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Chapter 5: Chapter 5

Problem 15

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Consider the problem of determining whether a Turing machine \(M\) on an input \(w\) ever attempts to move its head left at any point during its computation on \(w\). Formulate this problem as a language and show that it is decidable.

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The given problem can be formulated as a language L: \(L = \{ \langle M, w \rangle \mid\) M is a Turing machine that tries to move its head to the left at least once during the computation on input w\(\}\). To show that L is decidable, we constructed a Turing machine D that simulates M on input w, observing its moves. If D detects a leftward move, it accepts; if it reaches the end of the simulation without detecting a leftward move, it rejects. D satisfies both correctness and termination properties, proving that L is decidable.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Define the Language

Let's define a language L for the given problem: \(L = \{ \langle M, w \rangle \mid\) M is a Turing machine that tries to move its head to the left at least once during the computation on input w\(\}\). The goal is to show that L is decidable.

Step 2: Construct a Decider for L

To do this, we will need to construct a Turing machine D that decides L. The Turing machine D works as follows for a given input \(\langle M, w \rangle\): 1. Simulate M on w. 2. During the simulation, observe all the moves of M's head. 3. If at any point, we find a move where M's head moves left, immediately "accept" since this is a positive instance of the language. 4. If the simulation of M on w halts and no leftward move was detected, "reject" since it is a negative instance of the language.

Step 3: Analyze the Decider's Properties

In order to prove that the constructed Turing machine D is indeed a decider for L, we need to examine the following properties: - **Correctness**: For every \(\langle M, w \rangle\), D accepts if and only if M makes a move to the left on input w. This is true in our construction. D executes M on w and accepts once it finds the leftward move. If it finishes simulating M on w without finding any leftward move, it rejects the input. - **Termination**: For every \(\langle M, w \rangle\), D halts. D always halts. This is because either it finds the left head move during the simulation and immediately "accepts," or it finishes simulating M on w, meaning it has exhausted all possible computations of M on w without detecting a leftward move, and "rejects." Since D satisfies both the correctness and termination properties, D is indeed a decider for L.

Step 4: Conclusion

We have successfully formulated the given problem as a language L and shown that L is decidable by constructing a Turing machine D that decides L. D accepts an input when it detects a leftward move during the computation of M on w, and rejects otherwise, after exhausting all possible states.

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Most popular questions from this chapter

Chapter 5

$$ \text { Show that } A \text { is decidable iff } A \leq_{\mathrm{m}} 0^{*} 1^{*} \text {. } $$
Chapter 5

Say that a variable \(A\) in CFG \(G\) is necessary if it appears in every derivation of some string \(w \in G\). Let $N E C E S S A R Y_{\mathrm{CFG}}=\\{\langle G, A\rangle \mid A$ is a necessary variable in \(G\\}\). a. Show that \(N E C E S S A R Y_{\mathrm{CFG}}\) is Turing-recognizable. b. Show that \(N E C E S S A R Y_{\mathrm{CFG}}\) is undecidable.
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Consider the problem of determining whether a two-tape Turing machine ever writes a nonblank symbol on its second tape during the course of its computation on any input string. Formulate this problem as a language and show that it is undecidable.
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Let \(\Gamma=\\{0,1, \sqcup\\}\) be the tape alphabet for all TMs in this problem. Define the busy beaver function $B B: \mathcal{N} \longrightarrow \mathcal{N}\( as follows. For each value of \)k\(, consider all \)k$-state TMs that halt when started with a blank tape. Let \(B B(k)\) be the maximum number of 1s that remain on the tape among all of these machines. Show that \(B B\) is not a computable function.
Chapter 5

A useless state in a Turing machine is one that is never entered on any input string. Consider the problem of determining whether a Turing machine has any useless states. Formulate this problem as a language and show that it is undecidable.
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