在回答另一个Stack Overflow问题时,我偶然发现了一个有趣的子问题。对6个整数的数组进行排序的最快方法是什么?

因为问题层次很低:

我们不能假设库是可用的(而且调用本身也有开销),只有纯C 为了避免清空指令管道(这有非常高的成本),我们可能应该最小化分支、跳转和其他类型的控制流中断(比如隐藏在&&或||序列点后面的那些)。 空间是有限的,最小化寄存器和内存的使用是一个问题,理想情况下,就地排序可能是最好的。

实际上,这个问题是一种Golf,其目标不是最小化源长度,而是最小化执行时间。我称之为“Zening”代码,就像Michael Abrash在《Zen of code optimization》一书及其续集中所使用的那样。

至于为什么它有趣,有几个层面:

示例简单,易于理解和测量,不需要太多的C技能 它显示了对问题选择好的算法的影响,也显示了编译器和底层硬件的影响。

下面是我的参考(简单的,不是优化的)实现和测试集。

#include <stdio.h>

static __inline__ int sort6(int * d){

    char j, i, imin;
    int tmp;
    for (j = 0 ; j < 5 ; j++){
        imin = j;
        for (i = j + 1; i < 6 ; i++){
            if (d[i] < d[imin]){
                imin = i;
            }
        }
        tmp = d[j];
        d[j] = d[imin];
        d[imin] = tmp;
    }
}

static __inline__ unsigned long long rdtsc(void)
{
  unsigned long long int x;
     __asm__ volatile (".byte 0x0f, 0x31" : "=A" (x));
     return x;
}

int main(int argc, char ** argv){
    int i;
    int d[6][5] = {
        {1, 2, 3, 4, 5, 6},
        {6, 5, 4, 3, 2, 1},
        {100, 2, 300, 4, 500, 6},
        {100, 2, 3, 4, 500, 6},
        {1, 200, 3, 4, 5, 600},
        {1, 1, 2, 1, 2, 1}
    };

    unsigned long long cycles = rdtsc();
    for (i = 0; i < 6 ; i++){
        sort6(d[i]);
        /*
         * printf("d%d : %d %d %d %d %d %d\n", i,
         *  d[i][0], d[i][6], d[i][7],
         *  d[i][8], d[i][9], d[i][10]);
        */
    }
    cycles = rdtsc() - cycles;
    printf("Time is %d\n", (unsigned)cycles);
}

生的结果

随着变体的数量越来越多,我将它们都收集到一个测试套件中,可以在这里找到。在Kevin Stock的帮助下,实际使用的测试没有上面展示的那么简单。您可以在自己的环境中编译和执行它。我对不同目标架构/编译器上的行为很感兴趣。(好了,伙计们,把它放在答案里,我将+1一个新结果集的每个贡献者)。

一年前,我把答案给了Daniel Stutzbach(高尔夫),因为他是当时最快的解决方案(排序网络)的来源。

Linux 64位,gcc 4.6.1 64位,Intel Core 2 Duo E8400, -O2

Direct call to qsort library function : 689.38 Naive implementation (insertion sort) : 285.70 Insertion Sort (Daniel Stutzbach) : 142.12 Insertion Sort Unrolled : 125.47 Rank Order : 102.26 Rank Order with registers : 58.03 Sorting Networks (Daniel Stutzbach) : 111.68 Sorting Networks (Paul R) : 66.36 Sorting Networks 12 with Fast Swap : 58.86 Sorting Networks 12 reordered Swap : 53.74 Sorting Networks 12 reordered Simple Swap : 31.54 Reordered Sorting Network w/ fast swap : 31.54 Reordered Sorting Network w/ fast swap V2 : 33.63 Inlined Bubble Sort (Paolo Bonzini) : 48.85 Unrolled Insertion Sort (Paolo Bonzini) : 75.30

Linux 64位,gcc 4.6.1 64位,Intel Core 2 Duo E8400, -O1

Direct call to qsort library function : 705.93 Naive implementation (insertion sort) : 135.60 Insertion Sort (Daniel Stutzbach) : 142.11 Insertion Sort Unrolled : 126.75 Rank Order : 46.42 Rank Order with registers : 43.58 Sorting Networks (Daniel Stutzbach) : 115.57 Sorting Networks (Paul R) : 64.44 Sorting Networks 12 with Fast Swap : 61.98 Sorting Networks 12 reordered Swap : 54.67 Sorting Networks 12 reordered Simple Swap : 31.54 Reordered Sorting Network w/ fast swap : 31.24 Reordered Sorting Network w/ fast swap V2 : 33.07 Inlined Bubble Sort (Paolo Bonzini) : 45.79 Unrolled Insertion Sort (Paolo Bonzini) : 80.15

我包括了-O1和-O2的结果,因为令人惊讶的是,在一些程序中,O2的效率低于O1。我想知道什么具体的优化有这种效果?

对建议解决方案的评论

插入排序(丹尼尔·斯图茨巴赫)

正如预期的那样,最小化分支确实是一个好主意。

排序网络(丹尼尔·斯图茨巴赫)

比插入排序好。我想知道主要的效果是不是避免外部循环。我试着通过展开插入排序来检查,确实我们得到了大致相同的数字(代码在这里)。

排序网络(保罗R)

迄今为止最好的。我用来测试的实际代码在这里。目前还不知道为什么它的速度几乎是其他排序网络实现的两倍。参数传递?快速max ?

排序网络12 SWAP与快速交换

根据Daniel Stutzbach的建议,我将他的12交换排序网络与无分支快速交换相结合(代码在这里)。它确实更快,到目前为止最好的,只有很小的利润率(大约5%),因为可以使用更少的交换。

同样有趣的是,无分支交换似乎比在PPC架构上使用if的简单交换效率低得多(4倍)。

调用库qsort

To give another reference point I also tried as suggested to just call library qsort (code is here). As expected it is much slower : 10 to 30 times slower... as it became obvious with the new test suite, the main problem seems to be the initial load of the library after the first call, and it compares not so poorly with other version. It is just between 3 and 20 times slower on my Linux. On some architecture used for tests by others it seems even to be faster (I'm really surprised by that one, as library qsort use a more complex API).

等级次序

Rex Kerr proposed another completely different method : for each item of the array compute directly its final position. This is efficient because computing rank order do not need branch. The drawback of this method is that it takes three times the amount of memory of the array (one copy of array and variables to store rank orders). The performance results are very surprising (and interesting). On my reference architecture with 32 bits OS and Intel Core2 Quad E8300, cycle count was slightly below 1000 (like sorting networks with branching swap). But when compiled and executed on my 64 bits box (Intel Core2 Duo) it performed much better : it became the fastest so far. I finally found out the true reason. My 32bits box use gcc 4.4.1 and my 64bits box gcc 4.4.3 and the last one seems much better at optimizing this particular code (there was very little difference for other proposals).

更新:

正如上面公布的数字所示,这种效果在gcc的后续版本中仍然得到了增强,Rank Order的速度始终是其他任何替代版本的两倍。

用重新排序的交换对网络进行排序

The amazing efficiency of the Rex Kerr proposal with gcc 4.4.3 made me wonder : how could a program with 3 times as much memory usage be faster than branchless sorting networks? My hypothesis was that it had less dependencies of the kind read after write, allowing for better use of the superscalar instruction scheduler of the x86. That gave me an idea: reorder swaps to minimize read after write dependencies. More simply put: when you do SWAP(1, 2); SWAP(0, 2); you have to wait for the first swap to be finished before performing the second one because both access to a common memory cell. When you do SWAP(1, 2); SWAP(4, 5);the processor can execute both in parallel. I tried it and it works as expected, the sorting networks is running about 10% faster.

使用简单交换对网络进行排序

One year after the original post Steinar H. Gunderson suggested, that we should not try to outsmart the compiler and keep the swap code simple. It's indeed a good idea as the resulting code is about 40% faster! He also proposed a swap optimized by hand using x86 inline assembly code that can still spare some more cycles. The most surprising (it says volumes on programmer's psychology) is that one year ago none of used tried that version of swap. Code I used to test is here. Others suggested other ways to write a C fast swap, but it yields the same performances as the simple one with a decent compiler.

“最佳”代码如下:

static inline void sort6_sorting_network_simple_swap(int * d){
#define min(x, y) (x<y?x:y)
#define max(x, y) (x<y?y:x) 
#define SWAP(x,y) { const int a = min(d[x], d[y]); \
                    const int b = max(d[x], d[y]); \
                    d[x] = a; d[y] = b; }
    SWAP(1, 2);
    SWAP(4, 5);
    SWAP(0, 2);
    SWAP(3, 5);
    SWAP(0, 1);
    SWAP(3, 4);
    SWAP(1, 4);
    SWAP(0, 3);
    SWAP(2, 5);
    SWAP(1, 3);
    SWAP(2, 4);
    SWAP(2, 3);
#undef SWAP
#undef min
#undef max
}

如果我们相信我们的测试集(是的,它很差,它的唯一好处是简短,简单,易于理解我们所测量的内容),那么一个排序的结果代码的平均循环次数低于40个循环(执行6个测试)。这使得每次交换平均为4个周期。我称之为惊人的快。还有其他可能的改进吗?


当前回答

Try 'merging sorted list' sort. :) Use two array. Fastest for small and big array. If you concating, you only check where insert. Other bigger values you not need compare (cmp = a-b>0). For 4 numbers, you can use system 4-5 cmp (~4.6) or 3-6 cmp (~4.9). Bubble sort use 6 cmp (6). Lots of cmp for big numbers slower code. This code use 5 cmp (not MSL sort): if (cmp(arr[n][i+0],arr[n][i+1])>0) {swap(n,i+0,i+1);} if (cmp(arr[n][i+2],arr[n][i+3])>0) {swap(n,i+2,i+3);} if (cmp(arr[n][i+0],arr[n][i+2])>0) {swap(n,i+0,i+2);} if (cmp(arr[n][i+1],arr[n][i+3])>0) {swap(n,i+1,i+3);} if (cmp(arr[n][i+1],arr[n][i+2])>0) {swap(n,i+1,i+2);}

最初的韩剧 9 8 7 6 5 4 3 2 10 0 89 67 45 23 01…Concat两个排序的列表,列表长度= 1 6789 2345 01…Concat两个排序的列表,列表长度= 2 23456789 01…Concat两个排序的列表,列表长度= 4 0123456789……Concat两个排序的列表,列表长度= 8

JS代码

function sortListMerge_2a(cmp) { var step, stepmax, tmp, a,b,c, i,j,k, m,n, cycles; var start = 0; var end = arr_count; //var str = ''; cycles = 0; if (end>3) { stepmax = ((end - start + 1) >> 1) << 1; m = 1; n = 2; for (step=1;step<stepmax;step<<=1) //bounds 1-1, 2-2, 4-4, 8-8... { a = start; while (a<end) { b = a + step; c = a + step + step; b = b<end ? b : end; c = c<end ? c : end; i = a; j = b; k = i; while (i<b && j<c) { if (cmp(arr[m][i],arr[m][j])>0) {arr[n][k] = arr[m][j]; j++; k++;} else {arr[n][k] = arr[m][i]; i++; k++;} } while (i<b) {arr[n][k] = arr[m][i]; i++; k++; } while (j<c) {arr[n][k] = arr[m][j]; j++; k++; } a = c; } tmp = m; m = n; n = tmp; } return m; } else { // sort 3 items sort10(cmp); return m; } }

其他回答

我认为你的问题有两个方面。

The first is to determine the optimal algorithm. This is done - at least in this case - by looping through every possible ordering (there aren't that many) which allows you to compute exact min, max, average and standard deviation of compares and swaps. Have a runner-up or two handy as well. The second is to optimize the algorithm. A lot can be done to convert textbook code examples to mean and lean real-life algorithms. If you realize that an algorithm can't be optimized to the extent required, try a runner-up.

I wouldn't worry too much about emptying pipelines (assuming current x86): branch prediction has come a long way. What I would worry about is making sure that the code and data fit in one cache line each (maybe two for the code). Once there fetch latencies are refreshingly low which will compensate for any stall. It also means that your inner loop will be maybe ten instructions or so which is right where it should be (there are two different inner loops in my sorting algorithm, they are 10 instructions/22 bytes and 9/22 long respectively). Assuming the code doesn't contain any divs you can be sure it will be blindingly fast.

我知道这是一个老问题。

但我刚刚写了一个不同的解,我想分享一下。 只使用嵌套的MIN MAX,

它的速度并不快,因为它每个都要用114个, 可以简单地降低到75吗,就像这样-> pastebin

但它不再是单纯的最小最大值了。

什么可能工作是做最小/最大对多个整数一次与AVX

PMINSW参考

#include <stdio.h>

static __inline__ int MIN(int a, int b){
int result =a;
__asm__ ("pminsw %1, %0" : "+x" (result) : "x" (b));
return result;
}
static __inline__ int MAX(int a, int b){
int result = a;
__asm__ ("pmaxsw %1, %0" : "+x" (result) : "x" (b));
return result;
}
static __inline__ unsigned long long rdtsc(void){
  unsigned long long int x;
__asm__ volatile (".byte 0x0f, 0x31" :
  "=A" (x));
  return x;
}

#define MIN3(a, b, c) (MIN(MIN(a,b),c))
#define MIN4(a, b, c, d) (MIN(MIN(a,b),MIN(c,d)))

static __inline__ void sort6(int * in) {
  const int A=in[0], B=in[1], C=in[2], D=in[3], E=in[4], F=in[5];

  in[0] = MIN( MIN4(A,B,C,D),MIN(E,F) );

  const int
  AB = MAX(A, B),
  AC = MAX(A, C),
  AD = MAX(A, D),
  AE = MAX(A, E),
  AF = MAX(A, F),
  BC = MAX(B, C),
  BD = MAX(B, D),
  BE = MAX(B, E),
  BF = MAX(B, F),
  CD = MAX(C, D),
  CE = MAX(C, E),
  CF = MAX(C, F),
  DE = MAX(D, E),
  DF = MAX(D, F),
  EF = MAX(E, F);

  in[1] = MIN4 (
  MIN4( AB, AC, AD, AE ),
  MIN4( AF, BC, BD, BE ),
  MIN4( BF, CD, CE, CF ),
  MIN3( DE, DF, EF)
  );

  const int
  ABC = MAX(AB,C),
  ABD = MAX(AB,D),
  ABE = MAX(AB,E),
  ABF = MAX(AB,F),
  ACD = MAX(AC,D),
  ACE = MAX(AC,E),
  ACF = MAX(AC,F),
  ADE = MAX(AD,E),
  ADF = MAX(AD,F),
  AEF = MAX(AE,F),
  BCD = MAX(BC,D),
  BCE = MAX(BC,E),
  BCF = MAX(BC,F),
  BDE = MAX(BD,E),
  BDF = MAX(BD,F),
  BEF = MAX(BE,F),
  CDE = MAX(CD,E),
  CDF = MAX(CD,F),
  CEF = MAX(CE,F),
  DEF = MAX(DE,F);

  in[2] = MIN( MIN4 (
  MIN4( ABC, ABD, ABE, ABF ),
  MIN4( ACD, ACE, ACF, ADE ),
  MIN4( ADF, AEF, BCD, BCE ),
  MIN4( BCF, BDE, BDF, BEF )),
  MIN4( CDE, CDF, CEF, DEF )
  );


  const int
  ABCD = MAX(ABC,D),
  ABCE = MAX(ABC,E),
  ABCF = MAX(ABC,F),
  ABDE = MAX(ABD,E),
  ABDF = MAX(ABD,F),
  ABEF = MAX(ABE,F),
  ACDE = MAX(ACD,E),
  ACDF = MAX(ACD,F),
  ACEF = MAX(ACE,F),
  ADEF = MAX(ADE,F),
  BCDE = MAX(BCD,E),
  BCDF = MAX(BCD,F),
  BCEF = MAX(BCE,F),
  BDEF = MAX(BDE,F),
  CDEF = MAX(CDE,F);

  in[3] = MIN4 (
  MIN4( ABCD, ABCE, ABCF, ABDE ),
  MIN4( ABDF, ABEF, ACDE, ACDF ),
  MIN4( ACEF, ADEF, BCDE, BCDF ),
  MIN3( BCEF, BDEF, CDEF )
  );

  const int
  ABCDE= MAX(ABCD,E),
  ABCDF= MAX(ABCD,F),
  ABCEF= MAX(ABCE,F),
  ABDEF= MAX(ABDE,F),
  ACDEF= MAX(ACDE,F),
  BCDEF= MAX(BCDE,F);

  in[4]= MIN (
  MIN4( ABCDE, ABCDF, ABCEF, ABDEF ),
  MIN ( ACDEF, BCDEF )
  );

  in[5] = MAX(ABCDE,F);
}

int main(int argc, char ** argv) {
  int d[6][6] = {
    {1, 2, 3, 4, 5, 6},
    {6, 5, 4, 3, 2, 1},
    {100, 2, 300, 4, 500, 6},
    {100, 2, 3, 4, 500, 6},
    {1, 200, 3, 4, 5, 600},
    {1, 1, 2, 1, 2, 1}
  };

  unsigned long long cycles = rdtsc();
  for (int i = 0; i < 6; i++) {
    sort6(d[i]);
  }
  cycles = rdtsc() - cycles;
  printf("Time is %d\n", (unsigned)cycles);

  for (int i = 0; i < 6; i++) {
    printf("d%d : %d %d %d %d %d %d\n", i,
     d[i][0], d[i][1], d[i][2],
     d[i][3], d[i][4], d[i][5]);
  }
}

编辑: 受Rex Kerr的启发, 比上面的混乱快多了

static void sort6(int *o) {
const int 
A=o[0],B=o[1],C=o[2],D=o[3],E=o[4],F=o[5];
const unsigned char
AB = A>B, AC = A>C, AD = A>D, AE = A>E,
          BC = B>C, BD = B>D, BE = B>E,
                    CD = C>D, CE = C>E,
                              DE = D>E,
a =          AB + AC + AD + AE + (A>F),
b = 1 - AB      + BC + BD + BE + (B>F),
c = 2 - AC - BC      + CD + CE + (C>F),
d = 3 - AD - BD - CD      + DE + (D>F),
e = 4 - AE - BE - CE - DE      + (E>F);
o[a]=A; o[b]=B; o[c]=C; o[d]=D; o[e]=E;
o[15-a-b-c-d-e]=F;
}

排序使用cmp==0的4个项目。 cmp的数量是~4.34 (FF原生的有~4.52),但是比合并列表花费3倍的时间。但如果你有大数字或大文本,最好少做cmp操作。 编辑:修复bug

在线测试http://mlich.zam.slu.cz/js-sort/x-sort-x2.htm

function sort4DG(cmp,start,end,n) // sort 4
{
var n     = typeof(n)    !=='undefined' ? n   : 1;
var cmp   = typeof(cmp)  !=='undefined' ? cmp   : sortCompare2;
var start = typeof(start)!=='undefined' ? start : 0;
var end   = typeof(end)  !=='undefined' ? end   : arr[n].length;
var count = end - start;
var pos = -1;
var i = start;
var cc = [];
// stabilni?
cc[01] = cmp(arr[n][i+0],arr[n][i+1]);
cc[23] = cmp(arr[n][i+2],arr[n][i+3]);
if (cc[01]>0) {swap(n,i+0,i+1);}
if (cc[23]>0) {swap(n,i+2,i+3);}
cc[12] = cmp(arr[n][i+1],arr[n][i+2]);
if (!(cc[12]>0)) {return n;}
cc[02] = cc[01]==0 ? cc[12] : cmp(arr[n][i+0],arr[n][i+2]);
if (cc[02]>0)
    {
    swap(n,i+1,i+2); swap(n,i+0,i+1); // bubble last to top
    cc[13] = cc[23]==0 ? cc[12] : cmp(arr[n][i+1],arr[n][i+3]);
    if (cc[13]>0)
        {
        swap(n,i+2,i+3); swap(n,i+1,i+2); // bubble
        return n;
        }
    else    {
    cc[23] = cc[23]==0 ? cc[12] : (cc[01]==0 ? cc[30] : cmp(arr[n][i+2],arr[n][i+3]));  // new cc23 | c03 //repaired
        if (cc[23]>0)
            {
            swap(n,i+2,i+3);
            return n;
            }
        return n;
        }
    }
else    {
    if (cc[12]>0)
        {
        swap(n,i+1,i+2);
        cc[23] = cc[23]==0 ? cc[12] : cmp(arr[n][i+2],arr[n][i+3]); // new cc23
        if (cc[23]>0)
            {
            swap(n,i+2,i+3);
            return n;
            }
        return n;
        }
    else    {
        return n;
        }
    }
return n;
}

对于任何优化,最好的方法总是测试、测试、测试。我至少会尝试网络排序和插入排序。如果我要打赌,我会根据过去的经验把钱押在插入排序上。

你知道输入数据吗?某些算法在处理特定类型的数据时会表现得更好。例如,插入排序在已排序或几乎已排序的数据上执行得更好,因此如果几乎已排序的数据的概率高于平均水平,则插入排序将是更好的选择。

您发布的算法类似于插入排序,但看起来您以更多比较为代价最小化了交换的数量。但是,比较的代价要比交换高得多,因为分支会导致指令管道暂停。

这是一个插入排序实现:

static __inline__ int sort6(int *d){
        int i, j;
        for (i = 1; i < 6; i++) {
                int tmp = d[i];
                for (j = i; j >= 1 && tmp < d[j-1]; j--)
                        d[j] = d[j-1];
                d[j] = tmp;
        }
}

这是我如何建立一个排序网络。首先,使用该站点为适当长度的网络生成最小的SWAP宏集。将其包装在一个函数中得到:

static __inline__ int sort6(int * d){
#define SWAP(x,y) if (d[y] < d[x]) { int tmp = d[x]; d[x] = d[y]; d[y] = tmp; }
    SWAP(1, 2);
    SWAP(0, 2);
    SWAP(0, 1);
    SWAP(4, 5);
    SWAP(3, 5);
    SWAP(3, 4);
    SWAP(0, 3);
    SWAP(1, 4);
    SWAP(2, 5);
    SWAP(2, 4);
    SWAP(1, 3);
    SWAP(2, 3);
#undef SWAP
}

这是我对这个线程的贡献:一个包含唯一值的6成员int向量(valp)的优化1,4间隙壳排序。

void shellsort (int *valp)
{      
  int c,a,*cp,*ip=valp,*ep=valp+5;

  c=*valp;    a=*(valp+4);if (c>a) {*valp=    a;*(valp+4)=c;}
  c=*(valp+1);a=*(valp+5);if (c>a) {*(valp+1)=a;*(valp+5)=c;}

  cp=ip;    
  do
  {
    c=*cp;
    a=*(cp+1);
    do
    {
      if (c<a) break;

      *cp=a;
      *(cp+1)=c;
      cp-=1;
      c=*cp;
    } while (cp>=valp);
    ip+=1;
    cp=ip;
  } while (ip<ep);
}

在我的惠普dv7-3010so笔记本电脑上,双核Athlon M300 @ 2 Ghz (DDR2内存),它在165个时钟周期内执行。这是一个平均计算从计时每个独特的序列(6!/总共720)。使用OpenWatcom 1.8编译到Win32。这个循环本质上是一个插入排序,有16条指令/37字节长。

我没有一个64位的环境来编译。