Contradiction proof for inequality of P and NP?How can I argue that $3mathsfSATleq_p mathsfIndSet$ is polynomial in time?Problem with my proof that NP = coNP?Why does Schaefer's theorem not prove that P=NP?Logarithmic Randomness is Necessary for PCP TheoremTrouble seeing the contradiction in diagonalization proofIs 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?Find the flaw in the P != NP proofTesting algorithm for a modified sieve of EratosthenesFinding a complexity by solving inequality

Lock in SQL Server and Oracle

Do I have to worry about players making “bad” choices on level up?

Were there two appearances of Stan Lee?

Is thermodynamics only applicable to systems in equilibrium?

Build a trail cart

Pawn Sacrifice Justification

How to creep the reader out with what seems like a normal person?

Are Boeing 737-800’s grounded?

Why does processed meat contain preservatives, while canned fish needs not?

Historically, were women trained for obligatory wars? Or did they serve some other military function?

If Earth is tilted, why is Polaris always above the same spot?

When to use 1/Ka vs Kb

Has any spacecraft ever had the ability to directly communicate with civilian air traffic control?

Why does Bran Stark feel that Jon Snow "needs to know" about his lineage?

Possible to set `foldexpr` using a function reference?

Asahi Dry Black beer can

Binary Numbers Magic Trick

Modify locally tikzset

Why is current rating for multicore cable lower than single core with the same cross section?

What are the spoon bit of a spoon and fork bit of a fork called?

Does the EU Common Fisheries Policy cover British Overseas Territories?

Is it possible to Ready a spell to be cast just before the start of your next turn by having the trigger be an ally's attack?

What word means to make something obsolete?

Sci-fi novel series with instant travel between planets through gates. A river runs through the gates



Contradiction proof for inequality of P and NP?


How can I argue that $3mathsfSATleq_p mathsfIndSet$ is polynomial in time?Problem with my proof that NP = coNP?Why does Schaefer's theorem not prove that P=NP?Logarithmic Randomness is Necessary for PCP TheoremTrouble seeing the contradiction in diagonalization proofIs 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?Find the flaw in the P != NP proofTesting algorithm for a modified sieve of EratosthenesFinding a complexity by solving inequality













10












$begingroup$


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?










share|cite|improve this question











$endgroup$







  • 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 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
    2 days ago







  • 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
    2 days ago















10












$begingroup$


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?










share|cite|improve this question











$endgroup$







  • 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 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
    2 days ago







  • 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
    2 days ago













10












10








10


3



$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Discrete lizard

4,85311540




4,85311540










asked Apr 25 at 4:12









inverted_indexinverted_index

18816




18816







  • 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 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
    2 days ago







  • 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
    2 days ago












  • 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 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
    2 days ago







  • 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
    2 days ago







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
2 days ago





$begingroup$
@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
2 days ago





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
2 days ago




$begingroup$
@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
2 days ago










2 Answers
2






active

oldest

votes


















53












$begingroup$


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 $NP$.






share|cite|improve this answer











$endgroup$








  • 5




    $begingroup$
    "Polynomial time reductions aren't free" +1 for good wording.
    $endgroup$
    – Rick Decker
    Apr 25 at 12:46






  • 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
    yesterday


















8












$begingroup$

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 $k''>k'$.






share|cite|improve this answer









$endgroup$













    Your Answer








    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "419"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f108496%2fcontradiction-proof-for-inequality-of-p-and-np%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    53












    $begingroup$


    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 $NP$.






    share|cite|improve this answer











    $endgroup$








    • 5




      $begingroup$
      "Polynomial time reductions aren't free" +1 for good wording.
      $endgroup$
      – Rick Decker
      Apr 25 at 12:46






    • 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
      yesterday















    53












    $begingroup$


    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 $NP$.






    share|cite|improve this answer











    $endgroup$








    • 5




      $begingroup$
      "Polynomial time reductions aren't free" +1 for good wording.
      $endgroup$
      – Rick Decker
      Apr 25 at 12:46






    • 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
      yesterday













    53












    53








    53





    $begingroup$


    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 $NP$.






    share|cite|improve this answer











    $endgroup$




    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 $NP$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 25 at 5:49

























    answered Apr 25 at 4:38









    orlporlp

    6,53511127




    6,53511127







    • 5




      $begingroup$
      "Polynomial time reductions aren't free" +1 for good wording.
      $endgroup$
      – Rick Decker
      Apr 25 at 12:46






    • 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
      yesterday












    • 5




      $begingroup$
      "Polynomial time reductions aren't free" +1 for good wording.
      $endgroup$
      – Rick Decker
      Apr 25 at 12:46






    • 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
      yesterday







    5




    5




    $begingroup$
    "Polynomial time reductions aren't free" +1 for good wording.
    $endgroup$
    – Rick Decker
    Apr 25 at 12:46




    $begingroup$
    "Polynomial time reductions aren't free" +1 for good wording.
    $endgroup$
    – Rick Decker
    Apr 25 at 12:46




    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
    yesterday




    $begingroup$
    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
    yesterday











    8












    $begingroup$

    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 $k''>k'$.






    share|cite|improve this answer









    $endgroup$

















      8












      $begingroup$

      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 $k''>k'$.






      share|cite|improve this answer









      $endgroup$















        8












        8








        8





        $begingroup$

        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 $k''>k'$.






        share|cite|improve this answer









        $endgroup$



        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 $k''>k'$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 25 at 14:58









        David RicherbyDavid Richerby

        71k16109199




        71k16109199



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Computer Science Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f108496%2fcontradiction-proof-for-inequality-of-p-and-np%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Sum ergo cogito? 1 nng

            三茅街道4182Guuntc Dn precexpngmageondP