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);
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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},
$ `( k, z9 _0 j+ K6 x% Y/ t/ ^. b" R
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],
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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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},
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b[i]=b[i]-a[i,k]*b[k],
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i++
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},
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k++
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},
4 T% t G- A6 X1 ^
d=a[(n-1),n-1],
* N- ~7 v( t$ H6 @3 v% a9 n
if{ abs(d)+1.0==1.0,
; F9 w% E, j4 R6 m$ D9 w: t7 x
printff("fail\r\n"),
' o$ U& j. ]* m6 v
return(0)
( X3 P3 t8 o; s' W+ }4 A
},
3 E5 f, E( ?! Q% y" b- B$ S
b[n-1]=b[n-1]/d,
$ }4 m; G {% O7 @
i=n-2, while{i>=0,
; z0 y! C3 z8 A* w9 k4 Z" _) |) m
t=0.0,
+ r' z8 }( F! Z5 ~0 l: A- w
j=i+1, while{j<n, t=t+a[i,j]*b[j], j++},
! R: @" H+ M; {$ b
b[i]=b[i]-t,
! z# r8 s! U% n
i--
' W8 C6 G1 m: d% F, ^) F$ a8 T& g
},
) n3 S7 a( {. S1 F2 ? R7 h
js[n-1]=n-1,
& ` K) w* i2 l/ D! L' T
k=n-1, while{k>=0,
- C' A4 V" z4 ?$ Y. s
if{(js[k]!=k), t=b[k], b[k]=b[js[k]], b[js[k]]=t},
+ n. v$ d3 T, p, b8 L% ], o6 E" W! E
k--
9 ~5 _1 M S6 A) J) f8 t
},
& K5 C0 P/ V& Y! a
return(1)
1 }3 L8 p3 R8 O; c4 a
};
. _& v+ Y/ H1 u8 z( _
0 |1 D0 ?! g- B# N$ h; `
main(:i,a,b,aa,bb,t0)=
0 h; s& t8 @. Z* C7 Z, t% {
{
( Z" u! V+ e6 K: A- p, e2 j
oo{a=arrayinit{2,4,4 :
6 f2 K |( S# }3 H0 I: M5 f0 ~
0.2368,0.2471,0.2568,1.2671,
6 L. c$ V, X. c
0.1968,0.2071,1.2168,0.2271,
. U4 k! |8 [8 G6 C( J7 h
0.1581,1.1675,0.1768,0.1871,
/ q% l3 D) E; R6 S) _6 x g
1.1161,0.1254,0.1397,0.1490},
) J; a8 ^5 ~. O) D- z2 M; {
b=arrayinit{1,4 : 1.8471,1.7471,1.6471,1.5471},
6 o4 U- D7 B! u; l* L. l5 j4 q
aa=array[4,4], bb=array[4]
4 ]- t) B8 @4 L7 ^4 \3 K/ e9 d' {+ P
},
5 J6 p! i7 | k# p8 J, t
t0=clock(),
2 _; t5 }# V7 Y+ ]% s5 [: o
i=0, while{i<10000, aa.=a, bb.=b, agaus(aa,bb,4), i++},
2 |0 Z8 H W/ V; H/ M
outm[bb],
8 C ]4 L( I! j+ F
[clock()-t0]/1000
1 J6 S3 N( v+ m
};
复制代码
结果:
% N0 E. X- l2 {: O3 L
1.04058 0.987051 0.93504 0.881282
# c# i! F* ^ G' ~) _
" S2 M+ R' A& q- [( e6 i0 P
2.125
0 ^5 C; v, Q3 [4 Z: p) V$ d
6 \3 L5 [( x- n/ G; c D" ?( |
Forcal用函数sys::A()对数组元素进行存取:
!using["math","sys"];
; a6 `8 a/ T2 x) W4 V
agaus(a,b,n : js,l,k,i,j,is, d,t)=
L2 `5 a, i$ W' c9 o4 J
{
8 ^- ]- N8 k$ N9 U7 S
oo{ js=array(n)},
2 L- H! d' J. Z1 u9 J
l=1, k=0,
6 _% |- n( L c1 {6 l- I. t d- C. V4 K
while{ k<n-1,
4 h' G# C G7 W: E# \% v& t
d=0.0, i=k,
+ ~6 w# L. r: v. ?0 y
while{ i<n,
! k! F) g, h2 \) x
j=k, while{j<n,
/ F; ]. J! a+ y7 h! G
t=abs(A[a,i,j]),
5 E" ?, `3 c# N- f5 ~6 H4 A2 a
if{t>d, d=t, A[js,k]=j, is=i},
, z$ V' m" O3 k3 }
j++
$ F8 M0 Z" O0 K
},
/ }0 _3 m" w9 X0 @( V, U
i++
% M5 b# P9 n. q: j" m3 Y
},
" ?. e2 `2 p& q: I
which{ d+1.0==1.0, l=0,
# k- b: Q1 _9 n4 V
{ if{ (A[js,k]!=k),
1 T. x8 M: H4 P
i=0, while{i<n,
/ m" G& f0 H5 {: U. r8 ~
t=A[a,i,k], A[a,i,k]=A[a,i,A[js,k]], A[a,i,A[js,k]]=t,
; q% `: V5 X$ i+ e& b/ i& J& n
i++
% V3 E/ M5 n2 \, A) o1 d
}
# Y! F% R3 F- h; |. K# w1 H
},
( f$ c: r) Y( H) r! ^- F$ L( i
if{ (is!=k),
4 r, U4 `7 D) Y- v% p
j=k, while{j<n,
1 a, c) {5 e0 a- r% c8 T
t=A[a,k,j], A[a,k,j]=A[a,is,j], A[a,is,j]=t,
" W% G5 L3 `& U) Y
j++
! y5 x5 E4 |3 f3 E' Z5 s5 F
},
/ C" \7 m6 ^+ I+ m6 t7 @, y
t=A[b,k], A[b,k]=A[b,is], A[b,is]=t
% X0 c _/ E; l, N- I* b% b
}
% Y v; O5 R. ^9 v, r5 e
}
8 U: Z! `" o7 t5 I7 ?7 M
},
" u9 M$ O0 T# R M
if{ (l==0),
, n3 e& O% S+ {+ T( c% f) E2 T
printff("fail\r\n"),
+ ^6 Q3 a2 U) D1 g, s, A' s K
return(0)
& O- o' \- I9 @+ D3 M2 X/ o) G
},
, n/ ^' H) b+ Q6 f+ B
d=A[a,k,k],
3 q) G8 v5 B6 T7 _. a
j=k+1, while {j<n, A[a,k,j]=A[a,k,j]/d, j++},
# n3 f- Z; i* Z; A
A[b,k]=A[b,k]/d,
* [+ p& `2 j0 y9 p
i=k+1, while {i<n,
* C* }1 I& ^; b5 K, B2 q$ X% W0 h
j=k+1, while{j<n,
( w+ {9 s- O1 J: M* Z5 e% E
A[a,i,j]=A[a,i,j]-A[a,i,k]*A[a,k,j],
! e9 g+ C* m" K3 I
j++
; t* s( ]% t. `* k
},
5 h, h" o/ n5 I- d. I) l
A[b,i]=A[b,i]-A[a,i,k]*A[b,k],
9 G, ?% s4 r& r% ?# g
i++
- ?$ \ O6 T- F7 W2 i- A
},
& F* U" b- ^6 ~& \1 T* {1 H" X+ l# P
k++
" s) m: k0 g: b% L& [% f
},
0 N) G; \; n0 x, V0 P
d=A[a,(n-1),n-1],
5 ]9 o; d$ j% k
if{ abs(d)+1.0==1.0,
& ^$ u3 r+ a, T1 A S0 u$ u7 u
printff("fail\r\n"),
* I0 y: G# z2 M/ k3 b. F" H
return(0)
! s) Q# ^& }* _7 O# p/ c
},
% e2 g( n5 K1 s R$ G9 V4 c f
A[b,n-1]=A[b,n-1]/d,
! ]( k3 x& N! m
i=n-2, while{i>=0,
: X1 M9 P4 d9 Y% m( A& ~2 z% j
t=0.0,
- i* s: Q9 I5 Z0 ? h6 ?% i* p
j=i+1, while{j<n, t=t+A[a,i,j]*A[b,j], j++},
7 c0 H% U2 z3 x. k5 J! ^
A[b,i]=A[b,i]-t,
! ?) i6 z9 B$ Q( g
i--
1 f* |* r# M2 R9 }- Q: A. @
},
: r9 y0 n" w6 Z0 D' h1 e) J b
A[js,n-1]=n-1,
% M* z. d9 ?( C& S9 D7 I
k=n-1, while{k>=0,
7 U* M3 r7 ?6 D; d. ~2 W
if{(A[js,k]!=k), t=A[b,k], A[b,k]=A[b,A[js,k]], A[b,A[js,k]]=t},
$ ]* X5 N& i- m+ ~1 t
k--
/ h+ o" c) ]5 q. I
},
, n0 }; L4 F4 M* i9 r
return(1)
b1 R: @) S# B. [( z. p
};
! e P" r6 f$ T5 L$ J: |1 ~
0 X! A3 b! x8 f. W- Y
main(:i,a,b,aa,bb,t0)=
5 b: S h" k+ l+ Z0 e) K
{
5 t. Q% B" y1 ]8 Z j% Q: H4 S
oo{a=arrayinit{2,4,4 :
+ `% {! J' V# R' l0 L9 Z
0.2368,0.2471,0.2568,1.2671,
. H& C0 y, { u; I7 Z
0.1968,0.2071,1.2168,0.2271,
$ m5 i9 v, W* p0 I% e) q
0.1581,1.1675,0.1768,0.1871,
4 H0 q( I. J3 \1 X" E+ E
1.1161,0.1254,0.1397,0.1490},
J( z$ X' D( V* a: {6 I; ?
b=arrayinit{1,4 : 1.8471,1.7471,1.6471,1.5471},
+ g0 ~& K$ @, _; o2 i7 B0 X4 \
aa=array[4,4], bb=array[4]
+ o( D& N9 I- n1 f
},
7 Z3 z @6 S7 ?
t0=clock(),
^" c: h0 K7 v2 r$ `! q
i=0, while{i<10000, aa.=a, bb.=b, agaus(aa,bb,4), i++},
: l" t0 n: y0 s9 m7 |
outm[bb],
2 H' B% A' j) s& ?! H0 f
[clock()-t0]/1000
% b$ F; b; I' u* E. h% `
};
复制代码
结果:
9 B; b4 E5 ]2 W- C8 m
1.04058 0.987051 0.93504 0.881282
! f2 c, j" M2 }+ }/ T/ S" [2 z" \
) X2 i# M1 _3 g- ` l
1.454
/ n8 u! G# h- T+ I/ v: L* F& Y
. L" i& w8 P( n( c b; n% L8 |
----------
+ @: J! i: _8 |9 ]) q
1 L" e% m& g' n! A$ ?
可以看出C/C++、matlab、Forcal耗时之比为 1 :25:68 (Forcal不使用函数sys::A())。
, w: c' G1 |* s1 o; Z1 {
可以看出C/C++、matlab、Forcal耗时之比为 1 :25:47 (Forcal使用函数sys::A())。
3 A* e' B0 N- Y
$ Z2 S% V% ?8 E' S2 D5 \
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
9 P6 D+ e! k/ @" Z
1 B7 e6 O, z4 y( S+ g, U$ |% ~
C/C++代码:
#include "stdafx.h"
; A/ ]# x7 V% C7 G) U2 m6 ^
#include <stdio.h>
" h7 a9 w+ F% I/ [) y! e9 V- D g: R
#include <stdlib.h>
; `( ] A/ d. Z6 [- E
#include "time.h"
6 X [' g& h9 q: C
#include "math.h"
. t- ^& \# S/ x
. g0 [1 ] `9 E6 k* |6 y6 j; F$ s
double simp1(double x,double eps);
$ t- G; J! \3 p" m0 A. D
void fsim2s(double x,double y[]);
- Z8 J" y3 d/ W8 h# `8 L
double fsim2f(double x,double y);
1 c% y( C' F+ Q3 u u1 V2 V
: x4 ]6 i9 q7 y! q& ]1 ~: \5 c' a2 q: K
double fsim2(double a,double b,double eps)
9 W R9 r' {4 s& _
{
% y& g+ t3 |- N9 x. h W
int n,j;
9 s: j) N+ j4 Y" M. k5 z7 t- L7 r& }
double h,d,s1,s2,t1,x,t2,g,s,s0,ep;
$ Y. L: ?# F5 u( j4 ~
0 p# w2 h' S2 X
n=1; h=0.5*(b-a);
% ~' l- b8 a6 U
d=fabs((b-a)*1.0e-06);
/ P# w2 ?( x$ K+ W' M9 d3 X9 o1 O
s1=simp1(a,eps); s2=simp1(b,eps);
7 f2 \* _1 s0 v& C
t1=h*(s1+s2);
% I! ?% N1 m- T- W0 O5 d
s0=1.0e+35; ep=1.0+eps;
! v2 h; t+ u* B; Y# j
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
1 e: Y E3 `8 S1 C" y6 B$ k
{
& E- A) F9 y$ n+ h3 g8 |
x=a-h; t2=0.5*t1;
; z5 U' V5 T- c$ a! U+ j7 y
for (j=1;j<=n;j++)
4 T {" a7 I( S
{
9 Y" A. Y# Q( z4 V4 j9 [) Q
x=x+2.0*h;
" g2 ~, L- w( f7 a) _% d3 `
g=simp1(x,eps);
, F0 I7 X! }4 h- r4 A
t2=t2+h*g;
, @7 ]$ ~) k/ r& v8 S& V$ w# A
}
, A! g0 C2 ]. v3 ~
s=(4.0*t2-t1)/3.0;
; H- r, P# x! i- E G9 W
ep=fabs(s-s0)/(1.0+fabs(s));
7 ^2 l/ y- l1 {, Q$ |& E) ]9 y
n=n+n; s0=s; t1=t2; h=h*0.5;
' w+ t) }: }) F2 p! k
}
& G. h& Y) _. k" ~: b* G2 @% i" @* m
return(s);
# T; k1 Q* @$ R! u3 r" d }' ]
}
; q% B q$ O' F: t# e1 t
& X4 ^2 d! z8 T- z, A
double simp1(double x,double eps)
- n: M4 ]+ r- u, u) V0 i( D
{
& y2 r5 D- u$ I2 h
int n,i;
$ u# b" i2 ?$ w; F; J* \: c; f
double y[2],h,d,t1,yy,t2,g,ep,g0;
5 P! G3 T- U8 B8 L: C. }
( w8 f- ?/ U. i6 C- G% S& L0 u! l
n=1;
8 y7 _( G, w. W! ]4 z s! t) X" Z$ }
fsim2s(x,y);
. o- J/ j8 R* ^8 A$ A
h=0.5*(y[1]-y[0]);
, y5 F3 E' w3 p9 `
d=fabs(h*2.0e-06);
& D6 K8 w; e/ |* P. K
t1=h*(fsim2f(x,y[0])+fsim2f(x,y[1]));
' Q3 p8 A" h! P/ ~+ n
ep=1.0+eps; g0=1.0e+35;
" ^7 v M! h- Y8 |3 ~, U
while (((ep>=eps)&&(fabs(h)>d))||(n<16))
1 \+ u' @9 c( _
{
2 }0 \& s: R% X( a; S1 b+ B
yy=y[0]-h;
( I- M* f6 d0 m, R
t2=0.5*t1;
; b5 O4 D, \+ Q, ]
for (i=1;i<=n;i++)
# n9 Q, K* F5 P5 D0 w+ V
{
* G0 L! V5 } }) L
yy=yy+2.0*h;
; A* O) E1 v- V H5 Q! L
t2=t2+h*fsim2f(x,yy);
' p" j, p( p+ ]
}
4 C- ~! T" R2 t5 h7 E6 u! U
g=(4.0*t2-t1)/3.0;
4 P7 O/ [% j- b9 V/ g9 L
ep=fabs(g-g0)/(1.0+fabs(g));
/ J; E2 @9 e4 f
n=n+n; g0=g; t1=t2; h=0.5*h;
6 f6 w; y3 f! c# U
}
/ o4 ]. R0 ^; x- b
return(g);
" O4 W! z3 D0 ~+ V4 _5 j
}
- c$ D3 r6 \! f$ S$ n- G; f
p( O* ~5 s* a3 h) |/ t1 @
void fsim2s(double x,double y[])
6 o' u r. L4 r b* I7 k& z
{
* T, |8 ` R" o3 Z4 Q
y[0]=-sqrt(1.0-x*x);
7 B: n% [9 p& }9 c1 f
y[1]=-y[0];
: l' P3 \. t0 k
}
# }2 n, Y8 Z$ N: ]2 ~7 L
: x7 `3 K3 u# L0 T
double fsim2f(double x,double y)
# ^: c: s3 Z" ^- V& l% p( _
{
. d- [5 N" a% ^4 B. ^# h5 R
return exp(x*x+y*y);
. `( \- w9 ?0 }% b
}
3 r% I" `! s5 f7 g3 {
2 P4 k; K: ?3 _. L
int main(int argc, char *argv[])
6 Z, @$ j3 g6 y, k% Z N+ `& J, r7 I& h
{
, }( i4 z4 \6 L$ T! g% v
int i;
4 W/ q) \8 r1 U6 `
double a,b,eps,s;
7 Y/ T7 T4 y7 u. j
clock_t tm;
4 o. t" t8 a# P+ h
# D. L+ w, V' @! d+ I' ]$ v" x
a=0.0; b=1.0; eps=0.0001;
- \! B6 T; H& c" L
tm=clock();
' y& c* U, p" {9 G8 [6 ^
for(i=0;i<100;i++)
- k" a) [ H6 D, i% f3 u
{
8 r& e9 o4 Z9 @: ~0 ^
s=fsim2(a,b,eps);
* e- o2 a3 {+ x( @. @) y* R' E
}
. y1 h# D+ g2 {6 d
printf("s=%e , 耗时 %d 毫秒。\n", s, (clock()-tm));
3 E8 T# d$ B# Q8 p! W! c Y
}
复制代码
结果:
* i6 i L, T( t$ Y/ v) k$ z
s=2.698925e+000 , 耗时 78 毫秒。
$ P1 v( c; @7 p1 q ?/ j: ~% H% O
& g, X* f7 }' k$ y% C
-------
6 ^; A, I) ^9 W* `. b. ?
8 {' j6 [+ ~7 U y
matlab代码:
%file fsim2.m
7 K, J4 V! ~: n0 W# v( J4 _
function s=fsim2(a,b,eps)
7 S! ~6 B! |) O# `5 H9 @( u3 u
n=1; h=0.5*(b-a);
( K1 ]7 }0 C1 C8 [: d. X
d=abs((b-a)*1.0e-06);
; K8 i1 F3 l% ^3 b
s1=simp1(a,eps); s2=simp1(b,eps);
6 K1 d/ X; P2 E' G; y. g. j
t1=h*(s1+s2);
5 ]& `+ I1 T9 n% r
s0=1.0e+35; ep=1.0+eps;
$ O' f8 K0 i8 H* x2 m5 i* o
while ((ep>=eps)&&(abs(h)>d))||(n<16),
2 ~- `( y; K. |
x=a-h; t2=0.5*t1;
" R0 G9 j" s8 Z, }! u
for j=1:n
& d; o& N) t, z% {9 x
x=x+2.0*h;
$ R# ^& \! s$ \. K! z
g=simp1(x,eps);
7 v$ g! U& n, ~$ @" U& X2 H% }" |
t2=t2+h*g;
, v2 g2 {$ [' ^( n- A
end
- [0 X& ]+ [4 N. X. e' u" F" w% }
s=(4.0*t2-t1)/3.0;
2 p/ {$ g3 N& L9 j* M- |
ep=abs(s-s0)/(1.0+abs(s));
) J$ B; h3 ~. [- W8 C. |
n=n+n; s0=s; t1=t2; h=h*0.5;
9 n8 \; e- [9 e( p5 t& W
end
) X, \1 {. L4 [
end
0 ~& ~$ w, U, Q0 l: q9 v
' _+ n/ R0 g d( r) J% r# W
function g=simp1(x,eps)
6 t2 W) `# }( G
n=1;
2 ?$ O) y' L, G" }+ L
[y0,y1]=f2s(x);
3 E2 ?+ b; B5 }. z# b$ Q6 b% X7 o
h=0.5*(y1-y0);
1 t; ? E7 i) M, H3 D0 t
d=abs(h*2.0e-06);
* a) A1 c5 Z+ M8 k
t1=h*(f2f(x,y0)+f2f(x,y1));
4 B8 t! T0 w1 E
ep=1.0+eps; g0=1.0e+35;
1 M5 K! M8 p5 d. D* Q. y# L/ [
while (((ep>=eps)&&(abs(h)>d))||(n<16))
: s" l* C( J0 ]8 q
yy=y0-h;
9 e+ {8 R6 L, b! t* L8 f2 h
t2=0.5*t1;
5 l; G, _1 \2 s9 f7 f/ w) H$ F7 P2 ]6 ~
for i=1:n
: a! M% P4 y3 G" c w! X5 L: p. D
yy=yy+2.0*h;
4 {* m9 Q0 c- H- n p
t2=t2+h*f2f(x,yy);
$ u; x0 b/ V$ R, ]) d3 o( X D6 A
end
2 | M& U3 \9 W1 M2 Q
g=(4.0*t2-t1)/3.0;
( A5 T5 ~+ d! x3 C/ z/ e; _
ep=abs(g-g0)/(1.0+abs(g));
0 S% |! V L5 K3 j, Y
n=n+n; g0=g; t1=t2; h=0.5*h;
3 T7 v2 V5 q$ a8 o* M3 p
end
1 L: K. e2 w% J! P
end
& i6 E" x, g$ U, H+ g- A3 x
( T! @" e: q& O. ~
%file f2s.m
( l5 M! i8 p$ h' t, o- g' v5 }
function [y0,y1]=f2s(x)
4 K6 o3 x C. G% N: p6 W8 x6 v
y0=-sqrt(1.0-x*x);
9 O# Y7 d7 e: Z6 _% q
y1=-y0;
* p D) ]( e8 U9 ?$ M: Y( k
end
! Q0 \/ A; _; M. m9 J' l* ~
7 ~2 i1 h2 t+ p5 B9 }
%file f2f.m
7 d8 i* ]9 X# p2 y* `
function c=f2f(x,y)
0 _0 v9 x7 K$ Q$ P" H
c=exp(x*x+y*y);
8 o8 W. a7 v6 H
end
" ~# J: l5 J2 m) K$ G
8 {+ a+ e7 `; [. c, l% [
%%%%%%%%%%%%%
, l0 Q1 d8 P/ w7 y: J9 R5 V
3 L \' N# N( X9 O& l, M6 E
>> tic
" f+ X! w! _" ]4 q5 G( t j
for i=1:100
* Q* I# Z$ |) G6 G/ b: G
a=fsim2(0,1,0.0001);
R# {, ]5 ?0 V, c
end
, h" T! I8 y6 R% K0 u" q1 j9 a
a
% s6 u* ?2 q0 g2 S% [
toc
) z1 y% K0 N: V% M a* L
; [* z: s: ~2 D1 z' O+ t+ W
a =
1 _ U- M8 z" M' Z
+ q. X! o8 ` M* ^5 N. C' O
2.6989
; e1 K0 {, w2 ^* z, ~
8 W. n1 a! R) I" g$ A
Elapsed time is 0.995575 seconds.
复制代码
-------
# r1 B2 _2 Y4 [! z" O% \# g
9 [* I- r6 d4 k/ ~. O( N
Forcal代码:
fsim2s(x,y0,y1)=
$ u" z0 v) |- I. C7 q
{
& p2 e; _ D- Q! X d6 ?* S
y0=-sqrt(1.0-x*x),
3 D: ^+ z) Q s: n5 R
y1=-y0
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};
. Y6 y _+ @; ^- A* F1 i
fsim2f(x,y)=exp(x*x+y*y);
4 q! D6 L3 U+ s
//////////////////
" b3 ^, _0 I. v C. R3 d
simp1(x,eps : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
* W1 G! I$ T6 Y4 m# K- i/ I& H8 l8 M
{
! l. @6 D' l2 `/ K& _
n=1,
8 P; l) S* d" U; h `
fsim2s(x,&y0,&y1),
+ L$ R# _: r. ~1 }$ d! R, t* I
h=0.5*(y1-y0),
4 H. Q2 S1 V7 U
d=abs(h*2.0e-06),
( v$ V9 l. f) l' y
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
2 d5 ~* w& l, a9 o& W. u
ep=1.0+eps, g0=1.0e+35,
: H9 N; R4 M4 t" p7 K
while {((ep>=eps)&(abs(h)>d))|(n<16),
# s2 p0 {8 O$ Z2 s, ? D4 @ @
yy=y0-h,
# D0 s+ o0 r4 A: u
t2=0.5*t1,
2 C) d1 L0 g9 C7 l( K
i=1, while{i<=n,
5 A1 a* e5 v6 p' t
yy=yy+2.0*h,
& Y* ~% ^7 \1 g5 y& [5 O/ e8 d- g
t2=t2+h*fsim2f(x,yy),
' Y" z7 K) C: M! H8 N
i++
. t5 t1 @: w4 t3 O
},
7 ?. j9 l1 |9 g9 [
g=(4.0*t2-t1)/3.0,
& ]8 v1 o2 n) e3 t
ep=abs(g-g0)/(1.0+abs(g)),
( W1 f7 X- u/ y$ G+ b
n=n+n, g0=g, t1=t2, h=0.5*h
/ g3 {1 H" O! R- `
},
' {( U# O) ]+ X: a- L: j
g
3 j' U3 p/ e, G* {$ J [+ r. p
};
' G; k( I" a1 @% g7 \
3 x+ A8 I7 C4 X8 H+ t' F: J
fsim2(a,b,eps : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
2 ^5 d& a. ^) ?5 X8 k; L# K
{
: u$ e/ G' g6 C% d. T' A& N$ R
n=1, h=0.5*(b-a),
- a" H% A/ g# t1 u% h- w
d=abs((b-a)*1.0e-06),
: z; q( X* x5 G' `" T6 I4 }
s1=simp1(a,eps), s2=simp1(b,eps),
8 S F9 S" x& k" `4 U
t1=h*(s1+s2),
) N3 r: N @2 A( J0 o( @5 H9 x
s0=1.0e+35, ep=1.0+eps,
0 [9 d2 h- G+ k$ b- }7 L1 ]! T
while {((ep>=eps)&(abs(h)>d))|(n<16),
6 ` B0 W% h; M$ r
x=a-h, t2=0.5*t1,
8 F; S2 r; x3 o+ ]: X5 F' G
j=1, while{j<=n,
& A6 ^3 ]2 ` w) p" Q% |, y% _
x=x+2.0*h,
5 m" O6 K" j/ x8 p0 [
g=simp1(x,eps),
?/ l3 m$ `! h" n5 \4 y4 z
t2=t2+h*g,
R: \8 V) `/ U: n( e- u3 Q
j++
$ B" G- [$ L# Q) k! K9 N
},
6 Q- b+ s( x4 v% O/ L1 i
s=(4.0*t2-t1)/3.0,
5 h' ~/ V, ~# g/ \; a
ep=abs(s-s0)/(1.0+abs(s)),
: j7 q+ t! l) d& q* \) d o3 I
n=n+n, s0=s, t1=t2, h=h*0.5
6 c$ n4 f0 u. E/ A& f$ g0 J
},
0 J* Y- d* X( p8 i. l
s
; m6 u! c$ g0 I* ^
};
1 z1 B. j, x5 |* a' H
/ r0 L' f# i N9 U1 t
//////////////////
9 ~" U, S1 L4 K% g1 G
- K; U7 {! k& p! ?# Z4 K6 j0 b
mvar:
6 ]: |/ G$ \, k+ S. ]3 q3 ]/ h
t0=sys::clock(),
, ]3 ^* d B) o5 _: b
i=0, while{i<100, a=fsim2(0,1,0.0001), i++}, a;
3 c- S) k9 _& r8 s
[sys::clock()-t0]/1000;
复制代码
结果:
0 e0 R3 c" T) }0 e
2.698925000624303
6 a% [ r d) J2 C/ P0 h
0.328
, \. u3 p) ~7 B4 I) s2 |
4 P1 T5 N! N$ o. f, X) ^( O
---------
2 g( W# M' B* f4 Y! {
3 e8 i. R, l. Z& ~ Q0 i2 t+ E' t
本例C/C++、matlab、Forcal运行耗时之比为 1:12.7:4.2 。
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l+ x6 W0 ]9 i
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。
$ X; y p5 }/ `7 e# g( {/ \ z
: ^0 h* X R4 B* |+ ]
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
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) H) `" p, { Z+ F$ I
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。
6 v6 o8 X1 R* g6 y8 \: z
% A8 b, t8 p4 P+ R: F
不再给出C/C++代码,因其效率不会发生变化。
8 ~- L* y1 y4 q7 R2 Q
$ u$ W2 S) R/ z( j* Z+ i C4 k
Matlab代码:
%file fsim2.m
0 t% ~4 Z& N. U- Z
function s=fsim2(a,b,eps,fsim2s,fsim2f)
* G! C, w) }: U# D1 \
n=1; h=0.5*(b-a);
* k( r& C$ y1 h0 q! m: v
d=abs((b-a)*1.0e-06);
5 v: e9 m$ T2 v2 c8 s0 v/ m
s1=simp1(a,eps,fsim2s,fsim2f); s2=simp1(b,eps,fsim2s,fsim2f);
2 X. f5 M3 `4 ]5 M" J* h
t1=h*(s1+s2);
( N' V* a! E9 y
s0=1.0e+35; ep=1.0+eps;
. @! o% N+ _) z) J9 r5 U
while ((ep>=eps)&&(abs(h)>d))||(n<16),
3 C& |1 m3 R4 S# p% v* p
x=a-h; t2=0.5*t1;
+ w* g( T5 H9 p( m3 \
for j=1:n
% T! _' G: Y. C) I9 G& N5 {+ c* y' I
x=x+2.0*h;
9 Y! y% F2 m0 q6 u$ n
g=simp1(x,eps,fsim2s,fsim2f);
" Z2 d7 a T8 T3 R4 g! @8 c! ~
t2=t2+h*g;
% w9 I6 x: p/ l1 t
end
8 v' A5 E% w- O* v4 x! b1 y/ W6 `
s=(4.0*t2-t1)/3.0;
6 }8 M% x" |2 J4 ]1 u9 X
ep=abs(s-s0)/(1.0+abs(s));
: C7 o0 Y8 G2 @. e
n=n+n; s0=s; t1=t2; h=h*0.5;
* h5 e7 ~( L+ X1 T' W" m3 d
end
9 D( ^$ k6 K; F
end
3 t: E3 G* t6 r8 A0 u( R
" \1 `6 C/ u+ q5 }5 A6 S
function g=simp1(x,eps,fsim2s,fsim2f)
9 C) @5 s! z( y; s. V2 F
n=1;
; g5 h; `7 f; w
[y0,y1]=fsim2s(x);
I7 E; m. O5 m- H, Q
h=0.5*(y1-y0);
! \& w$ M! }9 S' V2 A* f5 w- Z
d=abs(h*2.0e-06);
9 ]) z8 e) }& w* A
t1=h*(fsim2f(x,y0)+fsim2f(x,y1));
* K1 B" E" h# z3 d2 N r
ep=1.0+eps; g0=1.0e+35;
2 R4 v {8 P% a: G* @( {. T, I( I
while (((ep>=eps)&&(abs(h)>d))||(n<16))
/ K* k# C$ |, i: K1 l6 T6 A
yy=y0-h;
8 b0 k) v0 U1 V6 G% S* t9 Y. N6 x- x! F
t2=0.5*t1;
- E' A P, f( N6 O7 E. B
for i=1:n
- M+ S8 U( p L
yy=yy+2.0*h;
1 D1 N) q1 x r5 q0 I% `. A- B
t2=t2+h*fsim2f(x,yy);
# h, D: P. ^5 l) K" o" p6 C8 b! |
end
2 n% g1 r+ b+ h* f1 S" x1 M2 Q
g=(4.0*t2-t1)/3.0;
; |, D0 y* J- [3 z1 w0 m
ep=abs(g-g0)/(1.0+abs(g));
, O0 F& @- V6 l
n=n+n; g0=g; t1=t2; h=0.5*h;
" Y5 C3 {- {6 i: N7 r% C8 d
end
4 c& G& j0 A Q
end
& M ?0 c% G$ G
- B! h0 ]' G( ~4 m
%file f2s.m
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function [y0,y1]=f2s(x)
0 L. y' F; R# g: ]1 d/ H+ J C
y0=-sqrt(1.0-x*x);
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y1=-y0;
9 h/ Y+ X. f0 A/ U8 J" z6 @; c# I
end
- C" h8 Q/ `. I
) z6 B, x5 T9 N
%file f2f.m
) b8 b/ {. A- z# v+ h" I% ^/ @
function c=f2f(x,y)
1 _% n- x4 U8 E. r
c=exp(x*x+y*y);
8 @' a) G) l+ l" W, X- `
end
& a/ O( U2 H6 l: R# X$ J. G4 }7 a
$ W8 R6 I' }9 X3 p# D
%%%%%%%%%%%%%%%%
: L7 b% e6 ^- G, ], C
9 G% ?7 x% y" a, ^1 N' N \
>> tic
3 U. y3 c7 {. h, h
for i=1:100
" X& i; q7 O v \. ~
a=fsim2(0,1,0.0001,@f2s,@f2f);
% R6 }5 T2 _2 ?4 d8 D/ f1 n- a
end
. @' Z# U2 E- [, _% l' j# T- F
a
1 `3 r3 F" W2 }7 e ]* Y( ]
toc
- j9 F6 G* }9 P
) b8 y, {/ j* n# Y/ h
a =
8 v: U) S9 ]! z: \$ t) t% ]! l
+ H# Y q! d: }3 i+ y: w; y! U' t) `. J
2.6989
3 w; N/ S' R# A/ }. I5 D
% G8 B0 R# v" t+ |3 }8 M5 X7 y2 s, _
Elapsed time is 1.267014 seconds.
复制代码
--------
# g' ~+ {+ }3 T5 B) J
/ j1 [& z- s' Z7 C$ i& s2 O4 }
Forcal代码:
simp1(x,eps,fsim2s,fsim2f : n,i,y0,y1,h,d,t1,yy,t2,g,ep,g0)=
& P6 ?* `% X( ~# m% A
{
8 n4 L7 K F/ B/ r/ s. Q
n=1,
" o* [; n/ o, o, \( J3 Z- b2 F
fsim2s(x,&y0,&y1),
$ I) w/ b; c! v3 M+ V# _% L
h=0.5*(y1-y0),
( j2 f# T" O4 i2 T2 y
d=abs(h*2.0e-06),
# n* M2 W' v ~0 j/ e
t1=h*(fsim2f(x,y0)+fsim2f(x,y1)),
: V8 T# b8 F/ z v7 o3 V4 X
ep=1.0+eps, g0=1.0e+35,
4 h/ N& u) J( ?$ P0 a) ~1 B6 W
while {((ep>=eps)&(abs(h)>d))|(n<16),
# v3 g! d$ X' I8 `5 u
yy=y0-h,
" u2 U+ E B3 G" p% c+ K; s6 ^1 @
t2=0.5*t1,
- U; D5 v! ?0 F9 H5 H
i=1, while{i<=n,
# z& C$ T x/ D2 h
yy=yy+2.0*h,
* [- N: b$ p- |' }* u! i' T
t2=t2+h*fsim2f(x,yy),
7 Z. d3 i2 w" p
i++
! y! F4 ]+ ?+ D5 C7 W* n
},
+ k7 d$ \2 k0 Y! g
g=(4.0*t2-t1)/3.0,
2 C c- f# w7 {
ep=abs(g-g0)/(1.0+abs(g)),
; n2 X) \& n' c6 E
n=n+n, g0=g, t1=t2, h=0.5*h
& v/ F/ D" W' `' d' o
},
3 O; J' R5 }( `) C! X# T1 ]
g
5 K9 q0 `/ D8 B% O& F! S* Q
};
" }% |9 c! E: `" V2 S) Y5 i
) h& D+ Z' N2 ?! @! L
fsim2(a,b,eps,fsim2s,fsim2f : n,j,h,d,s1,s2,t1,x,t2,g,s,s0,ep)=
% ~9 u$ Q' g" d* a3 d2 `
{
/ K# K7 u* R( P0 T7 ~3 O. t$ I
n=1, h=0.5*(b-a),
( P) M8 f) V; Q0 N! u& }! e7 u
d=abs((b-a)*1.0e-06),
# D3 [ V: G4 g8 l8 P
s1=simp1(a,eps,fsim2s,fsim2f), s2=simp1(b,eps,fsim2s,fsim2f),
0 z5 c. s- q& u* ~ { I4 a
t1=h*(s1+s2),
# h* J/ H8 p# j9 o
s0=1.0e+35, ep=1.0+eps,
% e" _2 O- D; J! y+ @( ~+ m
while {((ep>=eps)&(abs(h)>d))|(n<16),
) J: g z) s ~# o+ X
x=a-h, t2=0.5*t1,
1 I2 x; {6 v- m k6 t3 g! O
j=1, while{j<=n,
" N& [; Z F) X2 I# @& K
x=x+2.0*h,
r0 p4 D" Q# J$ R- B
g=simp1(x,eps,fsim2s,fsim2f),
: a6 X3 L) w+ U7 r( d" H5 O9 s% f8 t
t2=t2+h*g,
$ u+ }3 Q5 L. r7 @7 ^
j++
7 F# R- n* a: D8 g! r+ x' U
},
7 L9 Z( p! l7 u2 r' a- Y
s=(4.0*t2-t1)/3.0,
7 r& r6 t- F: h9 I; f' B
ep=abs(s-s0)/(1.0+abs(s)),
# L4 e c/ ~: ?4 C. V
n=n+n, s0=s, t1=t2, h=h*0.5
# S; ]; E* }2 A( Q1 W
},
& a( p/ P/ R& G2 b
s
% u" P) h# u6 O6 i/ |% X
};
$ o4 a) r2 G2 g8 r
3 z0 b4 m/ z; F- g$ l ^, [4 T
//////////////////
& L) p. c n9 O7 i1 k
8 k3 j1 l, s5 ], ]5 R
f2s(x,y0,y1)=
- N/ b( e$ W/ u3 {. i
{
3 N3 Q& c! M9 e/ p6 e! X
y0=-sqrt(1.0-x*x),
8 e9 \& a5 @9 _
y1=-y0
: u/ z' s+ \9 |$ d* A2 x9 |* }" x
};
. ~! P n' J% p n
f2f(x,y)=exp(x*x+y*y);
3 Y) P8 _" T9 C/ ^0 O- V
9 F* p- ~+ |* r, D$ |
mvar:
, O4 @3 p( Y0 ^7 B1 f$ \
t0=sys::clock(),
- K9 {' O0 q( X- l3 s9 g
i=0, while{i<100, a=fsim2(0,1,0.0001,HFor("f2s"),HFor("f2f")), i++}, a;
8 \2 k! E# \) v. h; ^( y9 D; u
[sys::clock()-t0]/1000;
复制代码
结果:
6 A4 z; M8 V& i& N
2.698925000624303
! x7 _' t% z6 j" X' k
0.844
: S u# `9 A) a9 W9 T1 g
6 S% u" f/ ]5 M! W& Q) {4 d: n1 D
--------
) q$ a( y, ?; \; D! G) T
* K2 I" Y- J; y( u, H* x& Y
本例matlab与Forcal耗时之比为1.267014 :0.844。Forcal仍有优势。
( P. L4 p% @* X5 N
' Y. H# V8 l' \' W: v/ G# ]3 V
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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