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

Answer 1

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))