我正在研究Sedgewick C部分第1-4部分的算法 on p172中的Shell类型。
我使用size (数组的长度),而不是l和r (开始和结束);所以我的代码是
int i,j,h;
int key;
for( h=1;h<=(size-1)/9;h=h*3+1);
for(;h>0;h/=3)
{
for(i=h;i<size;i++)
{
key=num[i];
j=i;
while(j>=h&&key>num[j-h];j-=h)
{
num[j]=num[j-h];
}
num[j]=key;
}
}这一切我都知道。我读过TAOCP。我知道1,4,13,…是最好的序列(可比)。但在这个位置上,我的代码
for(h=1;h<size;h=h*3+1);我的问题是:他为什么写h<(size-1)/9
“/9”是什么意思?
发布于 2015-01-13 03:26:46
循环:
for (h = 1; h < size; h = h * 3 + 1)
;大多数情况下会超调数组的大小。可选回路使间隙保持在范围内。
您可以通过这样一个琐碎的测试程序看到这一点:
#include <stdio.h>
static inline int hs_gap9(int size)
{
int h;
for (h = 1; h <= (size - 1) / 9; h = h * 3 + 1)
;
return h;
}
static inline int hs_gap3(int size)
{
int h;
for (h = 1; h < size; h = h * 3 + 1)
;
return h;
}
int main(void)
{
for (int i = 1; i < 100; i++)
printf("Size: %3d; gap9 = %d; gap3 = %d\n", i, hs_gap9(i), hs_gap3(i));
return 0;
}样本输出:
Size: 1; gap9 = 1; gap3 = 1
Size: 2; gap9 = 1; gap3 = 4
Size: 3; gap9 = 1; gap3 = 4
Size: 4; gap9 = 1; gap3 = 4
Size: 5; gap9 = 1; gap3 = 13
Size: 6; gap9 = 1; gap3 = 13
Size: 7; gap9 = 1; gap3 = 13
Size: 8; gap9 = 1; gap3 = 13
Size: 9; gap9 = 1; gap3 = 13
Size: 10; gap9 = 4; gap3 = 13
Size: 11; gap9 = 4; gap3 = 13
Size: 12; gap9 = 4; gap3 = 13
Size: 13; gap9 = 4; gap3 = 13
Size: 14; gap9 = 4; gap3 = 40
Size: 15; gap9 = 4; gap3 = 40
Size: 16; gap9 = 4; gap3 = 40
…
Size: 34; gap9 = 4; gap3 = 40
Size: 35; gap9 = 4; gap3 = 40
Size: 36; gap9 = 4; gap3 = 40
Size: 37; gap9 = 13; gap3 = 40
Size: 38; gap9 = 13; gap3 = 40
Size: 39; gap9 = 13; gap3 = 40
Size: 40; gap9 = 13; gap3 = 40
Size: 41; gap9 = 13; gap3 = 121
Size: 42; gap9 = 13; gap3 = 121
Size: 43; gap9 = 13; gap3 = 121
…
Size: 97; gap9 = 13; gap3 = 121
Size: 98; gap9 = 13; gap3 = 121
Size: 99; gap9 = 13; gap3 = 121如您所见,“gap3”算法返回一个比数组大小更大的h初始值。“gap9”算法返回小于数组大小的h初始值。这在循环上节省了一些开销(保存外部循环的一次迭代,中间循环在第一个循环中退出,而不接触内循环。
https://stackoverflow.com/questions/27914025
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