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
- And 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 \(L(M) = L(M')\).
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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 \(\forall q \in K, E(q)=\{p \in K:(q,e)\vdash_M^*(p,e)\}\)
\(\delta(Q,a) = \bigcup_{q\in Q}\bigcup_{p:(q,a,p)\in\Delta}E(p)\) 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 \(A^*\).
创建日期: 2024年9月10日 11:37:04