L2 m : m is a tm and l m is infinite
WebOct 15, 2024 · TM = { M is a TM and L(M)= } –It is undecidable! •EQ TM = {(M1,M2) M1,M2 are TMs and L(M1)=L(M2)} •Instead of setting up a reduction from A TM we can use other undecidableproblems such as E TM –Assume towards contradiction R is a decider for EQ TM –Construct a decider S for E TM such that on input where M is a TM 1. WebProblem 4 (10 points) Let L2 = {M is a TM and L(M) = 2}. In other words, Ly consists of all encodings of turing machines that accept exactly 2 strings. Show that L2 is …
L2 m : m is a tm and l m is infinite
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WebClaim 1. If a language L and its complement L are both semi-decidable, then L is decidable. Proof. Let M L be a TM accepting L, and let M L be a TM accepting L . On input x, run both TMs \in parallel", until one of them accepts. (At some nite point in time, one of the machines must accept as every input x is either in L or in L .) If M WebTM m INFINITE, where INFINITE TM = fh M ij is a T uring mac hine and accepts in nitely man yw ords g A TM = f j M is a TM, w isaw ord, and accepts g. Solution: Since A TM is undecidable and w e pro v ed already that m INFINITE, then w e kno w that INFINITE TM is undecidable. Note that FINITE the complemen t of. Hence, FINITE TM has to b ...
WebTM. If M does not accept w, then L(M 2) is L(00∗11∗), so M 2 ̸∈S TM. Hence, M 2 belongs to S TM if and only if M accepts w, so a solution for S TM can be used to solve A TM; i.e., A TM reduces to S TM. Because S is assumed to decide S TM, the TM A decides A TM because stage 3 of the TM A accepts M,w if and only if S accepts M 2 . But we ... WebProof: Let M1 and M2 be TM’s for L1 and L2. We show there is a TM M that recognizes L1 U L2 by giving a high-level description of nondeterministic TM M. • Construction: Let M = “On input w: 1. Nondeterministically guess i = 1 or 2, and check if w is accepted by Mi by running Mi on w. If Mi accepts w, accept.”
WebThe following TM M decides L = L1 intersection L2: Let M = "on input string w: Run M1 and M2 in parallel (or one after the other). If both M1 and M2 accept w, then accept w. If either … WebINFINITETM = {〈M〉 M is a TM and L (M) is an infinite language}. Is it co-Turing-recognizable? prove your answer. This problem has been solved! You'll get a detailed …
Weblet M2 be a TM that decides L2. The following TM M decides L = L1 intersection L2: Let M = "on input string w: Run M1 and M2 in parallel (or one after the other). If both M1 and M2 accept w, then accept w. If either one rejects w, then reject w. Let's prove that the Turing machine M above decides the language
WebTM = { M is a TM and L(M)=Φ} (p. 217) Theorem 5.2 E TM is undecidable Assume R decides E TM, i.e. given as input, R accepts if L(M) is empty rejects if L(M) is not Use R to construct an S that decides A TM as follows Given any , first convert M to M 1 as follows On any input x, If x != w, M 1 rejects dr. brent hardin with oxford urology assocWebTM = fhMijM is a TM and L(M) is regulargis undecidable. Proof. Let R be a TM that decides REGULAR TM and construct TM S to decide A TM. S = \On input hM;wi, where M is a TM and w is a string: 1.Construct the following TM M 2. 2. … encased rebel armorWebdecider for the language L1. 2. Run M2 on input w. Again, the computation is guaranteed to halt. 3. If M1 accepted, and M2 rejected, then accept the string w, else reject. b. Let L1 and … encased skill booksWebJan 1, 2024 · This is the empty set, since every L (M) has an infinite number of TMs that accept it." The other answers are correct, and there are other ways to prove that every … encased rpg transportWebTM. If M does not accept w, then L(M 2) is L(00∗11∗), so M 2 ̸∈S TM. Hence, M 2 belongs to S TM if and only if M accepts w, so a solution for S TM can be used to solve A TM; i.e., A … dr brent hamilton nephrologyWebNov 9, 2005 · then M1 will write a nonblank, overwrite the nonblank with a blank and then accept w. Now we can create our decider for ATM. S = “On input , where M is a TM 1. Create M1 as described above 2. Run the decider D on input 3. If D accepts accept 4. If D rejects reject” Since D is a decider, S is also a decider. encased skill pointsWebP = {< M > M is a TM and 1011 ∈ L (M)}. Use Rice’s Theorem to prove the undecidability of the following language. P = {< M > M is a TM and 1011 ∈ L (M)}. Expert Answer 100% (2 ratings) Rice's Theorem: If P is a non-trivial property, and the language holding the property, Lp , is recognized by Turing machine M, then Lp= { encased shotgun