How to compute a 3D Morton number (interleave the bits of 3 ints)

Question

I\'m looking for a fast way to compute a 3D Morton number. This site has a magic-number based trick for doing it for 2D Morton numbers, but it doesn\'t seem obvious how to e

Answer 1

Good timing, I just did this last month!

The key was to make two functions. One spreads bits out to every-third bit. Then we can combine three of them together (with a shift for the last two) to get the final Morton interleaved value.

This code interleaves starting at the HIGH bits (which is more logical for fixed point values.) If your application is only 10 bits per component, just shift each value left by 22 in order to make it start at the high bits.

/* Takes a value and "spreads" the HIGH bits to lower slots to seperate them.
   ie, bit 31 stays at bit 31, bit 30 goes to bit 28, bit 29 goes to bit 25, etc.
   Anything below bit 21 just disappears. Useful for interleaving values
   for Morton codes. */
inline unsigned long spread3(unsigned long x)
{
  x=(0xF0000000&x) | ((0x0F000000&x)>>8) | (x>>16); // spread top 3 nibbles
  x=(0xC00C00C0&x) | ((0x30030030&x)>>4);
  x=(0x82082082&x) | ((0x41041041&x)>>2);
  return x;
}

inline unsigned long morton(unsigned long x, unsigned long y, unsigned long z)
{
  return spread3(x) | (spread3(y)>>1) | (spread3(z)>>2);
}