Register

Michael Sipser Edition 3 Exercise 4 Question 5 (Page No. 211)

137 views

Please log in or register to answer this question.

Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
0 answers
1
admin asked Oct 19, 2019
211 views
Michael Sipser Edition 3 Exercise 5 Question 24 (Page No. 240)
Let $J = \{w \mid \text{either $w = 0x$ for some $x \in A_{TM},$ or $w = 1y\:$ for some $y \in \overline{A_{TM}}\:\:$}\}$. Show that neither $J$ nor $\overline{J}$ is Turing-recognizable.
Let $J = \{w \mid \text{either $w = 0x$ for some $x \in A_{TM},$ or $w = 1y\:$ for some $y \in \overline{A_{TM}}\:\:$}\}$. Show that neither $J$ nor $\overline{J}$ is Tur...
User Avatar
0 votes
0 votes
0 answers
3
admin asked Oct 17, 2019
174 views
Michael Sipser Edition 3 Exercise 4 Question 11 (Page No. 211)
Let $INFINITE_{PDA} = \{\langle{ M \rangle} \mid \text{M is a PDA and L(M) is an infinite language}\}$. Show that $INFINITE_{PDA}$ is decidable.
Let $INFINITE_{PDA} = \{\langle{ M \rangle} \mid \text{M is a PDA and L(M) is an infinite language}\}$. Show that $INFINITE_{PDA}$ is decidable.
User Avatar
0 votes
0 votes
0 answers
4
admin asked Oct 17, 2019
187 views
Michael Sipser Edition 3 Exercise 4 Question 10 (Page No. 211)
Let $INFINITE_{DFA} = \{\langle{ A \rangle} \mid \text{ A is a DFA and L(A) is an infinite language}\}$. Show that $INFINITE_{DFA}$ is decidable.
Let $INFINITE_{DFA} = \{\langle{ A \rangle} \mid \text{ A is a DFA and L(A) is an infinite language}\}$. Show that $INFINITE_{DFA}$ is decidable.
User Avatar
Link Copied!