Reverse the order of bits in a bit array

问题

I have a long sequence of bits stored in an array of unsigned long integers, like this

struct bit_array
{
    int size; /* nr of bits */
    unsigned long *array; /* the container that stores bits */
}

I am trying to design an algorithm to reverse the order of bits in *array. Problems:

size can be anything, i.e. not necessarily a multiple of 8 or 32 etc, so the first bit in the input array can end up at any position within the unsigned long in the output array;
the algorithm should be platform-independent, i.e. work for any sizeof(unsigned long).

Code, pseudocode, algo description etc. -- anything better than bruteforce ("bit by bit") approach is welcome.

回答1:

My favorite solution is to fill a lookup-table that does bit-reversal on a single byte (hence 256 byte entries).

You apply the table to 1 to 4 bytes of the input operand, with a swap. If the size isn't a multiple of 8, you will need to adjust by a final right shift.

This scales well to larger integers.

Example:

11 10010011 00001010 -> 01010000 11001001 11000000 -> 01 01000011 00100111

To split the number into bytes portably, you need to use bitwise masking/shifts; mapping of a struct or array of bytes onto the integer can make it more efficient.

For brute performance, you can think of mapping up to 16 bits at a time, but this doesn't look quite reasonable.

回答2:

I like the idea of lookup table. Still it's also a typical task for log(n) group bit tricks that may be very fast. Like:

unsigned long reverseOne(unsigned long x) {
  x = ((x & 0xFFFFFFFF00000000) >> 32) | ((x & 0x00000000FFFFFFFF) << 32);
  x = ((x & 0xFFFF0000FFFF0000) >> 16) | ((x & 0x0000FFFF0000FFFF) << 16);
  x = ((x & 0xFF00FF00FF00FF00) >> 8)  | ((x & 0x00FF00FF00FF00FF) << 8);
  x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4)  | ((x & 0x0F0F0F0F0F0F0F0F) << 4);
  x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2)  | ((x & 0x3333333333333333) << 2);
  x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1)  | ((x & 0x5555555555555555) << 1);
  return x;
}

The underlying idea is that when we aim to reverse the order of some sequence we may swap the head and tail halves of this sequence and then separately reverse each of halves (which is done here by applying the same procedure recursively to each half).

Here is a more portable version supporting unsigned long widths of 4,8,16 or 32 bytes.

#include <limits.h>

#define ones32 0xFFFFFFFFUL
#if (ULONG_MAX >> 128)
#define fill32(x) (x|(x<<32)|(x<<64)|(x<<96)|(x<<128)|(x<<160)|(x<<192)|(x<<224))
#define patt128 (ones32|(ones32<<32)|(ones32<<64) |(ones32<<96))
#define patt64  (ones32|(ones32<<32)|(ones32<<128)|(ones32<<160))
#define patt32  (ones32|(ones32<<64)|(ones32<<128)|(ones32<<192))
#else
#if (ULONG_MAX >> 64)
#define fill32(x) (x|(x<<32)|(x<<64)|(x<<96))
#define patt64  (ones32|(ones32<<32))
#define patt32  (ones32|(ones32<<64))
#else
#if (ULONG_MAX >> 32)
#define fill32(x) (x|(x<<32))
#define patt32  (ones32)
#else
#define fill32(x) (x)
#endif
#endif
#endif

unsigned long reverseOne(unsigned long x) {
#if (ULONG_MAX >> 32)
#if (ULONG_MAX >> 64)
#if (ULONG_MAX >> 128)
  x = ((x & ~patt128) >> 128) | ((x & patt128) << 128);
#endif
  x = ((x & ~patt64) >> 64) | ((x & patt64) << 64);
#endif
  x = ((x & ~patt32) >> 32) | ((x & patt32) << 32);
#endif
  x = ((x & fill32(0xffff0000UL)) >> 16) | ((x & fill32(0x0000ffffUL)) << 16);
  x = ((x & fill32(0xff00ff00UL)) >> 8)  | ((x & fill32(0x00ff00ffUL)) << 8);
  x = ((x & fill32(0xf0f0f0f0UL)) >> 4)  | ((x & fill32(0x0f0f0f0fUL)) << 4);
  x = ((x & fill32(0xccccccccUL)) >> 2)  | ((x & fill32(0x33333333UL)) << 2);
  x = ((x & fill32(0xaaaaaaaaUL)) >> 1)  | ((x & fill32(0x55555555UL)) << 1);
  return x;
}

回答3:

In a collection of related topics which can be found here, the bits of an individual array entry could be reversed as follows.

unsigned int v;     // input bits to be reversed
unsigned int r = v; // 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

The reversal of the entire array could be done afterwards by rearranging the individual positions.

回答4:

You must define what is the order of bits in an unsigned long. You might assume that bit n is corresponds to array[x] & (1 << n) but this needs to be specified. If so, you need to handle the byte ordering (little or big endian) if you are going to use access the array as bytes instead of unsigned long.

I would definitely implement brute force first and measure whether the speed is an issue. No need to waste time trying to optimize this if it is not used a lot on large arrays. An optimized version can be tricky to implement correctly. If you end up trying anyway, the brute force version can be used to verify correctness on test values and benchmark the speed of the optimized version.

回答5:

The fact that the size is not multiple of sizeof(long) is the hardest part of the problem. This can result in a lot of bit shifting.

But, you don't have to do that if you can introduce new struct member:

struct bit_array
{
    int size; /* nr of bits */
    int offset; /* First bit position */
    unsigned long *array; /* the container that stores bits */
}

Offset would tell you how many bits to ignore at the beginning of the array.

Then you only only have to do following steps:

Reverse array elements.
Swap bits of each element. There are many hacks for in the other answers, but your compiler might also provide intrisic functions to do it in fewer instructions (like RBIT instruction on some ARM cores).
Calculate new starting offset. This is equal to unused bits the last element had.

回答6:

I would split the problem into two parts.

First, I would ignore the fact that the number of used bits is not a multiple of 32. I would use one of the given methods to swap around the whole array like that.

pseudocode:

for half the longs in the array:
    take the first longword;
    take the last longword;
    swap the bits in the first longword
    swap the bits in the last longword;

    store the swapped first longword into the last location;
    store the swapped last longword into the first location;

and then fix up the fact that the first few bits (call than number n) are actually garbage bits from the end of the longs:

for all of the longs in the array:
    split the value in the leftmost n bits and the rest;
    store the leftmost n bits into the righthand part of the previous word;
    shift the rest bits to the left over n positions (making the rightmost n bits zero);
    store them back;

You could try to fold that into one pass over the whole array of course. Something like this:

for half the longs in the array:
    take the first longword;
    take the last longword;
    swap the bits in the first longword
    swap the bits in the last longword;

    split both value in the leftmost n bits and the rest;

    for the new first longword:
        store the leftmost n bits into the righthand side of the previous word;
        store the remaining bits into the first longword, shifted left;

    for the new last longword:
        remember the leftmost n bits for the next iteration;
        store the remembered leftmost n bits, combined with the remaining bits, into the last longword;

    store the swapped first longword into the last location;
    store the swapped last longword into the first location;

I'm abstracting from the edge cases here (first and last longword), and you may need to reverse the shifting direction depending on how the bits are ordered inside each longword.

来源：https://stackoverflow.com/questions/33910399/reverse-the-order-of-bits-in-a-bit-array

标签

bitarray