[求助]求教大家,平方根辨识问题?

liljx_2008 · 发表于 2006-3-22 23:44

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近期遇到一个问题,在一篇论文上看到 A*X=0,其中A为已知的数据阵,X为向量.论文上说求X用平方根法辨识, 但不知是怎么求得? 在此请教各位大侠

[此贴子已经被cdwxg于2006-5-28 0:32:08编辑过]

liljx_2008 · 发表于 2006-3-23 21:07

非常感谢yejet的解答. 万分感谢!!!!!!! 以后有问题再向你请教.[em01][em01]

[此贴子已经被作者于2006-3-23 22:32:36编辑过]

liljx_2008 · 发表于 2006-3-23 22:56

刚看完yejet提供的资料,想问这个资料摘自何处? 可能是我现在一下子没看明白,所以我还想向你请教一下,如何辨识的问题,即A*X=0(其中A为n*n已知矩阵,X为一个列向量),如何求X.

yejet · 发表于 2006-4-7 19:54

<DIV class=quote>以下是引用liljx_2008在2006-3-23 22:56:16的发言： 刚看完yejet提供的资料,想问这个资料摘自何处? 可能是我现在一下子没看明白,所以我还想向你请教一下,如何辨识的问题,即A*X=0(其中A为n*n已知矩阵,X为一个列向量),如何求X.</DIV>
 ？？？？？？上面不是说得很清楚吗？

tongji · 发表于 2006-4-14 16:04

是呀，文章中的变量说明也不全面

huhust · 发表于 2006-4-15 11:13

yejet, 您有相应的程序吗？或者推荐一个相似的程序也行，我急用。先谢了！

[此贴子已经被作者于2006-4-15 11:17:03编辑过]

yejet · 发表于 2006-4-15 19:54

线性方程组的求解---平方根法及改进平方根法 
平方根法主要用于求解对称正定矩阵方程：
首先要提到的是有个关于正定矩阵的定理，说的是若A为n阶地称正定矩阵，则存在一个实的非奇异下三角矩阵L，使A=LL’（L’表示L的对称矩阵）
根据这个前提，在结合前面的LU分解算法，便有了这里的平方根算法：

 

平方根方法:
/*


squareRootMethod, coded by EmilMathew 05/9/10, you can modify and use these code as you wish , but there is no guarantee that it can fit all you need.


*/


void squareRootMethod(Type** inMatrixArr,Type* bList,Type* xAnsList,int size)


{


/*Maths Reason:


L*U=A*x=b


When you meet a duiCheng and zheng ding matrix:


it could be pation like this:


l11 l11 l21 ... ln1


l21 l22 * l22 ... ln2


.... ... ...


ln1 ln2 ... lnn lnn





i=j:


lij=sqrt(aij-sigma_k1...j-1(ljk^2))





i>j:


lij=(aij-sigma_k1...j-1(lik*ljk))/ljj





the steps below is very easy :


L*y=b;


U*x=y;





Enjoy!:)


by EmilMatthew


05/9/10.


*/


Type** l_Matrix,* yAnsList;


Type tmpData;


int i,j;





/*pointer data assertion*/


assertF(inMatrixArr!=NULL,"in squareRootMethod,matrixArr is NULL\n");


assertF(bList!=NULL,"in squareRootMethod,bList is NULL\n");


assertF(xAnsList!=NULL,"in squareRootMethod,xAnsList is NULL\n");





/*correct pass in matrix assertion*/


assertF(duiChengMatrixCheck(inMatrixArr,size),"in squareRootMethod,the pass in matrix is not dui cheng\n");





/*Mem Apply*/


listArrMemApply(&yAnsList,size);


twoDArrMemApply(&l_Matrix,size,size);





assertF(l_Matrix!=NULL,"in squareRootMethod,l_Matrix is null\n");


assertF(yAnsList!=NULL,"in squareRootMethod,yAnsList is null\n");





/*Core Program*/


for(i=0;i

for(j=0;j<=i;j++)


{


if(i==j)


{


tmpData=sumSomeRowPower(l_Matrix,j,0,j-1,2);


// printf("tmpData:%f\n",tmpData);


l_Matrix[j]=(float)sqrt(inMatrixArr-tmpData);


}


else


{


l_Matrix[j]=(inMatrixArr[j]-sumTwoRowBy(l_Matrix,i,j,0,j-1))/l_Matrix[j][j];


}


}


for(i=0;i

yAnsList=(bList-sumArr_JKByList_K(l_Matrix,yAnsList,i,0,i-1))/l_Matrix;





for(i=size-1;i>=0;i--)


xAnsList=(yAnsList-sumArr_KJByList_K(l_Matrix,xAnsList,i,i+1,size-1))/l_Matrix;





twoDArrMemFree(&l_Matrix,size);


free(yAnsList);


}





平方根算法的计算量约为普通三角分解算法的一半，但由于这里要用到开平方，效率不是很高，所以，便有了为效率而存在的改进版平方根算法:


改进的平方根算法:


/*


enhancedSquareRootMethod, coded by EmilMathew 05/9/10, you can modify and use these code as you wish , but there is no guarantee that it can fit all you need.


*/





如下:


void enhancedSquareRootMethod(Type** inMatrixArr,Type* bList,Type* xAnsList,int size)


{


Type** l_Matrix,** t_Matrix;


Type* yAnsList,* dList;





int i,j;





/*pointer data assertion*/


assertF(inMatrixArr!=NULL,"in enhancedSquareRootMethod,matrixArr is NULL\n");


assertF(bList!=NULL,"in enhancedSquareRootMethod,bList is NULL\n");


assertF(xAnsList!=NULL,"in enhancedSquareRootMethod,xAnsList is NULL\n");





/*correct pass in matrix assertion*/


assertF(duiChengMatrixCheck(inMatrixArr,size),"in enhancedSquareRootMethod,the pass in matrix is not dui cheng\n");





/*Mem Apply*/


listArrMemApply(&yAnsList,size);


listArrMemApply(&dList,size);


twoDArrMemApply(&l_Matrix,size,size);


twoDArrMemApply(&t_Matrix,size,size);





assertF(t_Matrix!=NULL,"in enhancedSquareRootMethod,t_Matrix is null\n");


assertF(l_Matrix!=NULL,"in enhancedSquareRootMethod,l_Matrix is null\n");


assertF(yAnsList!=NULL,"in enhancedSquareRootMethod,yAnsList is null\n");





for(i=0;i

l_Matrix=1;





for(j=0;j

{


dList[j]=inMatrixArr[j][j]-sumArr1_IKByArr2_JK(t_Matrix,l_Matrix,j,j,0,j-1);


for(i=j+1;i

{


t_Matrix[j]=inMatrixArr[j]-sumArr1_IKByArr2_JK(inMatrixArr,l_Matrix,i,j,0,j-1);


l_Matrix[j]=t_Matrix[j]/dList[j];


}


}








for(i=0;i

yAnsList=bList-sumArr_JKByList_K(l_Matrix,yAnsList,i,0,i-1);





for(i=size-1;i>=0;i--)


xAnsList=yAnsList/dList-sumArr_KJByList_K(l_Matrix,xAnsList,i,i+1,size-1);





/*mem free*/


twoDArrMemFree(&t_Matrix,size);


twoDArrMemFree(&l_Matrix,size);


free(yAnsList);


free(dList);


}





测试程序:


/*Square Method Algorithm test program*/


#include "Global.h"


#include "Ulti.h"


#include "MyAssert.h"


#include "Matrix.h"


#include


#include


#include


#include


#include








char *inFileName="inputData.txt";


/*


input data specification


len,


a00,a01,...,a0n-1,b0;


.....


an-10,an-11,...,an-1n-1,bn-1;


*/








char *outFileName="outputData.txt";


#define DEBUG 1





void main(int argc,char* argv[])


{


FILE *inputFile;/*input file*/


FILE *outputFile;/*output file*/





double startTime,endTime,tweenTime;/*time callopsed info*/





/*The read in data*/


int len,methodIndex;


Type** matrixArr;


Type* bList,* xAnsList;








int i,j;/*iterator index*/





/*input file open*/


if(argc>1)strcpy(inFileName,argv[1]);


assertF((inputFile=fopen(inFileName,"rb"))!=NULL,"input file error");


printf("input file open success\n");





/*outpout file open*/


if(argc>2)strcpy(outFileName,argv[2]);


assertF((outputFile=fopen(outFileName,"wb"))!=NULL,"output file error");


printf("output file open success\n");





fscanf(inputFile,"size=%d;\r\n",&len);


fscanf(inputFile,"method=%d;\r\n",&methodIndex);





/*Memory apply*/


matrixArr=(Type**)malloc(sizeof(Type*)*len);


for(i=0;i

matrixArr=(Type*)malloc(sizeof(Type)*len);





bList=(Type*)malloc(sizeof(Type)*len);


xAnsList=(Type*)malloc(sizeof(Type)*len);





/*Read info data*/


for(i=0;i

{


for(j=0;j

fscanf(inputFile,"%f,",&matrixArr[j]);


fscanf(inputFile,"%f;",&bList);


}








/*Check the input data*/


showArrListFloat(bList,0,len);


show2DArrFloat(matrixArr,len,len);





#if DEBUG


printf("\n*******start of test program******\n");


printf("now is runnig,please wait...\n");


startTime=(double)clock()/(double)CLOCKS_PER_SEC;


/******************Core program code*************/


switch(methodIndex)


{


case 1:


enhancedSquareRootMethod(matrixArr,bList,xAnsList,len);


printf("after the enhancedSquareRootMethod:the ans x rows is:\n");


fprintf(outputFile,"after the enhancedSquareRootMethod:the ans x rows is:(from x0 to xn-1)\r\n");


break;





case 2:





squareRootMethod(matrixArr,bList,xAnsList,len);


printf("after the SquartRootPationMethod:the ans x rows is:\n");


fprintf(outputFile,"after the SquartRootPationMethod:the ans x rows is:(from x0 to xn-1)\r\n");


break;





default:


printf("input method index error\n");


break;


}


showArrListFloat(xAnsList,0,len);


outputListArrFloat(xAnsList,0,len,outputFile);


/******************End of Core program**********/


endTime=(double)clock()/(double)CLOCKS_PER_SEC;


tweenTime=endTime-startTime;/*Get the time collapsed*/


/*Time collapsed output*/


printf("the collapsed time in this algorithm implement is:%f\n",tweenTime);


fprintf(outputFile,"the collapsed time in this algorithm implement is:%f\r\n",tweenTime);


printf("\n*******end of test program******\n");


#endif


for(i=0;i

free(matrixArr);


free(matrixArr);





free(xAnsList);


free(bList);





printf("program end successfully,\n you have to preess any key to clean the buffer area to output,otherwise,you wiil not get the total answer.\n");


getchar();/*Screen Delay Control*/


return;


}





测试结果:


平方根法:


test1:


输入:


size=3;


5,-4,1,1;


-4,6,-4,2;


1,-4,6,-3;


输出:


after the SquartRootPationMethod:the ans x rows is:(from x0 to xn-1)


1.00000 1.00000 0.00000





改进平方根法:


test1:


输入:


size=3;


method=1;


5,-4,1,1;


-4,6,-4,2;


1,-4,6,-3;





输出:


after the enhancedSquareRootMethod:the ans x rows is:(from x0 to xn-1)


1.00000 1.00000 0.00000





Test2:


after the enhancedSquareRootMethod:the ans x rows is:(from x0 to xn-1)


2.00000 1.00000 -1.00000





网上关于这个主题的相关参考:


http://jpkc.ecnu.edu.cn/gdds/xsxz/ZhangGengYun.htm


http://www.ascc.net/pd-man/linpack/node14.html

yejet · 发表于 2006-4-15 19:57

<DIV class=quote>以下是引用tongji在2006-4-14 16:04:39的发言： 是呀，文章中的变量说明也不全面</DIV>
 具体的你可依找本数值算法的书看看 比如丁丽娟的《数值计算方法》

多情清秋 · 发表于 2006-6-2 08:34

liljx_2008、huhust各加威望1点，yejet加威望2点
多情清秋 06.6.2

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[控制理论] [求助]求教大家,平方根辨识问题?

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