sfl*_*lee 4 c++ algorithm optimization stl
我有两个int vectors喜欢a[100],b[100].
计算汉明距离的简单方法是:
std::vector<int> a(100);
std::vector<int> b(100);
double dist = 0;
for(int i = 0; i < 100; i++){
if(a[i] != b[i])
dist++;
}
dist /= a.size();
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我想问一下,有更快的方法在C++中进行此计算或如何使用STL执行相同的工作吗?
你要求更快的方式.这是一个令人尴尬的并行问题,因此,使用C++,您可以通过两种方式利用它:线程并行性和通过优化进行矢量化.
//The following flags allow cpu specific vectorization optimizations on *my cpu*
//clang++ -march=corei7-avx hd.cpp -o hd -Ofast -pthread -std=c++1y
//g++ -march=corei7-avx hd.cpp -o hd -Ofast -pthread -std=c++1y
#include <vector>
#include <thread>
#include <future>
#include <numeric>
template<class T, class I1, class I2>
T hamming_distance(size_t size, I1 b1, I2 b2) {
return std::inner_product(b1, b1 + size, b2, T{},
std::plus<T>(), std::not_equal_to<T>());
}
template<class T, class I1, class I2>
T parallel_hamming_distance(size_t threads, size_t size, I1 b1, I2 b2) {
if(size < 1000)
return hamming_distance<T, I1, I2>(size, b1, b2);
if(threads > size)
threads = size;
const size_t whole_part = size / threads;
const size_t remainder = size - threads * whole_part;
std::vector<std::future<T>> bag;
bag.reserve(threads + (remainder > 0 ? 1 : 0));
for(size_t i = 0; i < threads; ++i)
bag.emplace_back(std::async(std::launch::async,
hamming_distance<T, I1, I2>,
whole_part,
b1 + i * whole_part,
b2 + i * whole_part));
if(remainder > 0)
bag.emplace_back(std::async(std::launch::async,
hamming_distance<T, I1, I2>,
remainder,
b1 + threads * whole_part,
b2 + threads * whole_part));
T hamming_distance = 0;
for(auto &f : bag) hamming_distance += f.get();
return hamming_distance;
}
#include <ratio>
#include <random>
#include <chrono>
#include <iostream>
#include <cinttypes>
int main() {
using namespace std;
using namespace chrono;
random_device rd;
mt19937 gen(rd());
uniform_int_distribution<> random_0_9(0, 9);
const auto size = 100 * mega::num;
vector<int32_t> v1(size);
vector<int32_t> v2(size);
for(auto &x : v1) x = random_0_9(gen);
for(auto &x : v2) x = random_0_9(gen);
cout << "naive hamming distance: ";
const auto naive_start = high_resolution_clock::now();
cout << hamming_distance<int32_t>(v1.size(), begin(v1), begin(v2)) << endl;
const auto naive_elapsed = high_resolution_clock::now() - naive_start;
const auto n = thread::hardware_concurrency();
cout << "parallel hamming distance: ";
const auto parallel_start = high_resolution_clock::now();
cout << parallel_hamming_distance<int32_t>(
n,
v1.size(),
begin(v1),
begin(v2)
)
<< endl;
const auto parallel_elapsed = high_resolution_clock::now() - parallel_start;
auto count_microseconds =
[](const high_resolution_clock::duration &elapsed) {
return duration_cast<microseconds>(elapsed).count();
};
cout << "naive delay: " << count_microseconds(naive_elapsed) << endl;
cout << "parallel delay: " << count_microseconds(parallel_elapsed) << endl;
}
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请注意,我没有对矢量大小进行划分
我的机器的结果(显示它对于只有2个物理核心的机器没有太大帮助...):
$ clang++ -march=corei7-avx hd.cpp -o hd -Ofast -pthread -std=c++1y -stdlib=libc++ -lcxxrt -ldl
$ ./hd
naive hamming distance: 89995190
parallel hamming distance: 89995190
naive delay: 52758
parallel delay: 47227
$ clang++ hd.cpp -o hd -O3 -pthread -std=c++1y -stdlib=libc++ -lcxxrt -ldl
$ ./hd
naive hamming distance: 90001042
parallel hamming distance: 90001042
naive delay: 53851
parallel delay: 46887
$ g++ -march=corei7-avx hd.cpp -o hd -Ofast -pthread -std=c++1y -Wl,--no-as-needed
$ ./hd
naive hamming distance: 90001825
parallel hamming distance: 90001825
naive delay: 55229
parallel delay: 49355
$ g++ hd.cpp -o hd -O3 -pthread -std=c++1y -Wl,--no-as-needed
$ ./hd
naive hamming distance: 89996171
parallel hamming distance: 89996171
naive delay: 54189
parallel delay: 44928
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另外我看到自动矢量化没有效果,可能要检查装配......
有关矢量化和编译器选项的示例,请查看我的博文.
据观察,即使对于增量,使用 double 也非常慢。所以你应该在(递增)中使用 int for,然后使用 double 进行除法。
作为加速,我能想到的一种测试方法是使用 SSE 指令:
伪代码:
distance = 0
SSE register e1
SSE register e2
for each 4 elements in vectors
load 4 members from a in e1
load 4 members from b in e2
if e1 == e2
continue
else
check each 4 members individually (using e1 and e2)
dist /= 4
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在真实的(非伪代码)程序中,可以对此进行调整,以便编译器可以使用cmov指令而不是branches.
这里的主要优点是我们从内存中读取的数据减少了 4 倍。
缺点是我们之前的每 4 次检查都会有一次额外的检查。
根据如何通过cmoves或在汇编中实现这branches一点,对于在两个向量中具有许多相邻位置且具有相同值的向量来说,这可能会更快。
我真的不知道与标准解决方案相比它的性能如何,但至少值得测试。