Lecture 1 - Finite Automata¶

Example :¶

Given a weighted graph G :

What's the MST?
What's the weight of the MST?
Does G have a ST with wieght at most k(k is a given integer)? -- Decision Problem

Given a string w, is \(w\in L=\{encode(G,k) | G\ is\ a\ graph,that\ has\ a\ ST\ with\ weight\ at\ most\ k\}\)?

Decision Problem \(\Rightarrow\) A Laguanage
A Laguanage \(\Rightarrow\) A Formal Language
Therefore, a Decision Problem P \(\Leftrightarrow\) A Language L

Definition :¶

Alphabet¶

A finite set of symbols

String¶

A finite sequence of symbols from an alphabet \(\Sigma\)
concatenation: \(w_1w_2\)
exponentiation: \(w^n\)
empty string: \(e\)
reverse: \(w^R\)

Language¶

A set of strings over an alphabet \(\Sigma\)
concatenation: \(L_1L_2 = \{w_1w_2 | w_1\in L_1,w_2\in L_2\}\)
exponentiation: \(L^n = \{w^n | w\in L\}\)
\(A^0 = \{e\}\)
\(A^* = \bigcup_{i=0}^{\infty}A^i\)
\(A^+ = \bigcup_{i=1}^{\infty}A^i\)
\(\Sigma^* = \bigcup_{i=0}^{\infty}\Sigma^i\)
\(\Sigma^+ = \bigcup_{i=1}^{\infty}\Sigma^i\)
\(\Sigma^i = \{w | w\ is\ a\ string\ of\ length\ i\}\)
Language Reverse: \(L^R = \{w^R | w\in L\}\)
\(\{\}\) vs \(\{e \}\) : size of \(\{\}\) is 0, size of \(\{e\}\) is 1

Computable Model¶

Finite Automata¶

A finite automaton is a 5-tuple \((K,\Sigma,\delta,s,F)\) where:

\(K\) is a finite set of states
\(\Sigma\) is an alphabet
\(\delta:K\times\Sigma\rightarrow K\) is the transition function
\(s \in K\) is the start state
\(F\subseteq K\) is the set of accept states

A configuration of an FA is a pair \((q,w)\) where \(q\in Q\) and \(w\in\Sigma^*\) where \(q\) is the current state and \(w\) is the remaining input.

yeild in one step: \((q,w)\vdash_M(q',w')\) if \(w = aw'\) and \(\delta(q,a) = q'\)
yeild in zero or more steps: \((q,w)\vdash_M^*(q',w')\) if \((q,w)\vdash_M(q_1,w_1)\vdash_M(q_2,w_2)\vdash_M\cdots\vdash_M(q',w')\)

M accepts a language L if (two conditions):

M accepts all strings in L
Or M rejects all strings not in L
M accepts a UNIQUE Language
\(L(M) = \{w | M \ accepts\ w\}\)

A language is regular if it is accepted by a finite automaton.

Theorem¶

If A and B are regular languages, so is \(A\cup B\). Proof:

Let \(M_A\) and \(M_B\) be FAs that accept A and B respectively.
\(M_A = (K_A,\Sigma,\delta_A,s_A,F_A)\)
\(M_B = (K_B,\Sigma,\delta_B,s_B,F_B)\)
Construct a new FA \(M\) that accepts \(A\cup B\):
\(K_v = K_A\times K_B\)
\(s_v = (s_A,s_B)\)
\(F_v =(F_A\times K_B)\cup(K_A\times F_B)\)
\(\delta_v((q_A,q_B),a) = (\delta_A(q_A,a),\delta_B(q_B,a))\) for any \((q_A,q_B)\in K_A\times K_B\) and \(a\in\Sigma\)

Non-deterministic Finite Automata¶

next state is not uniquely determined by the current state and the input symbol
e-transitions: \(\delta(q,e)\)

A NFA is a 5-tuple \((K,\Sigma,\Delta,s,F)\) where:

K is a finite set of states
\(\Sigma\) is an alphabet
\(\Delta:K\times(\Sigma\cup\{e\})\times K\) is the transition relation
\(s\in K\) is the start state
\(F\subseteq K\) is the set of accept states

Acceptance of a string by a NFA:

M on input w accepts if \((s,w)\vdash_M^*(q,e)\) for some \(q\in F\)
M accepts a language L if M accepts every string w in L

Example: Construct a NFA that accepts the language \(\{w | w \in \{a,b\}^*\),the second symbol from the end of w is b\(\}\)

Theorem¶

Given a NFA M, there exists a DFA M' such that \(L(M) = L(M')\).
Given a DFA M, there exists a NFA M' such that \(L(M) = L(M')\).
DFA M simulates "tree-like" computation of NFA M.
A NFA \(M = (K,\Sigma,\Delta,s,F)\) can be simulated by a DFA \(M' = (K',\Sigma,\delta,s',F')\) where:
- \(K' = 2^K\)
- \(2^K = \{q_1,q_2,\cdots,q_n,\{q_1\},\{q_2\},\cdots,\{q_n\},\{q_1,q_2\},\cdots,\{q_1,q_2,\cdots,q_n\}\}\)
- \(F' = \{q\in K' | q\cap F \neq \emptyset\}\)
- \(\delta(q,a) = \bigcup_{p\in q}\Delta(p,a)\) for any \(q\in K'\) and \(a\in\Sigma\)
- \(s' = E(s)\) where \(\forall q \in K, E(q)=\{p \in K:(q,e)\vdash_M^*(p,e)\}\)
- \(\delta(Q,a) = \bigcup_{q\in Q}\bigcup_{p:(p,a,q)\in\Delta}E(p)\) for any \(Q\subseteq K'\) and \(a\in\Sigma\)
A language is regular if and only if it is accepted by a NFA.

Proof¶

If A and B are regular languages, so is \(AB\).

Let \(M_A\) and \(M_B\) be FAs that accept A and B respectively.

\(M_A = (K_A,\Sigma,\delta_A,s_A,F_A)\)

\(M_B = (K_B,\Sigma,\delta_B,s_B,F_B)\)

Construct a new FA \(M\) that accepts \(A\cdot B\):

\(M = (K,\Sigma,\delta,s,F)\) where

\(K = K_A\bigcup K_B\)

\(s = s_A\)

\(F = F_B\)

\(\delta(q,a) = \delta_A(q,a)\bigcup\delta_B(q,a)\bigcup\{(q,e,s_B):q\in F_A\}\)

If A is a regular language, so is \(A^+\).

If A is a regular language, so is \(A^*\).

Complemetary Material¶

Cantor's Theorem, Countable Sets, and String Representations of the Real Numbers¶

Cantor's Theorem¶

There does not exits a one-to-one function RtS: \(\mathbb{R}\rightarrow\{0,1\}^*\)
(equivalently) The Real numbers are uncountable.That is, there does not exist a one-to-one function \(f:\mathbb{N}\rightarrow\mathbb{R}\)

Proof:

See Book page 94

Boolean functions are uncountable¶

The set of all boolean functions(mapping \(\{0,1\}^*\rightarrow\{0,1\}\)) is uncountable. * Proof see book page 99

Equivalent conditions for countability¶

The set S is countable
There exits a one-to-one map from \(\mathbb{N}\) to S
There exits an onto map from \(\{0,1\}^*\) to S
There exits a one-to-one map from S to \(\{0,1\}^*\)
There exits a one-to-one map from S to \(\mathbb{N}\)
There exits an onto map from some countable set T to S
There exits a one-to-one map from S to some countable set T

Representing Objects Beyond Numbers¶

Lemma: Suppose that \(E: O \rightarrow \{0,1\}^*\) is a one-to-one function. Then there exists a function \(D:\{0,1\}^*\rightarrow O\) such that \(D(E(o)) = o\) for all \(o\in O\).

Proof: Let \(o_0\) be an arbitrary object in O. Let \(w_0 = E(o_0)\). Let \(w_0 = w_1w_2\cdots w_n\) be the string representation of \(w_0\). Let \(o_1 = D(w_1)\), \(o_2 = D(w_1w_2)\), \(\cdots\), \(o_n = D(w_1w_2\cdots w_n)\). Then \(D(w_0) = o_n\).

Finite Representations¶

if \(O\) is finite, then we can represent every object in \(O\) as a string of length at most some number \(n\) if and only if \(|O|\leq 2^{n+1}-1\).

Lemma: For every two finite sets S and T, there exists a one-to-one function \(E:S\rightarrow T\) if and only if \(|S|\leq|T|\).

Proof see book page 101

Prefix-Free encoding¶

Proof see book page 103

Making Representations prefix-free¶

Proof see book page 104

最后更新: 2024年11月11日 23:05:25
创建日期: 2024年9月10日 11:37:04