algorithm - 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 extend it to 3D.

So basically I have 3 10-bit numbers that I want to interleave into a single 30 bit number with a minimal number of operations.

question from:https://stackoverflow.com/questions/1024754/how-to-compute-a-3d-morton-number-interleave-the-bits-of-3-ints

Answer 1

You can use the same technique. I'm assuming that variables contain 32-bit integers with the highest 22 bits set to 0 (which is a bit more restrictive than necessary). For each variable x containing one of the three 10-bit integers we do the following:

x = (x | (x << 16)) & 0x030000FF;
x = (x | (x <<  8)) & 0x0300F00F;
x = (x | (x <<  4)) & 0x030C30C3;
x = (x | (x <<  2)) & 0x09249249;

Then, with x,y and z the three manipulated 10-bit integers we get the result by taking:

x | (y << 1) | (z << 2)

The way this technique works is as follows. Each of the x = ... lines above "splits" groups of bits in half such that there is enough space in between for the bits of the other integers. For example, if we consider three 4-bit integers, we split one with bits 1234 into 000012000034 where the zeros are reserved for the other integers. In the next step we split 12 and 34 in the same way to get 001002003004. Even though 10 bits doesn't make for a nice repeated division in two groups, you can just consider it 16 bits where you lose the highest ones in the end.

As you can see from the first line, you actually only need that for each input integer x it holds that x & 0x03000000 == 0.