space and games

December 9, 2008

What Makes a Hint Good?

Filed under: General — Peter de Blanc @ 1:40 am

Nick Hay, Marcello and I discussed this question a while ago: if you had a halting oracle, how could you use it to help you prove a theorem, such as the Riemann Hypothesis? Let’s say you are only allowed to ask one question; you get one bit of information.

Of course, you might simply ask if the Riemann Hypothesis is true. If you trust your oracle, this may be good enough for your purposes. But let’s suppose what we really care about is the proof. Maybe we want to win a prize for our proof, and they won’t accept a proof by oracle.

Clearly, a complete proof of the Riemann Hypothesis is longer than one bit, so we can’t ask the oracle to output the proof for us. So what can we do? Well, we could ask it for a hint. Maybe we have some conjecture (which we can call Conjecture A) that could be useful in proving the RH. But we don’t know if A is true, nor do we know if proving A would actually help us prove the RH. So we could ask the oracle if it’s worth our time to try to prove A.

What we really want to know is if we would save time by trying to prove A, and then trying to use A to prove the RH, compared to some other strategy. To figure out how much time it would take, we need some sort of model of how we write proofs. Here’s a really simple model: “mathematicians write proofs by brute force; they look at every possible proof, from short to long, until finding one which proves their goal hypothesis, at which point they halt.”

Let’s suppose this model is true (lol). Let’s define the “difficulty” of a theorem as the length of its shortest proof. If our proofs are written in binary, then to prove a theorem of difficulty D, it takes time t = O(2D).

Let’s say D(RH) is the difficulty of the Riemann Hypothesis, D(A) is the difficulty of Conjecture A, and D(RH | A) is the difficulty of proving the RH once we have proved A. Then a mathematician who attempted to prove A before proving the RH would take time tA,RH ~ 2D(A) + 2D(RH | A), while a mathematician who attempted to prove the RH directly would take time tRH ~ 2D(RH).

So we could ask the oracle:

True or False? 2D(A) + 2D(RH | A) < 2D(RH)

If the oracle returns True, then we would start thinking about how to prove A.

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