Lu与C/C++、Forcal、MATLAB、Python、Lua等各种语言的速度比较
以下比较均是在同一台计算机上进行的,在其他计算机上结果会有所不同。
1 Lu与C/C++、Forcal、Lua的数值计算速度比较
C/C++代码:
z=0.0;
for(x=0.0;x<=1.0;x=x+0.0011)
{
for(y=1.0;y<=2.0;y=y+0.0011)
{
z=z+cos(1.0-sin(1.2*pow(x+0.1,y/2.0-x)+cos(1.0-sin(1.2*pow(x+0.2,y/3.0-x))))-cos(1.0-sin(1.2*pow(x+0.3,y/4.0-x)))-cos(1.0-sin(1.2*pow(x+0.4,y/5.0-x)+cos(1.0-sin(1.2*pow(x+0.5,y/6.0-x))))-cos(1.0-sin(1.2*pow(x+0.6,y/7.0-x)))));
}
}
以上C/C++代码运行结果:z=19160.536601703152,耗时约1.328秒。
Forcal计算结果z=19160.536601703152,耗时约1.828秒。
Lua代码:
function f(x,y)
return math.cos(1-math.sin(1.2*(x+0.1)^(y/2-x)+math.cos(1-math.sin(1.2*(x+0.2)^(y/3-x))))-math.cos(1-
math.sin(1.2*(x+0.3)^(y/4-x)))-math.cos(1-math.sin(1.2*(x+0.4)^(y/5-x)+math.cos(1-math.sin(1.2*(x+0.5)^(y/6-x))))-
math.cos(1-math.sin(1.2*(x+0.6)^(y/7-x)))))
end
function z()
local t = os.clock()
local x=0
local y=0
local z=0
for x=0,1,0.0011 do
for y=1,2,0.0011 do
z=z+f(x,y)
end
end
io.write(z)
io.write(string.format(" Time Elapsed %f/n", os.clock() - t))
end
z()
Lua运行结果:
19160.536601703 Time Elapsed 3.234000
Lu代码:
main(:x,y,z,t)=
{
t=clock(), z=0.0, x=0.0,
while{x<=1.0,
y=1.0,
while{y<=2.0,
z=z+cos(1.0-sin(1.2*(x+0.1)^(y/2.0-x)+cos(1.0-sin(1.2*(x+0.2)^(y/3.0-x))))-cos(1.0-sin(1.2*(x+0.3)^(y/4.0-x)))-cos(1.0-sin(1.2*(x+0.4)^(y/5.0-x)+cos(1.0-sin(1.2*(x+0.5)^(y/6.0-x))))-cos(1.0-sin(1.2*(x+0.6)^(y/7.0-x))))),
y=y+0.0011
},
x=x+0.0011
},
o{"z=",z,",耗时约",[clock()-t]/1000.0,"秒。/r/n"}
};
Lu运行结果:
z=19160.536601703152,耗时约2.656秒。
2 八皇后问题
据测定,以下八皇后问题,Lu的运行速度约为C++的1/23,而Forcal的运行速度约为C++的1/10。
// 在运行不同的程序时,Lu的速度,从接近C++到只有C++速度的几十分之一。
// Lu的建议是:对运行时间较长的程序,如确有必要,设计成二级函数由Lu调用,从而获得接近C++速度的性能。
// Lu与C++是无缝链接的。故C++能实现的功能,借助二级函数,Lu完全可以实现。
// 但没有Lu支持的C++程序,将无法获得高效率地实时编译计算字符串表达式的功能。
// 据测定,以下八皇后问题,Lu的运行速度约为C++的1/23。
// 八皇后问题是一个古老而著名的问题,是回溯算法的典型例题。
// 该问题是19世纪著名的数学家高斯1850年提出:在8×8格的国际象棋盘上摆放8个皇后,使其不能互相攻击,即任意两个皇后都不能处于同一行、同一列或同一斜线上,问有多少种摆法。
// 高斯认为有76种方案。1854年在柏林的象棋杂志上不同的作者发表了40种不同的解,后来有人用图论的方法解出92种结果。
// 以下算法是从网上搜来的,该算法没有最终给出排列组合,仅仅给出有多少种组合,但是算法确实十分奥妙。
//Lu源程序
(::sum,upperlim)= sum=0, upperlim=1, SetStackMax(1000);
test(row, ld, rd : pos,p : sum,upperlim)=
{
which
{
row != upperlim,
{
pos = {upperlim && [!!(row||ld||rd)]},
while{ pos,
p = pos&&(-pos),
pos = pos -p,
test[row+p, (ld+p)<<1, (rd+p)>>1]
}
},
sum++
}
};
main(:tm,n:sum,upperlim)=
{
tm=clock(), n=15,
upperlim=(upperlim<<n)-1,
test(0,0,0),
o["Queens=",n,",sum=",sum,",耗时约",[clock()-tm]/1000.0,"秒。/r/n"]
};
Lu运行结果:
Queens=15,sum=2279184,耗时约136.703秒。
完成相同功能的C++程序:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
long sum=0,upperlim=1;
void test(long row, long ld, long rd)
{
if (row != upperlim)
{
long pos = upperlim & ~(row | ld | rd);
while (pos){
long p = pos& -pos;
pos -= p;
test(row+p, (ld+p)<<1, (rd+p)>>1);
}
}
else
sum++;
}
int main(int argc, char *argv[])
{
time_t tm;
int n=15;
if(argc!=1)n=atoi(argv[1]);
tm=time(0);
if((n<1)||(n>32))
{
printf(" heh..I can’t calculate that./n");
exit(-1);
}
printf("%d Queens/n",n);
upperlim=(upperlim<<n)-1;
test(0,0,0);
printf("Number of solutions is %ld, %d seconds/n", sum,(int)(time(0)-tm));
}
VC运行结果:
15 Queens
Number of solutions is 2279184, 6 seconds
3 Matlab与Lu普通函数调用效率
Matlab 2009a的测试代码:
f=@(x,y,z,t)x+y+z;
tic;
s=0;
for x=0:1000
for y=0:100
for z=0:100
s=s+f(x,y,z);
end
end
end
s
toc
s =
6.126720600000000e+009
Elapsed time is 9.546717 seconds.
将函数写成m文件后效率会提高,如下例:
%file xyz.m
function c=xyz(x,y,z)
c=x+y+z;
end
测试代码:
tic;
s=0;
for x=0:1000
for y=0:100
for z=0:100
s=s+xyz(x,y,z);
end
end
end
s
toc
s =
6.126720600000000e+009
Elapsed time is 4.724592 seconds.
Lu代码:
f(x,y,z)=x+y+z;
main(:t,s,x,y,z)=
t=clock(),
s=0,
x=0, while{x<=1000,
y=0, while{y<=100,
z=0, while{z<=100,
s=s+f(x,y,z),
z++
},
y++
},
x++
},
o{"s=",s,", time=",[clock()-t]/1000.};
Lu运行结果:
s=6126720600, time=4.015
4 Matlab与Lu的递归函数调用效率
以Fibonacci递归程序为例进行比较。
Matlab 2009a的Fibonacci函数定义:
function k=fib(n)
if n == 0
k=0;
return;
else if n == 1
k=1;
return;
else
k=fib(n - 1) + fib(n - 2);
return;
end
end
运行结果:
tic;
fib(30)
toc
ans =
832040
Elapsed time is 26.315245 seconds.
Lu的Fibonacci函数及代码:
SetStackMax(1000);
F(n)= which{
n == 0,
return(0),
n == 1,
return(1),
return [F(n - 1) + F(n - 2)]
};
main(:t)= t=clock(), o{"F(30)=",F(30),", time=",[clock()-t]/1000.};
Lu运行结果:
F(30)=832040, time=0.875
5 变步长辛卜生二重求积法
C/C++代码:
#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include "time.h"
#include "math.h"
double simp1(double x,double eps);
void fsim2s(double x,double y[]);
double fsim2f(double x,double y);
double fsim2(double a,double b,double eps)
{
int n,j;
double h,d,s1,s2,t1,x,t2,g,s,s0,ep;
n=1; h=0.5*(b-a);
d=fabs((b-a)*1.0e-06);
s1=simp1(a,eps); s2=simp1(b,eps);
t1=h*(s1+s2);
s0=1.0e+35; ep=1.0+eps;
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
{
x=a-h; t2=0.5*t1;
for (j=1;j<=n;j++)
{
x=x+2.0*h;
g=simp1(x,eps);
t2=t2+h*g;
}
s=(4.0*t2-t1)/3.0;
ep=fabs(s-s0)/(1.0+fabs(s));
n=n+n; s0=s; t1=t2; h=h*0.5;
}
return(s);
}
double simp1(double x,double eps)
{
int n,i;
double y[2],h,d,t1,yy,t2,g,ep,g0;
n=1;
fsim2s(x,y);
h=0.5*(y[1]-y[0]);
d=fabs(h*2.0e-06);
t1=h*(fsim2f(x,y[0])+fsim2f(x,y[1]));
ep=1.0+eps; g0=1.0e+35;
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
{
yy=y[0]-h;
t2=0.5*t1;
for (i=1;i<=n;i++)
{
yy=yy+2.0*h;
t2=t2+h*fsim2f(x,yy);
}
g=(4.0*t2-t1)/3.0;
ep=fabs(g-g0)/(1.0+fabs(g));
n=n+n; g0=g; t1=t2; h=0.5*h;
}
return(g);
}
void fsim2s(double x,double y[])
{
y[0]=-sqrt(1.0-x*x);
y[1]=-y[0];
}
double fsim2f(double x,double y)
{
return exp(x*x+y*y);
}
int main(int argc, char *argv[])
{
int i;
double a,b,eps,s;
clock_t tm;
a=0.0; b=1.0; eps=0.0001;
tm=clock();
for(i=0;i<100;i++)
{
s=fsim2(a,b,eps);
}
printf("s=%e , 耗时 %d 毫秒。/n", s, (clock()-tm));
}
C/C++结果:
s=2.698925e+000 , 耗时 78 毫秒。
-------
matlab 2009a代码:
%file fsim2.m
function s=fsim2(a,b,eps)
n=1; h=0.5*(b-a);
d=abs((b-a)*1.0e-06);
s1=simp1(a,eps); s2=simp1(b,eps);
t1=h*(s1+s2);
s0=1.0e+35; ep=1.0+eps;
while ((ep>=eps)&&(abs(h)>d))||(n<16),
x=a-h; t2=0.5*t1;
for j=1:n
x=x+2.0*h;
g=simp1(x,eps);
t2=t2+h*g;
end
s=(4.0*t2-t1)/3.0;
ep=abs(s-s0)/(1.0+abs(s));
n=n+n; s0=s; t1=t2; h=h*0.5;
end
end
function g=simp1(x,eps)
n=1;
[y0,y1]=f2s(x);
h=0.5*(y1-y0);
d=abs(h*2.0e-06);
t1=h*(f2f(x,y0)+f2f(x,y1));
ep=1.0+eps; g0=1.0e+35;
while (((ep>=eps)&&(abs(h)>d))||(n<16))
yy=y0-h;
t2=0.5*t1;
for i=1:n
yy=yy+2.0*h;
t2=t2+h*f2f(x,yy);
end
g=(4.0*t2-t1)/3.0;
ep=abs(g-g0)/(1.0+abs(g));
n=n+n; g0=g; t1=t2; h=0.5*h;
end
end
%file f2s.m
function [y0,y1]=f2s(x)
y0=-sqrt(1.0-x*x);
y1=-y0;
end
%file f2f.m
function c=f2f(x,y)
c=exp(x*x+y*y);
end
%%%%%%%%%%%%%
>> tic
for i=1:100
a=fsim2(0,1,0.0001);
end
a
toc
a =
2.6989
Elapsed time is 0.995575 seconds.
-------
Lu代码:
fsim2s(x,y0,y1)=
{
y0=-sqrt(1.0-x*x),
y1=-y0
};
fsim2f(x,y)=exp(x*x+y*y);
//////////////////
simp1(x,eps : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
{
n=1,
fsim2s(x,&y0,&y1),
h=0.5*(y1-y0),
d=abs(h*2.0e-06),
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
ep=1.0+eps, g0=1.0e+35,
while {((ep>=eps)&(abs(h)>d))|(n<16),
yy=y0-h,
t2=0.5*t1,
i=1, while{i<=n,
yy=yy+2.0*h,
t2=t2+h*fsim2f(x,yy),
i++
},
g=(4.0*t2-t1)/3.0,
ep=abs(g-g0)/(1.0+abs(g)),
n=n+n, g0=g, t1=t2, h=0.5*h
},
g
};
fsim2(a,b,eps : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
{
n=1, h=0.5*(b-a),
d=abs((b-a)*1.0e-06),
s1=simp1(a,eps), s2=simp1(b,eps),
t1=h*(s1+s2),
s0=1.0e+35, ep=1.0+eps,
while {((ep>=eps)&(abs(h)>d))|(n<16),
x=a-h, t2=0.5*t1,
j=1, while{j<=n,
x=x+2.0*h,
g=simp1(x,eps),
t2=t2+h*g,
j++
},
s=(4.0*t2-t1)/3.0,
ep=abs(s-s0)/(1.0+abs(s)),
n=n+n, s0=s, t1=t2, h=h*0.5
},
s
};
//////////////////
main(:t0,i,a)=
t0=clock(),
i=0, while{i<100, a=fsim2(0,1,0.0001), i++},
o{"a=",a,", time=",[clock()-t0]/1000.};
Lu结果:
a=2.6989250006243033, time=0.657
---------
本例C/C++、matlab、Lu运行耗时之比为 1:12.7:8.5 。
=========
以下将变步长辛卜生二重求积算法写成通用的函数。不再给出C/C++代码,因其效率不会发生变化。
注意函数fsim2中增加了两个参数,函数句柄fsim2s用于计算二重积分时内层的上下限,函数句柄fsim2f用于计算积分函数的值。
Matlab 2009a代码:
%file fsim2.m
function s=fsim2(a,b,eps,fsim2s,fsim2f)
n=1; h=0.5*(b-a);
d=abs((b-a)*1.0e-06);
s1=simp1(a,eps,fsim2s,fsim2f); s2=simp1(b,eps,fsim2s,fsim2f);
t1=h*(s1+s2);
s0=1.0e+35; ep=1.0+eps;
while ((ep>=eps)&&(abs(h)>d))||(n<16),
x=a-h; t2=0.5*t1;
for j=1:n
x=x+2.0*h;
g=simp1(x,eps,fsim2s,fsim2f);
t2=t2+h*g;
end
s=(4.0*t2-t1)/3.0;
ep=abs(s-s0)/(1.0+abs(s));
n=n+n; s0=s; t1=t2; h=h*0.5;
end
end
function g=simp1(x,eps,fsim2s,fsim2f)
n=1;
[y0,y1]=fsim2s(x);
h=0.5*(y1-y0);
d=abs(h*2.0e-06);
t1=h*(fsim2f(x,y0)+fsim2f(x,y1));
ep=1.0+eps; g0=1.0e+35;
while (((ep>=eps)&&(abs(h)>d))||(n<16))
yy=y0-h;
t2=0.5*t1;
for i=1:n
yy=yy+2.0*h;
t2=t2+h*fsim2f(x,yy);
end
g=(4.0*t2-t1)/3.0;
ep=abs(g-g0)/(1.0+abs(g));
n=n+n; g0=g; t1=t2; h=0.5*h;
end
end
%file f2s.m
function [y0,y1]=f2s(x)
y0=-sqrt(1.0-x*x);
y1=-y0;
end
%file f2f.m
function c=f2f(x,y)
c=exp(x*x+y*y);
end
%%%%%%%%%%%%%%%%
>> tic
for i=1:100
a=fsim2(0,1,0.0001,@f2s,@f2f);
end
a
toc
a =
2.6989
Elapsed time is 1.267014 seconds.
--------
Lu代码:
simp1(x,eps,fsim2s,fsim2f : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
{
n=1,
fsim2s(x,&y0,&y1),
h=0.5*(y1-y0),
d=abs(h*2.0e-06),
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
ep=1.0+eps, g0=1.0e+35,
while {((ep>=eps)&(abs(h)>d))|(n<16),
yy=y0-h,
t2=0.5*t1,
i=1, while{i<=n,
yy=yy+2.0*h,
t2=t2+h*fsim2f(x,yy),
i++
},
g=(4.0*t2-t1)/3.0,
ep=abs(g-g0)/(1.0+abs(g)),
n=n+n, g0=g, t1=t2, h=0.5*h
},
g
};
fsim2(a,b,eps,fsim2s,fsim2f : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
{
n=1, h=0.5*(b-a),
d=abs((b-a)*1.0e-06),
s1=simp1(a,eps,fsim2s,fsim2f), s2=simp1(b,eps,fsim2s,fsim2f),
t1=h*(s1+s2),
s0=1.0e+35, ep=1.0+eps,
while {((ep>=eps)&(abs(h)>d))|(n<16),
x=a-h, t2=0.5*t1,
j=1, while{j<=n,
x=x+2.0*h,
g=simp1(x,eps,fsim2s,fsim2f),
t2=t2+h*g,
j++
},
s=(4.0*t2-t1)/3.0,
ep=abs(s-s0)/(1.0+abs(s)),
n=n+n, s0=s, t1=t2, h=h*0.5
},
s
};
//////////////////
f2s(x,y0,y1)=
{
y0=-sqrt(1.0-x*x),
y1=-y0
};
f2f(x,y)=exp(x*x+y*y);
main(:t0,i,a)=
t0=clock(),
i=0, while{i<100, a=fsim2(0,1,0.0001,HFor("f2s"),HFor("f2f")), i++},
o{"a=",a,", time=",[clock()-t0]/1000.};
Lu运行结果:
a=2.6989250006243033, time=0.719
--------
本例matlab与Lu耗时之比为1.267014 :0.719。
6 Matlab与Lu动态内存管理效率
matlab 2009a代码:
clear all
tic
s=0;
for k=1:1000
for i=1:1000
a={2,2,2,2};
s=s+a{1};
end
end
s
toc
s =
2000000
Elapsed time is 5.030599 seconds.
Lu代码:
main(:a,t0,s,k,i)=
t0=clock(),
s=0,
k=0, while{k<1000,
i=0, while{i<1000, a=lu(2,2,2,2), s=s+a[1], i++},
k++
},
o{"s=",s,", time=",[clock()-t0]/1000.};
Lu运行结果:
s=2000000, time=0.703
7 Lu与VC动态数组存取效率测试
VC源程序:
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <math.h>
#include <time.h>
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
clock_t old,now;
__int64 i,k;
__int64 *p;
char ch;
old=clock();
for(i=0,k=0;i<=1000000;i++)
{
p=new __int64[5];
p[3]=i; k=k+p[3];
delete[] p;
}
now=clock();
cout<<"vc++:"<<setprecision(20)<<k;
cout<<" 运行时间:"<<now-old<<" 即: "<<(double)(now-old)/CLOCKS_PER_SEC<<"秒"<<endl<<endl;
cin>>ch;
return 0;
}
VC运行结果:
vc++:500000500000 运行时间:218 即: 0.218秒
Lu源程序:
(:i,k,p,t)=
{ t=clock(),i=0,k=0,
(i<=1000000).while
{
p=new[ints,5],
A[p,3]=i, k=k+A[p,3],
del[p], //实际上在这里不需要使用del函数,去掉此语句将更快
i++
},
o{"结果=",k,", 时间=",[clock()-t]/1000.,"秒。/r/n"}
};
Lu运行结果:
结果=500000500000, 时间=0.86秒。
在该例子中,Lu的效率为VC的0.86/0.218=3.95分之一。若将new和del这两个函数移到循环体的外边,Lu的效率如下:
VC源程序:
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <math.h>
#include <time.h>
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
clock_t old,now;
__int64 i,k;
__int64 *p;
char ch;
old=clock(); p=new __int64 [5];
for(i=0,k=0;i<=20000000;i++)
{
p[3]=i; k=k+p[3];
}
delete[] p; now=clock();
cout<<"vc++:"<<setprecision(20)<<k;
cout<<" 运行时间:"<<now-old<<" 即: "<<(double)(now-old)/CLOCKS_PER_SEC<<"秒"<<endl<<endl;
cin>>ch;
return 0;
}
VC运行结果:
vc++:200000010000000 运行时间:125 即: 0.125秒
Lu源程序:
(:i,k,p,t)=
{ t=clock(),p=new[ints,5],i=0,k=0,
(i<=20000000).while
{
A[p,3]=i, k=k+A[p,3],
i++
},
del[p], //实际上在这里不需要使用del函数
o{"结果=",k,", 时间=",[clock()-t]/1000.,"秒。/r/n"}
};
Lu运行结果:
结果=200000010000000, 时间=7.922秒。
在该例子中,Lu的速度为VC的7.922/0.125=63.376分之一。注意循环次数都增加到了20000000次。
8 一段有趣的程序
在这个网页看到一篇各种语言运行速度比较的文章,摘抄如下:
— Erik Wrenholt (erik -at- timestretch.com) 2005-09-20
Language Time Relative Speed
C gcc-4.0.1 0.05 seconds 1.00 x
ocaml compiled 3.09.2 0.05 seconds 1.00 x
SBCL 1.0.2 0.13 seconds 2.55 x
Java 1.4.2 0.40 seconds 8.00 x
Io 20070410 Vector 1.40 seconds 28.09 x
Lua 5.1 1.50 seconds 30.00 x
ocaml bytecode 3.09.2 3.76 seconds 75.15 x
Python 2.5.1 9.99 seconds 199.80 x
Ghostscript 8.51 11.66 seconds 233.12 x
Perl 5.8.6 Optimized 12.37 seconds 247.34 x
TCL 8.4 Optimized 16.00 seconds 320.00 x
Perl 5.8.6 21.75 seconds 435.00 x
PHP 5.1.4 23.12 seconds 462.40 x
Javascript SpiderMonkey v1.6 31.06 seconds 621.27 x
Ruby 1.8.4 34.31 seconds 686.18 x
Emacs Lisp 47.25 seconds 945.00 x
Applescript 71.75 seconds 1435.00 x
Io 20070410 85.26 seconds 1705.13 x
用以上网址提供的c代码与Lu比较,c代码改写为vs 2008可接受的形式,编译运行,结果如下:
#include "stdafx.h"
#include <math.h>
#include <time.h>
#define BAILOUT 16
#define MAX_ITERATIONS 1000
int mandelbrot(double x, double y)
{
double cr = y - 0.5;
double ci = x;
double zi = 0.0;
double zr = 0.0;
int i = 0;
while(1) {
i ++;
double temp = zr * zi;
double zr2 = zr * zr;
double zi2 = zi * zi;
zr = zr2 - zi2 + cr;
zi = temp + temp + ci;
if (zi2 + zr2 > BAILOUT)
return i;
if (i > MAX_ITERATIONS)
return 0;
}
}
int _tmain (int argc, _TCHAR* argv[]) {
clock_t old,now;
old=clock();
int x,y;
for (y = -39; y < 39; y++) {
printf("/n");
for (x = -39; x < 39; x++) {
int i = mandelbrot(x/40.0, y/40.0);
if (i==0)
printf("*");
else
printf(" ");
}
}
printf ("/n");
now=clock();
double query_time = ((double)(now-old))/CLOCKS_PER_SEC;
printf ("C Elapsed %0.2f/n", query_time);
return 0;
}
运行结果:
*
*
*
*
*
***
*****
*****
***
*
*********
*************
***************
*********************
*********************
*******************
*******************
*******************
*******************
***********************
*******************
*******************
*********************
*******************
*******************
*****************
***************
*************
*********
*
***************
***********************
* ************************* *
*****************************
* ******************************* *
*********************************
***********************************
***************************************
*** ***************************************** ***
*************************************************
***********************************************
*********************************************
*********************************************
***********************************************
***********************************************
***************************************************
*************************************************
*************************************************
***************************************************
***************************************************
* *************************************************** *
***** *************************************************** *****
****** *************************************************** ******
******* *************************************************** *******
***********************************************************************
********* *************************************************** *********
****** *************************************************** ******
***** *************************************************** *****
***************************************************
***************************************************
***************************************************
***************************************************
*************************************************
*************************************************
***************************************************
***********************************************
***********************************************
*******************************************
*****************************************
*********************************************
**** ****************** ****************** ****
*** **************** **************** ***
* ************** ************** *
*********** ***********
** ***** ***** **
* * * *
C Elapsed 0.25
Forcal用的时间与c++相比,为1.078:0.25=4.312:1。
Lua的代码及运行时间。
#!/usr/local/bin/lua
-- By Erik Wrenholt
local BAILOUT = 16
local MAX_ITERATIONS = 1000
function iterate(x,y)
local cr = y-0.5
local ci = x
local zi = 0.0
local zr = 0.0
local i = 0
while 1 do
i = i+1
local temp = zr * zi
local zr2 = zr*zr
local zi2 = zi*zi
zr = zr2-zi2+cr
zi = temp+temp+ci
if (zi2+zr2 > BAILOUT) then
return i
end
if (i > MAX_ITERATIONS) then
return 0
end
end
end
function mandelbrot()
local t = os.clock()
for y = -39, 38 do
for x = -39, 38 do
if (iterate(x/40.0, y/40) == 0) then
io.write("*")
else
io.write(" ")
end
end
io.write("/n")
end
io.write(string.format("Time Elapsed %f/n", os.clock() - t))
end
mandelbrot()
运行输出图形与前面相同,仅给出运行时间:
Time Elapsed 0.860000
Python的代码及运行时间。
#!/usr/local/bin/python
# by Daniel Rosengren
import sys, time
stdout = sys.stdout
BAILOUT = 16
MAX_ITERATIONS = 1000
class Iterator:
def __init__(self):
stdout.write('Rendering...')
for y in range(-39, 39):
stdout.write('/n')
for x in range(-39, 39):
i = self.mandelbrot(x/40.0, y/40.0)
if i == 0:
stdout.write('*')
else:
stdout.write(' ')
def mandelbrot(self, x, y):
cr = y - 0.5
ci = x
zi = 0.0
zr = 0.0
i = 0
while True:
i += 1
temp = zr * zi
zr2 = zr * zr
zi2 = zi * zi
zr = zr2 - zi2 + cr
zi = temp + temp + ci
if zi2 + zr2 > BAILOUT:
return i
if i > MAX_ITERATIONS:
return 0
t = time.time()
Iterator()
stdout.write('/nPython Elapsed %.02f' % (time.time() - t))
运行结果:
Python Elapsed 2.84
=======================
完成相同功能的Lu代码如下:
mandelbrot(x, y : cr,ci,zi,zr,i,temp,zr2,zi2) =
{
cr = y - 0.5,
ci = x,
zi = 0.0,
zr = 0.0,
i = 0,
(1).while {
i ++,
temp = zr * zi,
zr2 = zr * zr,
zi2 = zi * zi,
zr = zr2 - zi2 + cr,
zi = temp + temp + ci,
if [zi2 + zr2 > 16, return (i)],
if [i > 1000, return (0)]
}
};
main (:i,x,y,old,now) = {
old=clock(),
y = -39,
while{ y < 39,
o("/r/n"),
x = -39,
while{ x < 39,
i = mandelbrot(x/40.0, y/40.0),
which{ i==0,
o("*"),
o(" ")
},
x++
},
y++
},
now=clock(),
o["/r/nLu Elapsed ",(now-old)/1000.0]
};
运行时输出的图形与C程序相同,在演示程序DemoLu32.exe(这是个windows程序)中的运行时间为:
Lu Elapsed 5.657
从运行结果可看出,Lu用的时间与c++相比,为5.657:0.25=22.628:1。
因C/C++、Forcal、Lua等均使用控制台程序,故设计加载Lu并演示以上程序的C/C++控制台程序如下:
#include <windows.h>
#include <iostream>
#include <math.h>
#include "lu32.h"
#pragma comment( lib, "lu32.lib" )
using namespace std;
void _stdcall LuMessage(wchar_t *pch) //输出动态库信息,该函数注册到Lu,由Lu二级函数调用
{
wcout<<pch;
}
void main(void)
{
void *hFor; //表达式句柄
luINT nPara; //存放表达式的自变量个数
LuData *pPara; //存放输入自变量的数组指针
luINT ErrBegin,ErrEnd; //表达式编译出错的初始位置和结束位置
int ErrCode1,ErrCode2; //错误代码
void *v;
wchar_t mandelbrot[]=L"mandelbrot(x, y : cr,ci,zi,zr,i,temp,zr2,zi2) =/r/n{/r/n cr = y - 0.5,/r/n ci = x,/r/n zi = 0.0,/r/n zr = 0.0,/r/n i = 0,/r/n (1).while {/r/n i ++,/r/n temp = zr * zi,/r/n zr2 = zr * zr,/r/n zi2 = zi * zi,/r/n zr = zr2 - zi2 + cr,/r/n zi = temp + temp + ci,/r/n if [zi2 + zr2 > 16, return (i)],/r/n if [i > 1000, return (0)]/r/n }/r/n }";//字符串表达式
wchar_t mymain[]=L"mymain (:i,x,y,old,now) = {/r/n old=clock(),/r/n y = -39,/r/n while{ y < 39,/r/n o(/"/r/n/"),/r/n x = -39,/r/n while{ x < 39,/r/n i = mandelbrot(x/40.0, y/40.0),/r/n which{ i==0,/r/n o(/"*/"),/r/n o(/" /")/r/n },/r/n x++/r/n },/r/n y++/r/n },/r/n now=clock(),/r/n o[/"/r/nLu Elapsed /",(now-old)/1000.0]/r/n}"; //字符串表达式
if(!InitLu()) return; //初始化Lu
InsertKey("/0/0/0/0",4,luPubKey_User,LuMessage,NULL,NULL,1,v); //使Lu运行时可输出函数信息
ErrCode1=LuCom(mandelbrot,0,0,0,hFor,nPara,pPara,ErrBegin,ErrEnd); //编译表达式
ErrCode2=LuCom(mymain,0,0,0,hFor,nPara,pPara,ErrBegin,ErrEnd); //编译表达式
if(ErrCode1 || ErrCode2)
{
cout<<"表达式有错误!错误代码:"<<ErrCode1<<" "<<ErrCode2<<endl;
}
else
{
LuCal(hFor,pPara); //计算表达式的值
}
FreeLu(); //释放Lu
}
运行时输出的图形与C程序相同,运行时间为:
Lu Elapsed 2.063
从运行结果可看出,Lu用的时间与c++相比,为2.063:0.25=8.252:1。
9 Lu与Python比较的测试例子
Python程序:
Python code>>> def f(k):
n=0
i=0
while i <=k:
n=n+i
i=i+1
return n
>>> f(100000000)
5000000050000000
耗时约32秒(用秒表测量)。
Lu程序:
f(k:i,n)= i=0,n=0, while{i<=k, n=n+i++}, n;
(:t)= t=clock(), o{"f(100000000)=",f(100000000),", time=",[clock()-t]/1000.};
Lu运行结果:
f(100000000)=5000000050000000, time=14.797
======
另一个测试程序:
Python 程序:
Python code>>> def f(x,y):
return math.cos(1-math.sin(1.2*(x+0.1)**(y/2-x)+math.cos(1-math.sin(1.2*(x+0.2)**(y/3-x))))-math.cos(1-math.sin(1.2*(x+0.3)**(y/4-x)))-math.cos(1-math.sin(1.2*(x+0.4)**(y/5-x)+math.cos(1-math.sin(1.2*(x+0.5)**(y/6-x))))-math.cos(1-math.sin(1.2*(x+0.6)**(y/7-x)))))
>>> def g(m,n):
x=0
i=0
while i <=m:
j=0
while j <=n:
x=x+f(i,j)
j=j+0.01
i=i+0.01
return x
>>> g(12.3,7.89)
404005.29176341009
Python运行约10秒(用秒表测量)。
Lu程序:
f(x,y)=cos(1-sin(1.2*(x+0.1)^(y/2-x)+cos(1-sin(1.2*(x+0.2)^(y/3-x))))-cos(1-sin(1.2*(x+0.3)^(y/4-x)))-cos(1-sin(1.2*(x+0.4)^(y/5-x)+cos(1-sin(1.2*(x+0.5)^(y/6-x))))-cos(1-sin(1.2*(x+0.6)^(y/7-x)))));
g(m,n:i,j,x)=
x=0,i=0,
(i<=m).while{
j=0,
(j<=n).while{
x=x+f(i,j),
j=j+0.01
},
i=i+0.01
},
x;
main(:t)= t=clock(), o{"g(12.3,7.89)=",g(12.3,7.89),", time=",[clock()-t]/1000.};
Lu结果:
g(12.3,7.89)=404005.29176341009, time=3.43
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最近更新:2011年10月19日