Finding the Big O of a simple algorithm

Question

I'm given the pseudocode of an algorithm:

1     for i=1 to n          //n times
2         j=2^i             //n times
3         while j>=2        //2+3+....? times
4               j=j/2       //1+2+....? times

Where 'to' means less than or equal to, and where '^' means power of.

I'm quite the sure that the first two lines run n times, however I'm not quite sure how to compute the other two lines.

I know that line 3's first two values would be 2... then 3, but I'm not sure what to do next there.

Same goes along with line 4, the first two values being 1.. then 2, but then I have no idea how to continue.

I'm told that the answer is O(n^2), where I have to add up all the lines to get the actual result.

Any ideas?

That's not what I'm asking, rather how to get the values of each line (adding them, etc). Also the answer I'm given is 0(n^2) — TTEd, Dec 03 '15 at 19:14
I.e how many times does each line execute, then add them to get the Big O. — TTEd, Dec 03 '15 at 19:15
If the second line multiplied the power2 from the previous loop by 2, would that be a Small O? — Weather Vane, Dec 03 '15 at 19:16
If this is pseudocode, what does `2^i` mean? In C, it would be a bitwise xor. In pseudocode, it might mean 2 to the i power. — Fred Larson, Dec 03 '15 at 19:17

score 1 · Answer 1 · answered Dec 03 '15 at 19:16

The inner loop will run log2(j) times as j is halved each iteration. While j is 2^i, then log2(j) = i. I.e. the inner loop running i times. As i is doing from 1 to n, the total iterations will be 1+2+3+4+..+n, while it is a simple arithmetic series, whose sum is (n+1)n/2, which is O(n^2) as stated. (remark: I might have missed some +/-1 constants somewhere, but it won't affect the resulting complexity).

score 1 · Accepted Answer · answered Dec 03 '15 at 19:17

Well, this one is a bit weird.

Let's begin by pretending that every arithmetic operation takes the same amount of time. In the case, look at what this code does:

2         j=2^i             //n times
3         while j>=2        //2+3+....? times
4               j=j/2       //1+2+....? times

If you think about how loop (3) executes, it will keep dividing j by two on each iteration. If you think of j as a power of two, it decreases the exponent by one on each iteration. This can only happen O(i) times before j drops down to two. Consequently, this part code runs in time O(i), assuming that all arithmetic operations take the same amount of time.

The outer loop will run n times, so the work done here will be proportional to 1 + 2 + 3 + 4 + ... + n = Θ(n²).

The problem is that the assumption that all arithmetic operations take the same amount of time isn't really realistic. The number 2ⁿ grows really fast and will quickly overflow any fixed-width integer type. A more realistic assumption of the time bounds here is that each operation will take time proportional to the number of bits in the numbers. I'm going to ignore this case because I suspect this is for a class assignment, but it's worth keeping that in mind going forward.

score 0 · Answer 3 · answered Dec 03 '15 at 19:16

Break this into smaller steps

3         while j>=2        //2+3+....? times
4               j=j/2       //1+2+....? times

Above loop for some j=N runs for ~log(N) times

Since i varies from 1 to N, j varies from 2^1 to 2^N. Each while loop now executes ~log(2^i) times, which equals ~i times

Hope this helps

Finding the Big O of a simple algorithm

3 Answers3