Get an array of the bit positions within a 64-bit integer

Question

OK, it may sound a bit complicated, but this is what I\'m trying to do :

• Take e.g. 10101010101
• And return { 0, 2, 4, 6, 8, 10 }

Answer 1

Replace your firstBit function with an intrinsic using the BSF or BSR instruction for a massive speedup.

In gcc, it'd be __builtin_ffsll and __builtin_ctzll

With Visual C+, _BitScanForward and _BitScanReverse

Answer 2

Here is another suggestion that can be profiled (can be combined with other suggestions for further optimization). Note, the loop here is O(number of set bits).

vector<UINT> bits_set (UINT64 data) 
{
    UINT n;
    vector<UINT> res;
    res.reserve(64);
    for (n = 0; data != 0; n++, data &= (data - 1))
    {
        res.push_back(log2(data & ~(data-1)));
    }
    return res;
}

Answer 3

const size_t size = sizeof(U64)*8;
U64 val = 1;

vector<UINT> res;
res.reserve(size);

for ( size_t i = size; i > 0; --i ) {
  if ( ( val & bitboard ) != 0 ) {
    res.push_back(i);
  }
  val <<= 1;
}

Answer 4

Bit shifts are really cheap. Lookup tables cost cache space, and your lookup also have integer multiplication. Just brute-force will be faster than clever techniques, I expect.

vector<UINT> DQBitboard::bits(U64 bitboard)
{
    vector<UINT> res;
    res.reserve(64);
    uint_fast8_t pos = 1;

    do {
        if (bitboard & 1) res.push_back(pos);
        ++pos;
    } while (bitboard >>= 1);

    return res;
}

You can unroll the loop a little bit, that may make it faster yet.

The std::vector is the most expensive part by far. Consider using the bitboard directly. For example:

struct bitboard_end_iterator{};
struct bitboard_iterator
{
    U64 value;
    uint_fast8_t pos;

    bitboard_iterator(U64 bitboard) : value(bitboard), pos(0)
    {
        ++(*this);
    }
    UINT operator*() const { return pos + 1; }
    bool operator==( bitboard_end_iterator ) const { return pos == 64; }
    operator bool() const { return pos < 64; }
    bitboard_iterator& operator++()
    {
        while (U64 prev = value) {
            value >>= 1;
            ++pos;
            if (prev & 1) return *this;
        }
        pos = 64;
        return *this;
    }
};

Now you can write

for( bitboard_iterator it(bitboard); it; ++it )
    cout << *it;

and I think you'll get your list of bits.

Version 2:

class bitboard_fast_iterator
{
    U64 value;
    uint_fast8_t pos;

public:
    bitboard_fast_iterator(U64 bitboard = 0) : value(bitboard), pos(__builtin_ctzll(value)) {}
    UINT operator*() const { return pos + 1; }
    operator bool() const { return value != 0; }
    bitboard_iterator& operator++()
    {
        value &= ~(1ULL << pos);
        pos = __builtin_ctzll(value);
        return *this;
    }
};

Answer 5

I have been wondering whether it would work faster with the bst assembly instruction. So I tried 3 implementations and got the following results for 10 million iterations:

Your implementation (Dr.Kameleon) 1.77 seconds

The log2() implementation (icepack) 2.17 seconds

My assembly implementation (me) 0.16 seconds

Output:

bits version:
Function started at 0
           ended at 177
              spent 177 (1.770000 seconds)
c version:
Function started at 177
           ended at 394
              spent 217 (2.170000 seconds)
c version:
Function started at 394
           ended at 410
              spent 16 (0.160000 seconds)

One point about C/C++, static is horrendous. It is actually compiled in a list of CPU instructions (NOT what I would expect either!!!) Instead, use an array outside your function in a nameless namespace. That will have the expected effect. Although in assembly you can use the .long (or some other size) and then %rip to reference the data from the IP.

Note: once compiled, I do not see the size (n) being used in my assembly version so I'm not too sure whether the returned array is valid. Outside of that, the code itself becomes a loop of 5 assembly instructions, hence the tiny increase in speed (about x10).

The reason for log2() slowness is that it converts the number to an xmm register and then does call to another function. It then converts the xmm register back to the regular register.

#include <stdlib.h>
#include <stdio.h>
#include <inttypes.h>
#include <unistd.h>
#include <sys/times.h>
#include <string.h>
#include <math.h>
#include <vector>

using namespace std;

namespace
{
const int index64[64] = {
    63,  0, 58,  1, 59, 47, 53,  2,
    60, 39, 48, 27, 54, 33, 42,  3,
    61, 51, 37, 40, 49, 18, 28, 20,
    55, 30, 34, 11, 43, 14, 22,  4,
    62, 57, 46, 52, 38, 26, 32, 41,
    50, 36, 17, 19, 29, 10, 13, 21,
    56, 45, 25, 31, 35, 16,  9, 12,
    44, 24, 15,  8, 23,  7,  6,  5  };
const uint64_t debruijn64 = 0x07EDD5E59A4E28C2ULL;
}

int firstBit(uint64_t bitboard)
{
    return index64[((bitboard & -bitboard) * debruijn64) >> 58];  
}

vector<int> bits(uint64_t bitboard)
{
    vector<int> res;
    res.reserve(64);

    while(bitboard)
    {
        int first = firstBit(bitboard);
        res.push_back(first);

        bitboard &= ~(1ULL << first);
    }
    return res;
}



vector<int> bits_c(uint64_t bitboard)
{
    int n;
    vector<int> res;
    res.reserve(64);
    for (n = 0; bitboard != 0; n++, bitboard &= (bitboard - 1))
    {
        res.push_back(log2(bitboard & ~(bitboard - 1)));
    }
    return res;
}


vector<int> bits_asm(uint64_t bitboard)
{
    int64_t n(0);
    int res[64];
    asm(
    "bsf %[b], %%rax\n\t"
    "je exit\n\t"
    ".align 16\n"
"loop:\n\t"
    "mov %%eax, (%[r],%[n],4)\n\t"
    "btr %%rax, %[b]\n\t"
    "inc %[n]\n\t"
    "bsf %[b], %%rax\n\t"
    "je loop\n"
"exit:\n\t"
    : /* output */ "=r" (n)
    : /* input */ [n] "r" (n), [r] "r" (res), [b] "r" (bitboard)
    : /* state */ "eax", "cc"
    );
    return vector<int>(res, res + n);
}




class run_timer
{
public:
    run_timer()
    {
    }

    void start()
    {
        times(&f_start);
    }

    void stop()
    {
        times(&f_stop);
    }

    void report(const char *msg)
    {
        printf("%s:\n"
               "Function started at %ld\n"
               "           ended at %ld\n"
               "              spent %ld (%f seconds)\n",
               msg,
               f_start.tms_utime,
               f_stop.tms_utime,
               f_stop.tms_utime - f_start.tms_utime,
               (double)(f_stop.tms_utime - f_start.tms_utime)/(double)sysconf(_SC_CLK_TCK));
    }

    struct tms f_start;
    struct tms f_stop;
};


int main(int argc, char *argv[])
{
    run_timer t;

    t.start();
    for(int i(0); i < 10000000; ++i)
    {
        bits(rand());
    }
    t.stop();
    t.report("bits version");

    t.start();
    for(int i(0); i < 10000000; ++i)
    {
        bits_c(rand());
    }
    t.stop();
    t.report("c version");

    t.start();
    for(int i(0); i < 10000000; ++i)
    {
        bits_asm(rand());
    }
    t.stop();
    t.report("c version");

    return 0;
}

Compiled with g++ with this command line:

c++ -msse4.2 -O2 -o bits -c bits.cpp

Although you may think that the -msse4.2 could be the problem with the log2() version, I tried without it and log2() is slower then.

Btw, I do not recommend this method since it is not portable. Only the Intel based computers will understand those instructions.

Answer 6

I actually think the fastest, simplest method is simply to loop around, but if we pass in a vector instead of later making a copy, it should be a little faster.

void DQBitboard::bits(U64 bitboard, vector<UINT> &res)
{
    res.clear();   // Make sure vector is empty. Not necessary if caller does this!
    int bit = 0;
    while (bitboard)
    {
        if (bitboard & 1) 
            res.push_back(bit);
        bit++;
        bitboard >>= 1;
    }

    return res;
}

The multiply in the findfirst will cost a bit, so I doubt it's really worth it.