Time hierarchy theorem

In computational complexity theory, the time hierarchy theorems are important statements about time-bounded computation on Turing machines. Informally, these theorems say that given more time, a Turing machine can solve more problems. For example, there are problems that can be solved with n² time but not n time.

The time hierarchy theorem for deterministic multi-tape Turing machines was first proven by Richard E. Stearns and Juris Hartmanis in 1965.^[1] It was improved a year later when F. C. Hennie and Richard E. Stearns improved the efficiency of the Universal Turing machine.^[2] Consequent to the theorem, for every deterministic time-bounded complexity class, there is a strictly larger time-bounded complexity class, and so the time-bounded hierarchy of complexity classes does not completely collapse. More precisely, the time hierarchy theorem for deterministic Turing machines states that for all time-constructible functions f(n),

$\mathbf{DTIME}\left(o\left(\frac{f(n)}{\log f(n)}\right)\right) \subsetneq \mathbf{DTIME}(f(n))$.

The time hierarchy theorem for nondeterministic Turing machines was originally proven by Stephen Cook in 1972.^[3] It was improved to its current form via a complex proof by Joel Seiferas, Michael Fischer, and Albert Meyer in 1978.^[4] Finally in 1983, Stanislav Žák achieved the same result with the simple proof taught today.^[5] The time hierarchy theorem for nondeterministic Turing machines states that if g(n) is a time-constructible function, and f(n) = o(g(n)), then

$\mathbf{NTIME}(f(n)) \subsetneq \mathbf{NTIME}(g(n))$.

The analogous theorems for space are the space hierarchy theorems. A similar theorem is not known for time-bounded probabilistic complexity classes, unless the class also has advice.^[6]

Background

Both theorems use the notion of a time-constructible function. A function $f:\mathbb{N}\rightarrow\mathbb{N}$ is time-constructible if there exists a deterministic Turing machine such that for every $n\in\mathbb{N}$, if the machine is started with an input of n ones, it will halt after precisely f(n) steps. All polynomials with non-negative integral coefficients are time-constructible, as are exponential functions such as 2ⁿ.

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 f(n) but can be solved in worst-case deterministic time f(n)². In other words,

$\mathbf{DTIME}(f(n)) \subsetneq \mathbf{DTIME}\left (f(n)^2 \right).$

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

Note 2. Even more generally, it can be shown that if f(n) is time-constructible, then

$\mathbf{DTIME}\left(o\left(\frac{f(n)}{\log f(n)}\right)\right)\subsetneq \mathbf{DTIME}\left (f(n) \right).$

For example, there are problems solvable in time n² but not time n, since n is in

$o\left(\frac{n^2}{\log {n^2}}\right).$

Proof

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

To prove this, we first define a language as follows:

$H_f = \left\{ ([M], x)\ |\ M \ \mbox{accepts}\ x \ \mbox{in}\ f(|x|) \ \mbox{steps} \right\}.$

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 that first calculates f(|x|), then writes out a row of 0s of that length, and then uses 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, so

$H_f \in \mathbf{TIME}(f(m)^3).$

The rest of the proof will show that

$H_f \notin \mathbf{TIME}(f( \left\lfloor \tfrac{m}{2} \right\rfloor ))$

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 attempt to reach a contradiction.

If H_f is in this time complexity class, it means we can construct some machine K which, given some machine description [M] and input x, decides whether the tuple ([M], x) is in H_f within

$\mathbf{TIME}(f( \left\lfloor \tfrac{m}{2} \right\rfloor )).$

Therefore we can use this K to construct another machine, N, which takes a machine description [M] and runs K on the tuple ([M], [M]), and then accepts only if K rejects, and rejects if K accepts. If now 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

$\mathbf{TIME}(f( \left\lfloor \tfrac{m}{2} \right\rfloor )) = \mathbf{TIME}(f( \left\lfloor \tfrac{2n+1}{2} \right\rfloor )) = \mathbf{TIME}(f(n)).$

Now if we feed [N] as input into N itself (which makes n the length of [N]) and ask the question whether N accepts its own description as input, we get:

• If N accepts [N] (which we know it does in at most f(n) operations), this means that K rejects ([N], [N]), so ([N], [N]) is not in H_f, and thus 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

$H_f \notin \mathbf{TIME}(f( \left\lfloor \tfrac{m}{2} \right\rfloor )).$

Extension

The reader may have realised that the proof is simpler because we have chosen a simple Turing machine simulation for which we can be certain that

$H_f \in \mathbf{TIME}(f(m)^3).$

It has been shown^[7] that a more efficient model of simulation exists which establishes that

$H_f \in \mathbf{TIME}(f(m) \log f(m))$

but since this model of simulation is rather involved, it is not included here.

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 $\subsetneq$ EXPTIME $\subsetneq$ 2-EXP $\subsetneq$ ... and NP $\subsetneq$ NEXPTIME $\subsetneq$ 2-NEXP $\subsetneq$ ....

For example, $\mathbf{P} \subsetneq \mathbf{EXPTIME}$ since $\mathbf{P} \subseteq \mathbf{DTIME} (2^n)\subsetneq \mathbf{DTIME} (2^{2n}) \subseteq \mathbf{EXPTIME}$. Indeed $\mathbf{DTIME}\left(2^n\right) \subseteq \mathbf{DTIME}\left(o\left(\frac{2^{2n}}{2n}\right)\right) \subsetneq \mathbf{DTIME}(2^{2n})$ 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.

See also

• Space hierarchy theorem

References

1. Hartmanis, J.; Stearns, R. E. (1 May 1965). "On the computational complexity of algorithms". Transactions of the American Mathematical Society (American Mathematical Society) 117: 285–306. doi:10.2307/1994208. ISSN 0002-9947. JSTOR 1994208. MR 0170805.
2. Hennie, F. C.; Stearns, R. E. (October 1966). "Two-Tape Simulation of Multitape Turing Machines". J. ACM (New York, NY, USA: ACM) 13 (4): 533–546. doi:10.1145/321356.321362. ISSN 0004-5411.
3. Cook, Stephen A. (1972). "A hierarchy for nondeterministic time complexity". Proceedings of the fourth annual ACM symposium on Theory of computing. STOC '72. Denver, Colorado, United States: ACM. pp. 187–192. doi:10.1145/800152.804913.
4. Seiferas, Joel I.; Fischer, Michael J.; Meyer, Albert R. (January 1978). "Separating Nondeterministic Time Complexity Classes". J. ACM (New York, NY, USA: ACM) 25 (1): 146–167. doi:10.1145/322047.322061. ISSN 0004-5411.
5. Stanislav, Žák (October 1983). "A Turing machine time hierarchy". Theoretical Computer Science (Elsevier Science B.V.) 26 (3): 327–333. doi:10.1016/0304-3975(83)90015-4.
6. Fortnow, L.; Santhanam, R. (2004). "45th Annual IEEE Symposium on Foundations of Computer Science": 316. doi:10.1109/FOCS.2004.33. ISBN 0-7695-2228-9. |chapter= ignored (help)
7. Luca Trevisan, Notes on Hierarchy Theorems, U.C. Berkeley

• Michael Sipser (1997). Introduction to the Theory of Computation. PWS Publishing. ISBN 0-534-94728-X. Pages 310–313 of section 9.1: Hierarchy theorems.
• Christos Papadimitriou (1993). Computational Complexity (1st ed.). Addison Wesley. ISBN 0-201-53082-1. Section 7.2: The Hierarchy Theorem, pp. 143–146.