Deterministic Time Hierarchy Theorem

Theorem
Let $\map f n$ be a time-constructible function.

Then there exists a decision problem which:
 * can be solved in worst-case deterministic time $\map f {2 n + 1}^3$

but:
 * cannot be solved in worst-case deterministic time $\map f n$.

In other words, the complexity class $\map {\mathsf {DTIME} } {\map f n} \subsetneq \map {\mathsf {DTIME} } {\map f {2 n + 1}^3}$.

Proof
Let $H_f$ be a set defined as follows:


 * $H_f = \set {\tuple {\sqbrk M, x}: \text {$M$ accepts $x$ in $\map f {\size x}$ steps} }$

where:
 * $M$ is a (deterministic) Turing machine
 * $x$ is its input (the initial contents of its tape)
 * $\sqbrk M$ denotes an input that encodes the Turing machine $M$

Let $m$ be the size of $\left({\sqbrk M, x }\right)$.

We know that we can decide membership of $H_f$ by way of a (deterministic) Turing machine that:
 * $(1): \quad$ calculates $f \left({\size x}\right)$
 * $(2): \quad$ writes out a row of $0$s of that length
 * $(3): \quad$ uses this row of $0$s as a 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 $f \left({m}\right)^3$ operations, so:


 * $ H_f \in \mathsf{DTIME} \left({ f \left({m}\right)^3 }\right)$

Now assume:


 * $H_f \in \mathsf{DTIME} \left({ f \left({ \left\lfloor{ \dfrac m 2 }\right\rfloor }\right) }\right)$

Then we can construct some machine $K$ which:
 * given some machine description $\left[{M_K} \right]$ and input $x$
 * decides within $ \mathsf{DTIME} \left({ f \left({ \left\lfloor{ \dfrac m 2 }\right\rfloor }\right) }\right)$ whether $\left({ \left[{ M_K }\right], x }\right) \in H_f$.

Construct another machine $N$ which:
 * takes a machine description $\left[{M_N}\right]$
 * runs $K$ on $\left({ \left[{M_N}\right], \left[{M_N}\right] }\right)$
 * accepts only if $K$ rejects, and rejects if $K$ accepts.

Let $m_n$ be the length of $\left[{M_N}\right]$.

Then $m$ (the length of the input to $K$) is twice $m_n$ plus some delimiter symbol, so:


 * $ m = 2m_n + 1 $

$N$'s running time is thus:

Now consider the case $M_N = N$.

That is we feed $\left[{N}\right]$ as input into $N$ itself).

In this case $m_n$ is the length of $\left[{N}\right]$.


 * If $N$ accepts $\left[{N}\right]$ (which we know it does in at most $f \left( {m_n} \right)$ operations):
 * By the definition of $N$, $K$ rejects $\left({ \left[{N}\right], \left[{N}\right] }\right)$
 * Therefore, by the definition of $K$, $ \left({ \left[{N}\right], \left[{N}\right] }\right) \notin H_f $
 * Therefore, by the definition of $H_f$, $N$ does not accept $\left[{N}\right]$ in $f \left( {m_n} \right)$ steps -- a contradiction.


 * If $N$ rejects $\left[{N}\right]$ (which we know it does in at most $f \left( {m_n} \right)$ operations):
 * By the definition of $N$, $K$ accepts $\left({ \left[{N}\right], \left[{N}\right] }\right)$
 * Therefore, by the definition of $K$, $ \left({ \left[{N}\right], \left[{N}\right] }\right) \in H_f $
 * Therefore, by the definition of $H_f$, $N$ does accept $\left[{N}\right]$ in $f \left( {m_n} \right)$ steps -- a contradiction.

Therefore, $K$ does not exist, and so:


 * $H_f \notin \mathsf{DTIME}\left({f \left({\left\lfloor{\dfrac m 2}\right\rfloor}\right)}\right)$

Substituting $2n + 1$ for $m$, we get:


 * $H_f \notin \mathsf{DTIME} \left({f \left({n}\right)}\right)$

and, from the earlier result:


 * $H_f \in \mathsf{DTIME} \left({f \left({2n+1}\right)^3}\right)$