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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.

David Richerby
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Eran
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1 Answers1

1

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.

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Raphael
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  • 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 05 '14 at 08:29
  • @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 05 '14 at 08:31
  • I guess I should assume that $\sqrt{n}$ is of the form of $2^k$ – Eran Feb 05 '14 at 08:32
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    @Quaker: Why? Just use the exact number of iterations. (My answer includes a hint.) – Raphael Feb 05 '14 at 08:33
  • 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 05 '14 at 09:28
  • @Eran That is basically what $\dots$ + guess & proof means. See the linked question for detail. – Raphael Feb 05 '14 at 10:00
  • If I do as I said, I can see that the first iteration of $j$ does $0,...,\sqrt{n}$ iterations, the second does $0,...,\frac{\sqrt{n}}{2}$ and so on until it does $0,1$ so the sum looks like this: $\sum_{i=1}^{log(\sqrt{n})}(2^i+1) = \Theta(\sqrt{n})$, why am I getting different results? – Eran Feb 05 '14 at 10:13