计算机理论导引第三版答案第四章,《计算理论导引》第四章：可判定性-学习笔记

第四章：可判定性

4 Decidability

关于其他章节的内容，请点这：《计算理论导引》学习笔记

4.1 Decidable Languages

几个可判定的语言

acceptance problem for DFA

the acceptane problem for DFAs of testing whether a particular deterministic finite automaton accepts a given string can be expressed as a language\( A_{DFA}\).

\( A_{DFA} = \{\langle B,w \rangle | \text{ B is a DFA that accepts input string w}\}\).

The problem of testing whether a DFA B accepts an input w is the same as the problem of testing whether is a member of the language \( A_{DFA}\).

We simply need to present a TM M that decides \( A_{DFA}\).

M = “on input , where B is a DFA and w is string:

1. Simulate B on input w.

2. If the simulation ends in an accept state, accept. If it ends in a nonaccepting state, reject.”

acceptance problem for NFA

\( A_{NFA} = \{\langle B,w \rangle | \text{ B is a NFA that accepts input string w}\}\).

N = “on input , where B is a NFA and w is string:

1. Convert NFA B to an equivalent DFA C.

2. Run TM M on input .

2. If M accepts, accept; otherwise, reject.”

acceptance problem for REX

\( A_{REX} = \{\langle B,w \rangle | \text{ B is a regular expression that generates string w}\}\).

P = “On input , where R is a regular expression and w is a string:

1. Convert regular expression R to an equivalent NFA A.

2. Run TM N on input .

3. If N accepts, accept; if N rejects, reject.”

emptiness testing for DFA

\( E_{DFA} = \{\langle A \rangle | \text{ A is a DFA and } L(A) = \emptyset\}\).

T = “On input , where A is a DFA:

1. Mark the start state of A.

2. Repeat until no new states get marked:

3. Mark any state that has a transition coming ino it from any state that is already marked.

4. If no accept state is marked, accept; otherwise, reject.”

whether two DFAs recognize the same language

\( EQ_{DFA} = \{\langle A,B \rangle | \text{ A and B are DFA and } L(A) = L(B)\}\).

\( L(C) = (L(A) \cap \overline{L(B)}) \cup (\overline{L(A)} \cap L(B))\).

We can construct C form A and B with the sonstructions for proving the class of regular languages closed under complementation, union, and intersection.

\( L(C) = \emptyset \iff L(A) = L(B)\).

F = “On input , where A and B are DFAs:

1. Condtruct DFA C as described.

2. Run TM T on input .

3. If T accepts, accept. If T rejects, reject.”

acceptance problem for CFG

\( A_{CFG} = \{\langle G,w \rangle\} | \text{ G is a CFG that generates string w }\}\).

直接枚举会得到recognizer而不是decider

If G were in Chomsky normal form, any derivation of w has 2n-1 steps, where n is the length of w.

In that case, checking only derivations with 2n-1 steps to determine whether G generates w would be sufficient.

S = “on input , where G is a CFG and w is a string:

1. Convert G to an equivalent grammar in Chomsky normal form.

2. List all derivations with 2n-1 steps, where n is the length of w; except if n = 0, then instead list all derivations with one step.

3. If any of these deravation generate w, accept; if not, reject.”

emptiness testing for CFG

\( E_{CFG} = \{\langle G \rangle | \text{ G is a CFG and } L(G) = \emptyset\}\).

转化为可达性问题，目的地是terminal symbols，起点是start variable

R = “On input , where G is a CFG:

1. Mark all terminal symbols in G.

2. Repeat until no new variables get marked:

3. Mark any variable A where G has a rule \( A \longrightarrow U_1U_2\dots U_k\) and each symbol \( U_1,\dots,U_k\) has already been marked.

4. If the start variable is not marked, accept; otherwise, reject.

whether two CFGs generate the same language

\( EQ_{CFG} = \{\langle G,H \rangle | \text{ G and H are CFGs and } L(G) = L(H)\}\).

在这里，\( EQ_{DFA}\)的套路用不了了

因为，The class of context-free languages is not closed under complementation or intersection.(Exercise 2.2)

事实上，这是不可判定的

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