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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=============
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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);
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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
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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;
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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,
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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,
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{ if{ (js[k]!=k),
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i=0, while{i<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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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,
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j++
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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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},
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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],
% d/ m& |* `& t# g0 e: z9 E
j=k+1, while {j<n, a[k,j]=a[k,j]/d, j++},
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b[k]=b[k]/d,
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i=k+1, while {i<n,
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j=k+1, while{j<n,
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a[i,j]=a[i,j]-a[i,k]*a[k,j],
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j++
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},
- ?2 h9 t) Q F3 |* E/ o
b[i]=b[i]-a[i,k]*b[k],
& C- V6 ^! c. n1 z E
i++
" O$ \: j% K" u. d
},
4 i! i9 i7 N1 C" N6 _0 g6 B
k++
1 O8 R- l; u& z5 U8 p- ?; q) Y7 A0 u
},
; O9 g8 J3 k: l, @! M, U
d=a[(n-1),n-1],
+ A4 u6 o6 K. w% ?$ }2 x4 S: }
if{ abs(d)+1.0==1.0,
( v" E7 f: Q1 Y3 K( V& Y- R
printff("fail\r\n"),
, \! |2 Q, Y% { `6 x- J( A+ ^ Y
return(0)
. @9 D5 U. D Z' Q9 e
},
- ^3 N7 k: e1 G- r0 ?$ O+ k/ ~
b[n-1]=b[n-1]/d,
* r0 Y6 t* a5 D/ V
i=n-2, while{i>=0,
4 ?6 J( s, P: s3 Z
t=0.0,
0 }; Z8 |/ E7 r# f
j=i+1, while{j<n, t=t+a[i,j]*b[j], j++},
Y c; ~1 ~) L9 ^7 j! ?
b[i]=b[i]-t,
$ a% l" h4 f: K+ m- U4 _/ A) r2 X
i--
3 V! w/ T+ r q/ ~
},
2 j1 I) R1 s0 j6 D6 Y) _4 H" W9 C( L
js[n-1]=n-1,
0 j1 J4 e7 r N3 o. ~ I
k=n-1, while{k>=0,
5 g" l& v( I! [ I! T! ~( R3 _( k. q
if{(js[k]!=k), t=b[k], b[k]=b[js[k]], b[js[k]]=t},
# p( g" W/ E' J% B; Z+ v
k--
# n2 ~5 g+ f) ~5 ]
},
! k' S3 l, Y) ]9 X+ L: ?% D2 x1 c
return(1)
% t3 Y% n6 m4 t. _5 b/ m: L _
};
" @& h9 U9 X6 z- P7 P
2 S! w. U8 D# m# s. }5 N
main(:i,a,b,aa,bb,t0)=
; ?* j# M% s* `7 z
{
; ?6 C9 D: @ Y1 [) s8 ~+ I
oo{a=arrayinit{2,4,4 :
( {* W+ P, C- \3 ^- B
0.2368,0.2471,0.2568,1.2671,
" w: m8 b/ b4 J' E8 |4 D9 |
0.1968,0.2071,1.2168,0.2271,
, k" J7 {# Q0 p) r. |2 G1 b, I. W! P
0.1581,1.1675,0.1768,0.1871,
3 [) v; g" T' O. h
1.1161,0.1254,0.1397,0.1490},
/ c8 f* }- |/ R1 R, {/ \
b=arrayinit{1,4 : 1.8471,1.7471,1.6471,1.5471},
; x# ~% V% `; z1 g1 ^. ]
aa=array[4,4], bb=array[4]
+ i: O( N$ I- L' _9 _. V$ B
},
3 K- v& v6 X. r) d' l
t0=clock(),
9 D; Z( b; `2 e. j
i=0, while{i<10000, aa.=a, bb.=b, agaus(aa,bb,4), i++},
& s- i: `5 i, w6 L' M: Y# ?" S
outm[bb],
4 D& g% L" C6 G, {9 b
[clock()-t0]/1000
/ Y* V) B2 W1 p0 g2 [
};
复制代码
结果:
Y5 n/ R4 W1 x$ J; e' g# ]8 B
1.04058 0.987051 0.93504 0.881282
; o2 L1 \0 _& V% [* S
4 E: D4 R) Q8 i& i* Q" o
2.125
0 ^: X6 w+ W* `' ~ K
. W0 K2 i9 X" j/ M6 e
Forcal用函数sys::A()对数组元素进行存取:
!using["math","sys"];
: I0 @5 u4 ?& a, w
agaus(a,b,n : js,l,k,i,j,is, d,t)=
& `9 W' z. F9 ]$ |! R
{
& b! g7 \2 K6 S
oo{ js=array(n)},
1 ~$ i( ?6 b/ u8 I7 `" V2 S
l=1, k=0,
* Y+ [2 o8 O5 U0 W1 N7 g2 d* d& d
while{ k<n-1,
c Q2 ~5 i F, I6 Y y7 w
d=0.0, i=k,
0 Z) E: `( X0 v' ~; o
while{ i<n,
" E/ @: r! R; H. k4 g. Q
j=k, while{j<n,
7 P6 Q- B# G4 b* Z- J; N( i
t=abs(A[a,i,j]),
, [9 r$ x8 Z1 f& n# ]5 ^" F' R
if{t>d, d=t, A[js,k]=j, is=i},
( G) j z; F& }
j++
4 S* P0 ~0 W0 F- Z! ?
},
% M7 u* X3 p8 e+ h" X
i++
$ ? @' B( o; F3 @# V" L
},
; Q( y8 D8 u1 o3 f7 Z
which{ d+1.0==1.0, l=0,
2 H' G7 [0 r; i3 M- _- n
{ if{ (A[js,k]!=k),
7 Y; c8 M" ^' f' Z1 L
i=0, while{i<n,
# ?& V5 F$ z: Q, D% V" A# r5 C
t=A[a,i,k], A[a,i,k]=A[a,i,A[js,k]], A[a,i,A[js,k]]=t,
|. C( J% r- M8 d. b; u
i++
* H$ o0 Q: C t
}
% H8 J0 ~4 k& C" T g5 \
},
, E3 _4 t c" L* S
if{ (is!=k),
- a6 l8 @' o2 y- `/ y0 L
j=k, while{j<n,
: i: R- y& f9 x" G
t=A[a,k,j], A[a,k,j]=A[a,is,j], A[a,is,j]=t,
* w7 y$ l, ]- e/ M
j++
1 f' v1 }6 ^0 b
},
8 w0 p5 g$ i: a8 h" n
t=A[b,k], A[b,k]=A[b,is], A[b,is]=t
' R" z5 H% R, \ g b$ n* s
}
# |; s. P* i* o V, L
}
- y: e( P# o( X
},
, \/ | |7 e* |) `7 B0 | _
if{ (l==0),
4 ~! S! S0 F3 A+ Z2 K
printff("fail\r\n"),
/ ?& b8 P! V2 e0 q6 X( U
return(0)
1 `; s2 E9 H* ~: K* c0 p" N0 w: ^
},
3 C/ Z0 u, d: \7 a
d=A[a,k,k],
( [/ v9 u5 L8 v8 W0 E4 D# i4 S" o U
j=k+1, while {j<n, A[a,k,j]=A[a,k,j]/d, j++},
. {$ G4 o! ^* r) n
A[b,k]=A[b,k]/d,
" i# I( y9 R% c+ l
i=k+1, while {i<n,
# S7 f$ t; P2 [% U2 ^
j=k+1, while{j<n,
F2 N* l3 `( }+ ~
A[a,i,j]=A[a,i,j]-A[a,i,k]*A[a,k,j],
" W" l, e/ r( D9 |+ W/ u
j++
* }- j2 e0 a& h6 g. Y! T: x
},
! O* }+ v) {; p- |
A[b,i]=A[b,i]-A[a,i,k]*A[b,k],
6 e v9 a% q$ X" p, L
i++
' T i/ b8 M' F
},
. X; x7 c+ P# u m2 H9 v; c4 i' [6 t
k++
; h9 g% H9 f0 m, f% u
},
' Z6 t' P$ }. |! } v
d=A[a,(n-1),n-1],
7 y0 Q. g9 |' x5 M) U
if{ abs(d)+1.0==1.0,
; p7 e$ \7 W& E3 D8 n7 [
printff("fail\r\n"),
; l+ K7 y1 n u4 C3 ?
return(0)
, z# \6 b% x+ Z; ~6 L% Q9 j
},
. ? M+ P: x* d, T4 U+ [3 y
A[b,n-1]=A[b,n-1]/d,
' ^% S0 P1 ^% k4 j- S
i=n-2, while{i>=0,
, w. P5 F) |1 K9 b7 y, W
t=0.0,
+ v) p6 Q4 n" ]# h6 W$ _
j=i+1, while{j<n, t=t+A[a,i,j]*A[b,j], j++},
: }! b, B. n8 ^2 `& p8 I! @9 q
A[b,i]=A[b,i]-t,
& r/ s, I0 ? P1 l* k0 V
i--
; N+ \3 U. l8 z; Z7 o( p5 j% ?! }
},
% s/ U" u( Z9 y6 C4 K: ^
A[js,n-1]=n-1,
9 ~) |4 p+ y$ _ @ L
k=n-1, while{k>=0,
% n: K2 z1 `* l+ p
if{(A[js,k]!=k), t=A[b,k], A[b,k]=A[b,A[js,k]], A[b,A[js,k]]=t},
3 F: |. U4 i/ V: T
k--
; v( O( H- A/ P# v7 L& i' P8 f9 B
},
3 j" a/ I5 J3 e9 X
return(1)
2 S0 _8 S) G7 l! A
};
6 a/ ]5 q- l; ^
4 v- f# q* N% k; O/ t$ r* ?
main(:i,a,b,aa,bb,t0)=
/ o! r P0 r* z H% P
{
% Q- G" g: w4 Q0 ^; [% t
oo{a=arrayinit{2,4,4 :
3 \3 D8 s- @/ Q
0.2368,0.2471,0.2568,1.2671,
7 e% C5 t0 S" U) v
0.1968,0.2071,1.2168,0.2271,
. _9 U- C+ U4 z9 Q
0.1581,1.1675,0.1768,0.1871,
. Z( ^+ M/ K6 g4 h7 G, u. `
1.1161,0.1254,0.1397,0.1490},
1 T8 x- P) N# ~) N
b=arrayinit{1,4 : 1.8471,1.7471,1.6471,1.5471},
7 ~( A" H, _* G c. s( h+ Y, k
aa=array[4,4], bb=array[4]
0 l. z" T/ Q" x3 @2 r
},
! }' `* c) z# X7 I% ]) r9 c, O
t0=clock(),
1 c2 j! H- Y9 X4 n! K2 c! Z% d7 S
i=0, while{i<10000, aa.=a, bb.=b, agaus(aa,bb,4), i++},
$ e7 P( N, g! t- t+ f; F
outm[bb],
% e# p s) r6 t$ A+ f* w, P. t
[clock()-t0]/1000
+ P5 C$ p' P, x y# \
};
复制代码
结果:
& v( P2 \# g0 G3 P+ |
1.04058 0.987051 0.93504 0.881282
: B* v% k! x- p7 a: ^2 {
3 Y( i( b6 w) `6 F' A, _
1.454
+ N* O6 O+ |+ O ?& ^; K
1 s/ K \! ?% i
----------
' c* c/ |6 n! k+ d% q7 k
0 q% l, B K- b5 M) Q, V
可以看出C/C++、matlab、Forcal耗时之比为 1 :25:68 (Forcal不使用函数sys::A())。
, U0 y9 a' m- h
可以看出C/C++、matlab、Forcal耗时之比为 1 :25:47 (Forcal使用函数sys::A())。
9 X% g8 s8 j( r$ ~ E
9 ?' J; K, I$ o# z; _4 L5 |& 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
4 V; ]% U+ p N6 v J$ l
! l- K3 N8 |' Q0 _" L9 K
C/C++代码:
#include "stdafx.h"
7 u6 b5 ]. V+ V8 q0 s, ]
#include <stdio.h>
$ C5 o; Z& } R3 S2 g4 M9 J
#include <stdlib.h>
9 M9 N0 c3 N0 C. ]3 `& }" G$ B
#include "time.h"
h7 j( O6 o1 ^& F
#include "math.h"
c7 m7 p. i8 c
* g9 p* S; ^: X/ y- a# w: G
double simp1(double x,double eps);
8 O( D. c/ B7 c0 _
void fsim2s(double x,double y[]);
/ l9 m% [" }: d* }5 _
double fsim2f(double x,double y);
. T9 y% R7 i3 v- o1 v
8 I' ^& G+ J. |: \2 a# ]
double fsim2(double a,double b,double eps)
7 U, T3 f0 O, s( z. B
{
: l1 d2 l6 V# E; X0 E( J; c$ O
int n,j;
9 {5 j7 j- G* w( ?5 r
double h,d,s1,s2,t1,x,t2,g,s,s0,ep;
2 H3 l2 j) C+ q/ d/ F& U; W u6 T
9 B3 C8 K6 {6 e
n=1; h=0.5*(b-a);
# U2 \! F* g# |5 g$ w7 A$ w
d=fabs((b-a)*1.0e-06);
: F9 q B9 ~! N6 A! e0 |- m
s1=simp1(a,eps); s2=simp1(b,eps);
) O7 |1 B& `/ W* H1 ]
t1=h*(s1+s2);
/ s7 [8 c2 k0 t4 w5 F) }
s0=1.0e+35; ep=1.0+eps;
" l) F' m9 F3 A7 [ I# v1 s
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
9 \; l" O5 a! O) S9 U# z. n2 z
{
. \5 F5 W- \$ n. g9 ~8 J
x=a-h; t2=0.5*t1;
8 K' y! g" o! q9 H( M' E! J1 C
for (j=1;j<=n;j++)
7 S! f5 y) f+ B, ~
{
" X' e8 @+ g$ Y( j; x5 O
x=x+2.0*h;
, q$ r& o, |. @- o) I
g=simp1(x,eps);
% l" v( [1 D' j8 J) T/ Z) n
t2=t2+h*g;
+ S6 R) a- ^. z2 q
}
4 o! G% C5 T+ o6 Z8 E, [
s=(4.0*t2-t1)/3.0;
|& H! _1 p+ G
ep=fabs(s-s0)/(1.0+fabs(s));
5 h# d& s4 C9 Z" g+ r3 C
n=n+n; s0=s; t1=t2; h=h*0.5;
' y, }8 t" n, ^6 F* x
}
: I' E$ [+ y- f1 b# D% m' w
return(s);
) X0 [: {% ?# r0 Q! v/ ^0 }& `
}
# C6 c! E. r8 t; q! B- k- F
% A& v1 A7 q* b! i( i9 m
double simp1(double x,double eps)
% O0 B8 {! Z$ d6 r: Q
{
2 v- S) V/ S# m% h/ \& n
int n,i;
- M) e( |+ l) Z( X& Z- B
double y[2],h,d,t1,yy,t2,g,ep,g0;
- \) `7 e$ m. E9 |8 X; ~% u; B2 k
5 b+ r" K) i6 [ B1 k1 t% V
n=1;
% V6 G9 `% L: }: ^9 o7 o
fsim2s(x,y);
, F; ]4 D! d5 T% Q! ]" _" V
h=0.5*(y[1]-y[0]);
3 U9 ~7 c7 @- a5 a, g7 |2 n
d=fabs(h*2.0e-06);
" }& S5 x$ K% @. P
t1=h*(fsim2f(x,y[0])+fsim2f(x,y[1]));
: z8 ^4 w8 P& x6 T; a
ep=1.0+eps; g0=1.0e+35;
+ \2 V$ ~2 p s8 A7 |4 i9 j. v6 f4 d
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
+ _8 x- ^5 u% ~. U t- v' O& K; O( I
{
0 ]: T! C0 H9 ^2 p* m
yy=y[0]-h;
; {2 h2 x) o# g9 o# t
t2=0.5*t1;
2 x# a$ U6 E+ s, }
for (i=1;i<=n;i++)
. o0 E+ f1 Y; D) X! H4 |& o6 |
{
2 x0 ^! M( [7 |+ ]
yy=yy+2.0*h;
& u3 _1 H# G9 }2 {
t2=t2+h*fsim2f(x,yy);
) P3 Z4 I" r4 C) H6 E! i
}
) E! \+ [% q' H7 p$ J: v% u2 `
g=(4.0*t2-t1)/3.0;
5 I: q5 ?3 D2 i
ep=fabs(g-g0)/(1.0+fabs(g));
* a6 j r% v& \7 `' r4 v8 j0 x
n=n+n; g0=g; t1=t2; h=0.5*h;
" S: k! y5 ^# \
}
5 b' x5 a# H/ g9 s D5 a
return(g);
( F" {0 s# S8 b1 u: }
}
% c% u( V. Y: m# b& T& B; B
0 |$ s8 ?/ T/ |6 n0 j0 `
void fsim2s(double x,double y[])
' c5 m0 ?* b% O) C! S
{
6 e7 l% Q6 M- }) v3 `
y[0]=-sqrt(1.0-x*x);
3 c: G; X, s% `7 R8 K. V1 Y
y[1]=-y[0];
# w% ?4 k" @$ p' u
}
2 S4 B- D5 R3 o* _: G9 p
; A2 a3 V/ s" ^- i
double fsim2f(double x,double y)
3 f2 q. y# @1 ]3 d; t# m4 p3 O
{
, c, ?# G6 C- G
return exp(x*x+y*y);
, a9 Q* g" G |9 @! k2 K$ D
}
% H. A* f3 A/ ]# e5 |
1 Y" W8 }/ b/ \+ ^ }
int main(int argc, char *argv[])
R/ }6 U4 D9 |7 ?$ N! i
{
0 |+ l( A! v+ ^/ ]2 E7 `0 b
int i;
% Q9 T# ~* n7 m- v( D1 K
double a,b,eps,s;
0 q4 @- o4 S* m! L- v
clock_t tm;
$ y! P- `, k$ c8 n# N5 ?6 E
: z2 M. P: N" T# _$ f0 f) H
a=0.0; b=1.0; eps=0.0001;
1 t1 H( H, v/ }1 Z1 A- x3 ^
tm=clock();
0 I9 e/ d0 ^2 B9 R
for(i=0;i<100;i++)
/ Z$ P3 ^! | \) z8 q4 N) e& h; U$ T- i
{
# l4 B0 Y& K/ V! y' W; B; h
s=fsim2(a,b,eps);
& P; a- B \- B% f
}
% |8 j( J# D% u/ T" A
printf("s=%e , 耗时 %d 毫秒。\n", s, (clock()-tm));
3 ~& p* J& F; L8 O' e! I2 a
}
复制代码
结果:
# u) e4 k7 ^; ^6 z( c& A
s=2.698925e+000 , 耗时 78 毫秒。
( W U$ V9 R0 d2 @' @: E* B& \
1 C0 S' d0 G2 O/ p. _! c: z
-------
7 p6 Y6 f1 d! |
' j# c. W* U% {" g, O6 G! l% i
matlab代码:
%file fsim2.m
" w6 |) e! o: q( p- F
function s=fsim2(a,b,eps)
' |9 _9 \+ l* ~) c; H; w% l. _
n=1; h=0.5*(b-a);
, E+ [6 E# ?3 o7 p3 k
d=abs((b-a)*1.0e-06);
) U# V3 r/ d8 M7 R
s1=simp1(a,eps); s2=simp1(b,eps);
5 X+ U) w9 o; B1 d
t1=h*(s1+s2);
, W5 L/ P* H0 `9 m f) j% u) j8 O
s0=1.0e+35; ep=1.0+eps;
) B* l2 J1 \1 l# n5 i3 F4 s) Q
while ((ep>=eps)&&(abs(h)>d))||(n<16),
1 H8 d% }% X: k! t" V- ?% A1 R
x=a-h; t2=0.5*t1;
4 ~# I) S9 o! t9 z* R+ w
for j=1:n
; y2 M6 g3 y$ t2 x! f7 \" V
x=x+2.0*h;
+ p1 i8 G( g/ T& v; I
g=simp1(x,eps);
3 d( H' D, p+ H% A8 r0 P# P8 Z
t2=t2+h*g;
: {+ O! y- U, N/ H2 H) G* B
end
! j! b# `6 F( f! _& I: X
s=(4.0*t2-t1)/3.0;
! e& ^* u1 Q/ k* t
ep=abs(s-s0)/(1.0+abs(s));
8 ?7 B0 a/ w- F) ^
n=n+n; s0=s; t1=t2; h=h*0.5;
9 Y. ~( Z2 ?/ k, `2 [. L: i$ Z
end
0 ?4 B* G# g7 W% a9 D
end
' s' M9 p3 {4 x' F% i, s: t0 a
6 {3 r/ ~* J* x" X1 e+ f
function g=simp1(x,eps)
8 ]1 k/ o* m0 b! ~) @/ W
n=1;
' s) a) q! ?6 n6 R3 s& a
[y0,y1]=f2s(x);
7 s7 k5 \; B5 e" z8 ~% ]
h=0.5*(y1-y0);
4 u8 y+ q; V/ {4 H* [
d=abs(h*2.0e-06);
! y7 T& A8 t7 y6 G* e9 j
t1=h*(f2f(x,y0)+f2f(x,y1));
1 B9 ?3 i w D3 a
ep=1.0+eps; g0=1.0e+35;
: j) q$ ~) i3 I' W0 P
while (((ep>=eps)&&(abs(h)>d))||(n<16))
# Z9 M1 E( Y/ f8 V9 V
yy=y0-h;
! c) c1 v; N; F5 D7 y
t2=0.5*t1;
! W. | O2 Q$ D) G+ n
for i=1:n
; p3 ]! s, l9 R% e( H f& B
yy=yy+2.0*h;
5 r1 @ O; b( c6 i3 P3 b) ]
t2=t2+h*f2f(x,yy);
* ]8 g8 |7 w3 V/ V. |
end
* D" b1 ~( ^0 X. V. P$ k; P% j
g=(4.0*t2-t1)/3.0;
3 Y. D t5 w5 I# h& s( C
ep=abs(g-g0)/(1.0+abs(g));
+ N1 l: ]( M1 S9 e) K+ \
n=n+n; g0=g; t1=t2; h=0.5*h;
/ \% B# G P1 p, Q
end
3 X0 x. C9 n4 }( l1 s! }0 A7 |
end
h& K+ C7 }3 T
- i3 @" X: w" A+ i5 u
%file f2s.m
) O$ k# M" z' [
function [y0,y1]=f2s(x)
" ~# l' |/ d* z* Z$ V
y0=-sqrt(1.0-x*x);
+ P$ S. q5 C7 p/ K
y1=-y0;
5 v% I+ Q: {& O# f, C J
end
' E. l- C% ~' m
* m0 Y; p. ], w
%file f2f.m
3 R( i% @: [& n3 h7 ~% [
function c=f2f(x,y)
: j3 \4 U6 _6 ^2 K6 Z) t( B% s
c=exp(x*x+y*y);
% p( x$ O; \+ T( M g
end
5 |8 u( G/ u% i) D
- l9 u+ e- K% n# v; A# V5 f
%%%%%%%%%%%%%
5 x( i# n& d3 U6 h9 ]
0 R2 p+ t0 ^% ^1 w6 W% J/ Q$ k
>> tic
1 A# y. L- e; w& U
for i=1:100
1 v2 d( y: { V3 G
a=fsim2(0,1,0.0001);
6 U+ v) ?1 w/ u( y
end
8 Y$ e, S; b, A4 r* J- ]' s
a
9 \. _+ K; H' g1 _9 ]
toc
2 w& @4 P' c: h! v/ j( v, g [
# D- x5 ?# a5 q$ Z' `% T% K8 H
a =
7 e) p' B4 g# E. E$ {" p# I+ j
: s" R. R2 S4 s0 p+ o( S/ n
2.6989
0 ]/ R- u) `$ E% g# e j- _
2 _9 i/ ?# o. T! I( y! f
Elapsed time is 0.995575 seconds.
复制代码
-------
8 O: y2 m- [ _3 S6 [6 F0 E- Y3 q
1 A- \4 M9 o( L* X( f( ~! g6 ?
Forcal代码:
fsim2s(x,y0,y1)=
% {; {* b' d1 X
{
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y0=-sqrt(1.0-x*x),
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y1=-y0
/ _: `, S8 Q- C# o+ M) Z# C6 d
};
$ C9 S5 n7 C8 S( k& }
fsim2f(x,y)=exp(x*x+y*y);
$ s3 q _" t; D1 k( X" ?
//////////////////
2 [" G6 F# l& b7 p6 J/ I
simp1(x,eps : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
' |/ Q8 r" [; S" ]! L
{
1 v- r9 R' Q, @) O! i- J0 K: a
n=1,
( z& F. I2 ~# k4 B @
fsim2s(x,&y0,&y1),
7 E" O- i/ D7 S& s& l" k
h=0.5*(y1-y0),
: Q' X, P6 U) {6 ^. o
d=abs(h*2.0e-06),
; f0 a. W. P4 E9 J+ |2 X
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
- J. {- j* J; u) s7 c% w
ep=1.0+eps, g0=1.0e+35,
5 I+ j: @: H |) w* N/ @
while {((ep>=eps)&(abs(h)>d))|(n<16),
7 m2 g( y F& ~4 s
yy=y0-h,
2 B |4 w( F; G1 L+ ~7 x
t2=0.5*t1,
8 {3 u5 p9 N" w) j( o, N5 @( j4 i
i=1, while{i<=n,
, K6 ~4 t! g, T5 Q1 @9 @% R
yy=yy+2.0*h,
6 ~4 P z7 G3 m
t2=t2+h*fsim2f(x,yy),
3 o+ Z: [+ O; g( a1 q1 s4 d
i++
% A3 S( N$ C1 N( m* D" T' P: O. v
},
' g- a8 J2 @! n- U5 S- U: H
g=(4.0*t2-t1)/3.0,
& }6 ?7 G4 |7 Z1 n
ep=abs(g-g0)/(1.0+abs(g)),
4 l9 l' C$ d: D
n=n+n, g0=g, t1=t2, h=0.5*h
) t, I$ f$ C$ e. M2 p; b
},
1 i# ?8 Y9 l7 W( {# u( ] c
g
9 I( p w2 A) ^ Z* B3 E
};
5 M/ C* Y: G6 L
* e0 i2 M* u9 ^' `: ]6 b
fsim2(a,b,eps : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
7 N; r: v3 l& n( }2 Y& u) u
{
$ s/ c& l( {! C! y( H/ @1 s; ~
n=1, h=0.5*(b-a),
- m, m' l4 k+ x. j& _' b
d=abs((b-a)*1.0e-06),
) |. v9 J% J9 y8 t
s1=simp1(a,eps), s2=simp1(b,eps),
- s" u) n. w$ p J) y
t1=h*(s1+s2),
3 V/ F$ M% X6 m! q: O0 |! @
s0=1.0e+35, ep=1.0+eps,
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while {((ep>=eps)&(abs(h)>d))|(n<16),
- ^9 a( A$ x: n/ N
x=a-h, t2=0.5*t1,
# s6 C4 z$ \3 y0 ^$ I% t% {8 Q- T5 w
j=1, while{j<=n,
/ ]4 R4 V) G0 j6 Y. h
x=x+2.0*h,
6 I$ N5 ?9 q D
g=simp1(x,eps),
6 [, @2 D/ `+ f5 \- C) T) N
t2=t2+h*g,
9 x2 y" `- V( _& ]0 J- D
j++
6 {) [) w9 p6 X y4 c" b5 L6 ^
},
/ d! Q9 y- S5 C% `5 j
s=(4.0*t2-t1)/3.0,
( g5 H( P9 k! n( P% \2 Q' P: e
ep=abs(s-s0)/(1.0+abs(s)),
, l6 Q6 t* n" O
n=n+n, s0=s, t1=t2, h=h*0.5
4 q2 g( h- \0 ]" E e1 q" {
},
1 J, Z2 a; Q4 [
s
, d6 w8 N! {7 B7 O+ N6 u" g Z( B
};
. o6 V) }1 i5 z( J
5 H$ g4 S2 @ I; H" v' {& a
//////////////////
* u! |) E3 U5 r* b3 x/ Y/ S
. b/ Y1 N. \* |9 n
mvar:
8 A5 {2 I/ ~/ B8 n
t0=sys::clock(),
$ |+ P2 }( a! o3 |/ b8 n
i=0, while{i<100, a=fsim2(0,1,0.0001), i++}, a;
7 z2 ` R( ?# P5 g# X5 n. R5 e
[sys::clock()-t0]/1000;
复制代码
结果:
( j- S4 \! B" ]& k
2.698925000624303
; B; p! q- [, b& o# n
0.328
/ @5 i" z# \& h# f. n0 D
. V- t o7 h4 O. w/ l
---------
8 Y& b9 O% m/ `# m/ z! h0 l/ ?
* J$ |, s: u+ h5 Q: n8 t
本例C/C++、matlab、Forcal运行耗时之比为 1:12.7:4.2 。
# i- X0 d$ _5 G6 G' B
' r: C1 I4 Z6 ?$ t9 S& A1 X
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。
4 n6 m0 E; v% y. k4 |
8 u" }$ r0 u: _" R) I
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
# P' u% Y% g* M- T9 ^ t
3 D4 n$ P: f2 q `$ ~ R
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。
& d4 v6 E% h p/ I1 g, ?
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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)
& o+ `/ ]0 h; r: x) H
n=1; h=0.5*(b-a);
3 d: W* O, j! X4 }9 u0 N; s& @; R
d=abs((b-a)*1.0e-06);
/ @" T, R/ N% y1 Z- {. y
s1=simp1(a,eps,fsim2s,fsim2f); s2=simp1(b,eps,fsim2s,fsim2f);
, _$ b" j9 w8 u0 k o& ?" R' x
t1=h*(s1+s2);
& _3 Z5 \7 I) v) e! {
s0=1.0e+35; ep=1.0+eps;
0 e* m* g: j" R
while ((ep>=eps)&&(abs(h)>d))||(n<16),
, Y0 N; Q1 R2 X( R4 y' n
x=a-h; t2=0.5*t1;
7 z% a$ ~- \6 l& }8 M X
for j=1:n
4 g1 z! {" `3 _. A
x=x+2.0*h;
3 ?( b5 [& |0 O6 |3 J
g=simp1(x,eps,fsim2s,fsim2f);
' M! Z4 w' s; O2 e# z
t2=t2+h*g;
( F8 U I2 Q! X: z: i
end
8 y# e7 o/ t2 O7 e3 p
s=(4.0*t2-t1)/3.0;
. L# n1 y" |8 B" v$ \' g
ep=abs(s-s0)/(1.0+abs(s));
( y3 T. d6 O/ d% k, x+ }2 b
n=n+n; s0=s; t1=t2; h=h*0.5;
; F1 @6 r2 [( u8 z$ e
end
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end
: T: c/ K( V1 F% h. A/ S) R
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function g=simp1(x,eps,fsim2s,fsim2f)
( j& V U8 f0 u% M9 x
n=1;
5 o$ {* A6 R( f& P) x/ I9 Y
[y0,y1]=fsim2s(x);
* Z; x1 G% {4 ?4 g+ c
h=0.5*(y1-y0);
) D* Y3 a ~" U- G9 u& F7 n
d=abs(h*2.0e-06);
8 P$ a0 P' S+ ?; F
t1=h*(fsim2f(x,y0)+fsim2f(x,y1));
# o5 g0 S0 I6 _
ep=1.0+eps; g0=1.0e+35;
& e) Z6 ~; N) z* k, q4 Q: S6 i
while (((ep>=eps)&&(abs(h)>d))||(n<16))
, p+ g' _$ t2 W5 E6 k/ h5 J( ?7 E$ t0 T
yy=y0-h;
( T; ~" [) q6 d, c' |; h( V* d [
t2=0.5*t1;
, A) |+ j8 U; i) J( n
for i=1:n
4 t V0 }$ M# a. u5 a* x( ~7 [8 d; N
yy=yy+2.0*h;
# V8 q" P! ?: Q; d* S
t2=t2+h*fsim2f(x,yy);
3 ~& U! f) w# X4 B% S% P& Y
end
, b. x; Y7 _! ^* q& R4 O9 Q
g=(4.0*t2-t1)/3.0;
, c( w- M4 @/ S4 d4 _0 ?
ep=abs(g-g0)/(1.0+abs(g));
! R/ u3 ?, k g F8 Y# ?
n=n+n; g0=g; t1=t2; h=0.5*h;
8 t+ R& `( g7 m. Q: |+ w
end
" S z, }8 S7 k1 n" D
end
# e7 m" P- G. Q# Z
6 ~- k; S. d7 Y2 l. g% q
%file f2s.m
+ p, B+ B& f8 i( Z" B8 x/ u
function [y0,y1]=f2s(x)
; m( h9 q1 s) s; c3 |
y0=-sqrt(1.0-x*x);
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y1=-y0;
: d' M" n" z: H6 b3 p
end
" z* v5 h; r% q1 A2 `% i8 W! G
r% x( |( m6 H" D
%file f2f.m
+ ^$ `; E) e) O& B, V
function c=f2f(x,y)
2 t- J" ^2 r9 k" @
c=exp(x*x+y*y);
, N) k! m9 |+ X
end
* l5 M$ Q7 k1 c) W8 ^
. m6 x6 F& e/ u& H
%%%%%%%%%%%%%%%%
2 ^8 Y& u' l6 z1 e: w
+ J {! a) K& U5 G
>> tic
" A, C( ]1 S0 w3 O% E& G: F
for i=1:100
" K: |, |, d7 a
a=fsim2(0,1,0.0001,@f2s,@f2f);
8 i& c& Z2 o( i, ], b
end
! y' s/ Z% u1 `4 J0 s& L! J
a
, H6 H' l9 y1 e
toc
6 h" ~, I' J; r; p& c7 q
$ l1 ?) v# Q0 `! {0 f" X* r& U. I
a =
/ y3 y1 b3 [5 W' w9 R! b- B
E# d4 b3 P. q Z: W
2.6989
9 ?* j* A6 M4 F$ e, ]! q" K- ^3 P
: ] U' e. ~/ m) R+ ]
Elapsed time is 1.267014 seconds.
复制代码
--------
# D" t0 G% `% X5 b* L# s! {
* }$ Q) y( C( c8 ^, n" K6 T
Forcal代码:
simp1(x,eps,fsim2s,fsim2f : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
$ c: w/ z+ ]! _8 [* @
{
0 d$ |. \7 Q; Q# ^; a `& u
n=1,
3 y; M5 a; c3 M X) U6 y
fsim2s(x,&y0,&y1),
& O) w' \. V0 v8 b7 c2 c( u& W
h=0.5*(y1-y0),
$ D0 }0 ~) U8 E( H& \% N
d=abs(h*2.0e-06),
# T- E4 X* M9 X& }7 z& F
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
5 {' X1 C ^4 }5 N* {0 |
ep=1.0+eps, g0=1.0e+35,
& ]7 N. r% v+ S6 N" X
while {((ep>=eps)&(abs(h)>d))|(n<16),
- t1 @! E. n0 R. o( ?
yy=y0-h,
* J8 S/ t7 J9 }: Q$ T
t2=0.5*t1,
! W2 t) Z6 @! r1 b( x. x
i=1, while{i<=n,
$ |* w1 h- |; t! Y5 N1 A& ~, d) h
yy=yy+2.0*h,
- ^8 Y' z7 K' b* C, l0 V0 ?
t2=t2+h*fsim2f(x,yy),
, i" |- v4 e9 K8 w
i++
# j- l7 D( x. x
},
( W! [7 @4 p# \( Q
g=(4.0*t2-t1)/3.0,
. g* |) k; _4 f G
ep=abs(g-g0)/(1.0+abs(g)),
1 h# [- z& a+ S( a6 v
n=n+n, g0=g, t1=t2, h=0.5*h
% j3 s( J7 o" ]* r* B. ?4 B& ]
},
. F6 G9 v# N, U5 }! a
g
; d8 [) Q. i( r& g
};
: A7 g- o1 N* t% \
5 N- X5 C$ f/ P$ r" W4 W w
fsim2(a,b,eps,fsim2s,fsim2f : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
& H. H; z+ @- N6 O$ G! N7 w' J
{
/ G+ s2 w7 \2 W0 q& |2 g' o! n
n=1, h=0.5*(b-a),
: T& |7 M- ]6 s: I
d=abs((b-a)*1.0e-06),
5 N* K3 @4 b( j$ ]; i L
s1=simp1(a,eps,fsim2s,fsim2f), s2=simp1(b,eps,fsim2s,fsim2f),
/ y5 L- V! w! d( E
t1=h*(s1+s2),
h, P% R$ U2 N4 ]
s0=1.0e+35, ep=1.0+eps,
0 B" ]0 R; M3 I7 n$ b, P0 F9 o
while {((ep>=eps)&(abs(h)>d))|(n<16),
0 s% i( {, c: u# E# G( q
x=a-h, t2=0.5*t1,
/ y, K e, C$ Y w4 h
j=1, while{j<=n,
( w% n( {4 ~5 U2 H! N+ G
x=x+2.0*h,
2 w9 e; M5 N: c7 c4 U6 b/ m
g=simp1(x,eps,fsim2s,fsim2f),
% o& L" o" X" u3 X
t2=t2+h*g,
. `0 o+ x' [ S% C) P& J2 N
j++
- x* o2 U( Y! ~, U' L0 S4 j7 G E3 T
},
2 B, Q0 ~6 r! p' x5 y- L
s=(4.0*t2-t1)/3.0,
. r% u+ ?" p5 z( q4 v
ep=abs(s-s0)/(1.0+abs(s)),
" V9 _: B4 f, `9 d' U- f5 u+ M
n=n+n, s0=s, t1=t2, h=h*0.5
2 M1 Y, f0 k- j% ~; I; ]
},
0 r1 o; f* x7 t. Y
s
$ u' P+ o1 i# Y0 R) T
};
* O6 r. ]4 i' _+ Y
4 O T/ r' r/ \+ X
//////////////////
7 w& X" m8 g/ t# E g/ P
+ ]. Q- b1 M+ M, a# [' `- P! q
f2s(x,y0,y1)=
4 C0 x1 o \- v6 `+ |# A
{
! _) ?. ~' E G4 s$ M3 `: [# A$ k1 I
y0=-sqrt(1.0-x*x),
" j1 Y' v& W7 F) z
y1=-y0
h1 p9 O) h8 r; j
};
$ a6 `; i3 X8 }8 }& G4 F/ O; e
f2f(x,y)=exp(x*x+y*y);
0 I% T- N( `% w8 |- L
% C! E/ ]6 u' b7 l, Y9 y) I
mvar:
$ s5 n! l% f& Z, `
t0=sys::clock(),
; M; B/ k* N& i0 O D
i=0, while{i<100, a=fsim2(0,1,0.0001,HFor("f2s"),HFor("f2f")), i++}, a;
6 r. b& r }7 Y. ^1 C1 Z3 c
[sys::clock()-t0]/1000;
复制代码
结果:
) T/ a* F; Q$ \ I4 j- j
2.698925000624303
7 i0 `4 D# Q8 w) ~: B
0.844
& y# ?5 j1 q2 f' A# Y
# A$ T% |3 j+ Q3 x' W
--------
& k h7 B/ V) I* O* Q6 D* S
' y* j# v. H* V3 V
本例matlab与Forcal耗时之比为1.267014 :0.844。Forcal仍有优势。
& m4 I$ i* f; b# C i/ A B
1 a" Y2 V" m/ b1 M. J2 E
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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