Ordering on Time Complexity Classes
In Sipser, the time complexity class $mathrm{TIME}(f(n))$ with respect to a function $f colon mathbb{N} to mathbb{N}$ is defined to be the collection of all languages that are decidable by an $O(f(n))$ time Turing machine.
It is easy to give some examples of containment relations between complexity classes from this definition: $mathrm{TIME}(n) =mathrm{TIME}(2n)$ and $mathrm{TIME}(n) subseteq mathrm{TIME}(n^2)$.
Now I guess it is natural (?) to ask the following question: given any two such complexity-generating functions $f,g$, is it always the case that either $mathrm{TIME}(f(n)) subseteq mathrm{TIME}(g(n))$ or $mathrm{TIME}(f(n)) supseteq mathrm{TIME}(g(n))$? If the answer is yes (which would surprise me a little), this would define a total order on the space of functions $mathbb N^{mathbb N}$.
computational-complexity
asked Nov 20 at 18:20
P. Factor
31519
add a comment |
In Sipser, the time complexity class $mathrm{TIME}(f(n))$ with respect to a function $f colon mathbb{N} to mathbb{N}$ is defined to be the collection of all languages that are decidable by an $O(f(n))$ time Turing machine.
It is easy to give some examples of containment relations between complexity classes from this definition: $mathrm{TIME}(n) =mathrm{TIME}(2n)$ and $mathrm{TIME}(n) subseteq mathrm{TIME}(n^2)$.
Now I guess it is natural (?) to ask the following question: given any two such complexity-generating functions $f,g$, is it always the case that either $mathrm{TIME}(f(n)) subseteq mathrm{TIME}(g(n))$ or $mathrm{TIME}(f(n)) supseteq mathrm{TIME}(g(n))$? If the answer is yes (which would surprise me a little), this would define a total order on the space of functions $mathbb N^{mathbb N}$.
computational-complexity
asked Nov 20 at 18:20
P. Factor
31519
It's certainly possible to find monotonic functions $f, g$ such that neither $f = O(g)$ nor $g = O(f)$; for example, let $E(n)$ equal $1$ if $n$ is even and $0$ otherwise, and define $f(n) = 2^{2^{n+E(n)}}$ and $g(n) = 2^{2^{n + 1 - E(n)}}$. I don't know if there are languages in the corresponding complexity classes, though.
– Connor Harris
Nov 20 at 18:34
add a comment |
In Sipser, the time complexity class $mathrm{TIME}(f(n))$ with respect to a function $f colon mathbb{N} to mathbb{N}$ is defined to be the collection of all languages that are decidable by an $O(f(n))$ time Turing machine.
It is easy to give some examples of containment relations between complexity classes from this definition: $mathrm{TIME}(n) =mathrm{TIME}(2n)$ and $mathrm{TIME}(n) subseteq mathrm{TIME}(n^2)$.
Now I guess it is natural (?) to ask the following question: given any two such complexity-generating functions $f,g$, is it always the case that either $mathrm{TIME}(f(n)) subseteq mathrm{TIME}(g(n))$ or $mathrm{TIME}(f(n)) supseteq mathrm{TIME}(g(n))$? If the answer is yes (which would surprise me a little), this would define a total order on the space of functions $mathbb N^{mathbb N}$.
computational-complexity
asked Nov 20 at 18:20
P. Factor
31519
In Sipser, the time complexity class $mathrm{TIME}(f(n))$ with respect to a function $f colon mathbb{N} to mathbb{N}$ is defined to be the collection of all languages that are decidable by an $O(f(n))$ time Turing machine.
It is easy to give some examples of containment relations between complexity classes from this definition: $mathrm{TIME}(n) =mathrm{TIME}(2n)$ and $mathrm{TIME}(n) subseteq mathrm{TIME}(n^2)$.
Now I guess it is natural (?) to ask the following question: given any two such complexity-generating functions $f,g$, is it always the case that either $mathrm{TIME}(f(n)) subseteq mathrm{TIME}(g(n))$ or $mathrm{TIME}(f(n)) supseteq mathrm{TIME}(g(n))$? If the answer is yes (which would surprise me a little), this would define a total order on the space of functions $mathbb N^{mathbb N}$.
computational-complexity
computational-complexity
asked Nov 20 at 18:20
P. Factor
31519
asked Nov 20 at 18:20
P. Factor
31519
asked Nov 20 at 18:20
P. Factor
31519
asked Nov 20 at 18:20
P. Factor
31519
asked Nov 20 at 18:20
P. Factor
31519
31519
It's certainly possible to find monotonic functions $f, g$ such that neither $f = O(g)$ nor $g = O(f)$; for example, let $E(n)$ equal $1$ if $n$ is even and $0$ otherwise, and define $f(n) = 2^{2^{n+E(n)}}$ and $g(n) = 2^{2^{n + 1 - E(n)}}$. I don't know if there are languages in the corresponding complexity classes, though.
– Connor Harris
Nov 20 at 18:34
add a comment |
It's certainly possible to find monotonic functions $f, g$ such that neither $f = O(g)$ nor $g = O(f)$; for example, let $E(n)$ equal $1$ if $n$ is even and $0$ otherwise, and define $f(n) = 2^{2^{n+E(n)}}$ and $g(n) = 2^{2^{n + 1 - E(n)}}$. I don't know if there are languages in the corresponding complexity classes, though.
– Connor Harris
Nov 20 at 18:34
It's certainly possible to find monotonic functions $f, g$ such that neither $f = O(g)$ nor $g = O(f)$; for example, let $E(n)$ equal $1$ if $n$ is even and $0$ otherwise, and define $f(n) = 2^{2^{n+E(n)}}$ and $g(n) = 2^{2^{n + 1 - E(n)}}$. I don't know if there are languages in the corresponding complexity classes, though.
– Connor Harris
Nov 20 at 18:34
It's certainly possible to find monotonic functions $f, g$ such that neither $f = O(g)$ nor $g = O(f)$; for example, let $E(n)$ equal $1$ if $n$ is even and $0$ otherwise, and define $f(n) = 2^{2^{n+E(n)}}$ and $g(n) = 2^{2^{n + 1 - E(n)}}$. I don't know if there are languages in the corresponding complexity classes, though.
– Connor Harris
Nov 20 at 18:34
add a comment |
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It's certainly possible to find monotonic functions $f, g$ such that neither $f = O(g)$ nor $g = O(f)$; for example, let $E(n)$ equal $1$ if $n$ is even and $0$ otherwise, and define $f(n) = 2^{2^{n+E(n)}}$ and $g(n) = 2^{2^{n + 1 - E(n)}}$. I don't know if there are languages in the corresponding complexity classes, though.
– Connor Harris
Nov 20 at 18:34