> "It means minimizing use of nested loops, which exponentially increase statement count."
Nesting two loops has an n^2 cost and nesting 3 levels deep costs n^3. At no point does it ever cost 2^n or any x^n.
It's polynomial, not exponential.
Nesting two loops has an n^2 cost and nesting 3 levels deep costs n^3. At no point does it ever cost 2^n or any x^n.
It's polynomial, not exponential.
I know in reality the frequency of iterations varies considerably but for simplicity of discussion let's remove variability.
Say we have a loop with 1000 iterations. That is at minimum 1000 statements in the loop body plus expression overhead from the loop itself. If this loop is nested once with a same sized loop there are now 1,000,000 iterations plus some expression overhead per iteration. If it is nested twice deep there are now 1 billion iterations.
That example is exponential of 1000. Given that there is overhead associated with operation of a loop it is actually greater than exponential. It may not be quite so dramatic as a logarithmic growth curve though.
I completely concede that in reality loops vary in iteration count and so nested loops aren't likely perfectly exponential unless their iteration counts are identical. The increase of iterations from nesting loops does increase more dramatically than a simple multiplicative operation as the depth of loop nesting increases, such that the growth of total iterations is a curve on a graph. A polynomial growth operation when graphed should present a straight diagonal line without the presence of a third variable.
no it's not. you simply have the definitions mixed up. exponential slow down or speed up means a^x where x=1000. you are describing polynomial growth, i.e. x^a (where x=1000 and a=3)
Adding more nested loops is bad, but it's not exponential. It's polynomial.
As you nest more and more loops, the big O complexity goes from N to N^2 (quadratic) to N^3 (cubic) to N^4, etc... N^(any number) is polynomial. Exponential would be 2^N or 3^N or any number raised to the N.
And usually when you encounter O(some constant), it's meant as "of the order of magnitude of", i.e. somewhere between some constant/10 and some constant * 10. That isn't the definition used here, but seems to be the cause of most complaints about asymptotic analysis being misapplied when no asymptotic analysis was being done in the first place.
1000^depth is not a constant: the function here is "f(depth) = number of times the inner loop body executes". f(depth) = O(1000^depth).
And, "times" here is serving the same role as "comparisons" in "Mergesort takes O(n log n) comparisons": it's referring to the thing that f is actually counting.
Yes, O(…) is just a mathematical construct, so you can apply it the way you have to mean "the amount of work done for a given input size increases exponentially with respect to the number of nested loops iterating over the input"… but that's not how most people interpret it in the context of a computer algorithm. (The exception would be if the nesting depth were dynamically determined by an input parameter, but I don't think that's what anyone here is talking about.)
Anyway I'm not trying to argue, I agree with your point, but I think everyone here is talking past each other trying to say the same thing, ultimately due to imprecision in semantics.
If you have a sequence of programs with polynomial runtime, but which grows as N, N^2, N^3, N^4, ..., that's a textbook example of exponential growth. It is not generated by running a program on increasingly larger inputs, so there's nothing to be put in the exponential runtime complexity class, but it's exponential nonetheless.
No, it isn't. N, N^2, N^3, N^4, ... is polynomial, not exponential. Exponential would be X^N. Look at the graph on the Wikipedia page I linked to.
Bringing this back to the original comment about nesting loops, perhaps what you're describing is some sort of metaprogramming that dynamically generates increasingly deeply nested loops based upon the size of the input. If someone were to write that program then yes, it would be exponential.
In the far more common case of a programmer nesting a few loops, the resulting run-time would be polynomial, not exponential.
defines = 0
tests = 0
increments = 0
defines++
for (let i = 0; i < 5; i++) {
tests++
increments++
}
console.log(`defines: ${defines}, tests: ${tests}, increments: ${increments}`)
1 level will increment 5 (5^1) times
2 levels will increment 25 (5^2) times
3 levels will increment 125 (5^3) times
levels tests defines
1 15 1
2 30 6
3 155 31
N^2 is polynomial.
2^N is exponential.
https://stackoverflow.com/questions/4317414/polynomial-time-...
Anyway, there might be some confusion of terms, perhaps. Exponential time in the algorithmic complexity sense is any algorithm that takes 2^n operations to complete. If you’re talking about the number of iterations after you loop through n things n times, then that increase itself would not be exponential. But that delta in iterations is unrelated to exponential and polynomial time complexity.
Let's not forget there is overhead to loops. At a minimum let's assume there is a single statement in the loop body, an increment statement and a terminal condition. In a single loop of 1000 iterations there are 3000 things to evaluate. The math becomes:
(n * 3)^x
If a simple loop is nested twice (3 depths) there would be 1 billion iterations but about 27 billion evaluations. Would it be correct to say that is just slightly faster polynomial growth?
That's only true if your n is referring to the number of iterations of each loop. If your n instead refers to the number of nested loops, you will indeed get a runtime of O(2^n).
Seeing as in this context we were talking about varying the number of nested loops, it makes sense that we would define our variable n = number of nested loops.