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Contradiction proof for inequality of P and NP?
Problem with my proof that NP = coNP?Contradiction between best-case running time of insertion sort and $nlog n$ lower bound?Logarithmic Randomness is Necessary for PCP TheoremTrouble seeing the contradiction in diagonalization proofWhat is wrong with this conditional proof of P=NP?Is it always possible to have one part of the reduction?If everyone believes P ≠ NP, why is everyone sceptical of proof attempts for P ≠ NP?Finding a complexity by solving inequality
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $mathitSAT in P$ which itself then follows that $mathitSAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $mathitSAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof?
complexity-theory time-complexity p-vs-np
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
$endgroup$
add a comment
|
$begingroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $mathitSAT in P$ which itself then follows that $mathitSAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $mathitSAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof?
complexity-theory time-complexity p-vs-np
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
$endgroup$
2
$begingroup$
Please, write something like$mathitSAT$instead of$SAT$. As Leslie Lamport wrote in his original LaTeX book, the latter stands for S times A times T.
$endgroup$
– Oliphaunt
Apr 25 at 22:31
$begingroup$
Better yet, use thecomplexitypackage and simply writeSAT. (I guess that's not available on this stack, though.)
$endgroup$
– Oliphaunt
Apr 25 at 22:39
$begingroup$
@Oliphaunt Why not suggest an edit when you can improve the post? Although I must say that here the difference (if any) is a lot more subtle than I'd expect.
$endgroup$
– Discrete lizard♦
Apr 26 at 7:29
1
$begingroup$
@Discretelizard I often do, but it was "too much work" this time (i was / am on mobile). Entering all those $ and is finicky work. I chose to educate instead. (This decision may not have been entirely rational.)
$endgroup$
– Oliphaunt
Apr 26 at 11:38
add a comment
|
$begingroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $mathitSAT in P$ which itself then follows that $mathitSAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $mathitSAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof?
complexity-theory time-complexity p-vs-np
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
$endgroup$
I'm trying to argue that N is not equal NP using hierarchy theorems. This is my argument, but when I showed it to our teacher and after deduction, he said that this is problematic where I can't find a compelling reason to accept.
We start off by assuming that $P=NP$. Then it yields that $mathitSAT in P$ which itself then follows that $mathitSAT in TIME(n^k)$. As stands, we are able to do reduce every language in $NP$ to $mathitSAT$. Therefore, $NP subseteq TIME(n^k)$. On the contrary, the time hierarchy theorem states that there should be a language $A in TIME(n^k+1)$, that's not in $TIME(n^k)$. This would lead us to conclude that $A$ is in $P$, while not in $NP$, which is a contradiction to our first assumption. So, we came to the conclusion that $P neq NP$.
Is there something wrong with my proof?
complexity-theory time-complexity p-vs-np
complexity-theory time-complexity p-vs-np
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
edited Apr 26 at 7:29
Discrete lizard♦
5,3471 gold badge16 silver badges42 bronze badges
5,3471 gold badge16 silver badges42 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
asked Apr 25 at 4:12
inverted_indexinverted_index
1911 silver badge7 bronze badges
1911 silver badge7 bronze badges
2
$begingroup$
Please, write something like$mathitSAT$instead of$SAT$. As Leslie Lamport wrote in his original LaTeX book, the latter stands for S times A times T.
$endgroup$
– Oliphaunt
Apr 25 at 22:31
$begingroup$
Better yet, use thecomplexitypackage and simply writeSAT. (I guess that's not available on this stack, though.)
$endgroup$
– Oliphaunt
Apr 25 at 22:39
$begingroup$
@Oliphaunt Why not suggest an edit when you can improve the post? Although I must say that here the difference (if any) is a lot more subtle than I'd expect.
$endgroup$
– Discrete lizard♦
Apr 26 at 7:29
1
$begingroup$
@Discretelizard I often do, but it was "too much work" this time (i was / am on mobile). Entering all those $ and is finicky work. I chose to educate instead. (This decision may not have been entirely rational.)
$endgroup$
– Oliphaunt
Apr 26 at 11:38
add a comment
|
2
$begingroup$
Please, write something like$mathitSAT$instead of$SAT$. As Leslie Lamport wrote in his original LaTeX book, the latter stands for S times A times T.
$endgroup$
– Oliphaunt
Apr 25 at 22:31
$begingroup$
Better yet, use thecomplexitypackage and simply writeSAT. (I guess that's not available on this stack, though.)
$endgroup$
– Oliphaunt
Apr 25 at 22:39
$begingroup$
@Oliphaunt Why not suggest an edit when you can improve the post? Although I must say that here the difference (if any) is a lot more subtle than I'd expect.
$endgroup$
– Discrete lizard♦
Apr 26 at 7:29
1
$begingroup$
@Discretelizard I often do, but it was "too much work" this time (i was / am on mobile). Entering all those $ and is finicky work. I chose to educate instead. (This decision may not have been entirely rational.)
$endgroup$
– Oliphaunt
Apr 26 at 11:38
2
2
$begingroup$
Please, write something like
$mathitSAT$ instead of $SAT$. As Leslie Lamport wrote in his original LaTeX book, the latter stands for S times A times T.$endgroup$
– Oliphaunt
Apr 25 at 22:31
$begingroup$
Please, write something like
$mathitSAT$ instead of $SAT$. As Leslie Lamport wrote in his original LaTeX book, the latter stands for S times A times T.$endgroup$
– Oliphaunt
Apr 25 at 22:31
$begingroup$
Better yet, use the
complexity package and simply write SAT. (I guess that's not available on this stack, though.)$endgroup$
– Oliphaunt
Apr 25 at 22:39
$begingroup$
Better yet, use the
complexity package and simply write SAT. (I guess that's not available on this stack, though.)$endgroup$
– Oliphaunt
Apr 25 at 22:39
$begingroup$
@Oliphaunt Why not suggest an edit when you can improve the post? Although I must say that here the difference (if any) is a lot more subtle than I'd expect.
$endgroup$
– Discrete lizard♦
Apr 26 at 7:29
$begingroup$
@Oliphaunt Why not suggest an edit when you can improve the post? Although I must say that here the difference (if any) is a lot more subtle than I'd expect.
$endgroup$
– Discrete lizard♦
Apr 26 at 7:29
1
1
$begingroup$
@Discretelizard I often do, but it was "too much work" this time (i was / am on mobile). Entering all those $ and is finicky work. I chose to educate instead. (This decision may not have been entirely rational.)
$endgroup$
– Oliphaunt
Apr 26 at 11:38
$begingroup$
@Discretelizard I often do, but it was "too much work" this time (i was / am on mobile). Entering all those $ and is finicky work. I chose to educate instead. (This decision may not have been entirely rational.)
$endgroup$
– Oliphaunt
Apr 26 at 11:38
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
$endgroup$
1
$begingroup$
It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total.
$endgroup$
– gnasher729
Apr 27 at 11:21
$begingroup$
Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time).
$endgroup$
– cody
Apr 30 at 19:40
add a comment
|
$begingroup$
Suppose that $mathrm3SATinmathrmNTIME[n^k]$. By the nondeterministic version of the time hierarchy theorem, for any $r$, there is a problem $X_rinmathrmNTIME[n^r]$ that is not in $mathrmNTIME[n^r-1]$. This is an unconditional result that doesn't depend on any kind of assumption such as $mathrmPneqmathrmNP$
Choose any $r>k$. Suppose we have a deterministic reduction from $X_r$ to $mathrm3SAT$ that runs in time $n^t$. It produces a $mathrm3SAT$ instance of size at most $n^t$, which can be solved in time at most $(n^t)^k=n^tk$. By our choice of $X_r$, we must have $tk>r-1$, so $t>(r+1)/k$. This function grows without bound with $r$.
This means that there is no bound on how long it can take to reduce an arbitrary $mathrmNP$ problem to $mathrm3SAT$. Even if $mathrm3SATin mathrmP$, there's still no bound on how long those reductions can take. So, in particular, even if $mathrm3SATinmathrmDTIME[n^k']$ for some $k'$, we can't conclude that $mathrmNPsubseteqmathrmDTIME[n^k']$, or even $mathrmNPsubseteqmathrmDTIME[n^k'']$ for some $k''>k'$.
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
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2 Answers
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$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
$endgroup$
1
$begingroup$
It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total.
$endgroup$
– gnasher729
Apr 27 at 11:21
$begingroup$
Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time).
$endgroup$
– cody
Apr 30 at 19:40
add a comment
|
$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
$endgroup$
1
$begingroup$
It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total.
$endgroup$
– gnasher729
Apr 27 at 11:21
$begingroup$
Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time).
$endgroup$
– cody
Apr 30 at 19:40
add a comment
|
$begingroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
$endgroup$
Then it yields that $SAT in P$ which itself then follows that $SAT in TIME(n^k)$.
Sure.
As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP subseteq TIME(n^k)$.
No. Polynomial time reductions aren't free. We can say it takes $O(n^r(L))$ time to reduce language $L$ to $SAT$, where $r(L)$ is the exponent in the polynomial time reduction used. This is where your argument falls apart. There is no finite $k$ such that for all $L in NP$ we have $r(L) < k$. At least this does not follow from $P = NP$ and would be a much stronger statement.
And this stronger statement does indeed conflict with the time hierarchy theorem, which tells us that $P$ can not collapse into $TIME(n^k)$, let alone all of $NP$.
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
edited Apr 25 at 5:49
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
answered Apr 25 at 4:38
orlporlp
7,3361 gold badge11 silver badges28 bronze badges
7,3361 gold badge11 silver badges28 bronze badges
1
$begingroup$
It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total.
$endgroup$
– gnasher729
Apr 27 at 11:21
$begingroup$
Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time).
$endgroup$
– cody
Apr 30 at 19:40
add a comment
|
1
$begingroup$
It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total.
$endgroup$
– gnasher729
Apr 27 at 11:21
$begingroup$
Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time).
$endgroup$
– cody
Apr 30 at 19:40
1
1
$begingroup$
It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total.
$endgroup$
– gnasher729
Apr 27 at 11:21
$begingroup$
It's not only the time for the reduction itself. You could reduce to a make larger problem. If I can solve X in O (n^5), and I can reduce a problem in Y in O (n^6) to a O(n^3) sized instance of X, then I need O (n^15) in total.
$endgroup$
– gnasher729
Apr 27 at 11:21
$begingroup$
Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time).
$endgroup$
– cody
Apr 30 at 19:40
$begingroup$
Amusingly, this argument applies to PTIME-complete problems as well, e.g. HORNSAT, which is solvable in linear time (but not all problems in P are linear time).
$endgroup$
– cody
Apr 30 at 19:40
add a comment
|
$begingroup$
Suppose that $mathrm3SATinmathrmNTIME[n^k]$. By the nondeterministic version of the time hierarchy theorem, for any $r$, there is a problem $X_rinmathrmNTIME[n^r]$ that is not in $mathrmNTIME[n^r-1]$. This is an unconditional result that doesn't depend on any kind of assumption such as $mathrmPneqmathrmNP$
Choose any $r>k$. Suppose we have a deterministic reduction from $X_r$ to $mathrm3SAT$ that runs in time $n^t$. It produces a $mathrm3SAT$ instance of size at most $n^t$, which can be solved in time at most $(n^t)^k=n^tk$. By our choice of $X_r$, we must have $tk>r-1$, so $t>(r+1)/k$. This function grows without bound with $r$.
This means that there is no bound on how long it can take to reduce an arbitrary $mathrmNP$ problem to $mathrm3SAT$. Even if $mathrm3SATin mathrmP$, there's still no bound on how long those reductions can take. So, in particular, even if $mathrm3SATinmathrmDTIME[n^k']$ for some $k'$, we can't conclude that $mathrmNPsubseteqmathrmDTIME[n^k']$, or even $mathrmNPsubseteqmathrmDTIME[n^k'']$ for some $k''>k'$.
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
$endgroup$
add a comment
|
$begingroup$
Suppose that $mathrm3SATinmathrmNTIME[n^k]$. By the nondeterministic version of the time hierarchy theorem, for any $r$, there is a problem $X_rinmathrmNTIME[n^r]$ that is not in $mathrmNTIME[n^r-1]$. This is an unconditional result that doesn't depend on any kind of assumption such as $mathrmPneqmathrmNP$
Choose any $r>k$. Suppose we have a deterministic reduction from $X_r$ to $mathrm3SAT$ that runs in time $n^t$. It produces a $mathrm3SAT$ instance of size at most $n^t$, which can be solved in time at most $(n^t)^k=n^tk$. By our choice of $X_r$, we must have $tk>r-1$, so $t>(r+1)/k$. This function grows without bound with $r$.
This means that there is no bound on how long it can take to reduce an arbitrary $mathrmNP$ problem to $mathrm3SAT$. Even if $mathrm3SATin mathrmP$, there's still no bound on how long those reductions can take. So, in particular, even if $mathrm3SATinmathrmDTIME[n^k']$ for some $k'$, we can't conclude that $mathrmNPsubseteqmathrmDTIME[n^k']$, or even $mathrmNPsubseteqmathrmDTIME[n^k'']$ for some $k''>k'$.
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
$endgroup$
add a comment
|
$begingroup$
Suppose that $mathrm3SATinmathrmNTIME[n^k]$. By the nondeterministic version of the time hierarchy theorem, for any $r$, there is a problem $X_rinmathrmNTIME[n^r]$ that is not in $mathrmNTIME[n^r-1]$. This is an unconditional result that doesn't depend on any kind of assumption such as $mathrmPneqmathrmNP$
Choose any $r>k$. Suppose we have a deterministic reduction from $X_r$ to $mathrm3SAT$ that runs in time $n^t$. It produces a $mathrm3SAT$ instance of size at most $n^t$, which can be solved in time at most $(n^t)^k=n^tk$. By our choice of $X_r$, we must have $tk>r-1$, so $t>(r+1)/k$. This function grows without bound with $r$.
This means that there is no bound on how long it can take to reduce an arbitrary $mathrmNP$ problem to $mathrm3SAT$. Even if $mathrm3SATin mathrmP$, there's still no bound on how long those reductions can take. So, in particular, even if $mathrm3SATinmathrmDTIME[n^k']$ for some $k'$, we can't conclude that $mathrmNPsubseteqmathrmDTIME[n^k']$, or even $mathrmNPsubseteqmathrmDTIME[n^k'']$ for some $k''>k'$.
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
$endgroup$
Suppose that $mathrm3SATinmathrmNTIME[n^k]$. By the nondeterministic version of the time hierarchy theorem, for any $r$, there is a problem $X_rinmathrmNTIME[n^r]$ that is not in $mathrmNTIME[n^r-1]$. This is an unconditional result that doesn't depend on any kind of assumption such as $mathrmPneqmathrmNP$
Choose any $r>k$. Suppose we have a deterministic reduction from $X_r$ to $mathrm3SAT$ that runs in time $n^t$. It produces a $mathrm3SAT$ instance of size at most $n^t$, which can be solved in time at most $(n^t)^k=n^tk$. By our choice of $X_r$, we must have $tk>r-1$, so $t>(r+1)/k$. This function grows without bound with $r$.
This means that there is no bound on how long it can take to reduce an arbitrary $mathrmNP$ problem to $mathrm3SAT$. Even if $mathrm3SATin mathrmP$, there's still no bound on how long those reductions can take. So, in particular, even if $mathrm3SATinmathrmDTIME[n^k']$ for some $k'$, we can't conclude that $mathrmNPsubseteqmathrmDTIME[n^k']$, or even $mathrmNPsubseteqmathrmDTIME[n^k'']$ for some $k''>k'$.
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
answered Apr 25 at 14:58
David RicherbyDavid Richerby
75.4k17 gold badges117 silver badges209 bronze badges
75.4k17 gold badges117 silver badges209 bronze badges
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$begingroup$
Please, write something like
$mathitSAT$instead of$SAT$. As Leslie Lamport wrote in his original LaTeX book, the latter stands for S times A times T.$endgroup$
– Oliphaunt
Apr 25 at 22:31
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Better yet, use the
complexitypackage and simply writeSAT. (I guess that's not available on this stack, though.)$endgroup$
– Oliphaunt
Apr 25 at 22:39
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@Oliphaunt Why not suggest an edit when you can improve the post? Although I must say that here the difference (if any) is a lot more subtle than I'd expect.
$endgroup$
– Discrete lizard♦
Apr 26 at 7:29
1
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@Discretelizard I often do, but it was "too much work" this time (i was / am on mobile). Entering all those $ and is finicky work. I chose to educate instead. (This decision may not have been entirely rational.)
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– Oliphaunt
Apr 26 at 11:38