Worse case analysis?

Question

k = 0
for i = 1 to n do
    for j = n down to i do
        k = k + 1
        if j < 8 then
            k = k + jI am practicing analyzing worse case for algorithms and run across this interesting algorithm (at least interesting to me). What is the runtime of the above algorithm? I think it's N^2 but I am not so sure - looks kind of tricky. What is the value of k after the algorithm finishes? Instead of giving the answer, I would appreciate it very much if you could show me the process along with explanation in analyzing the above algorithm. Thanks guys/gals.

Answer 1

k = 0
for i = 1 to n do
for j = n down to i do
k = k + 1
if j < 8 then
k = k + jI am practicing analyzing worse case for algorithms
and run across this interesting algorithm (at least
interesting to me). What is the runtime of the above
algorithm? I think it's N^2 but I am not so sure -
looks kind of tricky. What is the value of k after
the algorithm finishes? Instead of giving the
answer, I would appreciate it very much if you could
show me the process along with explanation in
analyzing the above algorithm. Thanks guys/gals.Okay, to begin with, the first clue is that almost every time you have two nested
for loops, it usually O(n^2). But I'll show you a bit of logic for this one.
Consider the inner loop:
for j = n down to ialthough i does change every time through the outer loop, you can still say the following: The complexity is still linear. The larger n gets, the more often you have to pass through the loop.
The only way for this inner loop to be, say, constant, is for a condition i - n = k to always hold true for a given constant k. But since i changes each time through the outer loop, while n stays constant, this is clearly not the case. So the complexity of this loop is linear, or O(n).
Now the outer loop is easier to evaluate:
for i = 1 to n Yeah, now this is the most straight forward for loop you can get. This loop always has a complexity of O(n).
Then the total complexity is O(n) * O(n) = O(n^2).
Here's another easy way to think about it. I'm going to visually mark a "*" for each time the code in the inner loop is executed. It will look something like this:
Now, that's a triangle, right? The base is n, and the hight is n + 1. And the area of the triangle is (n * (n + 1))/2 = (n^2 + n) / 2.
Thus, the big oh of this formula is O(n^2), which is the complexity of your program.
Hope that helps clear things up for you.
- Adam