Bit hacking and modulo operation

Question

While reading this: http://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith64BitsDiv

I came to the phrase:

The last step, which in

Answer 1

The modulo operation does not give you the inverted bits per se, it is just a binning operation.

First Line : word expansion

b * 0x0202020202 = 01001010 01001010 01001010 01001010 01001010 0

The multiplication operation has a convolution property, which means it replicate the input variable several times (5 here since it's a 8-bit word).

First Line : reversing bits

That's the most tricky part of the hack. You have to remember that we are working on a 8-bit word : b = abcdefgh, where [a-h] are either 1 or 0.

b  * 0x0202020202 = abcdefghabcdefghabcdefghabcdefghabcdefgha
    & 10884422010 = a0000f000b0000g000c0000h000d00000000e0000

Last Line : word binning

Modulo has a peculiar property : 10 ≡ 1 (mod 9) so 100 ≡ 10*10 ≡ 10*1 (mod 9) ≡ 1 (mod 9).

More generally, for a base b, b ≡ 1 (mod b - 1) so for all number a ≡ sum(a_k*b^k) ≡ sum (a_k) (mod b - 1).

In the example, base = 1024 (10 bits) so

b ≡ a0000f000b0000g000c0000h000d00000000e0000 
  ≡ a*base^4 + 0000f000b0*base^3 + 000g000c00*base^2 + 00h000d000*base +00000e0000 
  ≡ a + 0000f000b0 + 000g000c00 + 00h000d000 + 00000e0000 (mod b - 1)
  ≡  000000000a
   + 0000f000b0 
   + 000g000c00 
   + 00h000d000 
   + 00000e0000 (mod b - 1)
 ≡   00hgfedcba (mod b - 1) since there is no carry (no overlap)