Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Show that the language {M: M is a Turing machine with L(M) to be a finite set} is undecidable. You need to establish its connection to Atm={<M,w>| Turing machine M accepts input w}
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- Prove that the language {1"0"1k|k>0, n > 0} is not regular. DO NOT use pumping lemma.arrow_forward10. Let SINGLETM = { <M> | M is a TM and for all s Î L(M), |s| = 1 }. Show that SINGLETM is undecidable by reductionfrom ATM, the acceptance problem for Turing machines. Do not use Rice’s Theorem.arrow_forwardPresent a language B such that B is complete for the class of Turing-decidable languges under ≤m reductions.arrow_forward
- Consider the language L = { |‡x such that ï ¤ L(M₁)~ L(M₂)}. Show that L is not decidable.arrow_forwardRecall a set A is countable if |N| ≥ |A|. Recall that Cantor’s theorem shows that P(N) = {X ⊆ N}is uncountable. Prove this.Next,show that F = {X ⊆ N | |X| < ∞} is countable, with an onto functiong : N → F that is computable.arrow_forward2. Let A be a Turing-recognizable language consisting of descriptions of Turing machines, i.e., A = {(M1), (M2),...}, where every M; is a decider. Prove that some decidable language D is not decided by any decider M, whose description appears in A.arrow_forward
- For a Turing machine M, (M) refers to the binary representation of M. For a Turing machine M, L(M) contains the set of all strings accepted by M. For a Turing machine M and an input x = {0,1}*, Steps(M, x) refers to the number of steps taken by M to execute on x before it halts. Here, one step of execution of M on x = one movement (left or right) of the tape head. For a Turing machine M and an input x = {0,1}*, we define the following: ReachCells(M,x) = {i : M reaches ith tape cell when M is executed on x} Informally, it contains all locations on the tape that are visited when M is ecuted on x. The leftmost location on the tape is the first tape cell, the location next to it is the second tape cell, and so on. A string w₁ is an anagram of w2 if w₁ can be obtained by rearranging the alphabets of w2. Formally, if w₁ is an n length string, wê is called an anagram of w₁ if there exists a permutation à on n elements such that π(w₁) = W2.arrow_forwardnot handwritten Prove that the following problem is not decidable: Given a Turing Machine M, does the machine reject every input. (i.e., does L(M) = ∅)?arrow_forwardProve the language L = {1n | m ∈ N ∧n = 2m} is non-regular. Explain each step of your reasoning with precision.arrow_forward
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