Deterministic Time Hierarchy Theorem

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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}$


Aiming for a contradiction, suppose:

$H_f \in \map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$

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:

\(\displaystyle \map {\mathsf{DTIME} } {\map f {\floor {\frac m 2} } }\) \(=\) \(\displaystyle \map {\mathsf{DTIME} } {\map f {\floor {\frac {2 m_n + 1} 2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\mathsf{DTIME} } {\map f {m_n} }\)


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:

$H_f \in \map {\mathsf{DTIME} } {\map f {2 n + 1}^3}$

$\blacksquare$