Question

#include <iostream>
using namespace std;

int g(float A[] , int L , int H)
{
   if (L==H)
      if (A[L] > 0.0)
         return 1;
      else 
         return 0;
   int M = (L+H)/2;

   return g(A,L,M)+ g(A,M+1,H);
}
int main (void)
{       
   float A[] = {-1.5 ,3.1,-5.2,0.0};
   g(A,0,3);

   system ("pause");
   return 0;
}

its asking me what is return by the function g and what the function does

here is what i got so far

first call is g(A , 0 ,3)
-total skip the IF statement and M = 1 since it its a int
-return g(A,1,3) + g(A,2 3)

second call
– g(A,1,3) skipp the if statement again
– M = 0;
– g(A,2 3) skip the if statement again
– M= 2;

third call
-g(A, 0,0,)
return 0
-g(A,3,3)
return 0;

so it just return 0?

and i am guessing it is dividing the middle value and some sort of binary search?

Answer 1

It’s a convoluted way to count how many numbers in the array is greater than 0. And if you try to run this in a compiler, the return value is 1 because the only number that is greater than 0 in the array is 3.1.

at first run:

{-1.5, 3.1, -5.2, 0.0}
   0    1     2    3
   L               H

then since L=0 and H=3, M = (0+3)/2 = 3/2 = 1 when you get to g(A, L, M) + g(A, M+1, H), you branch into two:

{-1.5, 3.1, -5.2, 0.0}
   0    1     2    3
   L               H
   L1   H1    L2   H2

let’s do the left part g(A, L1, H1) = g(A, 0, 1) first:

{-1.5, 3.1, -5.2, 0.0}
   0    1     2    3
   L               H
   L1   H1    L2   H2
   ^^^^^^^

again since L1=0, H1=1, and so M1 = (0+1)/2 = 1/2 = 0 and you branch into two again g(A, 0, 0) and g(A, 1, 1):

{-1.5,    3.1,    -5.2, 0.0}
   0       1        2    3
   L               H
   L1      H1      L2    H2
L11,H11 L12,H12

on the left part, since -1.5 <= 0 therefore g(A, L11, H11) = g(A, 0, 0) = 0, on the right part, since 3.1 > 0 therefore g(A, L12, H12) = g(A, 1, 1) = 1.

So therefore g(A, 0, 1) = g(A, 0, 0) + g(A, 1, 1) = 1.

Do the same with g(A, L2, H2), and you get that g(A, L, H) = g(A, L1, H1) + g(A, L2, H2) = 1 + 0 = 1.

@Nawaz had a good idea of visualizing this into a binary tree, basically you start with at the root of the tree:

{-1.5, 3.1, -5.2, 0.0}

At the second layer of iteration, you split the array into two:

     {-1.5, 3.1, -5.2, 0.0}
              /   \
             /     \
            /       \
           /         \
   {-1.5, 3.1}    {-5.2, 0.0}

At the third layer, you split again:

     {-1.5, 3.1, -5.2, 0.0}
              /   \
             /     \
            /       \
           /         \
   {-1.5, 3.1}    {-5.2, 0.0}
      /   \          /   \
     /     \        /     \
 {-1.5}   {3.1}  {-5.2}   {0.0}

At this point L==H so, we can evaluate the nodes:

     {-1.5, 3.1, -5.2, 0.0}
              /   \
             /     \
            /       \
           /         \
   {-1.5, 3.1}    {-5.2, 0.0}
      /   \          /   \
     /     \        /     \
 {-1.5}   {3.1}  {-5.2}   {0.0}
    |       |       |       |
    0       1       0       0

and to find the return values, we sum up:

     {-1.5, 3.1, -5.2, 0.0}
              /   \
             /     \
            /       \
           /         \
   {-1.5, 3.1}    {-5.2, 0.0}
      0+1=1          0+0=0

and lastly

     {-1.5, 3.1, -5.2, 0.0}
             1+0=1