# Runtime analysis of a recursive algorithm with a tricky amount of work per recursive call

I want to analyze the runtime of this algorithm:

int fun (int arr[], int n) {
int result = 1;
int i, j;

if (n == 1)
return 1;

else {
result = fun(arr, 2n/3);
for (i = 1; i <= sqrt(n); i=i*2);
for (j=0; j<sqrt(n)/i; j++)
result += arr[j];

return result;
}
}


I can see that the runtime recurrence should be something like

$\qquad\displaystyle T(n) = T\left(\frac{2n}{3}\right) + \Theta(X)$

where $X$ is the time of the extra operations per recursion.

I can also see that the extra operations are:

\qquad\begin{align*} \sum_{i=1}^{\log(\sqrt{n})} \sum_{j=0}^{\frac{\sqrt{n}}{i}}1 &= \sum_{i=1}^{\log(\sqrt{n})}\frac{\sqrt{n}}{i} \\ &= \sqrt{n} \cdot \sum_{i=1}^{\log(\sqrt{n})} \frac{1}{i} \\ &= \sqrt{n}\cdot \log(\log(\sqrt{n})) \end{align*}

So all in all:

\qquad\begin{align*} T(1) &= 1 \\ T(n) &= T\left(\frac{2n}{3}\right) + \sqrt{n}\cdot \log(\log(\sqrt{n})) \end{align*}

But I could not continue from here to solve this recursion.

I can see three issues with what you have.

1. There are some inaccurracies in your sums. The outer one needs rounding of the upper boundary, the inner needs a $-1$.

2. $\displaystyle \sum_{i=1}^{\log(\sqrt{n})} \frac{1}{i} \neq \log(\log(\sqrt{n}))$

The true value of the sum is $H_{\log(\sqrt{n})}$ (will change slightly if you fix the sums). It's true that the difference vanishes in $\Theta$ if you go that route, but better not write equality where it does not hold.

3. You dropped the recursion at the end! You should have

$\qquad \displaystyle T(n) = T(2/3 \cdot n) + \dots$

From there, unfold the recurrence:

\qquad\begin{align*} T(n) &= T(2/3 \cdot n) + f(n) \\ &= T(4/9 \cdot n) + f(2/3 \cdot n) + f(n) \\ &\dots \end{align*}

Spot a pattern, guess the solution and prove it correct by induction! This part is well covered by our reference question in case you have trouble.

• I did drop the $-1$ since I was speaking in terms of theta notation, but you're probably right, I should have write equality only if I used the theta on the LHS. I don't really understand what was the first issue you mentioned. What's wrong with the outer sum? I can see that the inner sum is equal to $\frac{\sqrt{n}}{i}+1$ if that's what you mean. How do you suggest attacking iterative functions? Write $\sum$ for each loop and understand it's boundaries or maybe unfold the sum in a table? – Eran Feb 5 '14 at 8:29
• @Quaker Translating loops into sums is what I'd do, no problem there. As for the outer sum, note that the number of iterations of any for-loop is a natural number but $\log(\sqrt{n})$ is not (always), so the sum can't be right. (All of these inaccurracies vanish in a $\Theta$ in the end, so some people may advise sloppiness for "clarity". I don't think that is fruitful, though.) – Raphael Feb 5 '14 at 8:31
• I guess I should assume that $\sqrt{n}$ is of the form of $2^k$ – Eran Feb 5 '14 at 8:32
• @Quaker: Why? Just use the exact number of iterations. (My answer includes a hint.) – Raphael Feb 5 '14 at 8:33
• Is it possible to solve this iterative analysis by writing a table of the values that $i$ gets in each iteration and then the values that $j$ gets depending on $i$ and summing it all up? Would it work? – Eran Feb 5 '14 at 9:28