How does that recursive function work? - c++

Question

Might be a very basic question but I just got stuck with it. I am trying to run the following recursive function:
//If a is 0 then return b, if b is 0 then return a,
//otherwise return myRec(a/2, 2*b) + myRec(2*a, b/2)
but it just gets stuck in infinite loop. Can anybody help me to run that code and explain how exactly that function works? I built various recursive functions with no problems but this one just drilled a hole in my head.
Thanks.
Here is what I tried to do:
#include<iostream>
int myRec(int a, int b){
if (a==0){
return b;
}
if (b==0){
return a;
}
else return myRec(a/2, 2*b) + myRec(2*a, b/2);
}
int main()
{
if (46 == myRec(100, 100)) {
std::cout << "It works!";
}
}

Answer 1

Well, let us mentally trace it a bit:
Starting with a, b (a >= 2 and b >= 2)
myRec(a/2, 2*b) + something
something + myRec(2*a', b'/2)
Substituting for a/2 for a' and 2*b for b', we get myRec(2*(a/2), (b*2)/2), which is exactly where we started.
Therefore we will never get anywhere.
(Note that I have left out some rounding here, but you should easily see that with this kind of rounding you will only round down a to the nearest even number, at which point it will be forever alternating between that number and half that number)

Answer 2

I think you are missing on some case logic. I last program in C ages ago so correct my syntax if wrong. Assuming numbers less than 1 will be converted to zero automatically...
#include<iostream>
int myRec(int a, int b){
// Recurse only if both a and b are not zero
if (a!=0 && b!=0) {
return myRec(a/2, 2*b) + myRec(2*a, b/2);
}
// Otherwise check for any zero for a or b.
else {
if (a==0){
return b;
}
if (b==0){
return a;
}
}
}
UPDATE:
I have almost forgot how C works on return...
int myRec(int a, int b){
if (a==0){
return b;
}
if (b==0){
return a;
}
return myRec(a/2, 2*b) + myRec(2*a, b/2);
}
VBA equivalent with some changes for displaying variable states
Private Function myRec(a As Integer, b As Integer, s As String) As Integer
Debug.Print s & vbTab & a & vbTab & b
If a = 0 Then
myRec = b
End If
If b = 0 Then
myRec = a
End If
If a <> 0 And b <> 0 Then
myRec = myRec(a / 2, 2 * b, s & "L") + myRec(2 * a, b / 2, s & "R")
End If
End Function
Sub test()
Debug.Print myRec(100, 100, "T")
End Sub
Running the test in Excel gives this (a fraction of it as it overstacks Excel):
T: Top | L: Left branch in myRec | R: Right branch in myRec
The root cause will be the sum of the return which triggers more recursive calls.
Repeating of the original values of a and b on each branch from level 2 of the recursive tree...

Answer 3

So MyRec(2,2) = MyRec(1,4) + MyRec(4,1)
And MyRec(1,4) = MyRec(.5,8) + MyRec(2,2)
So MyRec(2,2) = MyRec(.5,8) + MyRec(2,2) + MyRec(4,1)
Oops.
(The .5's will actually be zeroes. But it doesn't matter. The point is that the function won't terminate for a large range of possible inputs.)

Answer 4

Expanding on gha.st's answer, consider the function's return value as a sum of expressions without having to worry about any code.
Firstly, we start with myRec(a,b). Let's just express that as (a,b) to make this easier to read.
As I go down each line, each expression is equivalent, disregarding the cases where a=0 or b=0.
(a,b) =
(a/2, 2b) + (2a, b/2) =
(a/4, 4b) + (a, b) + (a, b) + (4a, b/4)
Now, we see that at a non-terminating point in the expression, calculating (a,b) requires first calculating (a,b).
Recursion on a problem like this works because the arguments typically tend toward a 'base case' at which the recursion stops. A great example is sorting a list; you can recursively sort halves of the list until a list given as input has <= 2 elements, which is trivial without recursion. This is called mergesort.
However, your myRec function does not have a base case, since for non-zero a or b, the same arguments must be passed into the function at some point. That's like trying to sort a list, in which half of the list has as many elements as the entire list.

Answer 5

Try replacing the recursion call with:
return myRec(a/2, b/3) + myRec(a/3, b/2);