Lecture 6¶
FA and PDA, as we mentioned in previous lectures, can only accept a few of the languages. Now we are going to give a stronger automata, the Turing machine.
Turing Machine¶
Definition
A deterministic Turing machine (DTM) is a 5-tuple \(M = (K, \Sigma, \delta, s, H)\), where
- \(K\) is a finite set of states.
- \(\Sigma\) is the tape alphabet, containing the left end symbol \(\triangleright\) the blank \(\sqcup\).
- \(s \in K\) is the initial state.
- \(H \subseteq K\) is a set of halting state.
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\(\delta\) is a transition function. It describes in the current state, when read a symbol, what will the next state be and the head action be (move or write).
\[ \delta: (K - H) \times \Sigma \rightarrow K \times (\{\leftarrow, \rightarrow\} \cup (\Sigma - \{\triangleright\})). \]-
Also, there is a limitation of \(\delta\).
\[ \forall q \in K - H, \exists p \in K,\ \ s.t.\ \delta(q, \triangleright) = (p, \rightarrow) \]
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Definition
A configuration of a Turing machine is a member of the set
Specially, \((q, \triangleright w \underline{a} u)\) is a halting configuration if \(q \in H\).
Definition
- Yields in one step: \((q_1, \triangleright w_1 \underline{a_1} u_1) \vdash_M (q_2, \triangleright w_2 \underline{a_2} u_2)\) if either of the following cases.
- Writing: \(\delta(q_1, a_1) = (q_2, a_2), w_2 = w_1, u_2 = u_1\).
- Moving Left: \(\delta(q_1, a_1) = (q_2, \leftarrow), w_1 = w_2 a_2, u_2 = a_1 u_1\) (Specially, \(u_2 = e\) if \(a_1 = \sqcup, u_1 = e\)).
- Moving Right: \(\delta(q_1, a_1) = (q_2, \rightarrow), w_2 = w_1 a_1, u_1 = a_2 u_2\) (Specially, \(u_1 = e\) if \(a_2 = \sqcup, u_2 = e\)).
- Yields in \(N\) steps: \((q_1, \triangleright w_1 \underline{a_1} u_1) \vdash_M^N (q_2, \triangleright w_2 \underline{a_2} u_2)\) if
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Yields: \((q_1, \triangleright w_1 \underline{a_1} u_1) \vdash_M^* (q_2, \triangleright w_2 \underline{a_2} u_2)\) if \((q_1, \triangleright w_1 \underline{a_1} u_1) \vdash_M \cdots \vdash_M (q_2, \triangleright w_2 \underline{a_2} u_2)\)
- either \((q_1, \triangleright w_1 \underline{a_1} u_1) = (q_2, \triangleright w_2 \underline{a_2} u_2)\).
- or \((q_1, \triangleright w_1 \underline{a_1} u_1) \vdash_M \cdots \vdash_M (q_2, \triangleright w_2 \underline{a_2} u_2)\) in some \(k\)(\(k \in \mathbb{N}^*\)) steps.
Now let's fix \(\Sigma\), we give some simple Turing Machine examples.
Example
Symbol writing machine
\(M_a = (\{s, h\}, \Sigma, \delta, s, \{h\})\) (\(a \in \Sigma - \{\triangleright\}\)), where
- \(\forall b \in \Sigma - \{\triangleright\}\), \(\delta(s, b) = (h, a)\).
- \(\delta(s, \triangleright) = \delta(s, \rightarrow)\).
Heading moving machine
\(M_\leftarrow = (\{s, h\}, \Sigma, \delta, s, \{h\})\), where
- \(\forall b \in \Sigma - \{\triangleright\}\), \(\delta(s, b) = (h, \leftarrow)\).
- \(\delta(s, \triangleright) = \delta(s, \rightarrow)\).
\(M_\rightarrow = (\{s, h\}, \Sigma, \delta, s, \{h\})\), where
- \(\forall b \in \Sigma\), \(\delta(s, b) = (h, \rightarrow)\).
We call \(M_a\), \(M_\leftarrow\) and \(M_\rightarrow\) basic machines, sometimes we abbreviate them to \(a\), \(L\), \(R\).
For simplicity, we introduce the following symbols.
--->
--->
And some abbreviations.
- \(R_\sqcup\) find the next blank symbol on the right.
- \(L_\sqcup\) find the next blank symbol on the left.
- \(R_{\overline{\sqcup}}\) find the next non-blank symbol on the right.
- \(L_{\overline{\sqcup}}\) find the next non-blank symbol on the left.
Example
Left-shifting Machine \(S_\leftarrow\)
Language Recognization¶
Definition
input alphabet: \(\Sigma_0 \subseteq (\Sigma - \{\triangleright, \sqcup\})\).
input configuration: \((s, \triangleright \underline{\sqcup} w)\).
For \(L(M) = \{ w \in \Sigma_0^*: (s, \triangleright \underline{\sqcup} w) \vdash_M^* (h, \dots), h \in H \}\), we say \(M\) semidecides \(L(M)\) and \(L(M)\) is recursively enumerable / recognizable.
Suppose a Turing machine \(M = (K, \Sigma_0, \delta, s, \{y, n\})\), we say \(M\) decides a language \(L \subseteq \Sigma_0\) if
- \(\forall w \in L, (s, \triangleright \underline{\sqcup} w) \vdash_M^* (y, \dots)\), we say \(M\) accepts \(w\).
- \(\forall w \in \Sigma_0^* - L, (s, \triangleright \underline{\sqcup} w) \vdash_M^* (n, \dots)\), we say \(M\) rejects \(w\).
and \(L\) is recursive / decidable.
Theorem
If \(L\) is recursive, then \(L\) must be recursively enumerable.
Function Computation¶
Definition
\(\forall w \in \Sigma_0^*\), if \((s, \triangleright \underline{\sqcup} w) \vdash_M^* (h, \triangleright \underline{\sqcup} y)\) for some \(h \in H\) and some \(y \in \Sigma_0^*\), we say \(y\) is the output of \(M\) on \(w\), denoted by \(y = M(w)\).
Let \(f : \Sigma_0^* \rightarrow \Sigma_0^*\), we say \(M\) computes \(f\) if
and \(f\) is recursive / computable.
Example
Show that \(\{a^nb^nc^n | n \ge 0\}\) is recursive.
创建日期: 2023.11.21 23:52:18 CST