Asymptotic Expected Running Time

Question

I'm having some trouble with an asymptotic analysis question. The problem asks for both the asymptotic worst case running time and the asymptotic expected running time of a function. Random(n) generates a random number between 1 and n with uniform distribution (every integer between 1 and n is equally likely.)

Func2(A, n)
/* A is an array of integers */
1 s ← A[1];
2 k ← Random(n);
3 if (k < log2(n)) then
4    for i ← 1 to n do
5       j ← 1;
6       while (j < n) do
7          s ← s + A[i] ∗ A[j];
8          j ← 2 ∗ j;
9       end
10   end
11 end
12 return (s);

I was wondering how line 3 (if (k < log2(n)) then) effects the expected running time of the function. I believe lines 4 - 10 run at worst case cn^2 time, but I am unsure how to derive the expected running time due to the if statement. Thanks for any help!

-Matt

Bernhard Barker · Answer 1 · 2013-02-04T15:20:40.757

0

A tip:

The running time of lines 4-10 is not O(n^2).

Consider the values of j for the while loop:

j = 1, 2, 4, 8, 16, ...

That's not O(n).

edited Feb 04 '13 at 15:20

answered Feb 04 '13 at 14:52

Bernhard Barker

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I apologize, lines 4-10 would run at nlog(n) time correct? The problem also asks for the Expected Running Time of the function given that Random(n) produces a random number between 1 and n with uniform distribution. Is it correct that, ET(n) = Pr(k < log(n))ET(k < log(n)) + Pr(k > log(n))ET(k > log(n)) – Marcus Koz Feb 04 '13 at 15:12
Yes, lines 4-10 would be worst case `n log n`. I haven't dealt a lot with expected running time, so your guess is probably better than mine. – Bernhard Barker Feb 04 '13 at 15:20

Andrew · Accepted Answer · 2013-02-04T15:58:17.437

0

For big n, k almost always bigger then log(n). For example for n=1024, log(1024)=10 Probability that you will execute cycle is P=log(n)/n So Time will be

(log(n)/n)*(n*log(n))+ O(RandomFunc(n))

So everything depend on O(Random(n)). If O(Random(n)) = O(n)

O(n)>O(log(n)^2) = O(n)

Lines 4-10 is O(nlog(n))

edited Feb 04 '13 at 15:58

answered Feb 04 '13 at 15:37

Andrew

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j is increasing by a multiple of 2 every time. Would that not indicate that (while j < n) will execute log(n) times? And I believe we are to make the assumption that Random(n) runs in constant time. – Marcus Koz Feb 04 '13 at 15:51

score 0 · Answer 3 · answered May 01 '14 at 16:43

0

Using Sigma Notation, you can methodically do:

enter image description here

answered May 01 '14 at 16:43

Mohamed Ennahdi El Idrissi

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Asymptotic Expected Running Time

3 Answers3