Sunday, December 22, 2013

Solution to the polynomial question

In answer to the question that was bugging me in my last post, It turns out that every function operating on values [0..N-1] can be represented as a polynomial.  The proof is that you can generate the polynomial from the function's truth-table. (I'm not sure if that's a proof that a real mathematician would accept, but it works for me.)

The basic building block for this is the following expression, which returns a value of f(a) for x = a, and a value of 0 for all other values of x:

To build the polynomial for a function that takes only one argument, you sum up the polynomials for all possible values of x. So, you can say:

From there, it is just a hop, skip and a jump to functions with multiple arguments. For f(x,y), you can do this hideous thing:

Not necessarily practical, but it does prove the point that every possible function (mod N) can be represented as a polynomial.

(Note added 12/23/2013--I just realized this only works if N is prime, otherwise we don't have a modular multiplicative inverse for all (a-i), (i-j), (h-j). That suggests that for non-prime values of N there might be some functions that couldn't be expressed as polynomials, which is interesting.)

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