7 ms·

Is P=NP an Ill Posed Problem?

Confusion 17y ago
Doesn't argue that "P=?NP" is an ill posed problem, but rather that solving it may not be very important, as the details matter at lot for the implications the result may have. There are a lot of solutions that would settle the problem, without having any influence on the practical state of affairs. Introduces the term 'SAT' at some point, without explaining what it means. Wikipedia isn't helpful.
- alextp 17y ago
  SAT is the boolean satisfiability problem, a simple NP-complete problem often used to prove the other problems are NP-complete. An instance of SAT is a boolean formula, and the answer to it is yes if and only if the formula can be true for some assignment of variables. 3SAT is the same problem, only with the formula in a conjunctive normal form and each clause having exactly three variables. Also, it seems the poster is really just trying to get attention. Complexity theorists do know all that and have spent a lot of time trying to see which postulated consequences of P?=NP are necessary and which are not. (If I were to argue it is an ill-posed question, I'd go more in the direction of CO-NP (which is the set of problems for which "NO" answers have short certificates), and why this really strong emphasys on "yes" certificates for NP problems) Doron Zeilberger's take on that ( http://www.math.rutgers.edu/~zeilberg/Opinion98.html http://www.math.rutgers.edu/~zeilberg/Opinion98.html ) is a lot more interesting. And, also, the "easy" equals "polynomial time" is (a) a good heuristic, if you assume computational power increases exponentially (hence, a problem will be solvable in the foreseeable future if all it needs is polynomially more processing power for larger instances); (b) it is an easy to study class, since it is composable; (c) not the only thing complexity theorists argue about (ie, P=NP means practical bounds for some problems and collapses a lot of classes that are not trivially seen to be equal, like AM and MA)
  - anatoly 17y ago
    I think your evaluation of the post may have been somewhat rushed. - The poster is a notable complexity theorist and is not "just trying to get attention". - Doron Zeilberger's "a lot more interesting" take is completely subsumed in Richard Lipton's blogpost, in that it mentions other possibilities beyond the basic "what if it's polynomial with a really huge exponent" scenario. - Assuming computational power to increase exponentially indefinitely is completely ridiculous, and without the "indefinitely" part this assumption no longer works to justify "easy = P". - It is trivial to say that if P=NP, then AM=MA, because both AM and MA are in the polynomial hierarchy, and if P=NP, the hierarchy collapses completely. - The most important point to be taken away from the blogpost is not just that "easy = P" is an assumption that's been held on too unreflectively and may not turn out to be true. It's that studying what this assumption means, how it could break down, and why it hasn't, by and large, broken down so far, may get us closer to understanding P ?= NP.
    - alextp 17y ago
      You have a point. My closeted complexity theorist persona thinks it would be a lot more interesting to study the special cases of P (not just LOG, but seeing how O(n) and O(n²) or "NO(n)" problems behave) instead of the most general cases.
- jrp 17y ago
  In general, I think blogs should omit introductions to such basic material. It makes more sense to collect the introductions in one place (usually textbooks, but certainly I would welcome someone blogging about intro to something -- see http://www.scottaaronson.com/democritus/ http://www.scottaaronson.com/democritus/ ). He could explain SAT, but then shouldn't he also explain what a randomized algorithm is, why P=NP is (as he says, possibly) relevant to circuit design, what P and NP are, etc. The problem with this approach is that it doesn't know where to stop (explaining what an algorithm is?).
  - Confusion 17y ago
    It's not the material that is the problem, it's the term. I understood the entire blogposting, except for that term. Now that someone explained it, I understand what it means and why it's important. I'm a physicist by training and a software engineer by profession; I think I qualify as the target audience for such a blogposting?
- pbhjpbhj 17y ago
  SAT is an abbreviation not an acronym. It stands for SATisfiability, as in http://en.wikipedia.org/wiki/Boolean_satisfiability_problem http://en.wikipedia.org/wiki/Boolean_satisfiability_problem . The blog on P=NP doesn't appear to mention it despite having 3 posts tagged as SAT. Sometimes we don't realise we're using jargon or we don't realise the wider audience we might be addressing. Edit: Oops, must refresh before re-joining a HN thread.
jganetsk 17y ago
I once heard Manuel Blum say that it could be shown that the result can be neither proven nor disproven given our framework of mathematics, similar to the Continuum Hypothesis. In this case, I would call P=NP an ill-posed problem. I think the OP makes great points in his article, but the title of the blog post is wrong. He really speaks of how the scope of P=NP is misconceived by most people.
- miloshh 17y ago
  Well, even if a given logical theory cannot decide P=NP, it is still a well-posed problem. Either there is a polynomial-time algorithm for SAT or there isn't - no middle-ground here. But I agree, the blog isn't about this issue at all.
  - jganetsk 17y ago
    Either there is a polynomial-time algorithm for SAT or there isn't - no middle-ground here. Statement of the continuum hypothesis: There is no set whose cardinality is strictly between that of the integers and that of the real numbers. You can provide your argument against the continuum hypothesis. No middle ground? There might not be a polynomial-time algorithm for SAT... but how would we know? You might not be able to prove that one does not exist. The absence of evidence is not the evidence of absence. And we could be eternally locked in this state. Middle ground exists, for constructivists, those who don't admit the law of the excluded middle (that any statement must be either true or false). And I think most theoretical computer scientists are of this persuasion.
    - miloshh 17y ago
      OK, I see what you mean, but I would argue that P=NP is a question of a very different nature than CH. CH is a statement about the existence of an abstract infinite set, so it is not that surprising that its truth depends on your very definition of a set, an elusive object with multiple possible meanings. On the other hand, P=NP is a statement about boolean formulas, Turing machines and polynomials. None of these are elusive, they are very cleanly defined, trivial concepts. So it's a very well-posed question. Of course, we could still not be able to prove the (non)existence of a polynomial algorithm for SAT in ZF set theory. In that case, too bad for ZF, but whether P=NP is still a meaningful question.
tome 17y ago
If David Hilbert woke up in 1000 years he wouldn't ask "Does P=NP?" but "what progress have we made on P=NP?". Once the Riemann Hypothesis is done with, it's done with, but P=NP or P!=NP opens up whole new lines of discussion. That said, P=NP could be independent of ZF[1], and if that was shown we'd be scratching our heads asking what P=NP actually means. [1] http://www.scottaaronson.com/papers/pnp.pdf http://www.scottaaronson.com/papers/pnp.pdf
dmg_83 17y ago
According to the Time Hierarchy theorem ( http://en.wikipedia.org/wiki/Time_hierarchy_theorem http://en.wikipedia.org/wiki/Time_hierarchy_theorem ), for each number k > 1, there exists a problem in P that can be decided (solved) in O(n^k), and not decided (solved) in O(n^j) for each j in [1, k). This was the first time I heard of this theorem (was googling for something I thought would weaken the author's point), and it really strengthens the authors argument - I thought his example of O(n^10) algorithms was sort of a fallacy but it appears not. If I'm properly understanding the Time Hierarchy Theorem, it is a very strong argument for what the author is suggesting (and somewhat disappointing for me, because I wish he were wrong).

Doesn't argue that "P=?NP" is an ill posed problem, but rather that solving it may not be very important, as the details matter at lot for the implications the result may have. There are a lot of solutions that would settle the problem, without having any influence on the practical state of affairs. Introduces the term 'SAT' at some point, without explaining what it means. Wikipedia isn't helpful.

alextp 17y ago
SAT is the boolean satisfiability problem, a simple NP-complete problem often used to prove the other problems are NP-complete. An instance of SAT is a boolean formula, and the answer to it is yes if and only if the formula can be true for some assignment of variables. 3SAT is the same problem, only with the formula in a conjunctive normal form and each clause having exactly three variables. Also, it seems the poster is really just trying to get attention. Complexity theorists do know all that and have spent a lot of time trying to see which postulated consequences of P?=NP are necessary and which are not. (If I were to argue it is an ill-posed question, I'd go more in the direction of CO-NP (which is the set of problems for which "NO" answers have short certificates), and why this really strong emphasys on "yes" certificates for NP problems) Doron Zeilberger's take on that ( http://www.math.rutgers.edu/~zeilberg/Opinion98.html http://www.math.rutgers.edu/~zeilberg/Opinion98.html ) is a lot more interesting. And, also, the "easy" equals "polynomial time" is (a) a good heuristic, if you assume computational power increases exponentially (hence, a problem will be solvable in the foreseeable future if all it needs is polynomially more processing power for larger instances); (b) it is an easy to study class, since it is composable; (c) not the only thing complexity theorists argue about (ie, P=NP means practical bounds for some problems and collapses a lot of classes that are not trivially seen to be equal, like AM and MA)
- anatoly 17y ago
  I think your evaluation of the post may have been somewhat rushed. - The poster is a notable complexity theorist and is not "just trying to get attention". - Doron Zeilberger's "a lot more interesting" take is completely subsumed in Richard Lipton's blogpost, in that it mentions other possibilities beyond the basic "what if it's polynomial with a really huge exponent" scenario. - Assuming computational power to increase exponentially indefinitely is completely ridiculous, and without the "indefinitely" part this assumption no longer works to justify "easy = P". - It is trivial to say that if P=NP, then AM=MA, because both AM and MA are in the polynomial hierarchy, and if P=NP, the hierarchy collapses completely. - The most important point to be taken away from the blogpost is not just that "easy = P" is an assumption that's been held on too unreflectively and may not turn out to be true. It's that studying what this assumption means, how it could break down, and why it hasn't, by and large, broken down so far, may get us closer to understanding P ?= NP.
  - alextp 17y ago
    You have a point. My closeted complexity theorist persona thinks it would be a lot more interesting to study the special cases of P (not just LOG, but seeing how O(n) and O(n²) or "NO(n)" problems behave) instead of the most general cases.
jrp 17y ago
In general, I think blogs should omit introductions to such basic material. It makes more sense to collect the introductions in one place (usually textbooks, but certainly I would welcome someone blogging about intro to something -- see http://www.scottaaronson.com/democritus/ http://www.scottaaronson.com/democritus/ ). He could explain SAT, but then shouldn't he also explain what a randomized algorithm is, why P=NP is (as he says, possibly) relevant to circuit design, what P and NP are, etc. The problem with this approach is that it doesn't know where to stop (explaining what an algorithm is?).
- Confusion 17y ago
  It's not the material that is the problem, it's the term. I understood the entire blogposting, except for that term. Now that someone explained it, I understand what it means and why it's important. I'm a physicist by training and a software engineer by profession; I think I qualify as the target audience for such a blogposting?
pbhjpbhj 17y ago
SAT is an abbreviation not an acronym. It stands for SATisfiability, as in http://en.wikipedia.org/wiki/Boolean_satisfiability_problem http://en.wikipedia.org/wiki/Boolean_satisfiability_problem . The blog on P=NP doesn't appear to mention it despite having 3 posts tagged as SAT. Sometimes we don't realise we're using jargon or we don't realise the wider audience we might be addressing. Edit: Oops, must refresh before re-joining a HN thread.

jganetsk 17y ago

I once heard Manuel Blum say that it could be shown that the result can be neither proven nor disproven given our framework of mathematics, similar to the Continuum Hypothesis. In this case, I would call P=NP an ill-posed problem. I think the OP makes great points in his article, but the title of the blog post is wrong. He really speaks of how the scope of P=NP is misconceived by most people.

miloshh 17y ago
Well, even if a given logical theory cannot decide P=NP, it is still a well-posed problem. Either there is a polynomial-time algorithm for SAT or there isn't - no middle-ground here. But I agree, the blog isn't about this issue at all.
- jganetsk 17y ago
  Either there is a polynomial-time algorithm for SAT or there isn't - no middle-ground here. Statement of the continuum hypothesis: There is no set whose cardinality is strictly between that of the integers and that of the real numbers. You can provide your argument against the continuum hypothesis. No middle ground? There might not be a polynomial-time algorithm for SAT... but how would we know? You might not be able to prove that one does not exist. The absence of evidence is not the evidence of absence. And we could be eternally locked in this state. Middle ground exists, for constructivists, those who don't admit the law of the excluded middle (that any statement must be either true or false). And I think most theoretical computer scientists are of this persuasion.
  - miloshh 17y ago
    OK, I see what you mean, but I would argue that P=NP is a question of a very different nature than CH. CH is a statement about the existence of an abstract infinite set, so it is not that surprising that its truth depends on your very definition of a set, an elusive object with multiple possible meanings. On the other hand, P=NP is a statement about boolean formulas, Turing machines and polynomials. None of these are elusive, they are very cleanly defined, trivial concepts. So it's a very well-posed question. Of course, we could still not be able to prove the (non)existence of a polynomial algorithm for SAT in ZF set theory. In that case, too bad for ZF, but whether P=NP is still a meaningful question.

tome 17y ago

If David Hilbert woke up in 1000 years he wouldn't ask "Does P=NP?" but "what progress have we made on P=NP?". Once the Riemann Hypothesis is done with, it's done with, but P=NP or P!=NP opens up whole new lines of discussion. That said, P=NP could be independent of ZF[1], and if that was shown we'd be scratching our heads asking what P=NP actually means. [1] http://www.scottaaronson.com/papers/pnp.pdf http://www.scottaaronson.com/papers/pnp.pdf

dmg_83 17y ago

According to the Time Hierarchy theorem ( http://en.wikipedia.org/wiki/Time_hierarchy_theorem http://en.wikipedia.org/wiki/Time_hierarchy_theorem ), for each number k > 1, there exists a problem in P that can be decided (solved) in O(n^k), and not decided (solved) in O(n^j) for each j in [1, k). This was the first time I heard of this theorem (was googling for something I thought would weaken the author's point), and it really strengthens the authors argument - I thought his example of O(n^10) algorithms was sort of a fallacy but it appears not. If I'm properly understanding the Time Hierarchy Theorem, it is a very strong argument for what the author is suggesting (and somewhat disappointing for me, because I wish he were wrong).