Why does a Turing machine take $n^k$ steps for computing an input?
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I was reading about Cook's Theorem for Turing machine. In its proof, it is said that the Turing machine would take at most $n^k$ steps (where $k$ is an integer and $k > 0$) to compute an input of length $n$.
This is probably assuming that the Turing machine does halt for the given input. It further says that we as there are at most $n^k$ steps, we don't need an infinite tape. A tape with $n^k$ elements is sufficient as the turning machine would not travel more than that.
Why do we say the Turing Machine needs at most $n^k$ steps?
turing-machines
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
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add a comment |
$begingroup$
I was reading about Cook's Theorem for Turing machine. In its proof, it is said that the Turing machine would take at most $n^k$ steps (where $k$ is an integer and $k > 0$) to compute an input of length $n$.
This is probably assuming that the Turing machine does halt for the given input. It further says that we as there are at most $n^k$ steps, we don't need an infinite tape. A tape with $n^k$ elements is sufficient as the turning machine would not travel more than that.
Why do we say the Turing Machine needs at most $n^k$ steps?
turing-machines
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
$endgroup$
add a comment |
$begingroup$
I was reading about Cook's Theorem for Turing machine. In its proof, it is said that the Turing machine would take at most $n^k$ steps (where $k$ is an integer and $k > 0$) to compute an input of length $n$.
This is probably assuming that the Turing machine does halt for the given input. It further says that we as there are at most $n^k$ steps, we don't need an infinite tape. A tape with $n^k$ elements is sufficient as the turning machine would not travel more than that.
Why do we say the Turing Machine needs at most $n^k$ steps?
turing-machines
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
$endgroup$
I was reading about Cook's Theorem for Turing machine. In its proof, it is said that the Turing machine would take at most $n^k$ steps (where $k$ is an integer and $k > 0$) to compute an input of length $n$.
This is probably assuming that the Turing machine does halt for the given input. It further says that we as there are at most $n^k$ steps, we don't need an infinite tape. A tape with $n^k$ elements is sufficient as the turning machine would not travel more than that.
Why do we say the Turing Machine needs at most $n^k$ steps?
turing-machines
turing-machines
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
edited Dec 17 '18 at 5:21
Hans Hüttel
3,1972921
3,1972921
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
asked Dec 17 '18 at 5:06
Sukumar GaonkarSukumar Gaonkar
61
61
add a comment |
add a comment |
1 Answer
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Cook's Theorem is a theorem in the theory of computational complexity. Here, we are only interested in Turing machines whose time complexity is well-defined. This can only be the case for Turing machines that always halt.
Cook's Theorem states that the language
$$mathrm{SAT} = { langle phi rangle mid phi ; text{is a satisfiable formula in propositional logic}}$$
is NP-complete. To show this, we must show that for any language $L in mathrm{NP}$, $L$ polynomial-time reduces to $mathrm{SAT}$.
But since $L in mathrm{NP}$, we know that there exists a nondeterministic Turing machine $M$ that decides $L$ and has polynomial time complexity. That is, we know there exists a $k > 0$ such that $M$ will never use more than $n^k$ steps for any input of size $k$. Within $n^k$ steps, $M$ can only visit $n^k$ cells on its tape.
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Cook's Theorem is a theorem in the theory of computational complexity. Here, we are only interested in Turing machines whose time complexity is well-defined. This can only be the case for Turing machines that always halt.
Cook's Theorem states that the language
$$mathrm{SAT} = { langle phi rangle mid phi ; text{is a satisfiable formula in propositional logic}}$$
is NP-complete. To show this, we must show that for any language $L in mathrm{NP}$, $L$ polynomial-time reduces to $mathrm{SAT}$.
But since $L in mathrm{NP}$, we know that there exists a nondeterministic Turing machine $M$ that decides $L$ and has polynomial time complexity. That is, we know there exists a $k > 0$ such that $M$ will never use more than $n^k$ steps for any input of size $k$. Within $n^k$ steps, $M$ can only visit $n^k$ cells on its tape.
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
$endgroup$
add a comment |
$begingroup$
Cook's Theorem is a theorem in the theory of computational complexity. Here, we are only interested in Turing machines whose time complexity is well-defined. This can only be the case for Turing machines that always halt.
Cook's Theorem states that the language
$$mathrm{SAT} = { langle phi rangle mid phi ; text{is a satisfiable formula in propositional logic}}$$
is NP-complete. To show this, we must show that for any language $L in mathrm{NP}$, $L$ polynomial-time reduces to $mathrm{SAT}$.
But since $L in mathrm{NP}$, we know that there exists a nondeterministic Turing machine $M$ that decides $L$ and has polynomial time complexity. That is, we know there exists a $k > 0$ such that $M$ will never use more than $n^k$ steps for any input of size $k$. Within $n^k$ steps, $M$ can only visit $n^k$ cells on its tape.
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
$endgroup$
add a comment |
$begingroup$
Cook's Theorem is a theorem in the theory of computational complexity. Here, we are only interested in Turing machines whose time complexity is well-defined. This can only be the case for Turing machines that always halt.
Cook's Theorem states that the language
$$mathrm{SAT} = { langle phi rangle mid phi ; text{is a satisfiable formula in propositional logic}}$$
is NP-complete. To show this, we must show that for any language $L in mathrm{NP}$, $L$ polynomial-time reduces to $mathrm{SAT}$.
But since $L in mathrm{NP}$, we know that there exists a nondeterministic Turing machine $M$ that decides $L$ and has polynomial time complexity. That is, we know there exists a $k > 0$ such that $M$ will never use more than $n^k$ steps for any input of size $k$. Within $n^k$ steps, $M$ can only visit $n^k$ cells on its tape.
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
$endgroup$
Cook's Theorem is a theorem in the theory of computational complexity. Here, we are only interested in Turing machines whose time complexity is well-defined. This can only be the case for Turing machines that always halt.
Cook's Theorem states that the language
$$mathrm{SAT} = { langle phi rangle mid phi ; text{is a satisfiable formula in propositional logic}}$$
is NP-complete. To show this, we must show that for any language $L in mathrm{NP}$, $L$ polynomial-time reduces to $mathrm{SAT}$.
But since $L in mathrm{NP}$, we know that there exists a nondeterministic Turing machine $M$ that decides $L$ and has polynomial time complexity. That is, we know there exists a $k > 0$ such that $M$ will never use more than $n^k$ steps for any input of size $k$. Within $n^k$ steps, $M$ can only visit $n^k$ cells on its tape.
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
answered Dec 17 '18 at 5:27
Hans HüttelHans Hüttel
3,1972921
3,1972921
add a comment |
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