Time hierarchy theorem

Proof overview

We need to prove that some time class TIME(g(n)) is strictly larger than some time class TIME(f(n)). We do this by constructing a machine which cannot be in TIME(f(n)), by diagonalization. We then show that the machine is in TIME(g(n)), using a simulator machine.

Deterministic time hierarchy theorem

Statement

Time Hierarchy Theorem. If f(n) is a time-constructible function, then there exists a decision problem which cannot be solved in worst-case deterministic time o(f(n)) but can be solved in worst-case deterministic time O(f(n)log f(n)). Thus

$${\displaystyle \mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(o(f(n)))⊊\mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f(n)\log f(n)\right).}$$ Equivalently, if ${\displaystyle f,g}$ are time-constructable, and ${\displaystyle f(n)\ln f(n)=o(g(n))}$, then

$${\displaystyle \mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(f(n))⊊\mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(g(n))}$$

Note 1. f(n) is at least n, since smaller functions are never time-constructible.

Example. ${\displaystyle \mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(n)⊊\mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(n(\ln n{)}^2)}$.

Proof

We include here a proof of a weaker result, namely that DTIME(f(n)) is a strict subset of DTIME(f(2n + 1)³), as it is simpler but illustrates the proof idea. See the bottom of this section for information on how to extend the proof to f(n)logf(n).

To prove this, we first define the language of the encodings of machines and their inputs which cause them to halt within f(|x|) steps: $${\displaystyle H_f=\left\{([M],x)|M\text{accepts}x\text{in}f(|x|)\text{steps}\right\}.}$$

Notice here that this is a time-class. It is the set of pairs of machines and inputs to those machines (M,x) so that the machine M accepts within f(|x|) steps.

Here, M is a deterministic Turing machine, and x is its input (the initial contents of its tape). [M] denotes an input that encodes the Turing machine M. Let m be the size of the tuple ([M], x).

We know that we can decide membership of H_f by way of a deterministic Turing machine R, that simulates M for f(x) steps by first calculating f(|x|) and then writing out a row of 0s of that length, and then using this row of 0s as a "clock" or "counter" to simulate M for at most that many steps. At each step, the simulating machine needs to look through the definition of M to decide what the next action would be. It is safe to say that this takes at most f(m)³ operations (since it is known that a simulation of a machine of time complexity T(n) for can be achieved in time ${\displaystyle O(T(n)\cdot |M|)}$ on a multitape machine, where |M| is the length of the encoding of M), we have that: $${\displaystyle H_f\in \mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f(m{)}^3\right).}$$

The rest of the proof will show that $${\displaystyle H_f\notin \mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f\left(\lfloor \frac{m}{2}\rfloor \right)\right)}$$

so that if we substitute 2n + 1 for m, we get the desired result. Let us assume that H_f is in this time complexity class, and we will reach a contradiction.

If H_f is in this time complexity class, then there exists a machine K which, given some machine description [M] and input x, decides whether the tuple ([M], x) is in H_f within $${\displaystyle \mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f\left(\lfloor \frac{m}{2}\rfloor \right)\right).}$$

We use this K to construct another machine, N, which takes a machine description [M] and runs K on the tuple ([M], [M]), ie. M is simulated on its own code by K, and then N accepts if K rejects, and rejects if K accepts.

If n is the length of the input to N, then m (the length of the input to K) is twice n plus some delimiter symbol, so m = 2n + 1. N's running time is thus $${\displaystyle \mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f\left(\lfloor \frac{m}{2}\rfloor \right)\right)=\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f\left(\lfloor \frac{2n+1}{2}\rfloor \right)\right)=\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f(n)\right).}$$

Now if we feed [N] as input into N' and ask the question whether N accepts its N' description as input, we get:

• If N accepts' [N' ] (which we know it does in at most f(n) operations since K halts on ([N ], [N']) in f(n) steps), this means that K rejects ([N'], [N']), so ([N], [N']) is not in H_f, and so by the definition of H_f, this implies that N does not accept [N'] in f(n) steps. Contradiction.

• If N rejects' [N'] (which we know it does in at most f(n) operations), this means that K accepts ([N'], [N']), so ([N ], [N']) is in H_f, and thus N does accept [N'] in f(n) steps. Contradiction.

We thus conclude that the machine K does not exist, and so $${\displaystyle H_f\notin \mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(f\left(\lfloor \frac{m}{2}\rfloor \right)\right).}$$

Extension

The reader may have realised that the proof gives the weaker result because we have chosen a simple Turing machine simulation for which we know that $${\displaystyle H_f\in \mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(f(m{)}^3).}$$

It is known that a more efficient simulation exists which establishes that $${\displaystyle H_f\in \mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(f(m)\log f(m)).}$$

Non-deterministic time hierarchy theorem

If g(n) is a time-constructible function, and f(n+1) = o(g(n)), then there exists a decision problem which cannot be solved in non-deterministic time f(n) but can be solved in non-deterministic time g(n). In other words, the complexity class NTIME(f(n)) is a strict subset of NTIME(g(n)).

Consequences

The time hierarchy theorems guarantee that the deterministic and non-deterministic versions of the exponential hierarchy are genuine hierarchies: in other words P ⊊ EXPTIME ⊊ 2-EXP ⊊ ... and NP ⊊ NEXPTIME ⊊ 2-NEXP ⊊ ....

For example, ${\displaystyle \mathsf{P}⊊\mathsf{E}\mathsf{X}\mathsf{P}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}}$ since ${\displaystyle \mathsf{P}\subseteq \mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(2^n)⊊\mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(2^{2n})\subseteq \mathsf{E}\mathsf{X}\mathsf{P}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}}$. Indeed, ${\displaystyle \mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(2^n\right)\subseteq \mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}\left(o\left(\frac{2^{2n}}{2n}\right)\right)⊊\mathsf{D}\mathsf{T}\mathsf{I}\mathsf{M}\mathsf{E}(2^{2n})}$ from the time hierarchy theorem.

The theorem also guarantees that there are problems in P requiring arbitrarily large exponents to solve; in other words, P does not collapse to DTIME(n^k) for any fixed k. For example, there are problems solvable in n⁵⁰⁰⁰ time but not n⁴⁹⁹⁹ time. This is one argument against Cobham's thesis, the convention that P is a practical class of algorithms. If such a collapse did occur, we could deduce that P ≠ PSPACE, since it is a well-known theorem that DTIME(f(n)) is strictly contained in DSPACE(f(n)).

However, the time hierarchy theorems provide no means to relate deterministic and non-deterministic complexity, or time and space complexity, so they cast no light on the great unsolved questions of computational complexity theory: whether P and NP, NP and PSPACE, PSPACE and EXPTIME, or EXPTIME and NEXPTIME are equal or not.

Sharper hierarchy theorems

The gap of approximately ${\displaystyle \log f(n)}$ between the lower and upper time bound in the hierarchy theorem can be traced to the efficiency of the device used in the proof, namely a universal program that maintains a step-count. This can be done more efficiently on certain computational models. The sharpest results, presented below, have been proved for:

• The unit-cost random-access machine
• A programming language model whose programs operate on a binary tree that is always accessed via its root. This model, introduced by Neil D. Jones is stronger than a deterministic Turing machine but weaker than a random-access machine.

For these models, the theorem has the following form:

If f(n) is a time-constructible function, then there exists a decision problem which cannot be solved in worst-case deterministic time f(n) but can be solved in worst-case time af(n) for some constant a (dependent on f).

Thus, a constant-factor increase in the time bound allows for solving more problems, in contrast with the situation for Turing machines (see Linear speedup theorem). Moreover, Ben-Amram proved that, in the above models, for f of polynomial growth rate (but more than linear), it is the case that for all ${\displaystyle \epsilon >0}$, there exists a decision problem which cannot be solved in worst-case deterministic time f(n) but can be solved in worst-case time ${\displaystyle (1+\epsilon )f(n)}$.