TM Variations and UTM MCQs: 10 Questions Explained

Multi-tape, nondeterministic and universal Turing machines look more powerful because they change the hardware picture or the possible execution paths. The recurring exam test is whether that change increases language-recognition power or only changes simulation efficiency. Usually it does not: extra tapes, extra heads and nondeterminism all leave the recognisable-language class exactly where it was, and only the pushdown cases genuinely move the boundary. Nine of the ten questions below come from previous-year papers, from GATE 2003 and 2011 through UGC NET, ISRO, Coal India, BEL and Bihar STET. Choose an option before you read the explanation. If the definitions feel loose, reset them with Turing Machines and Decidability: Halting Problem Proof, then come back. To place the topic in the wider syllabus, use the GATE category.

Calibrate the one distinction that controls the whole set

Separate recognition power from simulation cost before comparing models.

For a concrete two-tape configuration, let tape 1 be 1 0 [1] and tape 2 be 0 [1] 1, where brackets mark the heads. A single tape can encode both as # 1 0 [1] # 0 [1] 1 #. To reproduce one two-tape step, its single head scans the encoded region to read both marked symbols. Finite control records the applicable transition. The head then scans again to rewrite the required symbols and move both markers. The simulated step is possible, but it may take many single-tape moves.

So when an option calls two models "equivalent", ask what the word means. Equal language power and equal running time are different claims.

Questions 1-2: identify the basic machine and its standard variants

Question 1 (BEL 2023)

Which of the following is NOT part of the mechanical diagram of the Turing machine?

• A. Queue

• B. Finite control

• C. Input tape

• D. Read-write head

Correct answer: A. Queue.

Finite control stores the state, the tape stores symbols, and the read-write head observes or changes one cell. A queue is separate FIFO memory, not a component in this diagram. Queue automata can still exist.

Question 2 (Bihar STET 2025)

Which of the following is an extension to the basic model of Turing machine?

• A. Multi tape Turing machine

• B. Multi head Turing machine

• C. Nondeterministic Turing machine

• D. All of the above

Correct answer: D. All of the above.

Several tapes change storage, several heads change cell access, and nondeterminism permits several possible next moves. All are standard extensions. Each can be simulated by a deterministic single-tape TM for language recognition, possibly with extra time.

Questions 3-4: separate subset properties, determinism and expressive power

Question 3 (Coal India 2020)

Which of the following statement is false?

• A. Every regular language is also a context free language

• B. Every non deterministic Turing machine can be converted to an equivalent deterministic Turing machine

• C. Every subset of recursively enumerable set is recursive

• D. Every NFA can be converted into equivalent DFA

Correct answer: C. Every subset of recursively enumerable set is recursive.

Regular languages are context-free, deterministic TMs simulate nondeterministic branches, and subset construction converts an NFA to a DFA. C fails: Σ* is recursive, hence recursively enumerable, but it has non-recursive subsets. A set's property need not pass to every subset.

Question 4 (GATE 2011)

Which of the following pairs have DIFFERENT expressive power?

• A. Deterministic finite automata (DFA) and Non-deterministic finite automata (NFA)

• B. Deterministic push down automata (DPDA) and Non-deterministic push down automata (NPDA)

• C. Deterministic single tape Turing machine and Non-deterministic single tape Turing machine

• D. Single tape Turing machine and multi-tape Turing machine

Correct answer: B. Deterministic push down automata (DPDA) and Non-deterministic push down automata (NPDA).

DFA and NFA have equal power, as reviewed in Finite Automata MCQs: 10 Solved DFA and NFA (GATE). Deterministic TMs simulate nondeterministic branches, and one tape encodes several tapes. Only the PDA pair differs: NPDA covers all context-free languages, while DPDA covers only the deterministic context-free subset.

Questions 5-6: reason about bounded input and two-stack memory

Question 5 (ISRO 2016)

A FSM (finite state machine) can be considered to be a Turing machine of finite tape length,

• A. without rewinding capability and unidirectional tape movement

• B. rewinding capability and unidirectional tape movement

• C. without rewinding capability and bidirectional tape movement

• D. rewinding capability and bidirectional tape movement

Correct answer: A. without rewinding capability and unidirectional tape movement.

In this simplified analogy, an FSM reads input left to right without revisiting or rewriting earlier positions. That matches a finite input tape with one-way movement and no rewind. It does not define every bounded-tape machine.

Question 6 (UGC NET 2018)

A pushdown automata behaves like a Turing machine when the number of auxiliary memory is :

• A. 0

• B. 1

• C. 1 or more

• D. 2 or more

Correct answer: D. 2 or more.

Two stacks represent both tape sides. For ... □ 0 1 [1] 0 □ ..., the left stack is 0, 1 bottom to top, and the right is 1, 0, □ top to bottom, with 1 current. To write X and move right, replace 1 by X, move X left, and expose 0. One stack gives PDA power. Two unbounded stacks give TM power.

Questions 7-8: understand a universal machine and a nested encoding

Question 7 (Bihar STET 2025)

A universal Turing machine is a ________.

• A. Single tape Turing machine

• B. Two-tape Turing machine

• C. Reprogrammable Turing machine

• D. All of the above

Correct answer: C. Reprogrammable Turing machine.

On input ⟨M, 1011⟩, a UTM reads machine M's encoded description and input 1011, then simulates M. The program is data, so another encoding changes the computation without rebuilding the simulator. Tape count is an implementation choice, not the meaning of universal. Only option C names that property, so C is the answer.

Question 8 (GATE 2003)

Let Σ = {a, b, c, d, e} be an alphabet. Define g(a) = 3, g(b) = 5, g(c) = 7, g(d) = 9, and g(e) = 11. Let p_i denote the i-th prime number, with p_1 = 2. For a non-empty string s = a_1a_2...a_n, where each a_i ∈ Σ, define f(s) = ∏_{i=1}^{n} p_i^{g(a_i)}. For a non-empty sequence <s_1, ..., s_k> of strings from Σ+, define h(s_1, ..., s_k) = ∏_{i=1}^{k} p_i^{f(s_i)}. Which of the following numbers is the encoding h of a non-empty sequence of strings?

• A. 2⁷ · 3⁷ · 5⁷

• B. 2⁸ · 3⁸ · 5⁸

• C. 2⁹ · 3⁹ · 5⁹

• D. 2¹⁰ · 5¹⁰ · 7¹⁰

Correct answer: B. 2⁸ · 3⁸ · 5⁸.

For a, f(a) = p_1^{g(a)} = 2^3 = 8. Thus <a, a, a> gives h(a,a,a) = p_1^8 p_2^8 p_3^8 = 2^8 · 3^8 · 5^8, which is option B. The other three fail on arithmetic you can do in your head. Every g value is at least 3, so every f value is a multiple of 2^3 = 8, and neither 7 nor 9 is, which rules out A and C. D uses primes 2, 5 and 7, skipping p_2 = 3, but the outer product must run over consecutive primes from p_1.

Questions 9-10: write the k-tape transition and calculate the slowdown

Question 9 (UGC NET 2022)

The transition function δ in a multi-tape Turing machine is defined as:

• A. δ: 2Q × Γ^k → 2^Q × Γ^k × {L, R, S}^k

• B. δ: Q × Q × Γ^k → Q × Q × Γ^k × {L, R, S}^k

• C. δ: Q × Γ k → Q × Γ^k × {L, R, S}^k

• D. δ: Q × Γ^k × 2^Q → Q × Γ^k × 2^Q × {L, R, S}^k

Correct answer: C. δ: Q × Γ k → Q × Γ^k × {L, R, S}^k.

Option C is printed with a typesetting slip: k after Γ lost its exponent. The intended function is δ: Q × Γ^k → Q × Γ^k × {L, R, S}^k. For k = 3, q4 reading (0,1,□) may enter q7, write (1,1,0), and move (R,L,S). That is one state plus three input symbols, then one state, three symbols and three moves.

Question 10

It is known that a single-tape Turing Machine (TM) and a multi-tape TM are equivalent in terms of language recognition power. Suppose a computation can be done in O(n) time on a k-tape TM. What will be the corresponding time complexity on an equivalent single-tape TM?

• A. O(n)

• B. O(n²)

• C. O(nk)

• D. None of the above

Correct answer: B. O(n²).

Use O(t(n)^2): each of t(n) steps may scan a region of order t(n). With t(n) = n, this is O(n · n) = O(n²). At n = 100, picture 100 scans across about 100 cells, a 10,000-scale shape rather than 100.

For # 1 0 [1] # 0 [1] 1 #, the simulator scans for both marked symbols, records the transition, then scans again to update both tracks. Recognition power is preserved, step-for-step time is not.

To re-attempt Question 10 under set conditions, work through the TM Variations & UTM subtopic hub, which collects the whole variants set in one place.

Score the set by misconception, then choose the next practice step

If you missed Questions 3 or 4, redraw the regular, deterministic-CFL, CFL and recursively-enumerable relationships. For Question 6, replay the exact ... □ 0 1 [1] 0 □ ... two-stack move. For Question 8, recompute f(a) = 8 before reading the outer exponents. For Questions 9 or 10, write the input and output tuple sizes before choosing a complexity option.

The short version: extra tapes, extra heads and nondeterminism preserve recognition power and cost only time, while a second stack is what lifts a pushdown automaton to Turing power. If the halting argument underneath that still feels shaky, reread Turing Machines and Decidability: Halting Problem Proof. For a sequenced syllabus, continue with GATE Guidance by Sanchit Sir or NTA-UGC-NET Paper - 2. Then retake missed questions and explain each rejected option aloud.