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Here by «algorithm» I mean a (halting) Turing machine with finite alphabet and memory.

Is it possible to obtain by purely existential (i.e. non-constructive) means the existence of an algorithm performing some required task, while beeing unable to explicit it (i.e. write it down or describe it constructively) ?

I hope the question is neither too vague nor too ill-posed, as I'm not so acquainted with these concepts! Thank you for any insight on the matter, feel free to retag or plainly close this naïve question if necessary.

Edit: in view of the answers given so far I think I can better explain what I meant originally. My question is more along the lines of

Is there a task for which an algorithm is known to exist while at the same time it is not possible to recognize the said algorithm?

By this I mean that our innability to perform the recognition is not a consequence of our current lack of actual knowledge, but of an intrinsec formal impossibility to do so. In other word working out that any algorithm outputs what is expected by the task is undecidable by inspection of its content.

(Final) edit: I thank everybody who took the pain to answer my rookie's question... I would like to formulate some comments regarding the (legitimate) objections that has been raised. First we may assume that the formal system is ZFC, in order to fix ideas. My question was not about "mass problems" (although the examples presented below really are interesting), I was rather asking about a specific task, so it relates more to undecidability problems, as was pointed out by Mark Sapir.

Actually MyFavouriteProblem concerns effective computability of holomorphic dynamical systems satisfying MyFavoriteProperty, which were known to exists by some "abstract" argument and for which I gave a more "constructive" proof. It is doubtlessly a practical improvement (my PC can do the computing), but I wanted to know if from a theoretical point of view some progress really has been made (actually to settle a friendly argument with a colleague of mine). So the starting point to this question lies in algorithms performing computations with arbitrary floating-point precision on complex numbers, understood as Taylor coefficients of power series (needs some infinite memory here, I'm afraid, but I don't think it really matters).

  • Regarding Joel's answer: this argument would be a striking satisfactory answer if it is proved that knowing the value of $N$ is indecidable in ZFC. Maybe one can work out an example where $\pi$ is replaced by some other real number.

  • Regarding Goldstern's answer (and subsequent comment/answers): I don't so much think that the argument is a cheat than it does not answer my edited question (edit made after the initial answer). My point is the following. Either some open question $Q$ is decidable in ZFC, in which case we all agree that the program "Yes" or the program "No" answers $Q$, but by inspection of the programs we can decide which one performs the required task once we know which answer is a fact, which is not a formal issue anymore. Either $Q$ is indecidable in ZFC and we cannot prove, in ZFC, the existence of an algorithm answering it (though I'm not sure of this part of my argument). Now if we attach $Q$ to our axioms set then we are back to the previous case, were answering $Q$ has become explicitely tautological. So I believe Goldstern's example does not meet the requirement of my question.

  • Timothy's answer is great since it offers a very nice comprehensive summary of the issue as I perceive it. If the unofficial answer to my question is that pessimistic statement, then let it be. The sloppy mathematician in me is perfectly happy with matters as they are ;)

    To conclude, I think I should accept Peter Shor's answer, since he gives an example which, as I understand it, answers my query. I'll do that in a little while.

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This question largely duplicates mathoverflow.net/questions/14918/… . –  Emil Jeřábek Apr 5 '13 at 16:23
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See also mathoverflow.net/questions/48014/…. –  Joel David Hamkins Apr 5 '13 at 16:46
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The rough answer that all the answers have in common is that if some piece of information is guaranteed to be finite then there is a Turing machine that has this information preprocessed and can do anything algorithmic with this information. But if you don't know explicitly this information then you will not be able to explicitly write down the Turing Machine. –  Benjamin Steinberg Apr 6 '13 at 13:26
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I agree with Joel that we need to know more precisely what you intend by "intrinsic formal impossibility" to know something. If it just means undecidability in ZFC (or ZFC plus some specific large cardinals), then Goldstern's answer works if we take the question Q to be something like CH that is undecidable in such theories. But if you intend the possibilities for knowledge to extend beyond provability in such formal systems, then it's not so clear what those possibilities are, nor even whether there would be anything at all that is intrinsically unknowable. –  Andreas Blass Apr 6 '13 at 15:18
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This is a question for which an answer is obvious once it is made sufficiently precise, but every answer feels like a cheat if one hasn't gone through the process of clarifying the question. I guess that means that there is an algorithm to answer this question but there is no way to recognize this algorithm? –  François G. Dorais Apr 6 '13 at 17:41

8 Answers 8

up vote 9 down vote accepted

Yes.

We take a fixed diophantine equation in variables $y,x_1, x_2, \ldots, x_k$. Task: for an input $n \in \mathbb{Z}$, output either

  1. Five values of $y$ for which solutions exist to the equation.
  2. A solution to the equation for which $y=n$.
  3. "No", in which case there must be no solution with $y=n$.

It is clear that for any diophantine equation, there is a program which performs this task. If there are five values of $y$ for which solutions exist, you need only have the program output these for all inputs. If there are fewer than four values of $y$ for which solutions exist, once you know the solutions, then writing the program is trivial.

However, telling whether a solution exists to a diophantine equation is undecidable.

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I assume the index $n$ of the $x$-variables is not the same as the input $n$? The diophantine equation is fixed? –  Michael Bächtold Apr 6 '13 at 18:02
    
@Michael: that's right ... the diophantine equation is fixed. Let me fix the variables. –  Peter Shor Apr 6 '13 at 18:11
    
+1............. –  Benjamin Steinberg Apr 7 '13 at 2:03
    
It's too bad that I don't see how to make an algorithm like this based on the Mordell Conjecture (Falting's Theorem), as that would be quite a bit more natural. –  Peter Shor Apr 9 '13 at 20:11

Here is a pure existence proof of computability, where we show that a function is computable, not by exhibiting an explicit algorithm, but by providing a list of infinitely many algorithms, and proving that one of them computes the function; we just don't know which one.

(This answer is adapted from an answer I gave to Hans Stricker's question Non-constructive proofs of decidability?)

Let $f(n)=1$, if there are $n$ consecutive $7$s somewhere in the decimal expansion of $\pi$, and $f(n)=0$ otherwise. Is this a computable function?

Perhaps we might try naively to compute it like this: on input $n$, start to enumerate the digits of $\pi$, and look for $n$ consecutive $7$s. If found, then output $1$. But then we realize, of course: what if on a particular input, you have searched for 10 years, and still not found the instance? We don't seem justified in outputting $0$ quite yet, since perhaps we might still find the consecutive $1$s by searching a bit more.

Nevertheless, we can prove that the function is computable by a pure existence proof. Namely, either there are arbitrarily long strings of $7$s in $\pi$ or there is a longest string of $7$s of some length $N$. In the former case, the function $f$ is the constant $1$ function, which is definitely computable. In the latter case, $f$ is the function with value $1$ for all input $n\leq N$ and value $0$ for $n\gt N$, which for any fixed $N$ is also a computable function.

So we have proved that $f$ is computable in effect by providing an infinite list of programs and proving that one of them computes $f$, but we don't know which one exactly. Indeed, I believe it is an open question of number theory which case is the right one.

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Thanks a lot for the answer, but this function might one day «become» explicit if N is ever determined by number theoricists. It is not proved, I surmise, that this number is unreachable. Or is there some point I'm missing? –  Loïc Teyssier Apr 6 '13 at 9:18
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Loic, I don't think it is possible for you so easily to rule out this kind of answer---mine and Goldstern's---and retain the essence of your question. To my way of thinking, both of these answers (and others) illustrate the fundamental phenomenon, where we are able to prove something is computable, but are not able to exhibit an explicit algorithm. –  Joel David Hamkins Apr 6 '13 at 10:56
    
I didn't want to sound too dismissive. It's just that I felt the answers you refer to did not address the (unexpressed) nature of ny question. Maybe my edit clarifies what I meant. –  Loïc Teyssier Apr 6 '13 at 14:22

The question is a little vague but here is the answer to one possible interpretation of it. Let $R$ be a finite group presentation with finite generating set $X$, $u$ is a word in $X$. Suppose $G$ is generated by $X$, and finite, then we there is an algorithm to find out if $u=1$ in $X$ (the word problem in every finite group is decidable). But there is no way we can know if there exists a universal algorithm that given $R$ and $u$ tells if $u=1$ in every finite group generated by $X$ that satisfies $R$ (i.e. the uniform word problem in finite groups is undecidable, see Slobodskoĭ, A. M. Undecidability of the universal theory of finite groups. Algebra i Logika 20 (1981), no. 2, 207–230). In fact that situation is quite normal. Decidability of a mass problem means an algorithm to solve it exist, but often it is a pure existence statement.

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@Mark, I believe Slobodoskii's theorem doesn't say exactly what you want. It says you cannot decide give a finite presentation <X|R> whether a word in X is 1 in every X-generated finite group satisfying the relations R. Note <X|R> can present an infinite group. If you know a finite presentation presents a finite group, you can solve the yes part by enumerating all consequences of R and the no part enumerating finite groups. –  Benjamin Steinberg Apr 5 '13 at 18:17
    
Yes, I was not careful enough. I will change the answer. –  Mark Sapir Apr 5 '13 at 18:39
    
Thanks for your answer! Yet I don't feel it relates fully to my admitedly vague quetion, although its nature pleases me better than other answers do. Maybe my edit clarifies what I meant. –  Loïc Teyssier Apr 6 '13 at 14:19
    
I see. The new formulation is much less vague. I (or somebody else) need to think more about it. –  Mark Sapir Apr 6 '13 at 20:12
    
@Loïc: It may help if you formulate your question in a most standard way, as a "mass problem". What exactly is the problem which you want undecidable, i.e. what is the input set and when an input should be recognized? –  Mark Sapir Apr 6 '13 at 20:17

If you are still not satisfied with any of the current answers, then I think the trouble is that you have not thought through clearly in your own mind what it means for it to be impossible to know some particular mathematical fact.

The first condition that you want to be satisfied is clear enough. You want it to be provably true that

There exists some Turing machine that solves My Favorite Problem.

Since Turing machines can be enumerated, this is equivalent to saying that we can prove that at least one of the statements in the following list is true.

Turing machine 1 solves My Favorite Problem.
Turing machine 2 solves My Favorite Problem.
Turing machine 3 solves My Favorite Problem.
Turing machine 4 solves My Favorite Problem.
$\vdots$

The second condition that you want to be satisfied is that not only do we currently not know any particular statement in this list to be true, but there should be some "intrinsic formal impossibility" preventing us from ever knowing it.

The trouble is that it's not clear what you mean by "an intrinsic formal impossibility." Suppose for example that Turing machine 314159 happens to solve My Favorite Problem. What's to stop someone from just baldly asserting, "I know that Turing machine 314159 solves My Favorite Problem"? Well, of course, being mathematicians, we object to people making such an assertion unless they can prove it. So probably you want your second condition to look something like this:

For every n such that Turing machine n solves My Favorite Problem, it is impossible to prove that Turing machine n solves My Favorite Problem.

Now the trouble with this statement as it stands is that it is not entirely formal, because we have not specified the axioms we are allowed to appeal to in our proof. In "ordinary mathematical life," we normally do not bother explicitly saying what axioms we allow. As trained mathematicians, we recognize a valid proof when we see it and that's good enough for ordinary mathematical work. However, once you start trying to make statements about the impossibility of proving something, you can't get away with this sloppiness any more. You have to specify some set of axioms. Otherwise, there's no way to stop someone from taking "Turing machine 314159 solves My Favorite Problem" as an axiom.

O.K., so we have to specify some set of axioms. So how about ZFC, everybody's favorite whipping boy? That's a reasonable choice. However, as Andreas Blass pointed out in the comments, once you fix a set of axioms, then something like Goldstern's answer works. Now of course Goldstern's answer seems like a cheat. We wouldn't want to be accused of cheating, would we? Of course not. So, let's choose some other set of axioms. Whoops, no good…any reasonable set of axioms will leave some statement undecided, and Goldstern's cheat slips in the door. Dang it, there must be some way to phrase the question so that no cheating answer is possible. Mustn't there?

I don't think there is. As soon as you find yourself dealing with one specific statement such as "Turing machine 314159 solves My Favorite Problem," there's no way to avoid this kind of "cheat" when it comes to saying what it means for it to be impossible to prove it.

The only other avenue I see for you to try is this: Instead of saying that for every n it is impossible to prove that Turing machine n solves My Favorite Problem, you could try asking this:

There is no algorithm that, given n, decides whether Turing machine n solves My Favorite Problem.

Ah! That sounds more like it! Or does it? The trouble is, you can choose just about any algorithmic problem you like and it will be undecidable whether Turing machine n solves it. This is Rice's theorem. So this isn't really what you want either, since it has nothing to do with the inscrutable structure of My Favorite Problem; it rather has to do with the inscrutability of Turing machines in general.

At the end of the day, I think you just have to accept the uncomfortable fact that there is no way to get an answer that is better than the ones that have been proposed already. There are problems for which we can currently prove that there is an algorithm but we don't know any algorithm yet, and we can prove that there is no general procedure for turning existence proofs of algorithms into constructive proofs, but if you try to push harder and ask for an explicit example of My Favorite Problem for which an algorithm is intrinsically unknowable, then you're inevitably going to run into the above difficulties.

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Very nice. Thanks! –  Thomas Apr 8 '13 at 9:57
    
Thank you indeed for having laid out very clearly what are the issues involved. –  Loïc Teyssier Apr 8 '13 at 11:27

The Robertson-Seymour graph minor theorem shows that membership in any given minor-closed family of graphs can be checked in polynomial time. It even shows this can be done in $\mathcal{O}(n^3)$ time, but gives no bound on the size of the hidden constant.

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In fact, I believe there are fairly simple problems with no known algorithms: for example, can this graph be embedded into three-dimensional space so that it does not contain any knotted cycles? (There were certainly no explicit algorithms known for this a few years ago.) –  Peter Shor Apr 6 '13 at 16:25

Take any open question Q. ("Is the Riemann hypothesis true"? "Is P=NP?" etc)

There is an algorithm that will answer Q correctly.

It is trivial to prove (in classical logic) that one of the following programs works:

  • Program 1: Print "yes".
  • Program 2: Print "no".
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And one day,we will have proven RH,and constructed a algorithm. So,the question is really ambiguous –  XL _at_China Apr 5 '13 at 15:27
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My answer is so nonconstructive that not only is the algorithm given in a "non-constructive, purely existential" way, but even the task the algorithm has to perform is given in a "non-constructive, purely existential" way. (I think we agree that there will always be open questions.) –  Goldstern Apr 5 '13 at 16:00
    
I prefer Sapir's answer. This is not an algorithmic task. For example, replace P=NP by CH. Then it depends on your ambient model of ZFC which machine works. –  Benjamin Steinberg Apr 5 '13 at 18:32
    
@Goldstern,what do you think about the dovetailed computation? –  XL _at_China Apr 6 '13 at 11:21
    
Thank you for your answer. I did remove my previous comment because I did not understand fully both what you meant and what I sought :) I hope my edit correct that. –  Loïc Teyssier Apr 6 '13 at 14:42

I've already written an answer but in light of the "final edit" to the question, I believe that something more can be said. In particular, I want to address the following comment.

Actually MyFavouriteProblem concerns effective computability of holomorphic dynamical systems satisfying MyFavoriteProperty, which were known to exists by some "abstract" argument and for which I gave a more "constructive" proof. It is doubtlessly a practical improvement (my PC can do the computing), but I wanted to know if from a theoretical point of view some progress really has been made.

If I read this correctly, the previous proof showed that the problem was algorithmically decidable (in some model of real computation that allows manipulation of infinite-precision real numbers). If this is the case then there are (at least) two ways in which your new algorithm can be regarded as making "theoretical" progress.

  1. It could be a situation like the Robertson–Seymour graph minor theorem mentioned in Noah Stein's answer. Here there is a whole family of computational problems, and the theory proves that every member of the family is solvable in polynomial time; however, for any particular member of the family, there may be no known polynomial-time algorithm, since the theory does not yield an effective algorithm for finding the polynomial-time algorithm. Finding an explicit algorithm for a particular member of the family for which an explicit algorithm was not known would count as theoretical progress in anyone's book. This sort of situation is rare, so I suspect that your case is not of this type, but if it is, it would be very interesting!

  2. The earlier proof might have yielded an explicit algorithm in principle if you were willing to unpack the argument, but the computational complexity would have been horrendous. Your proof yields a vastly superior computational complexity. Some old-style mathematicians might not regard this as "theoretical" progress, but I think the dominant viewpoint nowadays is that improvements in computational complexity count as theoretical progress and not just practical progress. Indeed, computational improvements often provide a convenient yardstick for measuring theoretical progress: Your new proof surely uses new ideas, and the fact that it gives a better algorithm is a clear demonstration of the value of the new ideas. As other examples, Gowers's new proof of Szemerédi's theorem or Folsom, Kent, and Ono's new formula for the partition function come to mind. The value of the new ideas was obvious to experts, but even non-experts could tell that something major had happened because of the vast computational improvements that were obtained.

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Without claiming anything near as importamt as the examples you mention, my situation indeed pertains to the second situation you depict. Thanks again for the time you spent on my logic-redated musings, in particular for the clarity and accessiibility of your answers. –  Loïc Teyssier Apr 8 '13 at 16:12

The answer has to be No,since we can enumerate all algorithm.

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But you can't decide if your enumerated algorithms halt, let alone if they are equivalent to some given algorithm. –  Thomas Apr 5 '13 at 14:55
    
That was indeed the problem I raised, being aware of the enumerability of algorithms. –  Loïc Teyssier Apr 6 '13 at 9:05
    
@Kahle and Loice,We may dovetail the computation,Is this not an algorithm to perform the required task? –  XL _at_China Apr 6 '13 at 10:50
    
Actually,I do not understand what a Turing Machine with finite language means.And a Turing Machine usually has unlimited tape which is equivalent to memory of modern digital computer. –  XL _at_China Apr 6 '13 at 13:10
    
Is «finite alphabet» better? I amended my post accordingly. Yet I don't agree that our «modern digital computer», or any other actual computer ever, for that matter, has infinite memory. –  Loïc Teyssier Apr 6 '13 at 14:45

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