Given a 2D array, convert to Z-Order - c++

Question

Having trouble wrapping my head around this conversion. I want to recursively convert a 2D NxN matrix into its z-order version.
For example given array:
[ 1 2 ]
[ 3 4 ]
The Z-order is
[ 1 2 3 4]
What are the steps recursively for the z-order conversion?

Answer 1

The recursive way is simple:
visit top-left
visit top-right
visit bottom-left
visit bottom-right
In code
#include <iostream>
template<typename M, typename CBACK>
void zorder(const M& m, int y0, int x0, int size,
CBACK cback)
{
if (size == 1) {
// Base case, just one cell
cback(m[y0][x0]);
} else {
// Recurse in Z-order
int h = size/2;
zorder(m, y0, x0, h, cback); // top-left
zorder(m, y0, x0+h, h, cback); // top-right
zorder(m, y0+h, x0, h, cback); // bottom-left
zorder(m, y0+h, x0+h, h, cback); // bottom-right
}
}
void print(int x) {
std::cout << x << " ";
}
int main(int argc, const char *argv[]) {
int x[][4] = {{ 1, 2, 3, 4},
{ 5, 6, 7, 8},
{ 9, 10, 11, 12},
{13, 14, 15, 16}};
zorder(x, 0, 0, 4, print);
std::cout << std::endl;
return 0;
}
Output of this program is
1 2 5 6 3 4 7 8 9 10 13 14 11 12 15 16
Note that there is also another non-recursive approach: to visit the elements in z-order you can just iterate over a counter and take odd-bits as y and even-bits as x (counting bits from 0):
int zorder_x_of(int index) {
int x = 0;
for (int b=0,k=0; (1<<b) <= index; b+=2,k++) {
x += ((index & (1<<b)) != 0) << k;
}
return x;
}
int zorder_y_of(int index) {
return zorder_x_of(index>>1);
}
template<typename M, typename CBACK>
void zorder2(const M& m, int size, CBACK cback)
{
for (int i=0; i<size*size; i++) {
cback(m[zorder_y_of(i)][zorder_x_of(i)]);
}
}
Note:
In the above code samples I created a function that accepts a "callback" (named cback) that is something that will be called with the elements of the matrix, one at a time, in z-order.
To allow using both as matrix and as callback anything that supports double [] indexing and anything that can be called I used a C++ template.
In the main program as matrix I've used a bi-dimensional array of integers and a function, but the code would have compiled even for example with an std::vector< std::vector< double > > as matrix and an object instance of a class providing operator()(double) as callback.

Answer 2

A 2D matrix behaves internally as an 1D array i.e. it is already in "Z order".
Just iterate across the pointer that points to the first element up to NxM where N is the number of columns and M the number of rows.
example:
int arr[2][2] = {{2,4},{3,5}};
for (int i=0; i<2 * 2; ++i){
std::cout << *(&arr[0][0] + i); // or *(arr + i)
}