Decision problem that can be verified but not run in n^2 time












2












$begingroup$


A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?






time-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    In what model of computation?
    $endgroup$
    – Curtis F
    5 hours ago










  • $begingroup$
    @CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
    $endgroup$
    – while1fork
    5 hours ago






  • 4




    $begingroup$
    @while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
    $endgroup$
    – Rick Decker
    4 hours ago


















2












$begingroup$


A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?






time-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    In what model of computation?
    $endgroup$
    – Curtis F
    5 hours ago










  • $begingroup$
    @CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
    $endgroup$
    – while1fork
    5 hours ago






  • 4




    $begingroup$
    @while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
    $endgroup$
    – Rick Decker
    4 hours ago
















2












2








2





$begingroup$


A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?






time-complexity







share|cite|improve this question











$endgroup$




A much much weaker idea similar to the P=NP question, is there a decision problem that can be verified in $O(n^2)$ time, but it can be proven that there is no algorithm that decides it $O(n^2)$ time?






time-complexity




time-complexity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







while1fork

















asked 5 hours ago









while1forkwhile1fork

183




183












  • $begingroup$
    In what model of computation?
    $endgroup$
    – Curtis F
    5 hours ago










  • $begingroup$
    @CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
    $endgroup$
    – while1fork
    5 hours ago






  • 4




    $begingroup$
    @while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
    $endgroup$
    – Rick Decker
    4 hours ago




















  • $begingroup$
    In what model of computation?
    $endgroup$
    – Curtis F
    5 hours ago










  • $begingroup$
    @CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
    $endgroup$
    – while1fork
    5 hours ago






  • 4




    $begingroup$
    @while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
    $endgroup$
    – Rick Decker
    4 hours ago


















$begingroup$
In what model of computation?
$endgroup$
– Curtis F
5 hours ago




$begingroup$
In what model of computation?
$endgroup$
– Curtis F
5 hours ago












$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
5 hours ago




$begingroup$
@CurtisF if it matters then please write that up as an answer. I was under the impression that that sort of thing didn't matter within reason, or else how can one ask "does P=NP"?
$endgroup$
– while1fork
5 hours ago




4




4




$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
4 hours ago






$begingroup$
@while1fork. It does matter if you're giving precise timing estimates, like $O(n^2)$.
$endgroup$
– Rick Decker
4 hours ago












1 Answer
1






active

oldest

votes


















3












$begingroup$

The answer depends a lot on the model of computation.



On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.



On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.






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    1 Answer
    1






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    active

    oldest

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    active

    oldest

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    3












    $begingroup$

    The answer depends a lot on the model of computation.



    On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.



    On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The answer depends a lot on the model of computation.



      On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.



      On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The answer depends a lot on the model of computation.



        On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.



        On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.






        share|cite|improve this answer









        $endgroup$



        The answer depends a lot on the model of computation.



        On one extreme: in the decision tree model, many lower bounds can be proven. For instance, consider the 3SUM problem. You can verify an alleged solution to the 3SUM problem in $O(n)$ time, but it's conjectured that no algorithm can solve the 3SUM problem in $O(n^{2-epsilon})$ time for any $epsilon>0$. One can prove this holds in some version of the algebraic decision tree model; one can prove that any algorithm to solve it in this model must take $Omega(n^2)$ time. This provides a gap of $O(n^2)$ vs $O(n)$ for solving vs verifying solutions, if the verifier is limited to the algebraic decision tree model. This is a pretty limited result; in this case, the model means that we're restricting attention to algorithms of a particular form. So, it doesn't rule out the possibility of some other algorithm (that does something weird) being faster.



        On the other extreme: if we allow arbitrary (non-uniform) boolean circuits, then there is no explicitly stated function $f$ on $n$ bits where we can currently prove that every circuit for computing $f$ needs $ge 3.1n$ gates. In other words, this is saying that we have no clue how to prove lower bounds for circuits. Roughly speaking, we have no known result of a problem where we can prove that solving it requires circuits of size $omega(n)$. See, e.g., https://cstheory.stackexchange.com/a/8005/5038 and https://cstheory.stackexchange.com/q/21400/5038. So, given our current state of knowledge, we have no hope of proving a result like that if the model of computation is unrestricted boolean circuits.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        D.W.♦D.W.

        99.8k12122286




        99.8k12122286






























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