Question

Which one of the following statements is equivalent to the following assertion?
Turing machine \(M\) decides the language \(L \subseteq \{0,1\}^*\)

• A. Turing machine \(M\) halts on all input strings in \(\{0,1\}^*\)
• B. Turing machine \(M\) accepts all input strings in \(L\)
• C. Turing machine \(M\) rejects all input strings in \(\{0,1\}^* - L\)
• D. Turing machine \(M\) accepts all input strings in \(L\) and rejects all input strings in \(\{0,1\}^* - L\)

Hint:

Deciding a language needs BOTH correct acceptance on L and correct rejection on its complement; that combination alone forces halting everywhere too.

Answer 1

The Correct Option is D

Step 1: Recall what it means for a Turing machine to "decide" a language.
A Turing machine \(M\) decides a language \(L\) if two conditions both hold: \(M\) halts on every input string (it never loops forever), and \(M\)'s halting behavior correctly classifies every string, accepting exactly the strings in \(L\) and rejecting exactly the strings not in \(L\), that is, the strings in \(\{0,1\}^* - L\). A machine that only accepts every string of \(L\) without addressing the rest is merely a "recognizer," which is weaker than a "decider."

Step 2: Check option (A) against this definition.
Option (A) only says \(M\) halts on all strings in \(\{0,1\}^*\). Halting alone says nothing about whether \(M\) accepts exactly \(L\), a machine could halt on every string yet accept the wrong set of strings and still satisfy (A), so (A) alone is not equivalent to deciding \(L\), it is a necessary condition but not a sufficient one.

Step 3: Check option (B).
Option (B) only requires \(M\) to accept every string in \(L\); it says nothing at all about what happens on strings outside \(L\). A machine satisfying (B) could loop forever, or wrongly accept, on strings not in \(L\), and it would still satisfy (B), so (B) describes only "recognizing" \(L\), not deciding it.

Step 4: Check option (C).
Option (C) only requires \(M\) to reject every string in \(\{0,1\}^* - L\); it says nothing about what \(M\) does on strings that are actually in \(L\). A machine could loop forever, or wrongly reject, on strings inside \(L\) and still satisfy (C), so (C) alone is also insufficient.

Step 5: Check option (D).
Option (D) requires \(M\) to accept every string in \(L\) AND reject every string in \(\{0,1\}^* - L\). Since every string in \(\{0,1\}^*\) is either in \(L\) or in its complement, this covers every possible input string, and every covered string is explicitly required either to be accepted or rejected, both of which are halting outcomes. So this single statement forces \(M\) to halt on all inputs (nothing is left undefined) and to classify every string correctly, which is precisely the two-part definition of "decides" from Step 1.

Final Answer:
Only option (D) captures both required parts of decidability, correct acceptance on \(L\) and correct rejection on its complement, together implying halting on every input.
\[ \boxed{\text{Option D}} \]