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!



Does there exist a function (however complex) which given an input in the form of any problem which can be solved in a rigorous and non-random way can return the solution to that problem.
Your question seems to concern the issue of the computability of solutions of computable functions, and the larger context for such a question is the subject known as [computable analysis](http://en.wikipedia.org/wiki/Computable_analysis). 


Carl Mummert has [a very nice blog post concerning the following theorems](http://m6c.org/w/2012/02/computable-roots/), which I believe lie at the heart of your question.



> 
> 
> > 
> > Here are several interesting results from computable analysis:
> > 
> > 
> > **Theorem 1.** If $f$ is a computable function from $\mathbb{R}$ to $\mathbb{R}$ and $a$ is an isolated root of $f$, then α is computable.
> > 
> > 
> > **Corollary 2.** If $p(x)$ is a polynomial over $\mathbb{R}$ with computable coefficients, then every root of $p(x)$ is computable.
> > 
> > 
> > **Theorem 3.** There is a effective closed subset of $\mathbb{R}$ which is nonempty (in fact, uncountable) but which has no computable point.
> > 
> > 
> > **Theorem 4.** There is a computable function from $\mathbb{R}$ to $\mathbb{R}$ which has uncountably many roots but no computable roots.
> > 
> > 
> > 
> 
> 
> 


But your question inquires not for an algorithm for each function separately, but a uniform algorithm working with all equations to be solved. Here, there are various non-uniformity results that one can mention. 


For example, by [the MRDP theorem](http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem), there can be no computable algorithm that determines whether a given integer polynomial equation in several variables has a solution in the integers. 


But meanwhile, there of course can be a computational procedure that maps any given Diophantine equation to an integer solution of it, when there is a solution, for one may simply undertake exhaustive search. 


