Can you tell if you have escaped from a recursive definition?

Question

Most people define a function, f(n) on N recursively. I think I can calculate f(n) without dealing with f(n-r) for any 0 < r < n. How do I know that my method isn't still going through the same calculations needed to find f(n-1) (or whatever previous terms are required to find f(n) recursively) -- ?

1. If my method takes many fewer calculations than the recursive way of calculating it does that show that I am not relying on f(n-r) for any 0 < r < n? What would "many fewer" have to mean for this to be significant?

2. The number of calculations my method takes still depends on n, just like the recursive way of calculating f(n), does that alone mean that the methods are pretty much the same?

3. If my method takes the same number (or more) calculations than recursive way of calculating f(n) is there any other way of telling if my method is not, in some way, duplicating the recursive way of calculating f(n)?

Examples:

f(n) is recursively defined to be f(n) = f(n-1) + 1 and f(1) = 1 Then f(n) = n. Clearly, f(n) = n is a much faster way to find f(2876) rather than counting up from 1.

f(n) is recursively defined to be f(n) = f(n-1) + f(n-2) This is a linear recurrence and has a closed-form solution. $F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}={{\varphi^n-(-1/\varphi)^{n}} \over {\sqrt 5}}\,$ (from wikipedia)

$S(n,k)=kS(n−1,k)+S(n−1,k−1)$ with $S(n,n)=S(n,1)=1$ (Stirling numbers of the second kind) Almost seems like it's a linear recurrence... but we need to know about k-1. These numbers are defined as "the number of ways to partition a set of n objects into k groups" so, if I have written a few programs to find S(n,k) from that definition I want to know if I "must" find the values in the linear recurrence along the way...

But, I was trying to keep it more general to make it interesting?

One more example:

$C(n,k)=C(n−1,k)+C(n−1,k−1)$ with $C(n,n)=C(n,1)=1$ but $C(n,k) = \frac{n!}{k!(n-k!)}$, most people like the 2nd one better of you want to look at large values of n and k>1...

Answer 1

You inquire about comparing your algorithm to a given recursive algorithm, but the more fundamental question would seem to be how good is your algorithm just by itself?

There are numerous ways to measure the efficacy of a computational algorithm using the ideas of computational complexity. That is, you should measure the complexity of your algorithm by the intensely studied classes of P, NP, PSPACE, EXP, and so on. That is, if you have an algorithm to calculate a function f, the important thing to look at is where does your algorithm sit with respect to these complexity measures: is it polynomial time? exponential time? Is there a nondeterministic polynomial time algorithm? Is there a PSPACE algorithm?

You inquire about comparing your algorithm to a given recursive algorithm. For such a comparison, one should use the measures of complexity theory. If these two algorithms have the same computational complexity, then they are equivalent by these measures, even if your algorithm does not exactly amount to performing the same computation, and the two methods would be equivalent in terms of their computational cost. But if one algorithm finds itself in a lower complexity classification, then it will inevitably be superior in the general case.

Answer 2

I guess it's worth saying this in an answer: I don't think this is a meaningful question. Consider the function defined recursively by $f(0) = 1, f(n+1) = 2 f(n)$. Clearly $f(n) = 2^n$ for all $n$. You shouldn't consider this formula an "escape" from the recursive definition, for the very simple reason that the exponential function is usually defined by this very recursion! (One can, of course, do something incredibly silly like define $2^n$ to be $e^{n \ln 2}$. Whether you think this constitutes an "escape" from the recursive definition is up to you, but what it doesn't constitute is a fast method to compute powers of two.)

What you can ask for, instead, is an algorithm faster than the naive one above. There is, in fact, such an algorithm; it goes by the name binary exponentiation or exponentiation by squaring, and it basically works by replacing the recursion $f(n+1) = 2f(n)$ by a pair of recursions $$f(2n) = f(n)^2, f(2n+1) = 2f(n)^2.$$

Does this constitute an "escape" from the recursive definition? I don't know; it still requires that you compute some smaller powers of two, just not as many.

Answer 3

My answer will forcibly be vague, as your question is quite vague. One reminder: iteration is essentially equivalent to recursion (i.e. you can simulate one with the other). So either your method is made up of only arithmetic operations and branching (but no loops, recursion, throwing exceptions, gotos, i.e. no control structures), in which case there is no hidden recursion, OR it contains some control structure which may well be 'hiding' the equivalent of a recursive definition.

The number of operations isn't going to help you in a substantial way, at least not unless you can bound them independent from 'n'. If your number of computations still depends on 'n', then whether you get an improvement depends on the actual dependence on 'n'. You can write a bad recursive search in ordered data in $O(n)$ and a good one in $O(\ln n)$, so details really do matter.