The code is as follows:

```
Ans = 1;
while( N > 0)
{
Ans = Ans*2;
N = N/2;
}
```

I can not figure out how N = N/2 will effect the runtime. I tried to find a pattern by checking how many times the loop would run if N = 1 .. 7 but I didnt notice any pattern. I think I must be going about this the wrong way.

Answer:

If you are talking about the asymptotic notation, the Big-Oh, the complexity will be `O(logn)`

. Count the operations and you will see that they are in fact a logarithm of base two.

The asymptotic notation basically tells you that the algorithm doesn't run **slower** than the given function of complexity. Most of the calculations can be seen from the code right away, however in the more complex algorithms, a calculation of instructions is needed.

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