Non-constructive proofs of decidability?

When I teach computability, I usually use the following example to illustrate the point.

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

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

Nevertheless, we can prove that the function is computable as follows. Either there are arbitrarily long strings of $1$ in $\pi$ or there is a longest string of $1$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 $n\lt N$ and value $0$ for $n\geq N$, 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. In this sense, this example has a resemblence to Gerhard's examples.

The Robertson–Seymour theorem implies that every minor-closed family $F$ of finite graphs is decidable in time $O(n^3)$. However, it does not provide an explicit algorithm until one supplies an explicit finite list of forbidden minors that characterize $F$; the proof that such a list always exists is non-constructive.

There is the standard example involving Fermat's Last Theorem, except that we now have a good idea what the set is. So let's replace it with "the smallest positive integer n which is a multiple of 4 and for which no Hadamard matrix of order n exists, or 1 if Hadamard matrices of all possible orders exist." This defines a singleton set, which is decidable. You could argue that in principle it is constructive, whereas I would argue that since we still don't know maximal determinants for small orders less than 100, you and your putative great-grandchildren will not see a value for n, so you will have a hard time showing to me that a construction based on this definition exists, as there is no guarantee of termination of the construction.

Alternatively, any finite set which is arrived at by nonconstructive means (E.g. encodings of counterexamples to Frankl's union-closed families conjecture, where the presumed proof that there are only finitely many is nonconstructive) should count as an example.

Better answers will arise once a good notion of nonconstructive has been specified. As to such a notion, I'll leave the philosophical wrangling to others.

Gerhard "Ask Me About System Design" Paseman, 2010.02.10