Turing Machines Tm Practice Questions

For a Turing machine $M$ to "decide" a language $L \subseteq \{0,1\}^*$, it must always halt for every possible input string $w \in \{0,1\}^*$. Specifically, if $w \in L$, $M$ must halt and accept $w$. Conversely, if $w \notin L$ (meaning $w \in \{0,1\}^* - L$), $M$ must halt and reject $w$. , "Turing machine $M$ accepts all input strings in $L$ and rejects all input strings in $\{0,1\}^* - L$", is the only statement that correctly captures both conditions for deciding a language. Options A, B, and C describe only partial aspects of the definition.

To find the minimum number of states, we use state minimization techniques, looking for equivalent states in the provided transition table.

1. Initial Scan for Equivalence: We compare states based on their next states and outputs for inputs $X=0$ and $X=1$.

   • Compare states (B, D, E, H). Notice that for inputs $X=0$ and $X=1$, states B and E have the same next states (D for $X=0$, C for $X=1$) and outputs (0 for both $X=0$ and $X=1$). Therefore, B and E are equivalent and can be merged into a new state, say BE.
   • Similarly, states D and H have the same next states (G for $X=0$, A for $X=1$) and outputs (1 for $X=0$, 0 for $X=1$). So, D and H are equivalent and can be merged into DH.

2. Updated Transition Table (after first merge):

   • Original A, C, F, G remain.
   • New states BE (from B, E) and DH (from D, H) are introduced.
   • The table is updated reflecting these merged states.

3. Second Scan for Equivalence (with new table):

   • Now, compare the remaining states (A, C, F, G, BE, DH). Observe states A and C. For inputs $X=0$ and $X=1$, states A and C now map to F and BE respectively, and both have outputs (0, 0). Thus, A and C are equivalent and can be merged into AC.

4. Final Set of States:
   After these merges, the distinct states remaining are: AC, BE, DH, F, G.
   This gives a total of 5 unique, non-equivalent states.

The minimum number of states required to realize this FSM is 5.

[CORRECT_OPTION: 5]

Let's analyze each statement for undecidability:
(A) "Given two Turing machines M1 and M2 , decide if L(M1 ) = L(M2 )." This is the equivalence problem for Turing machines. It is known to be UNDECIDABLE. This is because to determine if two TMs accept the same language, one would have to compare their behavior on all possible inputs, which is an infinite task.

(B) "Given a Turing machine M , decide if L(M ) is regular." This is a property of the language of a Turing machine. This problem is UNDECIDABLE. According to Rice's Theorem, any non-trivial property of the language recognized by a Turing machine is undecidable. Being regular is a non-trivial property.

(C) "Given a Turing machine M , decide if M accepts all strings." This is also a non-trivial property of the language recognized by a Turing machine (specifically, if $L(M) = \Sigma^*$). By Rice's Theorem, this problem is UNDECIDABLE.

(D) "Given a Turing machine M , decide if M takes more than 1073 steps on every string." This statement describes a property about the behavior of the Turing machine, not solely its language, on a finite domain. If M takes more than 1073 steps on every string, this implies its behavior for all inputs. However, if we consider strings of length up to 1073. For strings of length $L \leq 1073$, we can simulate M for 1073 steps. If M halts within 1073 steps for any string of length $L \leq 1073$, then the property is false. If M runs for more than 1073 steps for any string of length $L \leq 1073$, it is true for those strings. This property, however, applies to every string (including infinite length strings).
This statement is DECIDABLE. The condition "M takes more than 1073 steps on every string" refers to the runtime behavior of the TM. For a string $w$, M runs for at least 1074 steps. This is a decidable property. We can simulate the TM M on any input $w$ for 1073 steps. If M halts within 1073 steps for any $w$, then the property is false. If it does not halt within 1073 steps for any $w$, then it might be true.
The crucial part is "on every string". We can construct a TM $M'$ which takes input $w$, simulates $M$ on $w$ for 1073 steps. If $M$ halts, $M'$ halts and rejects. If $M$ does not halt within 1073 steps, $M'$ halts and accepts. Then, the question is whether $L(M') = \Sigma^*$. This is equivalent to whether $M$ takes more than 1073 steps on every string. But deciding $L(M') = \Sigma^*$ is undecidable by Rice's theorem.

Let's re-examine (D) with a simpler perspective based on the solution:
The solution states (D) is DECIDABLE.
If a TM takes more than 1073 steps on every string, it means for any given string, if we run the TM for 1073 steps, it will not halt.
We can check this for all strings of length less than or equal to 1073. There are a finite number of such strings. For each, we can simulate M for 1073 steps.
If for any of these finite strings, M halts in less than 1073 steps, then the property is false.
What about strings longer than 1073? If M has not halted on a string of length $>1073$ in 1073 steps, it has read only a prefix of length at most 1073. Its behavior after 1073 steps cannot depend on anything beyond this prefix.
So, the property is equivalent to asking: "Does M take more than 1073 steps on all strings of length up to 1073?"
This is a finite problem, and thus DECIDABLE. For a fixed number of steps ($K=1073$), one can decide properties about TM behavior within $K$ steps for a finite set of inputs (all inputs that influence behavior up to $K$ steps).

So, statements (A), (B), and (C) are UNDECIDABLE. Statement (D) is DECIDABLE.
The question asks "Which of the following is/are undecidable?".
So, (A), (B), (C) are the correct choices.

For a Turing machine M, if it takes at most 2021 steps, it can read at most 2021 bits of input. The halting behavior of M is thus determined by a finite number of input strings (those with length up to 2021 bits).
We can simulate M on all these finite input strings for 2021 steps.
For $L_1$, we check if M takes more than 2021 steps on all inputs. For $L_2$, we check if M takes more than 2021 steps on some input.
Since this check can be completed in a finite amount of time by simulating M on all relevant finite inputs, both $L_1$ and $L_2$ are decidable.

Let's analyze each language based on Rice's Theorem, which states that any non-trivial property of the language recognized by a Turing machine is undecidable. A property is non-trivial if it is true for some Turing machines but false for others.

1. $L_1 = \{\left \langle M \right \rangle | L(M) = \varnothing\}$: This is the emptiness problem for Turing machines. It is a non-trivial property (some TMs accept an empty language, others do not), so by Rice's Theorem, $L_1$ is undecidable.
2. $L_2 = \{\left \langle M,w,q \right \rangle | M \text{ on input } w \text{ reaches state } q \text{ \in exactly 100 steps}\}$: This property is about the behavior of the Turing machine itself (reaching a specific state within a fixed number of steps), not about the language it recognizes. This can be simulated and checked, making $L_2$ decidable.
3. $L_3 = \{\left \langle M \right \rangle | L(M) \text{ is not recursive}\}$: This is a property of the language recognized by $M$ (whether it's recursive or not recursive). It's non-trivial because some TMs recognize recursive languages, and others recognize non-recursive languages. By Rice's Theorem, $L_3$ is undecidable.
4. $L_4 = \{\left \langle M \right \rangle | L(M) \text{ contains at least 21 members}\}$: This is a non-trivial property of the language recognized by $M$ (e.g., a TM accepting only "a" has less than 21 members, while a TM accepting $\Sigma^*$ has infinitely many). By Rice's Theorem, $L_4$ is undecidable.

Therefore, $L_1, L_3,$ and $L_4$ are undecidable.

The final answer is $\boxed{A}$

The decidability of the given problems is as follows:
(I) The membership problem for an unrestricted grammar (which is equivalent to a Turing Machine) is undecidable.
(II) According to Rice's theorem, any non-trivial property of the language accepted by a Turing Machine is undecidable. Whether a language is regular is a non-trivial property, hence this problem is undecidable.
(III) The equivalence problem, which is determining if two grammars generate the same language ($L(G_1) = L(G_2)$), is undecidable for context-free grammars and higher.
(IV) The problem of equivalence between an NFA (accepting a regular language) and a DPDA (accepting a deterministic CFL) is decidable.
Thus, problems I, II, and III are undecidable.

A function $f$ is computable if there is a Turing Machine (TM) that halts on every input $x$ and produces $f(x)$. A language $L$ is recursive if there is a TM (a decider) that halts on every input, accepting strings in $L$ and rejecting strings not in $L$.

1. If $f$ is computable, we can construct a decider for $L_f = \{x\#f(x)|x \in A^*\}$. Given an input $w$, the decider first checks if it has the form $x\#y$. If so, it computes $f(x)$ (which is guaranteed to halt) and compares the result with $y$. This process always halts, so $L_f$ is recursive.

2. If $L_f$ is recursive, we can construct a TM to compute $f(x)$. Given an input $x$, this TM can enumerate all possible strings $y \in B^*$ and for each, it uses the decider for $L_f$ to check if $x\#y \in L_f$. Since $f$ is a total function, this search is guaranteed to find the unique correct $y$ and halt. Thus, $f$ is computable.

Therefore, $f$ is computable if and only if $L_f$ is recursive.

For L1 and L2, membership can be decided by a Turing Machine that always halts. We can simulate the machine M for a finite number of steps on a finite set of inputs to check if it runs for at least 2016 steps. This simulation is guaranteed to terminate, making L1 and L2 recursive languages.

L3 is the language of Turing Machines that accept the empty string, $\varepsilon$. This is a well-known undecidable problem, meaning it is not recursive. Although it is recursively enumerable, it is not recursive. Therefore, L1 and L2 are recursive, but L3 is not.

I. True: The class of Turing decidable (recursive) languages is closed under complementation. If a Turing Machine (TM) decides a language L, another TM can be constructed to decide its complement by swapping the accept and reject states.
II. False: By definition, any language in NP is decided by a non-deterministic TM that halts on all inputs. Therefore, every language in NP is Turing decidable.
III. True: As explained for statement II, all languages in NP are a subset of the set of Turing decidable languages.
Thus, only statements I and III are true.

This question asks us to characterize the language $L = \{ \lt M \gt \mid M \text{ is a Turing machine that accepts a string of length 2014} \}$.

1. Recognizability (Recursively Enumerable): A Turing machine $U$ can be constructed to enumerate all strings over $\Sigma$, simulate $M$ on each string of length 2014, and if $M$ accepts any of them, then $U$ accepts $\lt M \gt$. This means $L$ is recursively enumerable.
2. Decidability: To be decidable, there must be a Turing machine that halts on all inputs, accepting strings in $L$ and rejecting strings not in $L$. However, a Turing machine $M$ might loop forever on some input string of length 2014. If $M$ doesn't accept any string of length 2014, and loops on some, we can't definitively say it doesn't accept any such string within a finite time. This is related to the Halting Problem, which is undecidable.
3. Since $L$ is recursively enumerable but not decidable, it is undecidable but recursively enumerable.

The language $L$ is recursively enumerable because we can simulate the Turing machine $M$ on all strings of length 2014. If $M$ accepts any of these strings, then $\lt M \gt$ is in $L$. However, $L$ is undecidable because $M$ might loop forever on some inputs of length 2014, preventing a decider from always halting.

Let's analyze each statement:

1. For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine. This statement is TRUE. It's a fundamental result in computability theory that non-deterministic Turing machines (NTMs) and deterministic Turing machines (DTMs) are equivalent in power. Any language recognized by an NTM can also be recognized by a DTM.

2. Turing recognizable languages are closed under union and complementation. This statement is FALSE. Turing recognizable languages (also known as recursively enumerable languages) are closed under union, but not under complementation. The complement of a Turing recognizable language is not necessarily Turing recognizable.

3. Turing decidable languages are closed under intersection and complementation. This statement is TRUE. Turing decidable languages (also known as recursive languages) are closed under all Boolean operations, including intersection and complementation.

4. Turing recognizable languages are closed under union and intersection. This statement is TRUE. Turing recognizable languages are indeed closed under both union and intersection.

Based on the analysis, only statement 2 is FALSE.