I was curious as to whether given any problem, lets say:

$x^2-1 = 0$

There exists a function that given this set of symbols as input return the exact set of symbols contained in the answer. In this case $x=-1,x=1$

Surely, given that the process of getting from the problem's set of symbols to the solution's is not random, there exists a function that does what I said above.

The method I came up with involves mapping the symbols to numbers as follows:

$ x \to 1$

$ uppower \to 2$

$ downpower \to 3$

$ - \to 4$

$ = \to 5$

$ , \to 6$

Then for the numbers 0 - 9 $(n\to n+7)$. And that should do for now.

Then to start approximating the function we can express the problem (using Godel encoding) with a unique number. In this case:

$2^1*3^2*5^9*7^3*11^4*13^8 = x_1$

And we can express the solution as:

$2^1*3^5*5^4*7^8*11^6*13^1*17^5*19^8 = y_1$

So $a = y_1$ and our first approximation for our universal function is:

$a = y$

Now we can continue with a second data point and solve it with the previous simultaneously - e.g $(bx_1+a = y_1$ and $ bx_2+a = y_2)$. Keep adding points and solving in the same style (for the amount of points you have $p$ solve that many polynomials with the highest power as $p-1$). Eventually it should give an accurate answer given any quadratic to solve.

If true then you could keep adding points outside the scope of quadratics to any non-random progression from one sequence of symbols to another, until it became a universal function.

If false, then why? I know there are an infinite amount of functions for any given set of points, but surely given enough data within a particular range this method is likely to get pretty close to it, and if it doesn't then the process is probably random. Or maybe i am wrong in my assumption that there should be a function mapping a non-random symbolic problem to its solution. Or maybe something in my method limits it. Either or another way I would be most interested to know, thanks, Reuben.

P.S. I didn't know what to tag this in so any suggestions would be welcome!