How many times is this loop body repeated?

Question

In my Object-Oriented Programming course we discussed a topic that I don't think he ever named, I've tried to find out what it's name is to find a proper way to solve these, but I have had no luck.

This is not homework, but a question for clarification about the process to solve this problem.

for I = (N + 2) downto -1
    for J = (I - 1) to (N + 4)
        // Code is run here

The question is "How many times is // Code is run here ran?"

Here is what I have tried to solve this:

1) I = (N + 2) , J = [(N + 2) - 1] from this (and what I remember) you use b - a - 1 to solve for the number of times executed, which gives us X = [(N + 2) - 1] - (N + 2) - 1 which can be simplified to X = -2

2) I = -1 , J = ((-1) - 1) and X = ((-1) - 1) - (-1) - 1 which simplifies to X = -2`

I'm getting lost on dealing with the second for loop and how to finish the problem. I know that we have to end up with an answer such as r(r + 1)/2

I just want to say that I have attempted to look for a name of this type of technique, but he simply called it "Code Counting" which didn't return any searches relating to this topic.

Thank you

EDIT: This course was in Java, so that is why I used the Java tag for this question, if anyone is curious.

EDIT2: To clarify, this was on a written exam , so we are expected to do this via pen-and-paper, I would like an explanation of how to solve this question as I have attempted it many times and still end up with the wrong answer.

Answer 1

Just look at the "code" and start counting logically. In the first iteration of the outer loop (called OL) you execute the inner loop (IL) (N + 4) - (N + 2 - 1) + 1 times = 4 times.

Explanation of the +1 : if we run the loop from -1 to 2, we in fact run it 4 times: -1, 0, 1, 2, which in math is `2 - (-1) + 1.

The next time I = N + 1 , therefore the IL runs (N + 4) - (N + 1 - 1) + 1 times = 5 times. Same goes for the next step and the step after that, the times the IL is executed increase by one each time : 4 + 5 + 6 + ... . The question remaining is how far we go.

The last step is I = -1 , there IL gets run (N + 4) - (-1 - 1) + 1 = N + 7 times.

The sum you are looking for therefore seems to be 4 + 5 + 6 + ... + (N + 6) + (N + 7) . Which in fact is something like r(r + 1)/2 with a few substractions and additions.

The above numbers assume the to boundaries to be inclusive.

Note: whenever you come up with some kind of a formular, choose the input parameter as something small (like 0 or 1) and verify that the formula works for that value.

Summing the values using the little gaussian formula r * (r + 1) / 2 we have r -> N + 7 . And therefore (N + 7) * (N + 8) / 2 . But then we count the 3, 2 and 1 as well, which are actually not in the above listing, we need to subtract them and come to the final solution of:

(N + 7) * (N + 8) / 2 - 6

Answer 2

The algorithm as shown in the question looks like the good old Basic syntax

for X down/to Y, that includes Y

The outer loop goes from n+2 to -1 , so the inner loop goes

n+1 to n+4 =>   4 iterations
...
 -2 to n+4 => n+7 iterations

Summing all of these, we get

n+3
 ∑ (4+i)  =  4(n+4) + (n+3)(n+4) / 2
i=0
          =  (n+11)(n+4) / 2

which is also equal to (N + 7)(N + 8) / 2 - 6