A pushdown automaton behaves like a Turing machine when the number of total memory is:
2 or more
The question asks about the conditions under which a Pushdown Automaton (PDA) can achieve the computational power of a Turing Machine (TM). Let's first understand what each type of automaton is and how they handle memory.
A standard Pushdown Automaton is a model of computation that consists of:
PDAs with a single stack are used to recognize context-free languages.
A Turing Machine is a more powerful model of computation that consists of:
Turing Machines are considered the most powerful computational model for algorithms and can recognize recursively enumerable languages.
Turing Machines are strictly more powerful than standard Pushdown Automata. There are languages that a TM can recognize but a standard PDA cannot (e.g., the language \(\{a^n b^n c^n \mid n \ge 0\}\)).
The key difference in power lies in the memory. A standard PDA's single stack has limitations compared to a TM's tape. What happens if we give a PDA more memory?
It turns out that giving a PDA additional stacks significantly increases its computational power. Specifically:
The question asks about the "number of total memory". In the context of comparing PDAs and TMs, this memory refers to the auxiliary storage beyond the finite control. For a PDA, this is typically measured by the number of stacks.
Let's look at the options:
| Option | Number of Memory (Stacks) | Computational Power | Behaves like a TM? |
|---|---|---|---|
| 0 | 0 | Finite Automaton (Regular Languages) | No |
| 1 | 1 | Standard PDA (Context-Free Languages) | No (less powerful than TM) |
| 1 or more | 1 or more | Starts with standard PDA (power of 1 stack), can become TM with 2+ stacks. The condition for *behaving like* a TM requires reaching that power level. | Only if it has 2 or more stacks. This option includes the case of only 1 stack, which is not enough. |
| 2 or more | 2 or more | Equivalent to a Turing Machine (Recursively Enumerable Languages) | Yes |
As established, a PDA with two or more stacks has the same computational power as a Turing Machine. Therefore, when the number of total memory (interpreted as stacks in this context) is 2 or more, a Pushdown Automaton behaves like a Turing Machine.
A standard Pushdown Automaton with one stack is less powerful than a Turing Machine. However, by adding a second stack, the PDA gains enough memory and flexibility to simulate the operations of a Turing Machine's tape. Any additional stacks beyond the second one do not increase the power further. Thus, a PDA becomes equivalent in power to a Turing Machine when it has 2 or more stacks.
| Automaton Type | Auxiliary Memory | Languages Recognized | Equivalent TM Power? |
|---|---|---|---|
| Finite Automaton (FA) | None (0 memory) | Regular Languages | No |
| Pushdown Automaton (PDA) | 1 Stack | Context-Free Languages | No |
| Pushdown Automaton (PDA) | 2 or more Stacks | Recursively Enumerable Languages | Yes |
| Turing Machine (TM) | Infinite Tape | Recursively Enumerable Languages | By definition |
The relationship between these automata and the languages they recognize is part of the Chomsky Hierarchy, which classifies formal languages based on their complexity:
Adding a second stack to a PDA elevates its power from recognizing Context-Free Languages (Type 2) to recognizing Recursively Enumerable Languages (Type 0), making it equivalent to a Turing Machine.
Consider the grammer S → SbS | a.
Consider the following statements:
The string abababa has
(A) two parse trees
(B) two left most derivations
(C) two right most derivations
Which of the following is correct?
Consider the following NPDA = ({q 0, q 1, q f}, {a, b}, {1, z}, δ, q 0, z, {q f})
δ(q 0, λ, z) = {(q f, z)}
δ(q 0, a, z) = {(q 1, 11z)}
δ(q 1, a, 1) = {(q 1, 111)}
δ(q 1, b, 1) = {(q 1, λ)}
δ(q 1, λ, z) = {(q f, z}
Which of the following Language L is accepted by NPDA?
What language is accepted by the pushdown automation
M = ({q 0, q 1, q2}, {a, b}, {a, b, z}, δ, q 0, z, {q 2})
with δ(q 0, a, a) = {(q 0, aa)}; δ(q 0, b, a) = {(q 0, ba)}
δ(q 0, a, b) = {(q 0, ab)}; δ(q 0, b, b) = {(q 0, bb)}
δ(q 0, a, z) = {(q 0, az)}; δ(q 0, b, z) = {(q 0, bz)}
δ(q 0, λ, b) = {(q 1, b)}; δ(q 0, λ, a) = {(q 1, a)}
δ(q 1, a, a) = {(q 1, λ)}; δ(q 1, b, b) = {(q 1, λ)}
δ(q 1, λ, z) = {(q 2, z)}?