Bit hack to generate all integers with a given number of 1s - bit

Question

I forgot a bit hack to generate all integers with a given number of 1s. Does anybody remember it (and probably can explain it also)?

Answer 1

From Bit Twiddling Hacks
Update Test program Live On Coliru
#include <utility>
#include <iostream>
#include <bitset>
using I = uint8_t;
auto dump(I v) { return std::bitset<sizeof(I) * __CHAR_BIT__>(v); }
I bit_twiddle_permute(I v) {
I t = v | (v - 1); // t gets v's least significant 0 bits set to 1
// Next set to 1 the most significant bit to change,
// set to 0 the least significant ones, and add the necessary 1 bits.
I w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
return w;
}
int main() {
I p = 0b001001;
std::cout << dump(p) << "\n";
for (I n = bit_twiddle_permute(p); n>p; p = n, n = bit_twiddle_permute(p)) {
std::cout << dump(n) << "\n";
}
}
Prints
00001001
00001010
00001100
00010001
00010010
00010100
00011000
00100001
00100010
00100100
00101000
00110000
01000001
01000010
01000100
01001000
01010000
01100000
10000001
10000010
10000100
10001000
10010000
10100000
11000000
Compute the lexicographically next bit permutation
Suppose we have a pattern of N bits set to 1 in an integer and we want the next permutation of N 1 bits in a lexicographical sense. For example, if N is 3 and the bit pattern is 00010011, the next patterns would be 00010101, 00010110, 00011001,00011010, 00011100, 00100011, and so forth. The following is a fast way to compute the next permutation.
unsigned int v; // current permutation of bits
unsigned int w; // next permutation of bits
unsigned int t = v | (v - 1); // t gets v's least significant 0 bits set to 1
// Next set to 1 the most significant bit to change,
// set to 0 the least significant ones, and add the necessary 1 bits.
w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the number of trailing zeros. If you are using Microsoft compilers for x86, the intrinsic is _BitScanForward. These both emit a bsf instruction, but equivalents may be available for other architectures. If not, then consider using one of the methods for counting the consecutive zero bits mentioned earlier.
Here is another version that tends to be slower because of its division operator, but it
does not require counting the trailing zeros.
unsigned int t = (v | (v - 1)) + 1;
w = t | ((((t & -t) / (v & -v)) >> 1) - 1);
Thanks to Dario Sneidermanis of Argentina, who provided this on November 28, 2009.

Answer 2

For bit hacks I like to refer to this page: Bit Twiddling Hacks.
Regarding your specific question, read the part entitled Compute the lexicographically next bit permutation.
Compute the lexicographically next bit permutation
Suppose we have a pattern of N bits set to 1 in an integer and we want the next permutation of N 1 bits in a lexicographical sense. For example, if N is 3 and the bit pattern is 00010011, the next patterns would be 00010101, 00010110, 00011001,00011010, 00011100, 00100011, and so forth. The following is a fast way to compute the next permutation.
unsigned int v; // current permutation of bits
unsigned int w; // next permutation of bits
unsigned int t = v | (v - 1); // t gets v's least significant 0 bits set to 1
// Next set to 1 the most significant bit to change,
// set to 0 the least significant ones, and add the necessary 1 bits.
w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the number of trailing zeros. If you are using Microsoft compilers for x86, the intrinsic is _BitScanForward. These both emit a bsf instruction, but equivalents may be available for other architectures. If not, then consider using one of the methods for counting the consecutive zero bits mentioned earlier.
Here is another version that tends to be slower because of its division operator, but it does not require counting the trailing zeros.
unsigned int t = (v | (v - 1)) + 1;
w = t | ((((t & -t) / (v & -v)) >> 1) - 1);
Thanks to Dario Sneidermanis of Argentina, who provided this on November 28, 2009.

Answer 3

To add onto #sehe's answer included below (originally from Dario Sneidermanis also at http://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation.)
#include <utility>
#include <iostream>
#include <bitset>
using I = uint8_t;
auto dump(I v) { return std::bitset<sizeof(I) * __CHAR_BIT__>(v); }
I bit_twiddle_permute(I v) {
I t = v | (v - 1); // t gets v's least significant 0 bits set to 1
// Next set to 1 the most significant bit to change,
// set to 0 the least significant ones, and add the necessary 1 bits.
I w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
return w;
}
int main() {
I p = 0b001001;
std::cout << dump(p) << "\n";
for (I n = bit_twiddle_permute(p); n>p; p = n, n = bit_twiddle_permute(p))
{
std::cout << dump(n) << "\n";
}
}
There are boundary issues with bit_twiddle_permute(I v). Whenever v is the last permutation, t is all 1's (e.g. 2^8 - 1), (~t & -~t) = 0, and w is the first permutation of bits with one fewer 1s than v, except when v = 000000000 in which case w = 01111111. In particular if you set p to 0; the loop in main will produce all permutations with seven 1's, and the following slight modification of the for loop, will cycle through all permutations with 0, 7, 6, ..., 1 bits set -
for (I n = bit_twiddle_permute(p); n>p; n = bit_twiddle_permute(n))
If this is the intention, it is perhaps worth a comment. If not it is trivial to fix, e.g.
if (t == (I)(-1)) { return v >> __builtin_ctz(v); }
So with an additional small simplification
I bit_twiddle_permute2(I v) {
I t = (v | (v - 1)) + 1;
if (t == 0) { return v >> __builtin_ctz(v); }
I w = t | ((~t & v) >> (__builtin_ctz(v) + 1));
return w;
}
int main() {
I p = 0b1;
cout << dump(p) << "\n";
for (I n = bit_twiddle_permute2(p); n>p; n = bit_twiddle_permute2(n)) {
cout << dump(n) << "\n";
}
}
The following adaptation of Dario Sneidermanis's idea may be slightly easier to follow
I bit_twiddle_permute3(I v) {
int n = __builtin_ctz(v);
I s = v >> n;
I t = s + 1;
I w = (t << n) | ((~t & s) >> 1);
return w;
}
or with a similar solution to the issue I mentioned at the beginning of this post
I bit_twiddle_permute3(I v) {
int n = __builtin_ctz(v);
I s = v >> n;
I t = s + 1;
if (v == 0 || t << n == 0) { return s; }
I w = (t << n) | ((~t & s) >> 1);
return w;
}