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Given two functions in PHP, say 

function f($n) {
    return $n;
}

function g($n) {
    return pow($n, (2/3));
}

How to check if a function f(n) is in Ω(g(n)), Θ(g(n)) or O(g(n)) in PHP?

What I tried so far:

$n = INF;

$A = f($n) / g($n);

if ($A == 0) {
    echo "f(n) = O(g(n))";
} elseif (is_infinite($A)) {
    echo "f(n) = Ω(g(n))";
} elseif ($A != 0) {
    echo "f(n) = Θ(g(n))";
}

Shouldn't that work?

user3419936
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  • What do you mean by *function f(n) is in Ω(g(n)), Θ(g(n)) or O(g(n)) *? The `f($n) / g($n)` just divide the result of each function – Javad Mar 14 '14 at 21:56
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    PHP isn't a symbolic math engine. Consider using Mathematica instead. – p.s.w.g Mar 14 '14 at 21:56
  • Maybe I did it in the wrong way, but I want to know whether f(n) is asymptotically bounded by g(n) by given constants. That is what I mean, when I ask whether f(n) is in Ω(g(n)), Θ(g(n)) or O(g(n)). In this specific example, I mean: The program should decide whether n is in Ω(n^(2/3)), Θ(n^(2/3)) or O(n^(2/3)). – user3419936 Mar 14 '14 at 22:00
  • @p.s.w.g - maybe Mathematica is better, but if I want to do it in PHP, do you think there is a way? – user3419936 Mar 14 '14 at 22:01
  • These are all abstract concepts, you can't determine them by calling the functions with a specific argument. – Barmar Mar 14 '14 at 22:01
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    Big-O is not even about the range of the function, it's about the complexity of the algorithm for computing it. – Barmar Mar 14 '14 at 22:04
  • @Barmar: So what you say is that this cannot be done in PHP? I'm wondering how Mathematica and other programs do. – user3419936 Mar 14 '14 at 22:06
  • I'm saying it can't be done in any regular programming language. Maybe Mathematica can do it because it's a symbolic math language, not a procedural programming language. – Barmar Mar 14 '14 at 22:11

2 Answers2

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Your basic idea is correct: you have to find the limit of f(n)/g(n) as n grows without bound. Unfortunately there is no easy way to compute the exact limit in PHP, since that requires symbolic computations which is best left to a computer algebra system such as Mathematica or Maxima.

You can approximate the limit by computing f(n)/g(n) for increasing values of n and seeing if you get a sequence that approaches a fixed value. For example: 

$n=1;
while ($n < 1e300) {
    $A = f($n)/g($n);
    echo $A, "\n";
    $n *= 1e12;
}

In this particular case the sequence of f(n)/g(n) seems to grow without bound, so the numerical evidence suggests that f(n) is in Ω(g(n)). This is not a proof though; symbolic methods are needed for that.

Joni
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Both the time and space requirements for both f() and g() are in Ω(1), Θ(1) and O(1).

Dwayne Towell
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  • Really? How can a function depending on a variable be constant? – user3419936 Mar 14 '14 at 22:23
  • Please read about Big-O notation here. http://en.wikipedia.org/wiki/Big_O_notation The general idea of Big-O, as mentioned in a comment above, is to characterize the running time (or memory usage) of an algorithm. Both of your functions have constant running time--that is the amount of time it takes to determine the answer is NOT dependent of the value of N. – Dwayne Towell Mar 14 '14 at 22:42
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    Big-O notation is not about running time or memory use, but about the behavior of functions. The first sentence from the Wikipedia article you link to: "In mathematics, big O notation describes the limiting behavior of a function when the argument tends towards a particular value or infinity, usually in terms of simpler functions." Algorithms come into play only when the functions you deal with happen to describe some characteristic of an algorithm, such as the number of comparisons made by a sort. In the case here it's wrong to say that f is in O(1) because f grows without bound. – Joni Mar 15 '14 at 07:19