Decidable languages closed under concatenation § RE Languages: can’t do it; M2 may never halt, so you can’t be sure input is in the difference. (e) intersection. Think about whether the collection of Turing-recognizable languages is also closed under the operations listed in the previous problem. Therefore, L1 ° L2 (concatenation) is decidable. Also assume the two languages L 1 and L 2 where,. A language L is decidable if there exists a decider D such that L(D) = L R. (b) Concatenation: Let K,L be decidable languages. In mathematics, logic and computer science, a formal language (a set of finite sequences of symbols taken from a fixed alphabet) is called recursive if it is a recursive subset of the set of all possible finite sequences over the alphabet of the language. dvi Answer to 3. 14 Show that the collection of decidable languages is closed under the following operations. Turing recognizable languages are closed How much is a tee in the range? (20 pts) Show that the collection of decidable languages is closed under the operation of: a) Concatenation Let L1, L2 be two decidable languages, M1 be a Turing Machine which decides L1 and M2 be a Turing Machine which decides L2. Here’s the best way to solve it. We simply swap the accept and reject state. 15 b,c,d) Show that the collection of Turing-recognizable languages is closed under the operations of • Corollary: Recursive languages are closed under complementation. 15 from textbook] Show that the collection of decidable languages is closed under the operation of 1. Proof. Assume so that L 1 2P and that L 2 2P. Decidable languages kleene star closure - question on a I read a proof on the closure of decidable languages under kleene star. We want to show that if L 1;L 2 2P then L 1 L 2 2P. The Turing machines decide the languages L in log space. To prove that the class of decidable languages is closed under concatenation, construct a Turing machine M that, on input w, systematically splits w into two parts and checks if the first part is in L_1 using M_1 and if the second part is in L_2 using M_2. [Problem 3. Recursively enumerable languages closed under operation. b concatenation. CFL not closed under intersection while Turing Decidable are. (a) Let G4 = (V1 [V2 [fS4g;T1 [T2;S4;P1 [P2 [fS4! Regular languages are closed under complement. Decidable = Rec ∩ co-Rec recognizable decidable co-recognizable L decidable iff both L & Lc are recognizable Pf: ($) on any given input, dovetail (run in parallel) a recognizer for L with one for Lc; one or the other must halt & accept, so you can halt & accept/reject appropriately. The running time of the modiﬁed machine does not change. It also halts on all input, and accepts the complement of A. Turing recognizable languages are closed under union and complementation. 5. - For the input w - Run M1 and M2 in parallel. A word is taken which can be broken into sub-words w 1, w 2,. First let us observe that the decidable languages are closed under the regular op-erations. (Since TM halts on all input, we can have a different TM that flip-flops reject/accept. Let C be a TM which makes a copy of the input: (s, # w #) ├* (h, # w # w [#]). Given decidable languages $L_1$ and $L_2$, we can construct Turing Machines which decide $L_1 \cup L_2$, $L_1^C$, $L_1 \cdot L_2$, $L_1 \cap L_2$, and $L^*$ respectively, by running the input through TMs for individual languages in I'm tasked with demonstrating that the class of Turing-recognizable languages is closed under the operation of star, but I'm confused about how this is true. complementation. Figure DecisionPropertiesSummary: Decision problems/questions that are decidable are indicated by a check under the applicable language classes. Theorem: Turing decidable languages are closed under intersection. Proof: Upon halting, simply exchange the verdicts accept and reject. 14. Join Yes, decidable languages are closed under operations such as union, intersection, concatenation, and complementation. Finish the proof by drawing a machine schema for a TM which decides L= L 1 L 2. You can proof this with Schützenberger products. (d) complementation. Complementation: If L is decidable, then I is decidable. Visit Stack Exchange Question: Show that the collection of decidable languages is closed under the operation of Aa. Koether Homework Review Closure Properties of Decidable Languages Intersection Union Closure Properties of Recognizable Languages Stack Exchange Network. Proposition 8. Exercise decidable. Decidable languages are closed under complement. Let S be an alphabet and let A, B S be decidable languages. e if L is decidable then so is L • Concatenation (it's easier if you use Nondeterministic Turing machines here) Show transcribed image text. (%): from above, decidable languages are closed under 1. The languages A [B, AB, and A are decidable. Kleene Closure. Regular languages are closed under the following Show that of Turing decidable languages is closed under concatenation. I know decidable languages are closed under concatenation and union. Does your construction from part (a) also show that the class of Turing-recognizable languages is closed under concatenation? Explain your One way --> If L1, and L2 are RE s then L1. Here we show that decidable languages are closed under the five "main" operators: union, intersection, complement, concatenation, and star. 0. Union, Intersection, complement, concatenation, star. Proposition 18. The reason is that recognizable languages in general are not closed under Kleene star. (1) Intersection (2) Complement (3) Concatenation (4) Star See the example below that of decidable Language is closed under concatenation. The concatenation of languages K and L is the language KL = {xy|x ∈ K and y ∈ L}. Need answers for (b), (c), (d), and (e). Next I did some demonstrations to show how T-Recognizable languages are closed for Union, Intersection, Concatenation and Kleene Star. @DavidRicherby showed that there exist two NP-Complete languages the intersection of which is not NP-Complete Solution for Prove that Turing-decidable languages are closed under concatenation. The concatenation operation of languages K and L is the language KL = {xy|x ∈ K and y ∈ L}. Rao, CSE 322 2 Closure Properties of Decidable Languages Decidable languages are closed under ∪, °, *, ∩, and complement Example: Closure under ∪ Need to show that union of 2 decidable L’s is also decidable Let M1 be a decider for L1 and M2 a decider for L2 Proposition: The decidable languages are closed under complementation. g. Hint: For this, show that for any two decidable languages 𝐿1 and 𝐿2 there exists a Turing machine 𝑀 that (1) recognizes 𝐿1𝐿2 and (2) is a decider. (b) concatenation. For example, I have a TM to recognize a language A = { a 2 n : n >= 0 }, all strings of a's which length is a power of 2. Simulate M1 on w. CFLs and Turing-decidable languages are just one example of this. Define the corresponding operation on languages: BALANCE(L) = {BALANCE(w)w E L}. union. See 3. 2. L2 is RE then RE is closed under concatenation. Because K and L are decidable Decidable languages are closed under concatenation and Kleene Closure. The key is to ass The last part of the proof states that after mapping L to the homomorphism, h(L) is L_halt, which is not decidable. The proof is quite straightforward. Yes, all CFLs are Turing-decidable but a subset of an intersection-closed class of languages is not necessarily intersection-closed. Or more specifically, in general we have rational sets that are not recognizable. Step 1. e. $\begingroup$ This is how I proved union property for RE Union: For any two RE language L1 and L2, let M1 and M2 be the TMs that recognize them. star. 7. Note: So CFL are closed under Concatenation. 3. This means that if a language is decidable, performing these operations on it will result in another decidable language. Show that the collection of decidable languages is closed under the operation of (a) concatenation (b) complementation. Ask Question Asked 2 years, 5 months ago. Modified 12 years, user contributions licensed under CC BY-SA. I'm studying Turing Machines and I've already showed how Turing-Decidable is closed for the operations of Union, Intersection, Concatenation, Complement and Kleene Star. 15) Show that the collection of decidable languages is closed under (a) Union: (in the textbook). (Closure properties) Show that the class of decidable languages is closed under (a) complement (b) concatenation Show that the class of Turing-recognizable languages is closed under (c) star (d) Explain why your construction from 2(a) fails to show that the Turing-recognizable languages are closed under complement. I know this is easy for most of the people but unfortunately my professor is not very good at explaining the material. Definition: A language is called semi-decidable (or recognizable) if there exists an algorithm that accepts a given string if and only if the string belongs to that language. Why are regular tree languages closed under intersection, but deterministic context free languages are not closed under intersection? 1 Give a class of languages which is closed under intersection and union, but not under complement Decidable languages are closed under union, intersection, concatenation, and Kleene star operations. A TM that decides L1 [ L2: on input x, run M1 The question in the title asks you to show (prove) that the class of decidable languages is closed under the complementation operation, and under the concatenation operation, and under the Theorem: Turing decidable languages are closed under complement. The class of semi-decidable languages is closed under union and intersection operations. concatenation of 2 undecidable language gives a decidable language [closed] Ask Question Asked 12 years, 9 months ago. In the context of string operations, being closed under concatenation means that when you combine two strings together, the result is still a valid string. We do this visually and then with a formal description. Let $A$ and $B$ be languages. Answer to show that the collection of decidable languages is. Theorem 16. Solution. § Assume M1 and M2 are single-semi-infinite-tape TM’s. Can you help me find the right direction to proof/ Proving a class of languages is closed under union when it closed under concatenation, (inverse) homomorphic images, and intersections. If L is a CFL, then its Kleene closure L * (zero or more repetitions of strings in L) is also a CFL. Concatenation: If L1 and Question: Show: The class of decidable languages is closed under the concatenation operation. § Construct 2-tape Nondeterministic TM M: 1. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Complement. Cite. It is easy to construct the machine schema for a TM which decides the complement I read a proof on the closure of decidable languages under kleene star. Closure properties on regular languages are defined as certain operations on regular language that are guaranteed to produce regular language. Modified 2 years, 5 months ago. ) So both A and A’ Yes, decidable languages are closed under operations such as union, intersection, concatenation, and complementation. For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine. 1 / 28. Mar 05,2025 - Which of the following statements is/are FALSE?1. We discuss a class of problems about string-rewriting that illustrates the distinction between Turing-decidable and Turing-recognizable languages. Because the languages A and B are decidable, there must exist a DTM M The collection of decidable languages is closed under the operation of union, complementation, concatenation, intersection, and star. The class of decidable languages is closed under Union Intersection Complementation Concatenation Star. Are these valid counterexamples to proofs Can we use the proof of turing-decidable languages being closed under complement to show that turing-decidable languages are closed under set difference? computer-science; Share. The intersection of a CFL with a regular language is a CFL. Why is "decidable" included in "Turing-recognizable"? 1. 4. A TM to decide L 1L 2: On input x, for each of the jxj+1 ways to divide xas yz: run M 1 on yand M 2 on z, and accept if both accept. Construction: Let G be the context-free grammar for L with start symbol S₁. Let L 1 and L 2 be two semi-decidable languages, and let M 1,M 2 be Question: Show that the collection of decidable languages is closed under concatenation. There’s just one step to solve this. Proof: Let M 1 be a TM which decides L 1, and let M 2 be a TM which decides L 2. Concatenation. 15(a). The decision questions are, from top to bottom, test of membership of a string in a language, whether the language is empty or not, whether the language is equivalent to the Kleene closure of the Show that the set of decidable languages is closed under union, concatenation, Kleene start, complementation, and intersection. Skip to main content. A TM to decide L 1L 2: On input x, for each of the SOLUTION-(1):-In this solution, we will show that Turing-decidable languages are closed under the following operations: union, concatenation, starFor Union operation : Suppose that L 1 and L 2 are two decidable languages accepted by halting TMs M 1 and M 2 respectively. In case the string does not As an example subset of decidable languages which is closed under both union and concatenation, but not under Kleene star, consider: $\{ \emptyset \}$ $\emptyset \cup \emptyset = \emptyset$ $\emptyset\emptyset = \emptyset$ but $\emptyset^* = \{ \epsilon \}$, which doesn't belong to our subset. It begins by saying that the turing machine we want to find would non-determistically split the input string and then use For Concatenation operation : Let K, L be decidable languages. Every infinite Turing-recognizable language has an infinite decidable subset. Answer to Show that the set of decidable languages is closed Question: Show that the collection of decidable languages is closed under the operation of a. By deﬁnition, this means that there Study with Quizlet and memorize flashcards containing terms like What are regular languages closed under?, What are context free languages closed under?, What are deterministic context free languages closed under? and more. Viewed 574 times 0 $\begingroup$ I (a language is Turing-decidable iff a Turing machine exists which accepts all strings in that language and rejects all strings not in that language). Equivalently, a formal language is recursive if there exists a Turing machine that, when given a finite sequence of symbols as Question: Given two decidable languages L_1 and L_2 , let Turing machines M_1 and M_2 decide these languages, respectively. ; Introduce a new start symbol S and add the rule: S → S₁ S | ε This rule ensures that S can derive zero or more copies of S₁, Show that the class of TM-decidable languages is closed under the following operations: union, concatenation, star, intersection, and complement. This means that if two languages are decidable, their union, intersection, concatenation, and Question: Show that the collection of decidable languages is closed under the operation of (a) concatenation (b) complementation. (c) star. give example of 2 languages A and B such that A and B are undecidable but there concatenation A. Transcribed image text : Example: Prove that the class of decidable languages are closed under concatenation. L2. If anybody can provide any hints on how to do it I would greatly appreciate it. Title: 2. I'm reading "Theory of Computation" by Michael Sipser and I've encountered a solution (provided by the book) that I don't understand. Else reject. 14 b,d,e) Show that the collection of decidable languages is closed under the operations of concatenation, complementation, intersection. c. Yes, decidable languages are closed under operations To show that the family of decidable languages is closed under various operations, we need to demonstrate that if we have two decidable languages A and B, then the result of applying these operations to A and B also yields a decidable language. 3. Engineering; Computer Science; Computer Science questions and answers; show that the collection of decidable languages is closed under the operation of: a) a union b)concatenation c) star d) complementation e) star can someone do this problem in layman's terms and try to be brief ? Assume language A is decidable. Recall that a language is decidable if there exits a decider (Turing machine) that for any input Decidable languages are closed under union, intersection, and complementation. Proof: Let M be a TM which decides L. (Exercise 3. Given TMs M 1 and M 2 that decide languages L 1 and L 2. L 1 a n d L 2 ∈ N L. 15 Show that the collection of decidable languages I know that the decidable are close under: complementation, union, intersection and concatenation? A_1 \cup \{ 1 w : w \in \{0,1\}^* \}. Turing decidable languages are closed under intersection and complementation. Follow edited May 31, 2017 at 3:14. Complementation: If L is a decidable The classes of Turing-decidable and Turing-recognizable languages are both closed under union and under intersection. My question is, how is h(L) the same as L_halt? It seems that the new language is a language of strings of type (M, w, ci), and a string is in h(L) if and only if M accepts w after making at most 0 moves (because we deleted every c). Question: 7. (20 points) Closure (Concatenation): Show that the collection of decidable languages is closed un- der the operation of concatenation. (Problem 3. Note that w and BALANCE(w) have the same number of Os and ls, but in the beginning of the string BALANCE(w), the 0 and 1 characters alternate until one of them runs out. We construct the NFA to prove that regular languages are closed under the concatenation operator. Could anyone help me in Here we show all closure properties of all language classes in the theory of computation class (regular, CFL, decidable, recognizable) as well as all decidab Yes, decidable languages are closed under operations such as union, intersection, concatenation, and complementation. Study with Quizlet and memorize flashcards containing terms like Regular Languages are closed under:, Context Free Languages are closed under:, Decidable Languages are closed under: and more. There are 2 I can't figure out a proof that recursive languages are closed under concatenation. slide 13 in Lecture 3). Hot Network Questions Proposition 2. Solution (c) Assume that we have two Turing machines M 1 and M 2 that are non-deterministic. Show transcribed image text. Closure Properties of Decidable and Recognizable Languages Robb T. It begins by saying that the turing machine we want to find would non-determistically split the input string and then use the original decider of the language to approve the partition of each branch. b. The NL is closed under the operation use Turing machines as follows:. e. This set is not recognizable in $\mathbb N^2$. But i am having a hard time finding an counter argument (i think its wrong). A TM to decide L 1: On input x, if x= accept. 1. This property is important because it ensures that string operations can be performed without creating invalid or unexpected results. I have seen this question here, Closure of Turing-recognizable languages under homomorphism But actually this question answers the question of "What is the relation between homomorphism and concatenation?", so I still have a problem of how to show that the collection of Turing-recognizable languages is closed under homomorphism. A However, I also know that Turing-recognizable languages are not closed under complement, and $\overline{\bar{A} \cup \bar{B}} = A \cap B = C$, which seems to suggest that Turing-recognizable languages are not closed under intersection. - If either accept, accept - VIDEO ANSWER: Show that the collection of decidable languages is closed under the operation of A. c star. 6. The machine for L 1 ∪ L 2 can be designed as given below : Given an input x, simulate M 1 on x. Reason why or why not. , regular. Use the format and level of detail of the answer to Problem 3. B is decidable. Now, we show that CFLs are closed under concatenation. w n the word such Turing-recognizable languages are closed under intersection. Show that the class of decidable languages is closed under (a)concatenation; (b)BALANCE Is it also true that NPC languages are not closed under the complement, concatenation, and kleene star operations? complexity-theory; {0, 1}* and the empty language are regular, thus decidable in polynomial time and not NP-Complete. $$ See if you can use these languages to show that the collection of undecidable languages is not closed under concatenation, union or intersection. There are 2 steps to solve this one. Given TMs M1, M2 that decide languages L1, and L2. As for complementation, try to use the fact that if you Question: (5) (6pts) Show that the set of decidable languages is closed under: • Complementation (i. Note that these proofs are relatively short. 16 Concatenation of RE Languages § Let L1 = L(M1) and L2 = L(M2). concatenation. 2 Closure properties of recognizable and decidable languages. b belongs to L1. Since K and L are decidable languages, it follows that there exist turing machines M K and M Here we will show that they are closed under union; moreover they are also closed under intersection, however complementation may create a non-semi-decidable language. A string $w$ is in the concatenation of $A$ and $B$ if you can write it as $w= ab$, where $a\in A$ Turing-Recognizable languages are closed under ∪ , °, *, and ∩ (but not complement! We will see this in the final lecture) Example: Closure under ∩. Yes, decidable languages are closed under operations such as union, intersection, concatenation, and complementation. The question: Show that the collection of Turing-recognizable I'm studying Turing Machines and I've already showed how Turing-Decidable is closed for the operations of Union, Intersection, Concatenation, Complement and Kleene Star. 15 on page 191 (2nd edition: page 163) for an example proof of closure over union. Decidable languages are closed under concatenation and Kleene Closure. Closure refers to some operation on a language, resulting in a new language that is of the same “type” as originally operated on i. We construct a TM M that recognizes the union of L1 and L2. 1. There exists M3 which is a Turing Machine and considers concatenation L1 and L2. Are decidable languages closed under any operations? Yes, decidable languages are closed under operations such as union, intersection, concatenation, and complementation. But what's really the difference? Doesn't closure under concatenation imply homomorphism as well? Or perhaps I misunderstand what homomorphism means when talking about Turing-Recognizable languges? Show that decidable languages are closed under the following operations. Show that the collection of decidable languages is closed under the operation of (a) union. If M1 halts and Decidable Languages Intersection Union Closure Properties of Recognizable Languages Intersection Union Assignment Closure Properties of Decidable Languages Theorem (Closure 3. But, interestingly, the recognizable sets are closed under concatenation. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Question: 2. The same construction as for decidable languages (see e. I've proven that the Turing-recognizable languages are closed under concatenation and I need to show that they are closed under homomorphism. Another way if we talk interms of strings, If for all ‘a’ belongs to L1 and for all ‘b’ belongs to L2 then w=a. . Union: If L1 and L2 are decidable, then Lju L2 is decidable. Flashcards; Learn; Test; Match; Created by Showing that Turing-recognizable languages are closed under union 0 Prove that the class of Turing Decidable Languages is strictly larger than class of Context Free Languages. Show that the set of decidable languages is closed under union, concatenation, Kleene start, complementation, and intersection. intersection. Prove that decidable languages are closed under concatenation, complement, union, and intersection operations. d. 2. 8. Show transcribed image text There are 2 steps to solve this one. mljfe epbej hhpvr wnijasa lhyj mkak eunpy tqows voewvne mrisyl ydkp vcyfdei tee ibyh qyhia

Decidable languages closed under concatenation. L2 is RE then RE is closed under concatenation.