#include
using namespace std;
int g(float A[] , int L , int H)
{
if (L==H)
if (A[L] > 0.0)
return 1;
else
retu
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