Prove that the following language L is decidable:L = {(M) : 3x € {0,1}*, 1000 ≤ ReachCells(M, x)}. 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 takenby M to execute on x before it halts. Here, one step of execution of M on x = one movement (left orright) 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. Theleftmost location on the tape is the first tape cell, the location next to it is the second tape cell, and soon.A string w₁ is an anagram of w2 if w₁ can be obtained by rearranging the alphabets of w2. Formally, ifw₁ is an n length string, wê is called an anagram of w₁ if there exists a permutation à on n elementssuch that π(w₁) = W2.

A Turing machine is an abstract mathematical model of computation introduced by Alan Turing in 1936. It consists of an infinite tape, a tape head that can read and write symbols on the tape, and a set of states that dictate the machine's behavior. The machine starts in an initial state, reads the symbol under the tape head, transitions to a new state based on the current state and symbol read, and moves the tape head left or right. Turing machines can perform various computational tasks and serve as a foundational concept in the theory of computation. They are used to explore the limits and capabilities of algorithms and form the basis for understanding computability and complexity in theoretical computer science.In the context of the above question, the language L is considered decidable if there exists a Turing machine that can determine whether a given Turing machine MM accepts some input xx such that 3x3x is in the set of reachable cells during the execution of MM. In other words, a language is decidable if there exists an algorithm (represented by a Turing machine) that can decide membership in that language for any given input.To prove that the language L is decidable, we need to show that there exists a Turing machine that decides L. The language L is defined as follows:L={(M)∣ there exists x∈{0,1}∗ such that 3x∈ReachCells(M,x)}L={(M)∣ there exists x∈{0,1}∗ such that 3x∈ReachCells(M,x)}Here, ReachCells(M,x) is the set of cells that are visited during the execution of Turing machine M on input x.To show decidability, we can construct a Turing machine T that decides L. The high-level idea of the proof is to simulate the execution of the Turing machine M on all strings x of length ∣x∣=n≤k where k is a fixed constant that we'll determine later. If 3x is in the set of reachable cells for any x, then (M) is in L. Otherwise, if 3x is not in the set of reachable cells for all x, then (M) is not in L.Here are the steps for constructing the decider T: Enumerate all strings x of length n≤k. For each x, simulate the execution of M on input x and record the set of cells visited during the simulation. If 3x is in the set of visited cells for any x, accept. Otherwise, if 3x is not in the set of visited cells for all x, reject.The constant k is determined by considering the maximum number of cells that M can visit in k steps. This is finite since the number of cells is finite, and M is deterministic. Therefore, we can set k to be the maximum number of steps needed for M to reach its maximum number of visited cells.Since the simulation and checking process is finite and deterministic, the Turing machine T will halt for any input (M), and thus L is decidable.The language L is decidable because there exists a Turing machine that can decide membership in L. The decider simulates the execution of Turing machine M on all possible inputs of a certain length and checks if 3x is in the set of reachable cells during the simulation. If 3x is found for any input x, then (M) is in L; otherwise, it is not. The simulation is guaranteed to halt since the number of possible inputs is finite. Therefore, L is decidable.

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Prove that the following language L is decidable: L = {(M) : 3x = {0,1}*, 1000 € ReachCells(M,x)}.

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Question

Prove that the following language L is decidable:
L = {(M) : 3x € {0,1}*, 1000 ≤ ReachCells(M, x)}.

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.

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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.
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