I am presented with the following code segment and asked to find the growth rate, which can be done by finding the number of times the variable sum is incremented:

sum = 0
for i = 1 to 2 * n
 for j = 1 to i * i
   for k = 1 to j

I tried making this into a function called run(n) in Python that returns sum and got run(1) = 11, run(2) = 192, run(3) = 1183, and run(4) = 4488. However, my book only presents 3 possible answers, those being $\Theta(n^4)$, $\Theta(n^5)$, and $\Theta(n^6)$.

I don't think my results match any of these choices. I might be confused as to what the question is asking as it says

present answers in terms of summations

or my code might be incorrect, but I am confused on what to do next. I am very new to algorithms, so any help would be appreciated.


2 Answers 2


The innermost loop increments sum exactly $j$ times. The middle loop runs $i^2$ times with increasing values of $j$ so it increments sum $\dfrac{i^2(i^2+1)}2$ times. And finally, the outer loop runs $2n$ times, for a total of $$\sum_{i=1}^{2n}\frac{i^2(i^2+1)}2=\frac{n (1 + 2 n) (1 + 4 n) (2 + 3 n + 6 n^2)}{15}$$ incrementations. [Thanks to Wolfram Alpha.]


Let $I$ be the number of times the for i is executed, $J$ the number of times for j, and $K$ the number of times for k is executed.

  1. $I$ is $\Theta(n)$ times
  2. $J$ is $\Theta(I \cdot I) = \Theta(n^2)$ times
  3. $K$ is $\Theta(J) = \Theta(n^2)$

The correct answer is the product of $I$, $J$ and $K$, i.e., $\Theta(n \cdot n^2 \cdot n^2) = \Theta(n^5)$.

  • $\begingroup$ That was a very clear explanation, thank you! $\endgroup$
    – user163191
    Commented Sep 26, 2023 at 9:27
  • $\begingroup$ Caution, this method can lead to false results. Take an outer loop for $i$ from $1$ to $n$, and inner loop from $1$ to $2^i$. You would say $\Theta(n)\cdot\Theta(2^n)=\Theta(n\cdot2^n)$ though the correct answer is $\Theta(2^n)$. $\endgroup$
    – user16034
    Commented Sep 26, 2023 at 9:56

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