Counting Primitive Operations & Calculating Big-O Notation

问题

I've written a piece of java code that when given an array (arrayX), works out the prefix averages of that array and outputs them in another array (arrayA). I am supposed to count the primitive operations and calculate the Big-O notation (which i'm guessing is the overall number of calculations). I have included the java code and what I believe to be the number of primitive operations next to each line, but I am unsure whether I have counted them correctly. Thanks in advance and sorry for my inexperience, i'm finding this hard to grasp :)

double [] arrayA = new double [arrayX.length]; *(3 Operations)*

    for (int i = 0; i < arrayX.length; ++i) *(4n + 2 Operations)* 
    {
        double sum = arrayX[i]; *(3n Operations)*

        for (int j = 0; j < i; ++j) *(4n^2 + 2n Operations)*
        {
            sum = sum + arrayX[j]; *(5n^2 Operations)*
        }
        arrayA[i] = sum/(i+1); *(6n Operations)*
    }
    return arrayA; *(1 Operation)*

Total number of operations: 9n^2 +15n + 6

回答1:

I don't think there's any standard definition of "what constitutes a primitive operation". I'm assuming this is a class assignment; if your instructor has given you detailed information about what operations to count as primitive operations, then go by that, otherwise I don't think he can fault you for any way you count them, as long as you have a reasonable explanation.

Regarding the inner loop:

for (int j = 0; j < i; ++j)

please note that the total number of times the loop executes is not n², but rather 0+1+2+...+(n-1) = n(n-1)/2. So your calculations are probably incorrect there.

Big-O notation is not really "the total number of calculations"; roughly speaking, it's a way of estimating how the number of calculations grows when n grows, by saying that the number of calculations is roughly proportional to some function of n. If the number of calculations is Kn² for any constant K, we say that the number of calculations is O(n²) regardless of what the constant K is. Therefore, it doesn't completely matter what you count as primitive operations. You might get 9n², someone else who counts different operations may get 7n² or 3n², but it doesn't matter--it's all O(n²). And the lower-degree terms (15n+6) don't count at all, since they grow more slowly than the Kn² term. Thus they aren't relevant to determining the appropriate big-O formula.

来源：https://stackoverflow.com/questions/26898148/counting-primitive-operations-calculating-big-o-notation