Serge Rogatch Serge Rogatch - 1 year ago 85
C++ Question

Interleave bits efficiently

I need to make

out of 2
interleaving the bits: if
, I need C=
. Is there a way to do this efficiently? So far I've got only the naive approach with a
loop of 32 iterations, where each iteration does

I guess there should be some mathematical trick like multiplying A and B by some special number which results in interleaving their bits with zeros in the resulting 64-bit number, so that what only remains is to
these products. But I can't find such multiplier.

Another potential way to go is a compiler intrinsic or assembly instruction, but I don't know of such.

Answer Source

NathanOliver's link offers the 16-bit -> 32-bit implementation:

static const unsigned int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF};
static const unsigned int S[] = {1, 2, 4, 8};

unsigned int x; // Interleave lower 16 bits of x and y, so the bits of x
unsigned int y; // are in the even positions and bits from y in the odd;
unsigned int z; // z gets the resulting 32-bit Morton Number.  
            // x and y must initially be less than 65536.

x = (x | (x << S[3])) & B[3];
x = (x | (x << S[2])) & B[2];
x = (x | (x << S[1])) & B[1];
x = (x | (x << S[0])) & B[0];

y = [the same thing on y]

z = x | (y << 1);

Which works by:

  1. leave the low 8 bits of x where they are. Move the high 8 bits up by 8;
  2. divide in half and do the same thing, this time leaving the low pairs of 4 bits where they are and moving the others up by 4;
  3. and again, and again.

I.e. it proceeds as:

-> 00000000abcdefgh 00000000ijklmnop
-> 0000abcd0000efgh 0000ijkl0000mnop
-> 00ab00cd00ef00gh 00ij00kl00mn00op
-> 0a0b0c0d0e0f0g0h 0i0j0k0l0m0n0o0p

And then combines the two inputs together.

As per my earlier comment, to extend that to 64 bits, just add an initial shift by 16 and mask by 0x0000ffff0000ffff, either because you can intuitively follow the pattern or as a divide-and-conquer step, turning the 32-bit problem into two non-overlapping 16-bit problems and then using the 16-bit solution.

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