Calculate n % 12

4

(Language is irrelevant)

n-((48&(11-n))>>2)

Woo! Got to 4.

11-n will ensure all of the high order bits are set if and only if n>= 12.

48&(11-n) == if n>11 then 48 else 0

(48&(11-n))>>2 == if n>11 then 12 else 0

n-((48&(11-n))>>2) is the answer

4

A solution with a lookup table (it looks up i ^ (i % 12)):

i ^ (0x1d4c000 >> (i & 0xfc) & 30)

4

Here's another solution with 4 operations:

i - ((0xffff >> (i - 12)) & 12)

It assumes that the count operand of bitshifts is implicitly taken mod 32, i.e. x >> -1 is the same as x >> 31.

5

Another approach, using a lookup table:

i - (16773120 >> i & 1) * 12

bash – 1

echo `seq 0 11` `seq 0 11` | awk '{print $(number+1)}'

e.g.

$ echo `seq 0 11` `seq 0 11` | awk '{print $(0+1)}'
0

$ echo `seq 0 11` `seq 0 11` | awk '{print $(11+1)}'
11

$ echo `seq 0 11` `seq 0 11` | awk '{print $(12+1)}'
0

$ echo `seq 0 11` `seq 0 11` | awk '{print $(23+1)}'
11