Voi*_*tar 20 algorithm binary function
我想找到n最低设置位的位置而不是最低设置位.(我不是在谈论n第四位的价值)
例如,说我有:
0000 1101 1000 0100 1100 1000 1010 0000
我想找到第4位设置.然后我希望它返回:
0000 0000 0000 0000 0100 0000 0000 0000
如果popcnt(v) < n,这个函数返回会有意义0,但是对于我这个案例的任何行为都是可以接受的.
如果可能的话,我正在寻找比循环更快的东西.
Juk*_*ela 24
现在PDEP,使用BMI2指令集非常容易.这是一个带有一些例子的64位版本:
#include <cassert>
#include <cstdint>
#include <x86intrin.h>
inline uint64_t nthset(uint64_t x, unsigned n) {
return _pdep_u64(1ULL << n, x);
}
int main() {
assert(nthset(0b0000'1101'1000'0100'1100'1000'1010'0000, 0) ==
0b0000'0000'0000'0000'0000'0000'0010'0000);
assert(nthset(0b0000'1101'1000'0100'1100'1000'1010'0000, 1) ==
0b0000'0000'0000'0000'0000'0000'1000'0000);
assert(nthset(0b0000'1101'1000'0100'1100'1000'1010'0000, 3) ==
0b0000'0000'0000'0000'0100'0000'0000'0000);
assert(nthset(0b0000'1101'1000'0100'1100'1000'1010'0000, 9) ==
0b0000'1000'0000'0000'0000'0000'0000'0000);
assert(nthset(0b0000'1101'1000'0100'1100'1000'1010'0000, 10) ==
0b0000'0000'0000'0000'0000'0000'0000'0000);
}
Voi*_*tar 10
事实证明,没有循环确实可以做到这一点.预先计算此问题的(至少)8位版本是最快的.当然,这些表占用了缓存空间,但几乎在所有现代PC场景中仍然应该有净加速.在此代码中,n = 0返回最小设置位,n = 1是第二个至少,等等.
使用__popcnt解决方案
有一个使用__popcnt内在的解决方案(你需要__popcnt非常快,或者通过一个简单的循环解决方案的任何性能增益都没有实际意义.幸运的是,大多数SSE4 +时代处理器都支持它).
// lookup table for sub-problem: 8-bit v
byte PRECOMP[256][8] = { .... } // PRECOMP[v][n] for v < 256 and n < 8
ulong nthSetBit(ulong v, ulong n) {
ulong p = __popcnt(v & 0xFFFF);
ulong shift = 0;
if (p <= n) {
v >>= 16;
shift += 16;
n -= p;
}
p = __popcnt(v & 0xFF);
if (p <= n) {
shift += 8;
v >>= 8;
n -= p;
}
if (n >= 8) return 0; // optional safety, in case n > # of set bits
return PRECOMP[v & 0xFF][n] << shift;
}
这说明了分而治之的方法是如何运作的.
一般解决方案
对于"通用"架构,还有一个解决方案 - 没有__popcnt.它可以通过8位块处理来完成.您还需要一个查找表来告诉您一个字节的popcnt:
byte PRECOMP[256][8] = { .... } // PRECOMP[v][n] for v<256 and n < 8
byte POPCNT[256] = { ... } // POPCNT[v] is the number of set bits in v. (v < 256)
ulong nthSetBit(ulong v, ulong n) {
ulong p = POPCNT[v & 0xFF];
ulong shift = 0;
if (p <= n) {
n -= p;
v >>= 8;
shift += 8;
p = POPCNT[v & 0xFF];
if (p <= n) {
n -= p;
shift += 8;
v >>= 8;
p = POPCNT[v & 0xFF];
if (p <= n) {
n -= p;
shift += 8;
v >>= 8;
}
}
}
if (n >= 8) return 0; // optional safety, in case n > # of set bits
return PRECOMP[v & 0xFF][n] << shift;
}
当然,这可以通过循环完成,但是展开的形式更快,循环的不寻常形式将使编译器不太可能自动为您展开它.
v-1具有零的位置v具有其最低有效"一"位,而所有更高有效位都相同.这导致以下功能:
int ffsn(unsigned int v, int n) {
for (int i=0; i<n-1; i++) {
v &= v-1; // remove the least significant bit
}
return v & ~(v-1); // extract the least significant bit
}
例如，适用于这种情况的bit-twiddling hacks版本是
unsigned int nth_bit_set(uint32_t value, unsigned int n)
{
const uint32_t pop2 = (value & 0x55555555u) + ((value >> 1) & 0x55555555u);
const uint32_t pop4 = (pop2 & 0x33333333u) + ((pop2 >> 2) & 0x33333333u);
const uint32_t pop8 = (pop4 & 0x0f0f0f0fu) + ((pop4 >> 4) & 0x0f0f0f0fu);
const uint32_t pop16 = (pop8 & 0x00ff00ffu) + ((pop8 >> 8) & 0x00ff00ffu);
const uint32_t pop32 = (pop16 & 0x000000ffu) + ((pop16 >>16) & 0x000000ffu);
unsigned int rank = 0;
unsigned int temp;
if (n++ >= pop32)
return 32;
temp = pop16 & 0xffu;
/* if (n > temp) { n -= temp; rank += 16; } */
rank += ((temp - n) & 256) >> 4;
n -= temp & ((temp - n) >> 8);
temp = (pop8 >> rank) & 0xffu;
/* if (n > temp) { n -= temp; rank += 8; } */
rank += ((temp - n) & 256) >> 5;
n -= temp & ((temp - n) >> 8);
temp = (pop4 >> rank) & 0x0fu;
/* if (n > temp) { n -= temp; rank += 4; } */
rank += ((temp - n) & 256) >> 6;
n -= temp & ((temp - n) >> 8);
temp = (pop2 >> rank) & 0x03u;
/* if (n > temp) { n -= temp; rank += 2; } */
rank += ((temp - n) & 256) >> 7;
n -= temp & ((temp - n) >> 8);
temp = (value >> rank) & 0x01u;
/* if (n > temp) rank += 1; */
rank += ((temp - n) & 256) >> 8;
return rank;
}
当在单独的编译单元中编译时，在 gcc-5.4.0 上使用-Wall -O3 -march=native -mtune=native英特尔酷睿 i5-4200u，产生
00400a40 <nth_bit_set>:
400a40: 89 f9 mov %edi,%ecx
400a42: 89 f8 mov %edi,%eax
400a44: 55 push %rbp
400a45: 40 0f b6 f6 movzbl %sil,%esi
400a49: d1 e9 shr %ecx
400a4b: 25 55 55 55 55 and $0x55555555,%eax
400a50: 53 push %rbx
400a51: 81 e1 55 55 55 55 and $0x55555555,%ecx
400a57: 01 c1 add %eax,%ecx
400a59: 41 89 c8 mov %ecx,%r8d
400a5c: 89 c8 mov %ecx,%eax
400a5e: 41 c1 e8 02 shr $0x2,%r8d
400a62: 25 33 33 33 33 and $0x33333333,%eax
400a67: 41 81 e0 33 33 33 33 and $0x33333333,%r8d
400a6e: 41 01 c0 add %eax,%r8d
400a71: 45 89 c1 mov %r8d,%r9d
400a74: 44 89 c0 mov %r8d,%eax
400a77: 41 c1 e9 04 shr $0x4,%r9d
400a7b: 25 0f 0f 0f 0f and $0xf0f0f0f,%eax
400a80: 41 81 e1 0f 0f 0f 0f and $0xf0f0f0f,%r9d
400a87: 41 01 c1 add %eax,%r9d
400a8a: 44 89 c8 mov %r9d,%eax
400a8d: 44 89 ca mov %r9d,%edx
400a90: c1 e8 08 shr $0x8,%eax
400a93: 81 e2 ff 00 ff 00 and $0xff00ff,%edx
400a99: 25 ff 00 ff 00 and $0xff00ff,%eax
400a9e: 01 d0 add %edx,%eax
400aa0: 0f b6 d8 movzbl %al,%ebx
400aa3: c1 e8 10 shr $0x10,%eax
400aa6: 0f b6 d0 movzbl %al,%edx
400aa9: b8 20 00 00 00 mov $0x20,%eax
400aae: 01 da add %ebx,%edx
400ab0: 39 f2 cmp %esi,%edx
400ab2: 77 0c ja 400ac0 <nth_bit_set+0x80>
400ab4: 5b pop %rbx
400ab5: 5d pop %rbp
400ab6: c3 retq
400ac0: 83 c6 01 add $0x1,%esi
400ac3: 89 dd mov %ebx,%ebp
400ac5: 29 f5 sub %esi,%ebp
400ac7: 41 89 ea mov %ebp,%r10d
400aca: c1 ed 08 shr $0x8,%ebp
400acd: 41 81 e2 00 01 00 00 and $0x100,%r10d
400ad4: 21 eb and %ebp,%ebx
400ad6: 41 c1 ea 04 shr $0x4,%r10d
400ada: 29 de sub %ebx,%esi
400adc: c4 42 2b f7 c9 shrx %r10d,%r9d,%r9d
400ae1: 41 0f b6 d9 movzbl %r9b,%ebx
400ae5: 89 dd mov %ebx,%ebp
400ae7: 29 f5 sub %esi,%ebp
400ae9: 41 89 e9 mov %ebp,%r9d
400aec: 41 81 e1 00 01 00 00 and $0x100,%r9d
400af3: 41 c1 e9 05 shr $0x5,%r9d
400af7: 47 8d 14 11 lea (%r9,%r10,1),%r10d
400afb: 41 89 e9 mov %ebp,%r9d
400afe: 41 c1 e9 08 shr $0x8,%r9d
400b02: c4 42 2b f7 c0 shrx %r10d,%r8d,%r8d
400b07: 41 83 e0 0f and $0xf,%r8d
400b0b: 44 21 cb and %r9d,%ebx
400b0e: 45 89 c3 mov %r8d,%r11d
400b11: 29 de sub %ebx,%esi
400b13: 5b pop %rbx
400b14: 41 29 f3 sub %esi,%r11d
400b17: 5d pop %rbp
400b18: 44 89 da mov %r11d,%edx
400b1b: 41 c1 eb 08 shr $0x8,%r11d
400b1f: 81 e2 00 01 00 00 and $0x100,%edx
400b25: 45 21 d8 and %r11d,%r8d
400b28: c1 ea 06 shr $0x6,%edx
400b2b: 44 29 c6 sub %r8d,%esi
400b2e: 46 8d 0c 12 lea (%rdx,%r10,1),%r9d
400b32: c4 e2 33 f7 c9 shrx %r9d,%ecx,%ecx
400b37: 83 e1 03 and $0x3,%ecx
400b3a: 41 89 c8 mov %ecx,%r8d
400b3d: 41 29 f0 sub %esi,%r8d
400b40: 44 89 c0 mov %r8d,%eax
400b43: 41 c1 e8 08 shr $0x8,%r8d
400b47: 25 00 01 00 00 and $0x100,%eax
400b4c: 44 21 c1 and %r8d,%ecx
400b4f: c1 e8 07 shr $0x7,%eax
400b52: 29 ce sub %ecx,%esi
400b54: 42 8d 14 08 lea (%rax,%r9,1),%edx
400b58: c4 e2 6b f7 c7 shrx %edx,%edi,%eax
400b5d: 83 e0 01 and $0x1,%eax
400b60: 29 f0 sub %esi,%eax
400b62: 25 00 01 00 00 and $0x100,%eax
400b67: c1 e8 08 shr $0x8,%eax
400b6a: 01 d0 add %edx,%eax
400b6c: c3 retq
当作为单独的编译单元编译时，在本机上计时是困难的，因为实际操作和调用无用函数一样快（也在单独的编译单元中编译）；本质上，计算是在与函数调用相关的延迟期间完成的。
它似乎比我建议的二分搜索略快，
unsigned int nth_bit_set(uint32_t value, unsigned int n)
{
uint32_t mask = 0x0000FFFFu;
unsigned int size = 16u;
unsigned int base = 0u;
if (n++ >= __builtin_popcount(value))
return 32;
while (size > 0) {
const unsigned int count = __builtin_popcount(value & mask);
if (n > count) {
base += size;
size >>= 1;
mask |= mask << size;
} else {
size >>= 1;
mask >>= size;
}
}
return base;
}
其中循环正好执行五次，编译为
00400ba0 <nth_bit_set>:
400ba0: 83 c6 01 add $0x1,%esi
400ba3: 31 c0 xor %eax,%eax
400ba5: b9 10 00 00 00 mov $0x10,%ecx
400baa: ba ff ff 00 00 mov $0xffff,%edx
400baf: 45 31 db xor %r11d,%r11d
400bb2: 66 0f 1f 44 00 00 nopw 0x0(%rax,%rax,1)
400bb8: 41 89 c9 mov %ecx,%r9d
400bbb: 41 89 f8 mov %edi,%r8d
400bbe: 41 d0 e9 shr %r9b
400bc1: 41 21 d0 and %edx,%r8d
400bc4: c4 62 31 f7 d2 shlx %r9d,%edx,%r10d
400bc9: f3 45 0f b8 c0 popcnt %r8d,%r8d
400bce: 41 09 d2 or %edx,%r10d
400bd1: 44 38 c6 cmp %r8b,%sil
400bd4: 41 0f 46 cb cmovbe %r11d,%ecx
400bd8: c4 e2 33 f7 d2 shrx %r9d,%edx,%edx
400bdd: 41 0f 47 d2 cmova %r10d,%edx
400be1: 01 c8 add %ecx,%eax
400be3: 44 89 c9 mov %r9d,%ecx
400be6: 45 84 c9 test %r9b,%r9b
400be9: 75 cd jne 400bb8 <nth_bit_set+0x18>
400beb: c3 retq
例如，在对二进制搜索版本的 95% 的调用中不超过 31 个周期，而在对 bit-hack 版本的 95% 的调用中不超过 28 个周期；在 50% 的情况下，两者都在 28 个周期内运行。（循环版本在 95% 的调用中最多需要 56 个周期，中位数最多为 37 个周期。）
要确定在实际现实世界代码中哪个更好，必须在现实世界任务中进行适当的基准测试；至少对于当前的 x86-64 架构处理器，完成的工作很容易隐藏在其他地方（如函数调用）产生的延迟中。