What does “Undecidable” mean, anyway

D'oh. I missed that.

I do find it sometimes can lead to less verbose code even in OOP, because it makes logic more focused and DRY. It certainly can greatly reduce the amount of bugs in large codebases.

The set of all possible mapping from all possible finite strings to booleans is definitely *not* countable.

What I (and the article) mean by a "function" `f : string -> boolean` here is any arbitrary assignment of a single boolean value to every possible string. Let's consider two examples:

1. Some "expressible" function like "f(s) returns True if the length of s is odd, and returns False otherwise".

2. Some "random" function where some magical process has listed out every possible string and then, for each string, flipped a coin and assigned that string True if the coin came up heads, and False if it came up tails and wrote down all the results in a infinite table and called that the function f.

The first type of "expressible" function is the type that we most commonly encounter and that we can implement as programs (i.e., a list of finitely many instructions to go from string to boolean).

Nearly all of the second type of function -- the "random" ones -- cannot be expressible using a program. The only way to capture the function's behavior is to write down the infinite look-up table that assigns each string a boolean value.

Now you are probably asking, "How do you know that the infinite tables cannot be expressed using some program?" and the answer is because there are too many possible infinite tables.

To give you an intuition for this, consider all the way we can assign boolean values to the strings "a", "b", and "c": there are 2^3 = 8. For any finite set X of n strings there will be 2^n possible tables that assign a boolean value to each string in X. The critical thing here is that 2^n is always strictly larger than n for all n >= 1.

This fact that there are more tables mapping strings to boolean than strings still holds even when there are infinitely many strings. What exactly we mean by "more" here is what Cantor and others developed. They said that a set A has more things than a set B if you consider all possible ways you can pair a thing from A with a thing from B there will always be things in A that are left over (i.e., not paired with anything from B).

Cantor's diagonalization argument applied here is the following: let S be the set of all finite strings and F be the set of all functions/tables that assigns a boolean to each element in S (this is sometimes written F = 2^S). Now suppose F was countable. By definition, that would mean that there is a pairing that assigns each natural number to exactly one function from F with none left over. The set S of finite strings is also countable so there is also a pairing from natural numbers to all elements of S. This means we can pair each element of S with exactly one element of F by looking up the natural number n assigned to s \in S and pairing s with the element f \in F that was also assigned to n. Crucially, what assuming the countability of F means is that if you give me a string s then there is always single f_s that is paired with s. Conversely, if you give me an f \in F there must be exactly one string s_f that is paired with that f.

We are going to construct a new function f' that is not in F. The way we do this is by defining f'(s) = not f_s(s). That is, f'(s) takes the input string s, looks up the function f_s that is paired with s, calculates the value of f_s(s) then flips its output.

Now we can argue that f' cannot be in F since it is different to every other function in F. Why? Well, suppose f' was actually some f \in F then since F is countable we know it must have some paired string s, that is, f' = f_s for some string s. Now if we look at the value of f'(s) it must be the same as f_s(s) since f' = f_s. But also f'(s) = not f_s(s) by the way we defined f' so we get that f_s(s) = not f_s(s) which is a contradiction. Therefore f' cannot be in F and thus our premise that F was countable must be wrong.

Another way to see this is that, by construction, f' is different to every other function f in F, specifically on the input value s that is paired with f.

Thus, the set F of all functions from strings to boolean must be uncountable.

Right.

But brute force solvers for bcrypt and chess are already "free" in the sense of thought, cleverness, insight. We already have the "everything" algorithm: iterate through all possible solutions in O(2^n) time and pick the best one.

A halting solver gains us nothing in these scenarios.

(The code for scoring a solution is the same code that would need to go into the halting solver. For bcrypt it's checking if the input matches, for chess it's the number of turns until checkmate.)