Name: Student ID: 0 CSE 322 Spring 2010: Take-Home
Show that every infinite prefix-closed context free language contains an every infinite Turing-recognizable language has an infinite decidable subset.
Answers to the CSCE 551 Final Exam April 30
https://cse.sc.edu/~fenner/csce551/final-ans.pdf
Harvard University
Nov 11 2014 (A) To prove that L is not Turing-recognizable
CS46 lab 8
Mar 20 2022 Show that every infinite Turing-recognizable language has an infinite decidable subset. 4. Computable functions.
Solutions for Problem Set 7
Solution: Let L be an infinite recursively enumerable language. characterization of decidable languages in terms of enumeration: L1 is decidable iff ...
Homework 4 – Due Wednesday March 18
https://cs-people.bu.edu/mbun/courses/332_S20/handouts/hw4.pdf
Practice Problems for Final Exam: Solutions CS 341: Foundations of
In each part below if you need to prove that the given language L is decidable
Enumerators
(Sipser 3.19) Show that every infinite Turing-recognizable language has an infinite decid- able subset. (Hint: use the result from the previous problem.).
CS660 Homework 1
Apr 8 2018 True or false: every infinite Turing-recognizable language has an infinite decidable subset. (This comes from problem 1 on Problem Set 6
Enumerators
(Sipser 3.18) Show that a language is decidable iff some enumerator (Sipser 3.19) Show that every infinite Turing-recognizable language has an infinite ...
CS1010: Theory of Computation - Brown University
Turing Recognizable & Decidable Languages The set of strings that a Turing Machine M accepts is the language of M denoted as 6(=)or the language recognized by M –A language 6is Turing-recognizableif some Turing machine recognizesit •I e There exists a TM =such that =halts in the accept state for all and only the strings ??6
CSE 322 Spring 2010: Take-Home Final Exam SOLUTIONS Where
Show that every infinite Turing-recognizable language has an infinite decidable subset (Hint: Use the result in (a) and the result you know regarding Turing-recognizable languages and enumerator TMs (Theorem 3 21 in the text)) Let A be an infinite Turing-recognizable language
CSE 431 Spring 2006 Assignment - University of Washington
1 Prove that a language is decidable if and only if there is an enumerator that enumerates it in lexicographic order (Hint: Handle the case where the language is ?nite separately from the case when it is in?nite ) 2 Use the above to show that any in?nite Turing-recognizable language contains an in?nite decidable subset 3
CS46 lab 8 - Swarthmore College
Show that there is a decidable language C consisting of Turing machine descriptions such that every machine described in B has an equivalent machine in C and every machine described in C has an equivalent machine in B 6 (extra challenge) A TM is a language consisting of descriptions of Turing machines and it is Turing-recognizable Why does
CSE 431 Spring 2007 Assignment - University of Washington
an in?nite decidable subset 4 Let INFINITE PDA = {hMi M is a PDA and L(M) is an in?nite language} Show that INFINITE PDA is decidable 5 Show that the set of complex numbers QUADRATIC-ROOT = {x ? C there are integers a 6= 0 b and c such that ax2+bx+c = 0} is countable 6 (Bonus) Let C be a language Prove that C is Turing
Homework 9 Solutions - New Jersey Institute of Technology
We showed in a previous homework that the class of Turing-recognizablelanguages is closed under union soEQCFGis Turing-recognizable Here are the details of a TMTthat recognizesEQCFG wheres1 s2 s3 is anenumeration of strings in??in string order: = “On input G1 G2 whereG1andG2are CFGs: 0 Check ifG1andG2are valid CFGs
Does every infinite Turing-recognizable language have an infinite decidable subset?
Show that every infinite Turing-recognizable language has an infinite decidable subset. (Hint: Use the result in (a) and the result you know regarding Turing- recognizable languages and enumerator TMs (Theorem 3.21 in the text)). Let A be an infinite Turing-recognizable language.
Is there an enumerator for every string in a Turing-recognizable language?
(Hint: Use the result in (a) and the result you know regarding Turing- recognizable languages and enumerator TMs (Theorem 3.21 in the text)). Let A be an infinite Turing-recognizable language. Then, there exists an enumerator E that enumerates all strings in A (in some order, possibly with repetitions).
Is E' a decidable subset of a?
Therefore, the language of E’ is also infinite. Finally, since E’ only prints strings in lexicographic order, its language is decidable as proved in (a). Thus, the language of E’ is an infinite decidable subset of A.
Why L1-L2 is not decidable?
We know that L2 is Turing- recognizable but not decidable. Now L1-L2is the complement of the language L2. If we have a recognizer for a language and its complement, then we have a decider for the language. This is a contradiction, since ATMis undecidable. Hence we cannot always build a recognizer for the language L1-L2.
