Deterministic Time Hierarchy Theorem

Theorem
Let $f(n)$ be a time-constructible function. Then, there exists a decision problem which can be solved in worst-case deterministic time $f\left({2n+1}\right)^3$ but not in worst-case deterministic time $f(n)$.

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

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


 * $H_f = \left\{ { \left({ \left[{M}\right], x}\right) : \text{$M$ accepts $x$ in $f \left({\left\vert{x}\right\vert}\right)$ steps} }\right\} $

where:
 * $M$ is a deterministic Turing machine
 * $x$ is its input (the initial contents of its tape)
 * $\left[{M}\right]$ denotes an input that encodes the Turing machine $M$

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

We know that we can decide membership of $H_f$ by way of a deterministic Turing machine that first calculates $f \left({ \left\vert{x}\right\vert }\right)$, 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 $f \left({m}\right)^3$ 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 $\left[ {M_N} \right]$ and runs $K$ on $\left({ \left[{M_N}\right], \left[{M_N}\right] }\right)$, and then 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:


 * $\mathsf{DTIME} \left({ f \left({ \left\lfloor{ \dfrac m 2 }\right\rfloor }\right) }\right) = \mathsf{DTIME} \left({ f \left({ \left\lfloor{ \dfrac{2m_n+1} 2 }\right\rfloor }\right) }\right) = \mathsf{DTIME} \left({ f \left({m_n}\right) }\right) $

Now consider the case $ M_N = N $ (i.e. we feed $\left[{N}\right]$ as input into $N$ itself), in which 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) $