Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The graph reconstruction conjecture claims that (barring trivial examples) a graph on n vertices is determined (up to isomorphism) by its collection of (n-1)-vertex induced subgraphs (again up to isomorphism).

The way it is phrased ("reconstruction") suggests that a proof of the conjecture would be a procedure, indeed an algorithm, that takes the collection of subgraphs and then ingeniously "builds" the original graph from these.

But based on some experience with a related conjecture (the vertex-switching reconstruction conjecture), I am led to wonder whether this is something that is simply true "by accident". By this I mean that it is something that is just overwhelmingly unlikely to be false ... there would need to be a massive coincidence for two non-isomorphic graphs to have the same "deck" (as the collection of (n-1)-vertex induced subgraphs is usually called). In other words, the only reason for the statement to be true is that it "just happens" to not be false.

Of course, this means that it could never actually be proved.. and therefore it would be a very poor choice of problem to work on!

My question (at last) is whether anyone has either formalized this concept - results that can't be proved or disproved, not because they are formally undecidable, but just because they are "true by accident" - or at least discussed it with more sophistication than I can muster.

EDIT: Apologies for the delay in responding and thanks to everyone who contributed thoughtfully to the rather vague question. I have accepted Gil Kalai's answer because he most accurately guessed my intention in asking the question.

I should probably not have used the words "formally unprovable" mostly because I don't really have a deep understanding of formal logic and while some of the "logical foundations" answers contained interesting ideas, that was not really what I was trying to get at.

What I was really trying to get at is that some assertions / conjectures seem to me to be making a highly non-obvious statement about combinatorial objects, the truth of which depends on some fundamental structural understanding that we currently lack. Other assertions / conjectures seem, again, to me, to just be saying something that we would simply expect to be true "by chance" and that we would really be astonished if it were false.

Here are a few unproved statements all of which I believe to be true: some of them I think should reflect structure and others just seem to be "by chance" (which is which I will answer later, if anyone is still interested in this topic).

(1) Every projective plane has prime power order

(2) Every non-desarguesian projective plane contains a Fano subplane

(3) The graph reconstruction conjecture

(4) Every vertex-transitive cubic graph has a hamilton cycle (except Petersen, Coxeter and two related graphs)

(5) Every 4-regular graph with a hamilton cycle has a second one

Certainly there is a significant chance that I am wrong, and that something that appears accidental will eventually be revealed to be a deep structural theorem when viewed in exactly the right way. However I have to choose what to work on (as do we all) and one of the things I use to decide what NOT to work on is whether I believe the statement says something real or accidental.

Another aspect of Gil's answer that I liked was the idea of considering a "finite version" of each statement: let S(n) be the statement that "all non-desarguesian projective planes of order at most n have a Fano subplane". Then suppose that all the S(n) are true, and that for any particular n, we can find a proof - in the worst case, "simply" enumerate all the projective planes of order n and check each for a Fano subplane. But suppose that the length of the shortest possible proof of S(n) tends to infinity as n tends to infinity - essentially there is NO OTHER proof than checking all the examples. Then we could never make a finite length proof covering all n. This is roughly what I would mean by "true by accident".

More comments welcome and thanks for letting me ramble!

share|improve this question
2  
Can't one ask the same question about any hard conjecture? –  Gjergji Zaimi Aug 25 '11 at 11:44
3  
It seems unlikely to me that the question "Has anyone has formalized this concept?" will produce an interesting answers. Probably this is not the case and the answer is simply "No: no one has formalized that concept". On the other hand, asking for a list of examples of similar-sounding conjectures (i.e. things that are true by accident - Goldbach's conjecture being the prototypical example) might be more fruitful. If you decide to change the question in the way I suggest, then please don't forget to make it community-wiki. –  André Henriques Aug 25 '11 at 11:46
4  
Assuming the conjecture is true, one could computably reconstruct the original graph from the subgraphs by brute force - enumerate all the graphs on $n$ vertices and then eliminate them one at a time if they do not have the correct collection of subgraphs. So the truth of the conjecture would imply the existence of the procedure, but that procedure only works because the conjecture is already known to be true, and the procedure itself would exist even if the conjecture is false, although in that case the procedure doesn't do what it is supposed to do. –  Carl Mummert Aug 25 '11 at 11:53
7  
Perhaps, when you try to formalize the concept "true by accident", you arrive at the concept "true but formally undecidable"? At least I am having difficulty seeing what the distinction between the two might be. –  Mark Grant Aug 25 '11 at 12:18
3  
What sort of graphs are we talking about? Directed, with loops, with multiple edges? I am asking out of idle curiosity. –  Andrej Bauer Aug 25 '11 at 13:15

6 Answers 6

up vote 12 down vote accepted

This is a very interesting (yet rather vague) question. Most answers were in the direction of mathematical logic but I am not sure this is the only (or even the most appropriate) way to think about it. The notion of coincidence is by itself very complicated. (See http://en.wikipedia.org/wiki/Coincidence ). One way to put it on rigurous grounds is using probabilistic/statistical framework. Indeed, as Timothy mentioned it is sometimes possible to give a probabilistic heuristic in support of some mathematical statement. But its is notorious statistical problem to try to determine aposteriori if some events represent a coincidence.

I am not sure that (as the OP assumes) if a statement is "true by accident" it implies that it can never be proved. Also I am not sure (as implied by most answers) that "can never be proved" should be interpreted as "does not follow from the axioms". It can also refers to situations where the statement admits a proof, but the proof is also "accidental" as the original statement is, so it is unlikely to be found in the systematic way mathematics is developed.

In a sense (as mentioned in quid's answer), the notion of "true by accident" is related to mathematics psychology. It is more related to the way we precieve mathematical truths than to some objective facts about them.

Regarding the reconstruction conjecture. Note that we can ask if the conjecture is true for graphs with at most million vertices. Here, if true it is certainly provable. So the logic issues disappear but the main issue of the question remains. (We can replace the logic distinctions by computational complexity distinctions. But still I am not sure this will cpature the essence of the question.) There is a weaker form of the conjecture called the edge reconstruction conjecture (same problem but you delete edges rather than vertices) where much is known. There is a very conceptual proof that every graph with n vertices and more than nlogn edges is edge-reconstructible. So this gives some support to the feeling that maybe vertex reconstruction can also be dealt with.

Finally I am not aware of a heuristic argument that "there would need to be a massive coincidence for two non-isomorphic graphs to have the same 'deck'" as the OP suggested. (Coming up with a convincing such heuristic would be intereting.) It is known that various graph invariants must have the same value on such two graphs.

share|improve this answer
    
I have accepted this as best matching the spirit of my question; additional comments are in the "Edit" above. –  Gordon Royle Sep 1 '11 at 2:05
4  
I'm reminded of Problem 3c in Chapter 3 of Richard Stanley's Enumerative Combinatorics: "Let $f(n)$ be the number of non-isomorphic $n$-element posets. Let $P$ denote the statement that infinitely many values of $f(n)$ are palindromes when written in base ten. Show that $P$ cannot be proved or disproved in Zermelo-Fraenkel set theory." The bit about formal unprovability is, in my opinion, a red herring. I think Stanley is just saying that he thinks that if $P$ is false (which it surely is), then it is false by accident. –  Timothy Chow Nov 4 '11 at 14:19
    
This is a good example. We do not expect "hidden structures" in the sequence f(n), or, more importantly in the sequence of primes, or even in the writings of Shakespeare, but we dont expect that we will be able to refute their existence either. –  Gil Kalai Nov 4 '11 at 15:17

Apart from your specific example, the idea of truth-by-accident has been studied in the context of formal first-order languages, which includes the language of graph theory, and in his dissertation, Kurt Gödel proved that the statements that happen to be true in all models of a first order theory $T$ are exactly the statements that are provable in $T$. This is his famous completeness theorem.

Thus, any statement expressible in the first-order theory of groups that happens to be true in all groups will be provable from the group axioms, and any statement expressible in the first-order statement of graphs that happens to be true in all graphs will be provable from the axioms of graph theory.

Your statement, however, does not seem to be expressible directly in the language of graph theory, since it also uses the concept of cardinality and of subgraphs, so the completeness theorem does not apply directly to it for the language of graphs. Rather, it is a statement of number theory, and the relevant models for this case would include all the standard and nonstandard models of arithmetic.

So the relevant conclusion would be that if the statement were not provable in the first-order Peano's axioms PA, then there is a nonstandard model of arithmetic having a bad (pseudo)finite graph.

But the particular form of the statement means that it has complexity $\Pi^0_1$, which means it is a universal statement quantifying over the natural numbers, and if any such statement is independent of PA, then it is true, because if it is true in any model, then because the standard model is an initial segment of all the others, it follows that it must be true in the standard model and hence true. This level of complexity is the same complexity as many of the interesting independent statements, including consistency statements.


Incidentally, this seems to be my 500th answer on mathoverflow. It's been a lot of fun, and I've surely learned a lot of mathematics!

share|improve this answer
    
I think you mean 'true' rather than 'provable in ZF' since Con(ZF) is a (hopefully true) $\Pi^0_1$ statement that is not provable in ZF (again, hopefully). –  François G. Dorais Aug 25 '11 at 12:57
    
Yes, I'll edit. –  Joel David Hamkins Aug 25 '11 at 12:58
1  
@Joel: I confess that I don't understand how this answers the OP's question. Are you saying that there aren't any statements that are true by accident? –  Timothy Chow Aug 25 '11 at 20:08
5  
Congrats on your 500th! –  François G. Dorais Aug 25 '11 at 20:13
1  
Timothy, yes, I am arguing that for statements in a first-order language, the (profound) fact that true-in-all-models is the same as provable means that there is no accidental truth. A statement in the language of graph theory is true in all graphs if and only if it is provable from the axioms of a graph. So we don't need another distinction besides provable, negated-provable, and independent. In the case of statements of arithmetic, such a view is intimately connected with a considerations of nonstandard models of arithmetic, but I maintain that it still answers the question. –  Joel David Hamkins Aug 25 '11 at 21:48

Yes, this is what constructivism is all about! In intuitionistic logic, the law of excluded middle doesn't generally hold, so it is not always possible to derive $A$ (i.e. $A$ is true) from $\lnot\lnot A$ (i.e. $A$ is not false).

The particular case you're considering is a form of Markov's Principle, which can be worded as if it is not the case that there is no example, then an example does exist. Symbolically, the rule is $$\lnot\forall x\lnot A(x) \to \exists x A(x),$$ where $A(x)$ is required to be decidable: $\forall x(A(x) \lor \lnot A(x))$. In constructive mathematics, existence is very strong — it is not acceptable to merely show that there must be an example, one needs to actually produce an example in some way or another. Markov's principle says that showing that there must be an example is enough to prove existence. Thus this principle is not generally accepted by most schools of constructivism, except in limited instances.

share|improve this answer
2  
What you describe is not Markov’s rule, but Markov’s principle. Markov’s rule is the corresponding derivation rule which states that when $\neg\forall x\neg A(x)$ is provable, then $\exists x\,A(x)$ is provable (under appropriate conditions on $A$). All usual constructive theories (for example, Heyting arithmetic) are closed under this rule, even though they typically do not prove Markov’s principle. Therefore, even in constructive setting, it would suffice to prove that there is no counterexample in order to establish the conjecture. –  Emil Jeřábek Aug 25 '11 at 12:22
    
You're absolutely right, I just fixed it. –  François G. Dorais Aug 25 '11 at 12:39

Probably the closest thing to what you're looking for is Chaitin's proof of the incompleteness theorem, which shows that for any formal system $S$, there is a constant $L$ such that the statement "$K(s) > L$" is unprovable in $S$ for all strings $s$ (here $K$ denotes Kolmogorov complexity). The vast majority of such statements are true "at random" because a random string will have high Kolmogorov complexity.

However, you didn't ask whether there exists a family of statements that can be regarded as being "true by accident"; you asked whether a specific statement (the graph reconstruction conjecture) can be regarded as being "true by accident." So Chaitin's incompleteness theorem doesn't quite address your question as stated.

It is certainly possible, in some situations, to construct a heuristic probabilistic model that predicts that certain things ought to be true just for "random reasons." Perhaps the most famous example is Cramér's random model for the primes, which can be used to give heuristic "proofs" of various number-theoretic conjectures; e.g., one can use the model to predict that there will be only finitely many primes with such-and-such a property, because the probability that a prime $p$ has the property decreases rapidly to zero as $p\to\infty$. If a conjecture is predicted by such a model, and also "happens" to be unprovable in your favorite axiomatic system for mathematics, then it might be tempting to say that it is "true by accident."

This kind of thing could happen, but we don't know of any good examples involving "naturally occurring" mathematical conjectures. So it's all pure speculation at this point. Certainly we have no objective evidence that any particular mathematical conjecture isn't worth working on for this reason. I think it would be interesting, though, if you could develop a heuristic probabilistic model for graph theory, in the spirit of Cramér's model, that could "predict" various well-known graph-theoretic conjectures.

share|improve this answer

The statement in question can be formalized in the language of Peano Arithmetic, and I will treat it as a statement in that language. A similar analysis works for any effective theory stronger than PA, such as ZFC.

Consider the set of all sentences in the language of PA; define an order relation $R$ so that $\phi \mathbin{R} \psi$ if $\phi \to \psi$ is provable in PA. This gives a pre-order; if we perform the usual equivalence class construction then the resulting algebra is a partial order called a Lindenbaum algebra (*).

Because the graph reconstruction conjecture corresponds to a sentence $G$ in PA, it corresponds to a particular node in this algebra.

  • If $G$ is provable in PA, then $G$ corresponds to the bottom element of the algebra
  • If $G$ is false, it corresponds to the top node of the algebra, but in this case we're not very worried about its provability
  • Otherwise, $G$ corresponds to some intermediate node of the algebra. In that case, we cannot prove $G$ from PA, but we can prove $G$ by assuming PA plus any axiom either in the equivalence class of sentences that forms $G$'s node or in the equivalence class of any node higher than $G$'s node.

In every case, unless $G$ is false, $G$ is amenable to proof, but the proof will have to assume axioms that are strong enough to prove the desired conclusion. There is no sentence which could "never actually be proved", although there are plenty of sentences that cannot be proved in PA, and false sentences can only be proved from false axioms. The question is simply which axioms are required to prove a particular sentence.


*: Traditionally, a "Lindenbaum algebra" or "Lindenbaum–Tarski algebra" should be defined with the dual ordering of the ordering I use. But the ordering in which $0=1$ corresponds to the top of the algebra matches better with the diagrams we create to illustrate relationships between different axiom systems, such as 1. People also use the reverse ordering in the context of set theory, where large cardinal axioms are sorted by consistency strength, e.g. 2.

share|improve this answer
1  
(I think you mean '≥' and not '>'.) –  François G. Dorais Aug 25 '11 at 12:41
1  
For the usual definition of the Lindenbaum algebra, you would mean neither $>$ nor $\geq$ but $\leq$. But your use of "bottom", "top", and "higher" indicates that you really intend $\geq$ and thus the dual of the usual definition. –  Andreas Blass Aug 25 '11 at 14:14
    
I do mean the algebra which puts $0=1$ at the top. I was only using '$>$' as an arbitrary relation symbol. I changed it to an $R$ to avoid entirely any confusion about whether it should have a horizontal line. –  Carl Mummert Aug 25 '11 at 14:23
1  
(The reason for making falsity the bottom in the standard definition is, of course, that once you really treat it as an algebra rather than just order, it is quite inconvenient to have $[\phi]\land[\psi]=[\phi\lor\psi]$ and $[\phi]\lor[\psi]=[\phi\land\psi]$, where $[\phi]$ denotes the equivalence class of a formula $\phi$.) –  Emil Jeřábek Aug 25 '11 at 14:40
    
@Emil: Yes; it's just the order that is of interest here. The algebra itself is often uninteresting as an algebra, as you explained in mathoverflow.net/questions/65851/lindenbaum-algebras-and-models/… . The point of the diagrams I linked at the end of my answer is to show interesting suborders obtained by picking a few recognizable nodes out of the order. –  Carl Mummert Aug 25 '11 at 14:47

This is a take on this question of a form different from the other answers, along the lines of the comments of Gjergji Zaimi and André Henriques.

No doubt in present day mathematics there are some conjectures or perhaps even results that have an 'accidental' feel, while others do not have this feel.

However, it seems to me that to a certain and I believe considerable extent this is a subjective impression, or perhaps subjective is not really the right term and I should rather say, this impression is informed by the current state of mathematics.

I do not have really good and precise historical examples at hand, but I think it is true that for example certain results on diophantine equations (finiteness results of solutions, say) a long time ago had a much more accidental feel. But now with the developement of Arithmetic Geometry (eg Mordell conj.) some of them are much better and conceptually understood and now feel natural or at least not accidental anymore.

Perhaps some of the results and conjectures that today look accidental will at some point in the future be natural consequence of theories yet to be developped. Indeed, I believe it to be a quite common pattern of progress in mathematics that developements begin with concrete and isolated results and questions, and then theories follow that explain the 'accidents' and 'miracles'.

Goldbach's conjecture got mentioned and gets mentioned frequently as accidental. Let us look at something not too far away.

The prime numbers contain infinitely many 3-term arithmetic progressions (van der Corput, 1930s). Accident, yes or no? Perhaps one could have thought so, when it was proved, but today the situation is different and it is widely conjectured (very recent results of Sanders achieve this for sets of just a slightly larger density) that any set of positive integers with the density of the primes has this property.

Who knows where the Theory of Set Addition (more generally Additive Combinatorics/Number Theory) will stand in some decades or centuries? Perhaps, then Goldbach conjecture will be a corollary of some natural theory.

In brief, I believe 'true by accident' (in the informal way I understand it, which seems not that far from the questioners intent) is to a considerable extent a time-dependent notion; it seems thus difficult for me to imagine that its spirit can be captured in a formal (time-independent) theory.

share|improve this answer
    
Maybe I am misunderstanding the OP's take on these things, but the way I read it, the fact that any set of positive integers with the density of the primes should have the same property is precisely the sort of "accident" the OP was talking about: given how ubiquitous the primes are, a suitable random model would tell you that the statement has a very low chance of failing. In other words, it is not true for a special reason, it is true because it has no reason to be false. –  Alex B. Aug 25 '11 at 18:14
1  
@Alex: However, the OP also says that accidental truths cannot be proved. This puts an additional spin on the topic. I think quid is right that one's feeling of what sorts of things "cannot be proved" are informed mainly by one's sense of what current mathematical technology is or is not capable of achieving. –  Timothy Chow Aug 25 '11 at 18:22
1  
@Timothy Yes, I wasn't trying to address that part at all, since the wonderful answers already given address the provability issue very nicely. I was just joining quid in trying to understand what exactly the OP has in mind when he talks about accidents. What I am trying to say is that even though a theorem is provable, it can feel like it's true for no special reason other than the absence of a reason to be false. And the example quid gave seems to me to exemplify exactly this situation. –  Alex B. Aug 25 '11 at 18:46
1  
"it is true because it has no reason to be false." Zagier had a similar comment, say about why $\pi$ is normal; if it isn't, there must be some deep reason, but if it is normal, the reason is just that it's random. For the other part, you can model many things by random sense (like $\mu(n)$ as random $0,\pm 1$ coefficients of a Dirichlet series, for RH, or Cramer's model of primes, for gaps), and the question should be whether the general randomness persists in the problem at hand. Another similar is Goldbach (partitio numerorum for general) as said. –  Junkie Aug 26 '11 at 1:26
1  
Junkie, thank you for the examples. In particular, the quote on $\pi$, which I in principle knew but forgot. Let us look a bit more than a century back on the transcendence of certain numbers (not pi itself, but in Hilbert 7th problem). Perhaps somebody could have said...since there would need to be a reason for them to be algebraic, and there is not, they will be transcendental, but no proof in sight (if I remeber well Hilbert was quite pessimistic in his prediction when 7th prob. would be solved.) Yet, 'suddenly' came Gelfond--Schneider and a 'reason' for transcendence of these numbers. –  quid Aug 26 '11 at 14:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.