question about recursive even sum program

Question

hello i need program which returns sum of all even number <= to given n number
for example if n=even (let say 10) sum will be 2+4+6+8+10 or if n is odd (let say 13) 2+4+6+8+10+12 please help it should be recursive algorithm

i have done myself  

Answer 1

Not that I like doing others' homework, but I'm interested whether I could solve that task quickly, so here's a C-like pseudocode which I haven't tested.

int evenSum( int number )
{
    if( number % 2 == 1 ) {
        return evenSum( number - 1 );
    }
    if( number == 0 ) {
        return 0;
    }
    return number + evenSum( number - 2 );
}

Answer 2

This algorithm is quite easy. Try decomposing the problem:

• How would you add all numbers <= some n recursively?
  • Have an appropriate base case (hint: either x = n or x = 0 depending on how you start),
  • Add the current number to the return value,
  • Call the function again (recursion, duh), add its result to the return value.
• Now, just add a test in every recursive step to see whether the number is even.

Answer 3

Algorithms only:

def sumeven (n):
    if n <= 0:
         return 0
    if n is even:
         return n + sumeven (n-1)
    return sumeven (n-1)

As with all recursive problems, break it down into:

• the terminating (lowest-level) value.
• all other values defined in terms of a "lower-level" value.

In the above case, the rules are:

f(0) is 0.
f(n) is f(n-1) for all odd values of n.
f(n) is n + f(n-1) for all even values of n.

This can be optimised to skip odd values (other than the first), if necessary, by changing that last rule to:

f(n) is n + f(n-2) for all even values of n.

giving the algorithm:

def sumeven (n):
    if n <= 0:
         return 0
    if n is odd:
        return sumeven (n-1)
    return n + sumeven (n-2)

Answer 4

It is relatively simple to write in c with a for loop

sum = 0;
for (int i=0; i<=n; i+=2)
   sum += i;

if you want a recursive function you could do something like this (again, in c)

int sum(int n){
   if (n==0) return 0;
   if (n%2) return sum(n-1);
   return n+sum(n-2);
}

This sounds very homework-ey, I do hope you are going to learn from this and other answers and not just copy and paste.

EDIT Thanks for that oezi, fully missed it.

Answer 5

Initially, the solution is :

int SumEven(int n)
{
   return (n/2)*(1+(n/2));
}

And to make it recursive:

int SumEven(int n,unsigned int y)  // y has completely no effect!!
{
   if(y==0)
       return (n/2)*(1+(n/2));
   else 
      return SumEvent(n,y-1)
}

                                            (^_^)

Answer 6

(defun sum-twos (n)
  "This will sum the even numbers between 0 and N (inclusive)"
  (if (= 1 (mod n 2))
      (sum-twos (- n 1))
      (if (> n 0)
          (+ n (sum-twos (- n 2)))
          (+ n 0))))