Efficient Algorithm for Bit Reversal (from MSB->LSB to LSB->MSB) in C
What is the most efficient algorithm to achieve the following:
0010 0000 => 0000 0100
The conversion is from MSB->LSB to LSB->MSB. All bits must be reversed; that is, this is not endianness-swapping.
Answer
NOTE: All algorithms below are in C, but should be portable to your language of choice (just don't look at me when they're not as fast :)
Options
Low Memory (32-bit int, 32-bit machine)(from here):
unsigned int
reverse(register unsigned int x)
{
x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8));
return((x >> 16) | (x << 16));
}
From the famous Bit Twiddling Hacks page:
Fastest (lookup table):
static const unsigned char BitReverseTable256[] =
{
0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0,
0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8,
0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4,
0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC,
0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2,
0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,
0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6,
0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,
0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,
0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9,
0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,
0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,
0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3,
0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,
0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7,
0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF
};
unsigned int v; // reverse 32-bit value, 8 bits at time
unsigned int c; // c will get v reversed
// Option 1:
c = (BitReverseTable256[v & 0xff] << 24) |
(BitReverseTable256[(v >> 8) & 0xff] << 16) |
(BitReverseTable256[(v >> 16) & 0xff] << 8) |
(BitReverseTable256[(v >> 24) & 0xff]);
// Option 2:
unsigned char * p = (unsigned char *) &v;
unsigned char * q = (unsigned char *) &c;
q[3] = BitReverseTable256[p[0]];
q[2] = BitReverseTable256[p[1]];
q[1] = BitReverseTable256[p[2]];
q[0] = BitReverseTable256[p[3]];
You can extend this idea to 64-bit ints, or trade off memory for speed (assuming your L1 Data Cache is large enough), and reverse 16 bits at a time with a 64K-entry lookup table.
Others
Simple
unsigned int v; // input bits to be reversed
unsigned int r = v & 1; // r will be reversed bits of v; first get LSB of v
int s = sizeof(v) * CHAR_BIT - 1; // extra shift needed at end
for (v >>= 1; v; v >>= 1)
{
r <<= 1;
r |= v & 1;
s--;
}
r <<= s; // shift when v's highest bits are zero
Faster (32-bit processor)
unsigned char b = x;
b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16;
Faster (64-bit processor)
unsigned char b; // reverse this (8-bit) byte
b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
If you want to do this on a 32-bit int, just reverse the bits in each byte, and reverse the order of the bytes. That is:
unsigned int toReverse;
unsigned int reversed;
unsigned char inByte0 = (toReverse & 0xFF);
unsigned char inByte1 = (toReverse & 0xFF00) >> 8;
unsigned char inByte2 = (toReverse & 0xFF0000) >> 16;
unsigned char inByte3 = (toReverse & 0xFF000000) >> 24;
reversed = (reverseBits(inByte0) << 24) | (reverseBits(inByte1) << 16) | (reverseBits(inByte2) << 8) | (reverseBits(inByte3);
Results
I benchmarked the two most promising solutions, the lookup table, and bitwise-AND (the first one). The test machine is a laptop w/ 4GB of DDR2-800 and a Core 2 Duo T7500 @ 2.4GHz, 4MB L2 Cache; YMMV. I used gcc 4.3.2 on 64-bit Linux. OpenMP (and the GCC bindings) were used for high-resolution timers.
reverse.c
#include <stdlib.h>
#include <stdio.h>
#include <omp.h>
unsigned int
reverse(register unsigned int x)
{
x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8));
return((x >> 16) | (x << 16));
}
int main()
{
unsigned int *ints = malloc(100000000*sizeof(unsigned int));
unsigned int *ints2 = malloc(100000000*sizeof(unsigned int));
for(unsigned int i = 0; i < 100000000; i++)
ints[i] = rand();
unsigned int *inptr = ints;
unsigned int *outptr = ints2;
unsigned int *endptr = ints + 100000000;
// Starting the time measurement
double start = omp_get_wtime();
// Computations to be measured
while(inptr != endptr)
{
(*outptr) = reverse(*inptr);
inptr++;
outptr++;
}
// Measuring the elapsed time
double end = omp_get_wtime();
// Time calculation (in seconds)
printf("Time: %f seconds\n", end-start);
free(ints);
free(ints2);
return 0;
}
reverse_lookup.c
#include <stdlib.h>
#include <stdio.h>
#include <omp.h>
static const unsigned char BitReverseTable256[] =
{
0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0,
0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8,
0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4,
0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC,
0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2,
0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,
0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6,
0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,
0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,
0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9,
0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,
0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,
0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3,
0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,
0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7,
0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF
};
int main()
{
unsigned int *ints = malloc(100000000*sizeof(unsigned int));
unsigned int *ints2 = malloc(100000000*sizeof(unsigned int));
for(unsigned int i = 0; i < 100000000; i++)
ints[i] = rand();
unsigned int *inptr = ints;
unsigned int *outptr = ints2;
unsigned int *endptr = ints + 100000000;
// Starting the time measurement
double start = omp_get_wtime();
// Computations to be measured
while(inptr != endptr)
{
unsigned int in = *inptr;
// Option 1:
//*outptr = (BitReverseTable256[in & 0xff] << 24) |
// (BitReverseTable256[(in >> 8) & 0xff] << 16) |
// (BitReverseTable256[(in >> 16) & 0xff] << 8) |
// (BitReverseTable256[(in >> 24) & 0xff]);
// Option 2:
unsigned char * p = (unsigned char *) &(*inptr);
unsigned char * q = (unsigned char *) &(*outptr);
q[3] = BitReverseTable256[p[0]];
q[2] = BitReverseTable256[p[1]];
q[1] = BitReverseTable256[p[2]];
q[0] = BitReverseTable256[p[3]];
inptr++;
outptr++;
}
// Measuring the elapsed time
double end = omp_get_wtime();
// Time calculation (in seconds)
printf("Time: %f seconds\n", end-start);
free(ints);
free(ints2);
return 0;
}
I tried both approaches at several different optimizations, ran 3 trials at each level, and each trial reversed 100 million random unsigned ints. For the lookup table option, I tried both schemes (options 1 and 2) given on the bitwise hacks page. Results are shown below.
Bitwise AND
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -o reverse reverse.c
mrj10@mjlap:~/code$ ./reverse
Time: 2.000593 seconds
mrj10@mjlap:~/code$ ./reverse
Time: 1.938893 seconds
mrj10@mjlap:~/code$ ./reverse
Time: 1.936365 seconds
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O2 -o reverse reverse.c
mrj10@mjlap:~/code$ ./reverse
Time: 0.942709 seconds
mrj10@mjlap:~/code$ ./reverse
Time: 0.991104 seconds
mrj10@mjlap:~/code$ ./reverse
Time: 0.947203 seconds
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O3 -o reverse reverse.c
mrj10@mjlap:~/code$ ./reverse
Time: 0.922639 seconds
mrj10@mjlap:~/code$ ./reverse
Time: 0.892372 seconds
mrj10@mjlap:~/code$ ./reverse
Time: 0.891688 seconds
Lookup Table (option 1)
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.201127 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.196129 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.235972 seconds
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O2 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 0.633042 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 0.655880 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 0.633390 seconds
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O3 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 0.652322 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 0.631739 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 0.652431 seconds
Lookup Table (option 2)
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.671537 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.688173 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.664662 seconds
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O2 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.049851 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.048403 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.085086 seconds
mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O3 -o reverse_lookup reverse_lookup.c
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.082223 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.053431 seconds
mrj10@mjlap:~/code$ ./reverse_lookup
Time: 1.081224 seconds
Conclusion
Use the lookup table, with option 1 (byte addressing is unsurprisingly slow) if you're concerned about performance. If you need to squeeze every last byte of memory out of your system (and you might, if you care about the performance of bit reversal), the optimized versions of the bitwise-AND approach aren't too shabby either.
Caveat
Yes, I know the benchmark code is a complete hack. Suggestions on how to improve it are more than welcome. Things I know about:
- I don't have access to ICC. This may be faster (please respond in a comment if you can test this out).
- A 64K lookup table may do well on some modern microarchitectures with large L1D.
- -mtune=native didn't work for -O2/-O3 (
ldblew up with some crazy symbol redefinition error), so I don't believe the generated code is tuned for my microarchitecture. - There may be a way to do this slightly faster with SSE. I have no idea how, but with fast replication, packed bitwise AND, and swizzling instructions, there's got to be something there.
- I know only enough x86 assembly to be dangerous; here's the code GCC generated on -O3 for option 1, so somebody more knowledgable than myself can check it out:
32-bit
.L3:
movl (%r12,%rsi), %ecx
movzbl %cl, %eax
movzbl BitReverseTable256(%rax), %edx
movl %ecx, %eax
shrl $24, %eax
mov %eax, %eax
movzbl BitReverseTable256(%rax), %eax
sall $24, %edx
orl %eax, %edx
movzbl %ch, %eax
shrl $16, %ecx
movzbl BitReverseTable256(%rax), %eax
movzbl %cl, %ecx
sall $16, %eax
orl %eax, %edx
movzbl BitReverseTable256(%rcx), %eax
sall $8, %eax
orl %eax, %edx
movl %edx, (%r13,%rsi)
addq $4, %rsi
cmpq $400000000, %rsi
jne .L3
EDIT: I also tried using uint64_t types on my machine to see if there was any performance boost. Performance was about 10% faster than 32-bit, and was nearly identical whether you were just using 64-bit types to reverse bits on two 32-bit int types at a time, or whether you were actually reversing bits in half as many 64-bit values. The assembly code is shown below (for the former case, reversing bits for two 32-bit int types at a time):
.L3:
movq (%r12,%rsi), %rdx
movq %rdx, %rax
shrq $24, %rax
andl $255, %eax
movzbl BitReverseTable256(%rax), %ecx
movzbq %dl,%rax
movzbl BitReverseTable256(%rax), %eax
salq $24, %rax
orq %rax, %rcx
movq %rdx, %rax
shrq $56, %rax
movzbl BitReverseTable256(%rax), %eax
salq $32, %rax
orq %rax, %rcx
movzbl %dh, %eax
shrq $16, %rdx
movzbl BitReverseTable256(%rax), %eax
salq $16, %rax
orq %rax, %rcx
movzbq %dl,%rax
shrq $16, %rdx
movzbl BitReverseTable256(%rax), %eax
salq $8, %rax
orq %rax, %rcx
movzbq %dl,%rax
shrq $8, %rdx
movzbl BitReverseTable256(%rax), %eax
salq $56, %rax
orq %rax, %rcx
movzbq %dl,%rax
shrq $8, %rdx
movzbl BitReverseTable256(%rax), %eax
andl $255, %edx
salq $48, %rax
orq %rax, %rcx
movzbl BitReverseTable256(%rdx), %eax
salq $40, %rax
orq %rax, %rcx
movq %rcx, (%r13,%rsi)
addq $8, %rsi
cmpq $400000000, %rsi
jne .L3