Question

For a Turing machine \(M\), \(\langle M \rangle\) denotes an encoding of \(M\). Consider the following two languages: \[ L_1 = \{\langle M \rangle \mid M \text{ takes more than } 2021 \text{ steps on all inputs}\} \] \[ L_2 = \{\langle M \rangle \mid M \text{ takes more than } 2021 \text{ steps on some input}\} \] Which one of the following options is correct?

• A. Both \(L_1\) and \(L_2\) are decidable.
• B. \(L_1\) is decidable and \(L_2\) is undecidable.
• C. \(L_1\) is undecidable and \(L_2\) is decidable.
• D. Both \(L_1\) and \(L_2\) are undecidable.

Hint:

Any language that asks whether a Turing machine halts within or exceeds a fixed constant number of steps is decidable, because the simulation space is finite.

Answer 1

The Correct Option is A

Step 1: Observation about the time bound.
The value 2021 is a fixed constant. For any Turing machine M, its behavior within the first 2021 computation steps on any input can be completely simulated and examined.

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Step 2: Decidability of language L₂.
Language L₂ asks whether there exists some input on which M runs for more than 2021 steps.

To decide this, we enumerate all possible input strings of length at most 2021 and simulate M on each input for at most 2021 steps.

If for at least one input the machine does not halt within 2021 steps, then ⟨M⟩ ∈ L₂. Otherwise, it is not.

Since both the number of inputs and the simulation steps are finite, L₂ is decidable.

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Step 3: Decidability of language L₁.
Language L₁ requires that M runs for more than 2021 steps on all inputs.

Again, simulate M on every input string of length up to 2021 for at most 2021 steps.

If M halts within 2021 steps for even one input, then ⟨M⟩ ∉ L₁. If it exceeds 2021 steps for every such input, then ⟨M⟩ ∈ L₁.

This procedure is finite and hence decidable.

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Step 4: Final conclusion.
Because both L₁ and L₂ can be decided using bounded, finite simulations, both languages are decidable.

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Final Answer: (A)