299 views

1 Answer

1 votes
1 votes

A) Computable Enumerable Languages

I'm guessing Computable Enumerable Languages are Recursively Enumerable Languages.

(I'm answering this according to a closure properties table given by Ravindrababu Ravula sir. Below is an explanation that I could find when I googled the same.)  

The Turing-recognizable (or recursively enumerable) languages are closed over homomorphism.

Consider the case where L is RE. Design a NTM M for H(L), as follows. Suppose w is the input to M. On a second tape, M guesses some string x over the alphabet of L, checks that H(x) = w, and simulates the TM for L on x, if so. If x is accepted, then M accepts w. We conclude that the RE languages are closed under homomorphism.

Source: University of Alabama Assignment 

Please correct me if I'm wrong. 

Related questions

1 votes
1 votes
1 answer
1
Souvik33 asked Dec 4, 2022
351 views
Consider the following statementS: $\left \{ a^{n}b^{n+k}|n\geq 0,k\geq 1 \right \} \cup \left \{a^{n+k}b^{n}|n\geq 0,k\geq 3 \right \}$ is DCFLThe above statement is:TRU...
3 votes
3 votes
2 answers
2
1 votes
1 votes
1 answer
3
srestha asked Apr 23, 2019
981 views
How many number of $DFA$ states(minimal DFA) required which accepts the language $L=\left \{ a^{n}:n=\text{3 or n>= 2m for all m>= 1} \right \}$ ___________Answer will be...
0 votes
0 votes
0 answers
4