়েছে এবং ধ্রুব নয়, এটা একরকম একটি তরঙ্গ মত আচরণ করতে হবে, এবং আমি কিভাবে আপনি সম্ভবত তা করতে পারে polynomials এবং সূচকীয় ফাংশন মতো summands অসীম নম্বর দিয়ে শেষ পর্যন্ত ছাড়া দেখতে পায় না একটি টেলর সম্প্রসারণ। কেউ কি আমাকে আলোকিত করতে পারেন?

— অ্যালেক্স টেন ব্রিংক
সূত্র

আপনি এই উপপাদ্যের নাম জানেন?

— আর্টেম কাজনাটচিভ

@ আর্টেমকাজনাটচিভ: নাহ, কোনও ধারণা নেই। দুর্ভাগ্যক্রমে উইকিপিডিয়া কোনও রেফারেন্স দেয় না :(

— অ্যালেক্স টেন ব্রিঙ্ক

আরও সাধারণভাবে: একটি নিয়মিত ভাষায় প্রদত্ত দৈর্ঘ্যের শব্দের সংখ্যা

— গিলস 'এস-অশুভ হওয়া বন্ধ করুন'

14

For your language, can you take $p_0(x) = 1/2$ , $\lambda_0 = 1$ , $p_1(x) = 1/2$ , $\lambda_1 = -1$ , and $p_i(x) = \lambda_i = 0$ for $i > 1$ ? The Wikipedia article doesn't say anything about the coefficients being either positive or integral. The sum for my choices is

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

which seems to be 1 for even $n$ , and 0 for odd $n$ . Indeed, a proof by induction seems straightforward.

— Patrick87
সূত্র

Ah yes, of course, I'd forgotten about alternating minus signs. Will upvote once the day is over - I've hit the vote cap.

— Alex ten Brink

No induction needed for that claim.

— Raphael

@Raphael True, but then again, that only makes my assertion all the more accurate.

— Patrick87

11

@Patrick87 gives a great answer for your specific case, I thought I would give a tip of how to find $s_L(n)$ in the more general case of any language $L$ that can be represented by an irreducible DFA (i.e. if it is possible to get to any state from any state). Note that your language is of this type.

Proof of theorem for irreducible DFAs

Let $D$ be the transition matrix of your $m$ -state DFA, since it is irreducible, the matrix is normal and has a full eigenbasis $|\lambda_1\rangle ... |\lambda_m\rangle$ . Let $|A\rangle$ be the accept vector: i.e. $\langle i | A \rangle$ is 1 if $i$ is an accept state, and 0 otherwise. WLOG assume that $|1\rangle$ is the initial state, and since we have a complete eigenbasis, we know that $|1\rangle = c_1|\lambda_1\rangle + ... + c_m|\lambda_m\rangle$ for some coefficients $c_1 ... c_m$ (note that $c_i = \langle \lambda_i | i \rangle$ ).

Now we can prove a restricted case of the theorem in the question (restricted to irreducible DFAs; as an exercise generalize this proof to the whole theorem). Since $D$ is the transition matrix $D|1\rangle$ is the vector of states reachable after reading any one character, $D^2|1\rangle$ is the same for two characters, etc. Given a vector $|x\rangle$ , $\langle A|x\rangle$ is simply the sum of the components of $|x\rangle$ that are accept states. Thus:

s L (n) = ⟨ A | D n | 1 ⟩ = ⟨ A | D n (c 1 | λ 1 ⟩ . . . c m | λ m ⟩) = c 1 λ n 1 ⟨ A | λ 1 ⟩ + . . . + c m λ n m ⟨ A | λ m ⟩ = ⟨ A | λ 1 ⟩ ⟨ λ 1 | 1 ⟩ λ n 1 + . . . + ⟨ A | λ m ⟩ ⟨ λ m | 1 ⟩ λ n m = p 1 λ n 1 + . . . + p m λ m m

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

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.

Application to specific question

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

D = (0110)

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

and accept vector:

A = (10)

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

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) = 1 2 + 1 2 (- 1) n

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

— Artem Kaznatcheev
সূত্র

Maybe post this here?

— Raphael

@Raphael that was asked while I was figuring out the proof and typing up my answer, so I did not know about it when I asked.

— Artem Kaznatcheev

6

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 .

$s_L(n) = x^T A^n y.$ (The vector

x $x$ is the characteristic vector of the start state, the vector

y $y$ is the characteristic vector of all accepting state, and

Aij $A_{ij}$ is equal to the number of transitions from state

i $i$ to state

j $j$ 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

(λ), (λ 0 1 λ), ⎛ ⎝ ⎜ λ 00 1 λ 0 01 λ ⎞ ⎠ ⎟, ⎛ ⎝ ⎜ ⎜ ⎜ λ 000 1 λ 00 01 λ 0 001 λ ⎞ ⎠ ⎟ ⎟ ⎟, \dots

$\begin{pmatrix} \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{pmatrix}, \ldots$ If

λ≠0 $\lambda \neq 0$ , then the

n $n$ th powers of these blocks are

(λ n), (λ n 0 n λ n - 1 λ n), ⎛ ⎝ ⎜ λ n 00 n λ n - 1 λ n 0 (n 2) λ n - 2 n λ n - 1 λ n ⎞ ⎠ ⎟, ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ λ n 000 n λ n - 1 λ n 00 (n 2) λ n - 2 n λ n - 1 λ n 0 (n 3) λ n - 3 (n 2) λ n - 2 n λ n - 1 λ n ⎞ ⎠ ⎟ ⎟ ⎟ ⎟, \dots

$\begin{pmatrix} \lambda^n \end{pmatrix}, \begin{pmatrix} \lambda^n & n\lambda^{n-1} \\ 0 & \lambda^n \end{pmatrix}, \begin{pmatrix} \lambda^n & n\lambda^{n-1} & \binom{n}{2} \lambda^{n-2} \\ 0 & \lambda^n & n\lambda^{n-1} \\ 0 & 0 & \lambda^n \end{pmatrix}, \begin{pmatrix} \lambda^n & n\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \binom{n}{3}\lambda^{n-3} \\ 0 & \lambda^n & n\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} \\ 0 & 0 & \lambda^n & n\lambda^{n-1} \\ 0 & 0 & 0 & \lambda^n \end{pmatrix}, \ldots$ Here's how we got to these formulas: write the block as

B=λ+N $B = \lambda + N$ . Successive powers of

N $N$ are successive secondary diagonals of the matrix. Using the binomial theorem (using the fact that

λ $\lambda$ commutes with

N $N$ ),

B n = (λ + n) N = λ n + n λ n - 1 N + (n 2) λ n - 2 N 2 + \dots .

$B^n = (\lambda + n)^N = \lambda^n + n \lambda^{n-1} N + \binom{n}{2} \lambda^{n-2} N^2 + \cdots.$ When

λ=0 $\lambda = 0$ , the block is nilpotent, and we get the following matrices (the notation

[n=k] $[n = k]$ is

1 $1$ if

n=k $n=k$ and

0 $0$ otherwise):

([n = 0]), ([n = 0] 0 [n = 1] [n = 0]), ⎛ ⎝ ⎜ [n = 0] 00 [n = 1] [n = 0] 0 [n = 2] [n = 1] [n = 0] ⎞ ⎠ ⎟, ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ [n = 0] 000 [n = 1] [n = 0] 00 [n = 2] [n = 1] [n = 0] 0 [n = 3] [n = 2] [n = 1] [n = 0] ⎞ ⎠ ⎟ ⎟ ⎟ ⎟

$\begin{pmatrix} [n=0] \end{pmatrix}, \begin{pmatrix} [n=0] & [n=1] \\ 0 & [n=0] \end{pmatrix}, \begin{pmatrix} [n=0] & [n=1] & [n=2] \\ 0 & [n=0] & [n=1] \\ 0 & 0 & [n=0] \end{pmatrix}, \begin{pmatrix} [n=0] & [n=1] & [n=2] & [n=3] \\ 0 & [n=0] & [n=1] & [n=2] \\ 0 & 0 & [n=0] & [n=1] \\ 0 & 0 & 0 & [n=0] \end{pmatrix}$

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) λ i (n) + \sum j c j [n = j],

$s_L(n) = \sum_i p_i(n) \lambda_i(n) + \sum_j c_j [n=j],$ for some complex

λi,cj $\lambda_i,c_j$ and complex polynomials

pi $p_i$ . In particular, for large enough

n $n$ ,

s L (n) = \sum i p i (n) λ i (n) .

$s_L(n) = \sum_i p_i(n) \lambda_i(n).$

— Yuval Filmus
সূত্র

Thank you for the general treatment! You should consider combining your answer with mine and posting it as a full answer to this question. I think it would be more helpful than the current answer there.

— Artem Kaznatcheev