I know with absolute certainty that this is the wrong runtime, but I just wanted to show you how I got to it.

for (i = 0; i <= n; i++) do  
     j = 2*i  
     while j <= n do  
         j = j + 1  
         print("Hello World")  

Then I get the runtime
$Θ(\sum_{i=1}^n 1 + n -2i)=Θ(n+n^2-n(n+1))=Θ(0)$??

  • $\begingroup$ The runtime of your inner loop isn’t 1+n-2i. It is max (1, 1+n-2i). That’s the fatal mistake in your calculation. $\endgroup$ – gnasher729 Sep 29 '20 at 8:58
  • $\begingroup$ I follow from $\Theta(\sum_{i=1}^n 1+n)$ to $\Theta(n+n^2)$. How do you get from $\Theta(\sum_{i=1}^n-2i)$ to $\Theta(−n(n+1))$? $\endgroup$ – greybeard Sep 30 '20 at 6:03

The result you state is clearly wrong, since the second line (for example) runs $n+1$ times. The problem is in your sum, which you don't explain how you got to.

Here is the code again, with line numbers:

1: for i in 0,...,n:
2:   j = 2*i
3:   while j ≤ n:
4:     j = j + 1
5:     print "hello world"
6:   end while
7: end for

The running time heavily depends on the model. However, assuming that $n \geq 0$, the running time will be proportional to the number of times that lines 2 and 4 are executed. Line 2 is executed $n+1$ times. As for line 4, for a given value of $i$, it is executed $\max(n-2i+1,0)$ times (for example, if $2i = n$ it is executed once). In total, it is executed this many times:

$$ \sum_{i=0}^n \max(n-2i+1,0). $$ In order to calculate this sum, it will be easier to consider separately the cases $n$ even and $n$ odd. If $n = 2m$ then the maximal $i$ for which $n-2i+1>0$ is $m$. Furthermore, the summand goes from $n+1$ down to $1$ in jumps of $2$; there are $m+1=n/2+1$ summands in total. Therefore the sum equals $$ 1 + 3 + 5 + \cdots + (n + 1) = \frac{(n/2+1)(1+(n+1))}{2} = \frac{(n+2)^2}{4} = \Theta(n^2). $$ If $n = 2m+1$ then the maximal $i$ for which $n-2i+1>0$ is also $m$. This time, the summand goes from $n+1$ down to $2$, and there are $m+1 = (n+1)/2$ summands. Therefore the sum equals $$ 2 + 4 + 6 + \cdots + (n+1) = \frac{((n+1)/2)(2+(n+1))}{2} = \frac{(n+1)(n+3)}{4} = \Theta(n^2). $$ In total, we get that the number of times that line 4 is executed is $\Theta(n^2)$ in both cases, and so $\Theta(n^2)$ overall (this is a subtle point, and requires a short argument).

We conclude that the total running time is $\Theta(n^2)$. (This may need the assumption $n>0$, depending on your exact definition; you can take $\Theta(n^2+1)$ if you want to be extra sure.)


Sum $\sum_{i=1}^n (1 + n -2i)$ is wrong calculation, because while loop produce steps only when $2i \leqslant n$.

  • $\begingroup$ So it's $\sum_{i=1}^n n -2i$? But don't I need to add the runtime $Θ(1)$ it takes to run each step for the while loop? $\endgroup$ – Jacob Toho Sep 29 '20 at 7:59
  • $\begingroup$ If you count each line (not operation), then you'll have only $2$ working lines inside for loop after $n-2i<0$. $1$ assignment and $1$ checking while condition. $\endgroup$ – zkutch Sep 29 '20 at 8:24
  • $\begingroup$ Jacob, it's not n - 2i. When i = n, it doesn't take minus n steps to perform the loop, it takes one step, which is an awful lot more than minus n. $\endgroup$ – gnasher729 Sep 29 '20 at 10:28

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