数学建模社区-数学中国
标题:
极限测试之Matlab与Forcal真实演练
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作者:
forcal
时间:
2011-8-4 08:15
标题:
极限测试之Matlab与Forcal真实演练
首先要说明的是本测试系列,除了比较编译效率外,其余所有运行效率的比较都是将matlab的首次运行排出在外,因matlab程序首次运行效率较低。理论上,Forcal程序任意次运行效率都是一样的。不过话又说回来,任意程序包含的函数,有些是需要多次运行的,而有些仅运行一次,甚至一次都不运行,故matlab函数首次运行效率较低应该是一个缺点。但如果说,matlab函数首次运行会对函数进行优化,以后运行效率会显著提高,则matlab函数首次运行效率较低就成了一个优点。
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本次演练要用matlab和Forcal实现两个实用函数并进行测试。当然,对脚本来说,这些函数用C/C++或Fortran来实现应是脚本的最佳选择。
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=============
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1、求解实系数方程组的全选主元高斯消去法:包含大量数组元素存取操作
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C/C++代码:
#include "stdafx.h"
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#include <stdio.h>
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#include <stdlib.h>
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#include "time.h"
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#include "math.h"
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int agaus(double *a,double *b,int n)
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{
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int *js,l,k,i,j,is,p,q;
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double d,t;
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js=new int[n];
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l=1;
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for (k=0;k<=n-2;k++)
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{
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d=0.0;
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for (i=k;i<=n-1;i++)
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{
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for (j=k;j<=n-1;j++)
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{
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t=fabs(a[i*n+j]);
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if (t>d) { d=t; js[k]=j; is=i;}
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}
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}
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if (d+1.0==1.0)
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{
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l=0;
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}
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else
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{
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if (js[k]!=k)
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{
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for (i=0;i<=n-1;i++)
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{
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p=i*n+k; q=i*n+js[k];
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t=a[p]; a[p]=a[q]; a[q]=t;
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}
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}
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if (is!=k)
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{
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for (j=k;j<=n-1;j++)
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{
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p=k*n+j; q=is*n+j;
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t=a[p]; a[p]=a[q]; a[q]=t;
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}
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t=b[k]; b[k]=b[is]; b[is]=t;
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}
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}
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if (l==0)
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{
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delete[] js; printf("fail\n");
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return(0);
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}
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d=a[k*n+k];
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for (j=k+1;j<=n-1;j++)
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{
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p=k*n+j; a[p]=a[p]/d;
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}
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b[k]=b[k]/d;
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for (i=k+1;i<=n-1;i++)
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{
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for (j=k+1;j<=n-1;j++)
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{
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p=i*n+j;
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a[p]=a[p]-a[i*n+k]*a[k*n+j];
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}
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b[i]=b[i]-a[i*n+k]*b[k];
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}
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}
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d=a[(n-1)*n+n-1];
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if (fabs(d)+1.0==1.0)
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{
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delete[] js; printf("fail\n");
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return(0);
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}
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b[n-1]=b[n-1]/d;
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for (i=n-2;i>=0;i--)
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{
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t=0.0;
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for (j=i+1;j<=n-1;j++)
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{
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t=t+a[i*n+j]*b[j];
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}
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b[i]=b[i]-t;
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}
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js[n-1]=n-1;
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for (k=n-1;k>=0;k--)
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{
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if (js[k]!=k)
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{
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t=b[k]; b[k]=b[js[k]]; b[js[k]]=t;
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}
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}
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delete[] js;
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return(1);
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}
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int main(int argc, char *argv[])
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{
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int i,j,k;
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double a[4][4]=
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{ {0.2368,0.2471,0.2568,1.2671},
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{0.1968,0.2071,1.2168,0.2271},
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{0.1581,1.1675,0.1768,0.1871},
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{1.1161,0.1254,0.1397,0.1490} };
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double b[4]={1.8471,1.7471,1.6471,1.5471};
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double aa[4][4],bb[4];
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clock_t tm;
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tm=clock();
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for(i=0;i<10000;i++)
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{
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for(j=0;j<4;j++)
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{
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for(k=0;k<4;k++)
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{
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aa[j][k]=a[j][k];
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}
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}
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for(j=0;j<4;j++)
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{
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bb[j]=b[j];
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}
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agaus((double *)aa,bb,4);
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}
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printf("循环 %d 次, 耗时 %d 毫秒。\n", i,(clock()-tm));
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for (i=0;i<=3;i++)
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{
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printf("x(%d)=%e\n",i,bb[i]);
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}
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}
复制代码
结果:
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循环 10000 次, 耗时 31 毫秒。
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x(0)=1.040577e+000
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x(1)=9.870508e-001
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x(2)=9.350403e-001
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x(3)=8.812823e-001
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---------
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matlab 2009a代码:
%file agaus.m
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function c=agaus(a,b,n)
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js=linspace(0,0,n);
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l=1;
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for k=1:n-1
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d=0.0;
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for i=k:n
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for j=k:n
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t=abs(a(i,j));
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if (t>d)
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d=t; js(k)=j; is=i;
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end
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end
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end
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if d+1.0==1.0
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l=0;
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else
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if js(k)~=k
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for i=1:n
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t=a(i,k); a(i,k)=a(i,js(k)); a(i,js(k))=t;
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end
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end
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if is~=k
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for j=k:n
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t=a(k,j); a(k,j)=a(is,j); a(is,j)=t;
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end
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t=b(k); b(k)=b(is); b(is)=t;
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end
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end
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if l==0
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printf('fail\n');
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c=[];
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return;
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end
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d=a(k,k);
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for j=k+1:n
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a(k,j)=a(k,j)/d;
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end
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b(k)=b(k)/d;
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for i=k+1:n
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for j=k+1:n
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a(i,j)=a(i,j)-a(i,k)*a(k,j);
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end
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b(i)=b(i)-a(i,k)*b(k);
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end
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end
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d=a(n,n);
% ~1 g4 `/ C4 l1 P* w! ?6 q
if abs(d)+1.0==1.0
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printf('fail\n');
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c=[];
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return;
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end
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b(n)=b(n)/d;
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for i=n-1:-1:1
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t=0.0;
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for j=i+1:n
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t=t+a(i,j)*b(j);
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end
% u1 | P! n3 u3 z. _2 g( l
b(i)=b(i)-t;
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end
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js(n)=n;
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for k=n:-1:1
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if js(k)~=k
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t=b(k); b(k)=b(js(k)); b(js(k))=t;
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end
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end
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c=b;
% d8 }1 @. I) X7 d
return;
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end
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a=[0.2368,0.2471,0.2568,1.2671;
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0.1968,0.2071,1.2168,0.2271;
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0.1581,1.1675,0.1768,0.1871;
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1.1161,0.1254,0.1397,0.1490] ;
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b=[ 1.8471,1.7471,1.6471,1.5471];
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tic
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for i=1:10000
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c=agaus(a,b,4);
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end
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c
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toc
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c =
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1.0406 0.9871 0.9350 0.8813
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Elapsed time is 0.762713 seconds.
复制代码
----------
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Forcal代码:
!using["math","sys"];
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agaus(a,b,n : js,l,k,i,j,is, d,t)=
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{
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oo{ js=array(n)},
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l=1, k=0,
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while{ k<n-1,
6 U, h& Y: D, }: [0 k
d=0.0, i=k,
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while{ i<n,
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j=k, while{j<n,
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t=abs(a[i,j]),
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if{t>d, d=t, js[k]=j, is=i},
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j++
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},
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i++
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},
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which{ d+1.0==1.0, l=0,
% ?' }' O! Y! C; C
{ if{ (js[k]!=k),
?" v a- l$ L* b
i=0, while{i<n,
5 K9 o* N- S% N- q
t=a[i,k], a[i,k]=a[i,js[k]], a[i,js[k]]=t,
* b% D* [# V9 m( R
i++
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}
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},
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if{ (is!=k),
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j=k, while{j<n,
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t=a[k,j], a[k,j]=a[is,j], a[is,j]=t,
$ \, M9 y& A. B4 i: D
j++
% l% z4 D+ e. a; W, D
},
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t=b[k], b[k]=b[is], b[is]=t
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}
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}
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},
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if{ (l==0),
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printff("fail\r\n"),
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return(0)
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},
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d=a[k,k],
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j=k+1, while {j<n, a[k,j]=a[k,j]/d, j++},
6 m4 u- j- e( a& `. L
b[k]=b[k]/d,
* ^: l" i" a0 L" ^) V; D. G; b
i=k+1, while {i<n,
w5 ?0 W! ?2 _( a
j=k+1, while{j<n,
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a[i,j]=a[i,j]-a[i,k]*a[k,j],
k/ ^7 s) ^" T: j" i3 `
j++
/ I+ ~& ^+ k+ g! h! a
},
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b[i]=b[i]-a[i,k]*b[k],
% |% e. `, R9 P+ w* W+ X" G
i++
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},
1 e1 n: w1 \0 _0 g5 Y/ {
k++
" T7 p' b, W/ n- |0 P/ F
},
: u5 b* @+ P0 T6 d/ x# O
d=a[(n-1),n-1],
! f1 s! o8 Q$ n2 l2 {8 V# M
if{ abs(d)+1.0==1.0,
5 Q( g- F8 z8 V$ e# v
printff("fail\r\n"),
' V$ L0 F( @6 h. z {
return(0)
$ g7 j+ q8 o; r2 w7 g" S( V
},
8 T7 l' |1 q0 M
b[n-1]=b[n-1]/d,
' Y. _2 k- i- F, P+ G" R } D
i=n-2, while{i>=0,
% C* ?- S3 {: m* q* h
t=0.0,
$ l g3 f( J0 V4 i$ z
j=i+1, while{j<n, t=t+a[i,j]*b[j], j++},
1 Y7 y) e( a( T+ t
b[i]=b[i]-t,
) b- U8 o4 S) y* C e( g( t* X; ?
i--
9 P' s: p( r) _5 @. N3 J
},
" ?0 f" }, r; f% v; }5 y! ^- Z
js[n-1]=n-1,
+ \3 [. |1 [9 n% ^0 |
k=n-1, while{k>=0,
; }7 F! i; f% h3 q0 J
if{(js[k]!=k), t=b[k], b[k]=b[js[k]], b[js[k]]=t},
9 R$ X& \, y- D6 R4 q4 n* N
k--
7 A/ z% ^* y) A, j7 b
},
/ u4 O% b* c1 l" E* c
return(1)
6 F' P, t5 \4 f
};
% m1 M& x4 ]7 N1 O! L' V: U
0 i- B* w4 q2 e2 ~
main(:i,a,b,aa,bb,t0)=
! Y6 I. ^! ~6 m( t" O
{
5 g, i9 f# C! E, C. [3 ~( \7 N
oo{a=arrayinit{2,4,4 :
3 Y3 ~" ?( \5 x) M
0.2368,0.2471,0.2568,1.2671,
- Y4 R! w% K9 @! a/ ?4 F% E
0.1968,0.2071,1.2168,0.2271,
8 j2 Z; H2 d1 ~! V3 u
0.1581,1.1675,0.1768,0.1871,
! a; D2 n. W& O* \3 ^3 v# T
1.1161,0.1254,0.1397,0.1490},
t# n7 y+ A8 j5 u
b=arrayinit{1,4 : 1.8471,1.7471,1.6471,1.5471},
2 d }5 |+ ^3 C' Q! A
aa=array[4,4], bb=array[4]
& e: d4 `2 `- a" S7 E+ A7 o
},
8 C! i7 I: X' z$ v( J% `
t0=clock(),
! f h2 r/ P- w$ i$ {4 x; y
i=0, while{i<10000, aa.=a, bb.=b, agaus(aa,bb,4), i++},
8 L; ~) H2 x8 A$ C; U
outm[bb],
7 i, g. ~8 v) O; j8 n2 \
[clock()-t0]/1000
. x. u, e% r4 Y* a7 |# Z# H' }9 S
};
复制代码
结果:
% ^+ Z+ s1 A. z4 D) @: f" U8 T- A
1.04058 0.987051 0.93504 0.881282
! a" F6 y- h' q) z5 i& Q8 z4 V
4 u b6 d. b/ ~, t3 F
2.125
9 N5 s. I; V1 h2 [; J" y) T
4 ^- b% Y/ @. o
Forcal用函数sys::A()对数组元素进行存取:
!using["math","sys"];
8 l4 F6 E) Y3 f' [( T" w; k5 W; ]7 p
agaus(a,b,n : js,l,k,i,j,is, d,t)=
) C! V; b5 ^/ }' i, G9 O3 e9 e
{
' N; v ?3 A" R# r
oo{ js=array(n)},
: J* g! Q G( u- C
l=1, k=0,
6 g2 v+ u; m2 `5 Y) r* M
while{ k<n-1,
( S4 H6 B* G( V$ s( h9 T
d=0.0, i=k,
! j- \# \' k( }/ M- j
while{ i<n,
& G j; \, m- h- v
j=k, while{j<n,
3 x. T+ W7 X/ H
t=abs(A[a,i,j]),
! h! w+ K( K$ |1 `; ]
if{t>d, d=t, A[js,k]=j, is=i},
3 e( }6 X( x4 E# U3 a
j++
3 ~, A7 E. O9 z% ]2 h
},
H/ ~3 N& u" D9 e. x
i++
6 N* ?( L$ A! l
},
3 ^9 m& Z# m4 [/ {8 y: s! s! ?
which{ d+1.0==1.0, l=0,
& f/ l. d: {2 Y# Z
{ if{ (A[js,k]!=k),
& n9 w; j$ M1 T1 I \$ e
i=0, while{i<n,
+ ]6 |1 p" A( z! M) k' {
t=A[a,i,k], A[a,i,k]=A[a,i,A[js,k]], A[a,i,A[js,k]]=t,
; E# s0 ~' c8 g: n4 C* b" c
i++
+ H0 g2 D* H5 a; |$ t2 H
}
3 q; A6 ] P* I$ S
},
4 v5 Z$ }6 A/ X3 v H
if{ (is!=k),
& p+ f5 j2 u& F9 y. ^' C4 ]2 X: q
j=k, while{j<n,
+ E' ^; `7 E* g+ ^9 s$ i; e
t=A[a,k,j], A[a,k,j]=A[a,is,j], A[a,is,j]=t,
8 a* B5 n. s, M; f D
j++
$ }* w- ^" Z/ H L) y
},
4 X; T& z L" l4 x- Q$ d, e% J- G
t=A[b,k], A[b,k]=A[b,is], A[b,is]=t
7 `, \9 ~- m! j! S- p+ O+ D: K
}
: @) J* `5 S9 q. J( i
}
* f$ \+ z+ W" @* G2 r- E
},
/ B; O& n0 R# M' m# i+ V
if{ (l==0),
3 w1 x8 \, d3 Q) ~7 k/ u) [7 N
printff("fail\r\n"),
. s/ s5 b. Y7 B& Q2 C
return(0)
% ?2 N6 Z& h4 K
},
/ W+ ]9 M# {% V" X/ F
d=A[a,k,k],
: e! B% f2 a4 }/ c) }
j=k+1, while {j<n, A[a,k,j]=A[a,k,j]/d, j++},
/ F1 K! k, P9 d9 f8 A# o
A[b,k]=A[b,k]/d,
* n) _2 a) w$ X
i=k+1, while {i<n,
/ L7 }' A% n; A' |2 J* f
j=k+1, while{j<n,
* F; e- F( [" @5 W% m" G
A[a,i,j]=A[a,i,j]-A[a,i,k]*A[a,k,j],
3 d" `' X- l# L7 @+ x- G) |+ e
j++
8 e+ u: i6 J# V* \- |$ @' M
},
* b9 L9 k- |. q/ y4 `
A[b,i]=A[b,i]-A[a,i,k]*A[b,k],
" t8 G7 B* W* i0 A6 d
i++
4 ?9 h) o* V* y2 y4 k, X
},
. g- b* i6 B% f5 i1 w
k++
H/ C. W6 t7 _8 J$ G9 z: k
},
: S8 X. \% A8 f: w: r9 Z
d=A[a,(n-1),n-1],
* [8 p( `+ o J6 K2 B5 j
if{ abs(d)+1.0==1.0,
$ ~# @9 r! z' e g
printff("fail\r\n"),
, \5 H! J4 r, B: G) h
return(0)
1 N1 d- s) T9 H# [$ `8 ?% V) Y
},
' w- ^( u# q" x; Z. B6 r
A[b,n-1]=A[b,n-1]/d,
& x. V, a& ]3 r# f
i=n-2, while{i>=0,
) n7 _) T6 g* s0 ^ k
t=0.0,
) ?7 l* n3 {) {: d" a/ C5 Q; y" `
j=i+1, while{j<n, t=t+A[a,i,j]*A[b,j], j++},
/ E1 B, i1 p3 q: D( z" ?6 m' W9 M& m
A[b,i]=A[b,i]-t,
) c# i" k0 Y$ U# j8 _( }
i--
& i* n- V& u8 `6 N4 M5 w7 q/ [
},
0 \9 }" t6 L) F" o& o5 ]
A[js,n-1]=n-1,
, |1 I7 s( _0 X6 ?. k& x
k=n-1, while{k>=0,
( x9 m+ q7 @ f* z9 @* s( Z# p
if{(A[js,k]!=k), t=A[b,k], A[b,k]=A[b,A[js,k]], A[b,A[js,k]]=t},
, T6 w' W1 g- J. N( @
k--
- z f6 ^' h# a6 |
},
: U, ^5 M7 {) s2 l8 }6 e$ S, S
return(1)
8 |7 f0 b: b4 O ~
};
% R3 {, J4 p6 N7 O/ ^3 b
& v$ j( P& q0 x3 w4 }2 J
main(:i,a,b,aa,bb,t0)=
7 u8 t% s% C5 Z5 V; [+ L- v3 x
{
8 a# b. L0 a* m! V, [8 R
oo{a=arrayinit{2,4,4 :
( s' U1 k; O) P+ @ r$ ^8 }
0.2368,0.2471,0.2568,1.2671,
5 a, f k: Z$ l4 z. i! f% } w
0.1968,0.2071,1.2168,0.2271,
8 n, F5 v- d( b* G! \
0.1581,1.1675,0.1768,0.1871,
# j! p' [0 [! l! Z6 Z* w# m6 V
1.1161,0.1254,0.1397,0.1490},
, J5 O8 {& A* U! q/ d$ } A
b=arrayinit{1,4 : 1.8471,1.7471,1.6471,1.5471},
; F3 _' x) X# Z7 W
aa=array[4,4], bb=array[4]
2 Q- x$ x( q; s1 _7 p) }, N
},
/ m* c M! s8 ?0 E# E1 F) w. W* K" V) V
t0=clock(),
' W% n* e$ D# ^" }
i=0, while{i<10000, aa.=a, bb.=b, agaus(aa,bb,4), i++},
3 r- F) U: X' Y5 x) q
outm[bb],
& }# z, c) F' }( @- G0 D6 h
[clock()-t0]/1000
/ s; J4 e' H9 |, x
};
复制代码
结果:
n' T4 P$ b* `% E& b' b$ r
1.04058 0.987051 0.93504 0.881282
6 j9 Z( l% _' h# H& I- J
" e/ L9 G) X+ g) ?0 b% S8 K8 ^
1.454
$ l. F; I! c) n( R! n
0 }+ B, c4 I4 _# I- S
----------
6 x1 N' e" ~' ]9 z; I0 ~1 P) Q
1 }: k c6 Y+ G; T; x
可以看出C/C++、matlab、Forcal耗时之比为 1 :25:68 (Forcal不使用函数sys::A())。
. q0 t# Y+ [: K
可以看出C/C++、matlab、Forcal耗时之比为 1 :25:47 (Forcal使用函数sys::A())。
/ h" k7 g" h9 i3 ]2 p
% g( } T. E9 S9 \' k# Z! }
本例Forcal耗时较长的原因在于本例程序含有大量的数组元素存取操作。
作者:
forcal
时间:
2011-8-4 08:44
2、变步长辛卜生二重求积法:没有数组元素操作
- a U' {' v% q: ^; W% n
4 }) Y M+ ~$ U% o+ ~- s2 ]
C/C++代码:
#include "stdafx.h"
# W. n; S0 N' i( b1 e0 I& e: L0 D
#include <stdio.h>
# k* q0 |- w9 q/ }* j2 n% m
#include <stdlib.h>
7 {: G9 |$ L {. ?! D* }
#include "time.h"
# I9 r3 c& d& E7 E2 Y5 M9 X |
#include "math.h"
1 ]3 y3 R7 Y$ A' \+ O5 F! j
7 c0 j( ^, f ]6 P4 M% m o
double simp1(double x,double eps);
( d. Y, c; U" x; M" {2 L
void fsim2s(double x,double y[]);
7 ^8 t- ~$ [& P6 M
double fsim2f(double x,double y);
: X7 J, R% x: w/ \0 z: M1 @) R& V
7 U& U: }$ G) n& L
double fsim2(double a,double b,double eps)
W7 x- u9 S0 |6 @, ?5 u
{
3 t; e _4 D+ N' e7 z; ?
int n,j;
, j4 m3 ], X0 c) ~' P7 \6 T
double h,d,s1,s2,t1,x,t2,g,s,s0,ep;
* b; |) ^1 C1 p
5 q7 l, \$ t! t9 x
n=1; h=0.5*(b-a);
' \9 n" s: D9 B& G) G8 [
d=fabs((b-a)*1.0e-06);
# {/ @9 I; Z3 x3 x( p k/ N S- x
s1=simp1(a,eps); s2=simp1(b,eps);
- s- Q2 ]. e6 a; ^
t1=h*(s1+s2);
* h% a1 m4 t' m7 l3 W, L- J' Q4 B! V
s0=1.0e+35; ep=1.0+eps;
$ } _" n' Q7 t5 e4 ^+ w! m) R" Y
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
) [# D ~5 r1 O0 J N
{
6 O" q! ?8 y! U6 j# t
x=a-h; t2=0.5*t1;
n5 R8 J% {6 f. K. P" k
for (j=1;j<=n;j++)
+ _5 M; p; h1 a2 N
{
7 F( R0 E' ^! o
x=x+2.0*h;
) ?! j4 w! P( u, e/ J
g=simp1(x,eps);
( b2 u# c3 ~- k Z+ S. H2 A( ?
t2=t2+h*g;
% C5 Z' R% @+ z- ^; ]
}
- G5 H1 j) C" K! |# Y/ @
s=(4.0*t2-t1)/3.0;
& y2 r5 a" M2 l7 _2 z
ep=fabs(s-s0)/(1.0+fabs(s));
$ y& H9 i% q" E, ?5 e! a0 N v. v
n=n+n; s0=s; t1=t2; h=h*0.5;
5 a4 C2 @9 _5 h
}
5 f2 S4 F2 L' v' S! T8 x1 l
return(s);
& P* d1 B; d6 Q, }
}
P, L. ^- a: M0 x) o2 V) Q- w
, X$ E K& d; I$ o! N, e
double simp1(double x,double eps)
, M2 P0 `* N$ S( U; h6 `8 P, q
{
- x; x# ?! k0 u2 N( U% {
int n,i;
9 |, a0 N: z+ d. b
double y[2],h,d,t1,yy,t2,g,ep,g0;
& E& Z+ J1 C9 @+ e$ @
% R3 `. Q$ l! j; r# q
n=1;
# L+ ]2 x1 v: M5 f: I; @, O
fsim2s(x,y);
3 ^. j, F( G1 U) m; f$ M
h=0.5*(y[1]-y[0]);
! c- Q; P3 {! W' a
d=fabs(h*2.0e-06);
7 f; J5 Y( Y: m& ?2 S* I
t1=h*(fsim2f(x,y[0])+fsim2f(x,y[1]));
- R1 c4 V+ {( j. d7 g* B3 l( a
ep=1.0+eps; g0=1.0e+35;
. n# a$ Y4 t$ S( h% B# U
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
9 n4 r5 K6 i b# X! U9 C6 Q. E
{
, a0 h! o) e( E$ c
yy=y[0]-h;
3 `: H) B9 M7 h
t2=0.5*t1;
/ p# G9 y3 W6 I# Z/ {
for (i=1;i<=n;i++)
0 w; d) ^$ m" |. [- F' z1 p
{
1 @* {. Z d$ \& Z) h# P( J) c1 h
yy=yy+2.0*h;
- L6 T+ l0 @% H- P
t2=t2+h*fsim2f(x,yy);
: H( g* J! T4 A0 X
}
# |* v( K F' R# F) P3 m$ v, `3 @
g=(4.0*t2-t1)/3.0;
4 j$ b, R3 F" ^/ ?
ep=fabs(g-g0)/(1.0+fabs(g));
7 Y: G) p- l s4 T& ?
n=n+n; g0=g; t1=t2; h=0.5*h;
! w1 c8 h/ v( \" s, M, {; n3 n
}
" t$ P4 E! x: M/ l" `8 h" Q
return(g);
: w/ n- ]4 s @% d9 i
}
( h( O8 [& U: |, ^
1 S; Y$ ]/ Y/ [
void fsim2s(double x,double y[])
/ i% {* w. l" K5 p; E
{
% S9 o- A! G Y+ x! q: J
y[0]=-sqrt(1.0-x*x);
( O& W" @. f H9 V8 y
y[1]=-y[0];
; T# W% O: c0 N. I
}
- M7 R1 L# b% p( d! x! P* {
/ o8 e; M( T4 r
double fsim2f(double x,double y)
, o* W# M* K) {8 f: o' M; q% B* G
{
7 c+ H: H8 ?& V, R
return exp(x*x+y*y);
% F6 G1 b' b, t3 H. K1 }
}
5 ~5 x4 U G4 y3 e0 P' D( E9 _; V3 Q
* G% Y4 g+ q: u8 F
int main(int argc, char *argv[])
2 R" V3 Q1 K, s- p
{
6 {; ?! r/ R4 L1 |6 W" }* F
int i;
% a; v) ~* R2 F7 N' X, Y3 `
double a,b,eps,s;
- y/ |7 l) j; Q/ ?1 X
clock_t tm;
* t1 G- P8 f# U- \) p) E6 ?. Q. x
/ Z( E+ w3 v* F, L: V( M
a=0.0; b=1.0; eps=0.0001;
6 q5 q7 a" A' s2 o
tm=clock();
) J6 x7 `$ w; M$ R+ t( o
for(i=0;i<100;i++)
0 {3 M9 W7 D3 u' x! E* g
{
: @, h& I. P- ?6 b$ `
s=fsim2(a,b,eps);
1 S, O! w! y4 [/ d/ B
}
0 r# |3 R* B' N, K+ w9 z# \
printf("s=%e , 耗时 %d 毫秒。\n", s, (clock()-tm));
# A' ?4 m# m% w0 K D" t. `, z/ t
}
复制代码
结果:
% v: H* m+ E0 W3 C7 ]" Q- {' N
s=2.698925e+000 , 耗时 78 毫秒。
3 b& G: ^. ~6 O+ T- ]& W- C7 ~. ?% S! y: D
) }# G! B6 `* a4 k
-------
" U( a5 F! D( K: K7 n& s
/ _* o4 k4 P( L& y. L
matlab代码:
%file fsim2.m
' n' u. _8 x! ?. W+ J1 ]
function s=fsim2(a,b,eps)
) t. g' v7 u/ t8 G- j
n=1; h=0.5*(b-a);
7 a6 d" o O" T. ^! V `
d=abs((b-a)*1.0e-06);
% [1 @" q, m" T/ @/ h
s1=simp1(a,eps); s2=simp1(b,eps);
( d- F/ y% Q- ?8 N( r P
t1=h*(s1+s2);
3 \& {: R4 n! _+ ?1 g* d3 u
s0=1.0e+35; ep=1.0+eps;
! g5 E/ ?0 e! x8 w+ A
while ((ep>=eps)&&(abs(h)>d))||(n<16),
$ W' @4 n8 N7 a. d1 a: Z4 Q
x=a-h; t2=0.5*t1;
: ]# E& e L- l/ A
for j=1:n
0 M' U+ N: [) b/ e8 I3 x9 T. C
x=x+2.0*h;
" Z, a! ]3 H, y j- e! z5 \2 L
g=simp1(x,eps);
( V+ c8 e+ x O% }/ \. }$ F
t2=t2+h*g;
8 ~9 c4 Y5 |1 Q3 X
end
) R1 j. L1 F" |# \( Z% e
s=(4.0*t2-t1)/3.0;
% U5 v' I/ H! k5 r
ep=abs(s-s0)/(1.0+abs(s));
; `" T3 K1 _ h
n=n+n; s0=s; t1=t2; h=h*0.5;
1 _: r5 f. r* Z9 W. J( p
end
& ^9 h2 z& x6 I0 O8 m2 G ~
end
" B# |& {2 b7 E5 r1 x
. B& k: _# _2 A9 l7 i; s1 Z
function g=simp1(x,eps)
, b. \9 {5 @% k3 G
n=1;
- ~ m5 ?0 h- C: ]5 B; d# [
[y0,y1]=f2s(x);
5 [& \3 B. v- M$ x
h=0.5*(y1-y0);
5 f1 S/ I, \9 N' a' z. r" F# y$ l2 ]
d=abs(h*2.0e-06);
) T4 J, y, f5 ` i# I4 C' F( p/ {
t1=h*(f2f(x,y0)+f2f(x,y1));
1 r3 i G5 Y6 ~+ R
ep=1.0+eps; g0=1.0e+35;
% }3 i9 a" ^, Q7 C& r% x
while (((ep>=eps)&&(abs(h)>d))||(n<16))
, v$ B N1 c" m$ L# \: A
yy=y0-h;
3 H) ~ ^# W6 u$ B
t2=0.5*t1;
# D, b; X* i2 m9 A. q' F; b/ N
for i=1:n
8 ]7 M: J* r1 e. F0 G7 g3 B
yy=yy+2.0*h;
& v$ _! t7 M7 H, X$ {! M" t6 n
t2=t2+h*f2f(x,yy);
( C' D2 I' S( C1 M( m6 P
end
) s1 @6 a5 J0 ?! s+ `9 v
g=(4.0*t2-t1)/3.0;
# l: I' R& K& _9 @' b
ep=abs(g-g0)/(1.0+abs(g));
2 w8 I! _) z; o M2 u
n=n+n; g0=g; t1=t2; h=0.5*h;
+ `3 ]* d4 @, ]5 s" z4 w; F
end
2 ~. d) l. p0 ~1 j. g$ ?0 Q
end
/ i9 _; O. b g6 L/ @
' t# t3 }; @3 i+ k/ }
%file f2s.m
, x) B- J& t: ~9 v$ s6 v( J
function [y0,y1]=f2s(x)
3 y" r5 r& i9 ~9 n* L. n
y0=-sqrt(1.0-x*x);
1 Y- ` G* U( V2 A4 |7 Z: k0 a
y1=-y0;
' m$ m3 E8 z7 v2 i
end
7 b! s c% M5 s4 a4 [3 P
) T/ k: T# T w& y1 U% t9 g" u
%file f2f.m
. o- ?2 ~* F/ ~4 w \1 j5 }- D
function c=f2f(x,y)
" T! _4 n, W2 |
c=exp(x*x+y*y);
2 G, b O3 d! {1 D
end
5 r% C, ~9 T o6 p7 k7 c3 W' B
. G3 C3 R( ]; I3 V5 L) M
%%%%%%%%%%%%%
' X z; L4 o# Y* I( j* Q
& Z* `" H6 k1 ]( a$ ?0 w& F4 l
>> tic
. h6 I, g5 R5 f+ T9 }4 ~
for i=1:100
7 V. w/ ?) l: ~: J, O0 Q6 e
a=fsim2(0,1,0.0001);
' B4 {: ^1 ~; L0 x) a
end
6 @- G- V. y: X" a9 R2 D i) H
a
- m" ?4 j [$ i" g' z
toc
( k* J* E; s+ C; U: z) {
; w: i g3 g- N# U4 a6 \
a =
+ k* D% F3 E( Y" E; f
# M, x Q3 g: x* H1 V7 G" N
2.6989
) x7 p0 N5 L7 O) h) T* x) A# y
- i& q& l# p0 d) ?
Elapsed time is 0.995575 seconds.
复制代码
-------
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Forcal代码:
fsim2s(x,y0,y1)=
& m) s0 L- h" e) ~7 R% s
{
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y0=-sqrt(1.0-x*x),
; q8 i4 C" g) V. v- n/ U4 t( |3 Q
y1=-y0
3 L* }% G' a* z6 H8 @2 Q
};
, t' v! u7 z0 z
fsim2f(x,y)=exp(x*x+y*y);
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//////////////////
7 q2 A9 u) m# U' x! P+ i
simp1(x,eps : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
% |3 v9 ] P8 r9 M4 q6 T' r
{
/ M; B; |' Q- f5 T2 D
n=1,
) _1 i" z0 w6 J2 R- M9 L
fsim2s(x,&y0,&y1),
) L; e- j8 J8 c" C
h=0.5*(y1-y0),
* L" D% x+ d- n2 c
d=abs(h*2.0e-06),
2 ^& ?9 q3 u9 h G( }& z$ i0 q- a
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
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ep=1.0+eps, g0=1.0e+35,
9 f2 y4 X# n0 e% J; Q2 i) G
while {((ep>=eps)&(abs(h)>d))|(n<16),
3 p& y" A6 q f; k4 C/ q9 Z
yy=y0-h,
/ j: C. |/ h, b- \, i
t2=0.5*t1,
- j! l }6 U6 ^% g1 N3 X
i=1, while{i<=n,
! B4 _# q* ^* ?2 m8 E
yy=yy+2.0*h,
1 L9 b+ v9 Q' h) n* i* k
t2=t2+h*fsim2f(x,yy),
) w* I4 _ @, m0 m+ B
i++
* D! m/ n d6 _) \" r4 T7 S. m
},
; g# J x' q/ Z8 ?2 r! c) \4 p- n
g=(4.0*t2-t1)/3.0,
$ T6 O7 G# y5 H g* p% Z' K' ^! _0 v
ep=abs(g-g0)/(1.0+abs(g)),
- P% _7 p0 b' f# `
n=n+n, g0=g, t1=t2, h=0.5*h
6 j2 T% V0 {4 H1 H
},
$ e$ j- ]. H, r; i/ d! ~9 h+ x
g
7 E% H4 y6 s! r( n# y
};
% s' M& h2 U/ n, ]; ]: J' H8 s
' l b4 c5 ?" B4 D3 Q
fsim2(a,b,eps : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
0 k8 K3 u: ?* E6 w% f; {
{
/ a1 y. f/ y* V4 K
n=1, h=0.5*(b-a),
: c0 V5 \0 g7 Z- j% E; y
d=abs((b-a)*1.0e-06),
& x0 n* K8 ~1 X+ Z! o* h$ \
s1=simp1(a,eps), s2=simp1(b,eps),
4 b; b0 b" T L6 {2 ^4 \
t1=h*(s1+s2),
/ Q) E# b" d6 x
s0=1.0e+35, ep=1.0+eps,
3 ]& i0 s% O( [# p1 w! Z
while {((ep>=eps)&(abs(h)>d))|(n<16),
V$ j! P9 d V7 j
x=a-h, t2=0.5*t1,
$ r3 i" l5 \) Z: h6 M: p
j=1, while{j<=n,
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x=x+2.0*h,
" {* ]: b% G+ l0 O# \! U- G
g=simp1(x,eps),
6 o# ?" b8 x: r G
t2=t2+h*g,
; w- `0 v! i6 {( x
j++
0 [ u# d7 N+ i- B# V! _
},
; m5 m, E: |6 G* l3 I
s=(4.0*t2-t1)/3.0,
/ U2 G& a# ~4 ?" z# J
ep=abs(s-s0)/(1.0+abs(s)),
4 F* x& G" K+ k" S8 M6 o; h* p1 ~
n=n+n, s0=s, t1=t2, h=h*0.5
& e" @8 @0 v8 J2 z- U
},
! m- L+ l) s2 }% C
s
- j) h5 N, b" N# p
};
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9 y8 X* [3 I2 d0 Q, C5 }
//////////////////
6 n1 S- K5 {& f, S: t% _! r
3 u! u& y3 l& q1 ] t& A( c
mvar:
) ?+ f6 C6 o8 q: l: H$ \% ? J
t0=sys::clock(),
* z/ v+ b6 L, v/ y8 f7 } d
i=0, while{i<100, a=fsim2(0,1,0.0001), i++}, a;
Z8 y/ R9 N# U# `0 }2 e% m4 E0 S \
[sys::clock()-t0]/1000;
复制代码
结果:
0 D5 ~) m) u, `" v( w+ z' [% i' n
2.698925000624303
i3 k' V3 n/ p# ]9 L b3 q# w
0.328
( G& b7 t- }7 G* _
2 }# W$ z) I) D" i& ?7 G( T
---------
) m8 A1 {9 q# o5 r6 p1 L
$ K6 k- C" a" H: W
本例C/C++、matlab、Forcal运行耗时之比为 1:12.7:4.2 。
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$ [( f. n" E: P4 U1 J# r z
本例matlab慢的原因应该在于有大量的函数调用。另外,matlab要求每一个函数必须存为磁盘文件真的很麻烦。
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本例还说明,对C/C++和Forcal脚本来说,C/C++的一条指令,Forcal平均大约用4.2条指令进行解释。
作者:
forcal
时间:
2011-8-4 09:00
3、变步长辛卜生二重求积法,写成通用的函数:没有数组元素操作
8 b* a2 T: ], R$ [/ n
/ `' H5 @$ V% [5 `; L1 A" ?' A
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。
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3 |& _: U: j/ \( g. \
不再给出C/C++代码,因其效率不会发生变化。
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Matlab代码:
%file fsim2.m
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function s=fsim2(a,b,eps,fsim2s,fsim2f)
5 k; p3 p! ]% C5 j: U7 J7 x
n=1; h=0.5*(b-a);
; _; y7 Y5 B: C' J
d=abs((b-a)*1.0e-06);
9 U# Y8 L- S- D3 x6 T1 D2 E2 t
s1=simp1(a,eps,fsim2s,fsim2f); s2=simp1(b,eps,fsim2s,fsim2f);
2 A2 B8 f& p5 }* Q4 L
t1=h*(s1+s2);
4 D; \6 C0 ~8 y6 @) c6 n
s0=1.0e+35; ep=1.0+eps;
( V9 U4 n) s" k1 S1 [6 j, R
while ((ep>=eps)&&(abs(h)>d))||(n<16),
. q' `% x" p4 f: h
x=a-h; t2=0.5*t1;
; `, A' O0 X, b v1 z
for j=1:n
o L8 _" H- U: K( z& a) S
x=x+2.0*h;
) G7 \' `. V8 [/ t5 @
g=simp1(x,eps,fsim2s,fsim2f);
% F3 [& h x& H5 F7 |0 k. n1 M# r
t2=t2+h*g;
_0 s4 M {! ^3 T+ K6 g% M T
end
# @, d5 H. L4 c' X
s=(4.0*t2-t1)/3.0;
; Q* \6 ^/ o4 F2 G0 Y/ c: L1 V
ep=abs(s-s0)/(1.0+abs(s));
# {4 c1 F4 y4 u9 A" E
n=n+n; s0=s; t1=t2; h=h*0.5;
$ D) v4 a; E& X6 o
end
! X: k2 b- w- w
end
% B f2 y& C% c
' p& ?( X6 O3 W* c
function g=simp1(x,eps,fsim2s,fsim2f)
" w: X% w, @0 a7 r# ? M
n=1;
9 d, c* ~* b$ x* ]
[y0,y1]=fsim2s(x);
3 f4 b3 o5 S! M+ j& I. z" [
h=0.5*(y1-y0);
% J! v8 w; X8 ]. ` X+ I3 W) O
d=abs(h*2.0e-06);
( U# @, a0 @2 |" J6 o: T: B
t1=h*(fsim2f(x,y0)+fsim2f(x,y1));
1 W* Y! Z0 S: K0 u
ep=1.0+eps; g0=1.0e+35;
! z4 D. u h1 T" \2 m4 S
while (((ep>=eps)&&(abs(h)>d))||(n<16))
. |" G v' E/ [ z# o
yy=y0-h;
# S4 ?. K$ s% t4 u8 M6 h
t2=0.5*t1;
# p6 |/ X: q* q4 S0 W* C: T
for i=1:n
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yy=yy+2.0*h;
1 x) W) _5 d, J- d2 W% t6 {4 j& ~
t2=t2+h*fsim2f(x,yy);
2 _4 ~1 k' Z! Q
end
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g=(4.0*t2-t1)/3.0;
+ }3 c( ^# _% w4 A) U" z( P
ep=abs(g-g0)/(1.0+abs(g));
) X$ A; p1 P- W, Z" o8 b
n=n+n; g0=g; t1=t2; h=0.5*h;
) y% q+ E& j; E4 {% g9 _$ |
end
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end
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9 x! |6 K& e* j
%file f2s.m
& ~2 c7 {" c* J
function [y0,y1]=f2s(x)
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y0=-sqrt(1.0-x*x);
' K0 ]% R- D0 G G$ ?
y1=-y0;
. q* k2 q0 A- P1 D/ S& A
end
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5 i1 ?5 W$ x4 k- l
%file f2f.m
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function c=f2f(x,y)
7 T. d9 ]+ U E* s0 b5 Q' u! O& c
c=exp(x*x+y*y);
4 T0 N5 c/ P% k: e l) |& b# y
end
! j; O. R! ]. w d2 T! @2 q
4 q2 \4 U2 _9 G. `4 W
%%%%%%%%%%%%%%%%
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- |& ^3 e5 L2 M6 y3 E- j, t
>> tic
- @: a- J/ V+ [5 K `/ R
for i=1:100
% s1 i1 ]$ y- {& T& Y
a=fsim2(0,1,0.0001,@f2s,@f2f);
' t9 T* Y8 h" ]
end
' ]* H$ _4 O) B8 ?5 O+ B% N
a
$ h0 g* F, c0 N; }3 Q
toc
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- \+ c1 z5 p# _# w: l- b$ D, n
a =
# }+ u/ c; K: p0 e+ I0 B+ X$ T
8 S ~7 m$ V, K3 V/ l* r- e# C
2.6989
1 M# a1 A" t! Q
7 Z1 \1 H" \9 c$ I! Y
Elapsed time is 1.267014 seconds.
复制代码
--------
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Forcal代码:
simp1(x,eps,fsim2s,fsim2f : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
3 s* p! j) d+ j# a: `) O
{
. H' b2 R! \2 k
n=1,
- q" O! u4 N2 e6 c1 W
fsim2s(x,&y0,&y1),
5 p7 I3 a) f4 x( P! t3 Y, v% O
h=0.5*(y1-y0),
, B; F0 i7 @, m6 V
d=abs(h*2.0e-06),
1 `4 ?# ~: c' J2 s
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
0 ~& }1 d3 n# A1 ]; o8 x5 ^
ep=1.0+eps, g0=1.0e+35,
+ M4 Q8 m9 Q0 p' e8 n; F- P8 z' ]* c
while {((ep>=eps)&(abs(h)>d))|(n<16),
* [1 u* \( c+ }! Q0 y: r7 A
yy=y0-h,
+ R' D& q, E4 G9 o$ ^
t2=0.5*t1,
, X$ ^3 a- K0 L$ y7 C- f, D6 O3 f
i=1, while{i<=n,
E" _ {( z, `
yy=yy+2.0*h,
& v. H) J1 [1 e. W
t2=t2+h*fsim2f(x,yy),
- x& ~3 W+ `7 B7 a+ E
i++
" Y2 n6 _- o1 X6 h+ X" U
},
- h# \- y7 Q5 @* T9 h$ `; J
g=(4.0*t2-t1)/3.0,
# F/ m! l: q/ L& Q/ L( w+ m
ep=abs(g-g0)/(1.0+abs(g)),
2 ~3 C0 h& b2 W1 B$ F
n=n+n, g0=g, t1=t2, h=0.5*h
+ o O# D" K! W
},
- H. s2 t' H* c* e
g
; k* b+ k# J% x) V3 d' k
};
/ K$ j6 U7 S9 d" Q; {2 x
% L& R$ p B8 i& _2 g, b
fsim2(a,b,eps,fsim2s,fsim2f : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
E) {4 D* r7 u9 ^7 Y: y8 D; E- L" n& W
{
; ?$ ^. ]: h8 U% _
n=1, h=0.5*(b-a),
2 l! [0 B$ D# f& G$ ]" t3 L+ S
d=abs((b-a)*1.0e-06),
0 F( G, ^: O* ?2 y" Q
s1=simp1(a,eps,fsim2s,fsim2f), s2=simp1(b,eps,fsim2s,fsim2f),
6 D! Z- \$ R6 p
t1=h*(s1+s2),
" z1 i _: B/ T+ c2 t
s0=1.0e+35, ep=1.0+eps,
# H4 N) m( u, q
while {((ep>=eps)&(abs(h)>d))|(n<16),
. W0 O5 y: }( L) ^2 i
x=a-h, t2=0.5*t1,
$ M/ _3 f8 V5 u: Q
j=1, while{j<=n,
2 F7 }7 f2 ? s* z5 J# Q
x=x+2.0*h,
/ f) m/ t% ~6 s- x# B
g=simp1(x,eps,fsim2s,fsim2f),
9 ~* G: Q9 N# w; C6 l5 q9 s* x: B; m
t2=t2+h*g,
& x2 N. ^1 i. y& v0 a" i# d5 J
j++
4 X' S/ I! x% L. h
},
& R' x- s7 a. w* T5 P* t
s=(4.0*t2-t1)/3.0,
- U* x, a4 h# s9 Q
ep=abs(s-s0)/(1.0+abs(s)),
; y3 v# P. G5 N+ w$ H
n=n+n, s0=s, t1=t2, h=h*0.5
# d* V& Q; a4 x5 S2 @. {+ m
},
4 o, ^/ Y5 ~: P1 C7 D3 g( z' ^6 I
s
$ v" e. p: f2 E3 I6 u; |6 F
};
: t* A% k$ k& s" o# Z) r
" \, \+ j9 [7 Z) @$ w, v. g# b
//////////////////
0 u/ O0 V$ q8 b% ^8 k7 J( S
/ y# _9 ^0 v/ H' i: g' A; r K
f2s(x,y0,y1)=
4 e: H, V2 x- y! i
{
" ^9 _# J; A7 R" ~: j' P! k
y0=-sqrt(1.0-x*x),
- ]" T/ j" q' G; I4 U
y1=-y0
1 i/ x5 F5 v7 T( R( C7 v
};
/ H( L& F3 }1 v5 q) o
f2f(x,y)=exp(x*x+y*y);
* ^1 \ W" z( O
3 c6 ]+ x, @ h o0 A
mvar:
5 p- m$ l0 P7 |6 |
t0=sys::clock(),
V, Z! |" P5 y4 v I+ e3 {2 t4 Z
i=0, while{i<100, a=fsim2(0,1,0.0001,HFor("f2s"),HFor("f2f")), i++}, a;
% w/ A/ A7 A& D7 W& D
[sys::clock()-t0]/1000;
复制代码
结果:
4 Q1 }/ s! s$ Y. D4 ~& D( X- C* h
2.698925000624303
8 e5 H7 [+ o0 N
0.844
' v5 T: _* j. H1 j1 e
8 M' T7 q/ ~+ d0 W! ?
--------
/ w7 U: C; v y" o, b& T
( r- n0 Q% x4 `) E' ~
本例matlab与Forcal耗时之比为1.267014 :0.844。Forcal仍有优势。
( \( j8 _; o9 ]; `) ?
( d, G+ H& v5 [
本例Forcal耗时增加的原因:在函数fsim2及simp1中要动态查找函数句柄fsim2s,fsim2f,并验证其是否有效,故效率下降了。
作者:
sxjm567
时间:
2012-7-30 01:48
百年不遇的好帖子,不得不顶
作者:
1055486706
时间:
2012-9-2 16:31
确实很不错
作者:
1055486706
时间:
2012-9-5 09:29
好深奥哦,肿么办?
欢迎光临 数学建模社区-数学中国 (http://www.madio.net/)
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