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 $\tuple {\sqbrk M, x}$.
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 $\map f m^3$ operations, so:
:$ H_f \in \map {\mathsf{DTIME} } {\map f m^3}$
:
:$H_f \in \map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$
Then we can construct some machine $K$ which:
:given some machine description $\sqbrk {M_K}$ and input $x$
:decides within $\map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$ whether $\tuple {\sqbrk {M_K}, x} \in H_f$.
Construct another machine $N$ which:
:takes a machine description $\sqbrk {M_N}$
:runs $K$ on $\left({ \sqbrk {M_N}, \sqbrk {M_N} }\right)$
:accepts only if $K$ rejects, and rejects if $K$ accepts.
Let $m_n$ be the length of $\sqbrk {M_N}$.
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 $\sqbrk N$ as input into $N$ itself).
In this case $m_n$ is the length of $\sqbrk N$.
* If $N$ accepts $\sqbrk N$ (which we know it does in at most $\map f {m_n}$ operations):
** By the definition of $N$, $K$ rejects $\tuple {\sqbrk N, \sqbrk N}$
** Therefore, by the definition of $K$, $\tuple {\sqbrk N, \sqbrk N} \notin H_f$
** Therefore, by the definition of $H_f$, $N$ does not accept $\sqbrk N$ in $\map f {m_n}$ steps -- a contradiction.
* If $N$ rejects $\sqbrk N$ (which we know it does in at most $\map f {m_n}$ operations):
** By the definition of $N$, $K$ accepts $\tuple {\sqbrk N, \sqbrk N}$
** Therefore, by the definition of $K$, $\tuple {\sqbrk N, \sqbrk N} \in H_f$
** Therefore, by the definition of $H_f$, $N$ does accept $\sqbrk N$ in $\map f {m_n}$ steps -- a contradiction.
Therefore, $K$ does not exist.
So, by Proof by Contradiction:
:$H_f \notin \map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$
Substituting $2 n + 1$ for $m$, we get:
:$H_f \notin \map {\mathsf{DTIME} } {\map f n}$
and, from the earlier result:
:$H_f \in \map {\mathsf{DTIME} } {\map f {2 n + 1}^3}$
Category:Complexity Theory
Category:Computer Science
Category:Named Theorems