計(jì)算N階矩陣的隨機(jī)一致性指標(biāo)RI

1、實(shí) 驗(yàn) 報(bào) 告（一）課程名稱 數(shù) 學(xué) 建 模 實(shí)驗(yàn)項(xiàng)目 Matlab基本操作及其實(shí)際應(yīng)用 實(shí)驗(yàn)環(huán)境 PC機(jī)、Matlab學(xué)院/班級(jí) 學(xué)號(hào)/姓名 指導(dǎo)教師 實(shí)驗(yàn)日期成績(jī)一、實(shí)驗(yàn)名稱：隨機(jī)一致性指標(biāo)求解二、實(shí)驗(yàn)?zāi)康模?）掌握用matlab求解隨機(jī)一致性指標(biāo)的方法2）加深對(duì)隨機(jī)一致性指標(biāo)概念的理解三、實(shí)驗(yàn)內(nèi)容：用matlab或C+編寫程序分別計(jì)算n=3-30時(shí)的n階矩陣的隨機(jī)一致性檢驗(yàn)指標(biāo)的值RI。/本組實(shí)驗(yàn)隨機(jī)數(shù)產(chǎn)生正互反矩陣,這個(gè)數(shù)目必須取的相當(dāng)大（超過(guò)1000000，所以此程序跑起來(lái)比較費(fèi)時(shí)），才比較接近標(biāo)準(zhǔn)答案 #include<iostream>#include<cmat

2、h>#include<ctime>#include<cstdlib>#include<iomanip>using namespace std;int getNameta(double a,int n);的最大void createMatrix(double a,int n);int getMax(double b,int n); /建立隨機(jī)正互反矩陣 /獲取b矩陣中的最大值 /獲取當(dāng)前隨機(jī)正互反矩陣void Mifa(double a,double b,int n);void main()數(shù)int all = 0; int tem = 0; int r

3、i = 0; int n; /冪法過(guò)程 /矩陣階數(shù) /建立的隨機(jī)正互反矩陣個(gè)int m = 3000; /所有最大的和 /最終的RI（為了計(jì)算方便，將實(shí)際的RI擴(kuò)大了100倍進(jìn)行計(jì)算了）cout << "Please enter the size of the matrix:" << endl; cout << "n = " ; cin >> n; if(n > 2&& n <= 30) for(int i =0;i<m;i+) double *a = new double

4、n*n; /聲明n*n的隨機(jī)正互反矩陣 for(int j =0;j<n*n;j+) aj = 1;createMatrix(a,n); /建立n*n的隨機(jī)正互反矩陣 tem = all/m; /計(jì)算得RI ri = (tem - n*100); ri = ri/(n-1); /計(jì)算得到的RIall += getNameta(a,n); /獲得最大 （乘以100后的結(jié)果）int baiwei = ri/100; int shiwei = (ri-100*baiwei)/10; int gewei = ri-100*baiwei-10*shiwei; cout << "

5、;The RI of the " << n <<" size matrix is " << baiwei << "." << shiwei << gewei else if(n =1) else cout << "The RI of the " << n <<" size matrix is " << 0 << endl; << endl; cout <&l

6、t; "The size of n you input is wrong" << endl;void createMatrix(double a,int n)doublesample17=1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,1.0/2.0,1.0/3.0,1.0/4.0,1.0/5.0,1.0/6.0,1.0/7.0,1.0/8.0,1.0/9.0;int getNameta(double a,int n) for(int i=0;i<n;i+) for(int j=i;j<n;j+) int t; t = ran

7、d()%17; if( j = i) else an*i+j = samplet; an*j+i = 1.0/samplet; an*i+j=1.0; int tem=0; double *b=new doublen; for(int i=0;i<n;i+) Mifa(a,b,n); int max=getMax(b,n); while(tem != max) int nameta; nameta = tem; return nameta; tem = max; for(int d = 0;d<n;d+) Mifa(a,b,n); max = getMax(b,n); bd = bd*100/max; bi=1.0/n;void Mifa(double a,double b,int n) int getMax(double b,int n) int a = 0; for(int i = 0;i<n;i+) double *c = new doublen; for(int r = 0;r<n;r+) for(int i =0;i<n;i+) for(int j = 0 ;j<n;j+) ci += an*i+j*bj; cr = 0.0; for(int k =0;k<n;k+) bk = ck; if(100