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# [Solved]: Runtime analysis of a recursive algorithm with a tricky amount of work per recursive call

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Problem Detail:

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.