在现代 CPU 上，二进制搜索在哪个 n 比线性搜索更快? [英] At which n does binary search become faster than linear search on a modern CPU?
问题描述
由于分支预测的奇妙之处，二进制搜索可能比对整数数组的线性搜索慢.在典型的台式机处理器上，在使用二分搜索更好之前，该数组必须有多大?假设该结构将用于多次查找.
Due to the wonders of branch prediction, a binary search can be slower than a linear search through an array of integers. On a typical desktop processor, how big does that array have to get before it would be better to use a binary search? Assume the structure will be used for many lookups.
推荐答案
我尝试了一点 C++ 基准测试，我很惊讶 - 线性搜索似乎占了上风，最多有几十个项目，我还没有找到案例对于这些尺寸，二进制搜索更好.也许 gcc 的 STL 没有很好地调整?但是——你会用什么来实现这两种搜索?-)这是我的代码，所以每个人都可以看看我是否做了一些愚蠢的事情，会严重扭曲时间......:
I've tried a little C++ benchmarking and I'm surprised - linear search seems to prevail up to several dozen items, and I haven't found a case where binary search is better for those sizes. Maybe gcc's STL is not well tuned? But then -- what would you use to implement either kind of search?-) So here's my code, so everybody can see if I've done something silly that would distort timing grossly...:
#include <vector>
#include <algorithm>
#include <iostream>
#include <stdlib.h>
int data[] = {98, 50, 54, 43, 39, 91, 17, 85, 42, 84, 23, 7, 70, 72, 74, 65, 66, 47, 20, 27, 61, 62, 22, 75, 24, 6, 2, 68, 45, 77, 82, 29, 59, 97, 95, 94, 40, 80, 86, 9, 78, 69, 15, 51, 14, 36, 76, 18, 48, 73, 79, 25, 11, 38, 71, 1, 57, 3, 26, 37, 19, 67, 35, 87, 60, 34, 5, 88, 52, 96, 31, 30, 81, 4, 92, 21, 33, 44, 63, 83, 56, 0, 12, 8, 93, 49, 41, 58, 89, 10, 28, 55, 46, 13, 64, 53, 32, 16, 90
};
int tosearch[] = {53, 5, 40, 71, 37, 14, 52, 28, 25, 11, 23, 13, 70, 81, 77, 10, 17, 26, 56, 15, 94, 42, 18, 39, 50, 78, 93, 19, 87, 43, 63, 67, 79, 4, 64, 6, 38, 45, 91, 86, 20, 30, 58, 68, 33, 12, 97, 95, 9, 89, 32, 72, 74, 1, 2, 34, 62, 57, 29, 21, 49, 69, 0, 31, 3, 27, 60, 59, 24, 41, 80, 7, 51, 8, 47, 54, 90, 36, 76, 22, 44, 84, 48, 73, 65, 96, 83, 66, 61, 16, 88, 92, 98, 85, 75, 82, 55, 35, 46
};
bool binsearch(int i, std::vector<int>::const_iterator begin,
std::vector<int>::const_iterator end) {
return std::binary_search(begin, end, i);
}
bool linsearch(int i, std::vector<int>::const_iterator begin,
std::vector<int>::const_iterator end) {
return std::find(begin, end, i) != end;
}
int main(int argc, char *argv[])
{
int n = 6;
if (argc < 2) {
std::cerr << "need at least 1 arg (l or b!)" << std::endl;
return 1;
}
char algo = argv[1][0];
if (algo != 'b' && algo != 'l') {
std::cerr << "algo must be l or b, not '" << algo << "'" << std::endl;
return 1;
}
if (argc > 2) {
n = atoi(argv[2]);
}
std::vector<int> vv;
for (int i=0; i<n; ++i) {
if(data[i]==-1) break;
vv.push_back(data[i]);
}
if (algo=='b') {
std::sort(vv.begin(), vv.end());
}
bool (*search)(int i, std::vector<int>::const_iterator begin,
std::vector<int>::const_iterator end);
if (algo=='b') search = binsearch;
else search = linsearch;
int nf = 0;
int ns = 0;
for(int k=0; k<10000; ++k) {
for (int j=0; tosearch[j] >= 0; ++j) {
++ns;
if (search(tosearch[j], vv.begin(), vv.end()))
++nf;
}
}
std::cout << nf <<'/'<< ns << std::endl;
return 0;
}
以及我对核心二人组的一些计时:
and my a couple of my timings on a core duo:
AmAir:stko aleax$ time ./a.out b 93
1910000/2030000
real 0m0.230s
user 0m0.224s
sys 0m0.005s
AmAir:stko aleax$ time ./a.out l 93
1910000/2030000
real 0m0.169s
user 0m0.164s
sys 0m0.005s
无论如何，它们是可重复的...
They're pretty repeatable, anyway...
OP 说:Alex，我编辑了你的程序，只用 1..n 填充数组，不运行 std::sort，并进行大约 1000 万次(mod 整数除法)搜索.在奔腾 4 上的 n=150 处，二进制搜索开始脱离线性搜索.抱歉图表颜色.
OP says: Alex, I edited your program to just fill the array with 1..n, not run std::sort, and do about 10 million (mod integer division) searches. Binary search starts to pull away from linear search at n=150 on a Pentium 4. Sorry about the chart colors.
二进制与线性搜索 http://spreadsheets.google.com/pub?key=tzWXX9Qmmu3_COpTYkTqsOA&oid=1&output=image
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