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题目
编写二维实数矩阵(非方阵)的相关运算函数,并在主函数测试各个函数的功能,并输出计算结果。具体要求如下:
1. 实现两个矩阵的加减法: A+B, A-B (注意维度大小匹配)
2. 实现两个矩阵的乘法: A*B (注意维度大小匹配)
3. 求给定矩阵的转置矩阵: A^T
4. 计算给定方阵的迹(主对角元素之和) \mathrm{Tr}(A)
5. 计算给定方阵的主对角元素之积
6. 计算给定矩阵的1-范数 (最大的行绝对值和)
7. 计算给定矩阵的 无穷-范数 (最大的列绝对值和)
8. 计算给定矩阵的 弗罗贝尼乌斯范数(Frobenius)范数 (定义为所有矩阵元素的平方和再开根号;提示:也可以利用矩阵A乘以其自身的转置矩阵A^T然后取所得方阵的迹进行计算)
一、实现方法
1.1 实现思路
题目要求为实现矩阵本身和矩阵与矩阵之间的一些数值计算和操作。总体思想为,使用两个二维数组a、b来记录题目所给的矩阵,并用row1、col1、row2、col2来记录两个矩阵的行数和列数。 c为记录输出矩阵的二维数组。采用循环的方式实现题目所要求的各个操作。
二、函数实现
2.1 两个矩阵的加减法
注:由于题目未指明上限,此处对于数组的定义采用
#define N 100
对于一个大小为100*100的数组,对于题目中所需要的全局数组,其大小
tot = 2 * sizeof(a);
为163216Byte,即0.15Mb,对于现代计算机来说非常小,而且可以满足绝大部分实际计算需要。
\begin{pmatrix}
a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3}\\
a_{3,1}&a_{3,2}&a_{3,3}\\
\end{pmatrix}+
\begin{pmatrix}
b_{1,1}&b_{1,2}&b_{1,3}\\
b_{2,1}&b_{2,2}&b_{2,3}\\
b_{3,1}&b_{3,2}&b_{3,3}\\
\end{pmatrix}=
\begin{pmatrix}
c_{1,1}&c_{1,2}&c_{1,3}\\
c_{2,1}&c_{2,2}&c_{2,3}\\
c_{3,1}&c_{3,2}&c_{3,3}\\
\end{pmatrix}
\begin{pmatrix}
a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3}\\
a_{3,1}&a_{3,2}&a_{3,3}\\
\end{pmatrix}-
\begin{pmatrix}
b_{1,1}&b_{1,2}&b_{1,3}\\
b_{2,1}&b_{2,2}&b_{2,3}\\
b_{3,1}&b_{3,2}&b_{3,3}\\
\end{pmatrix}=
\begin{pmatrix}
c_{1,1}&c_{1,2}&c_{1,3}\\
c_{2,1}&c_{2,2}&c_{2,3}\\
c_{3,1}&c_{3,2}&c_{3,3}\\
\end{pmatrix}
两个函数的加减法,采用一般的循环进行相加和相减,存入c中之后进行输出,代码如下。
其中,将a[][N+1],b[][N+1]作为传递数组,省去数组复制的时间,同时在函数中保留原数组的行数、列数信息,用于后续的操作。
void p(double a[][N + 1], double b[][N + 1], int row1, int col1, int row2, int col2) //矩阵加法
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
if (row1 != row2 || col1 != col2)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
c[i][j] = a[i][j] + b[i][j]; //答案存储矩阵
}
for (int i = 1; i <= row1; i++)//输出
{
for (int j = 1; j <= col2; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
void d(double a[][N + 1], double b[][N + 1], int row1, int col1, int row2, int col2) //矩阵减法
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
if (row1 != row2 || col1 != col2)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
c[i][j] = a[i][j] - b[i][j]; //答案存储矩阵
}
for (int i = 1; i <= row1; i++)//输出
{
for (int j = 1; j <= col2; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
首先,为了确保两个矩阵是可以进行相加和相减的,进行了一个判断,当两个矩阵大小一致才进行后续的操作。
2.2 两个矩阵的乘法
\begin{pmatrix}
a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3}\\
\end{pmatrix}*
\begin{pmatrix}
b_{1,1}&b_{1,2}\\
b_{2,1}&b_{2,2}\\
b_{3,1}&b_{3,2}\\
\end{pmatrix}=
\begin{pmatrix}
c_{1,1}&c_{1,2}\\
c_{2,1}&c_{2,2}\\
\end{pmatrix}
矩阵的点乘,根据定义,采用三个for循环进行嵌套实现,复杂度为O(n^3),对于我们定义的数组大小来说,可以满足一般需要,如果对于更大的矩阵,可以采用Strassen演算法优化,或是采用并行矩阵乘法。
同样的,需要判断矩阵a的列数和矩阵b的行数是否一致再进行后续操作。
void dot(double a[][N + 1], double b[][N + 1], int row1, int col1, int row2, int col2)//矩阵乘法
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
if (row2 != col1)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col2; j++)
for (int k = 1; k <= col1; k++)
c[i][j] += a[i][k] * b[k][j]; //答案存储矩阵
}
for (int i = 1; i <= row1; i++)//输出
{
for (int j = 1; j <= col2; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
2.3 矩阵的转置
\begin{pmatrix}
a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3}\\
\end{pmatrix}^T=
\begin{pmatrix}
a_{1,1}&a_{2,1}\\
a_{1,2}&a_{2,2}\\
a_{1,3}&a_{2,3}\\
\end{pmatrix}
void T(double a[][N + 1], int row1, int col1) //矩阵转置
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
c[j][i] = a[i][j];
}
for (int i = 1; i <= col1; i++)//输出
{
for (int j = 1; j <= row1; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
2.4 矩阵的迹
tr
\begin{pmatrix}
a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3}\\
a_{3,1}&a_{3,2}&a_{3,3}\\
\end{pmatrix}=
a_{1,1} + a_{2,2} + a_{3,3}
矩阵的迹的实现使用一个循环直接求对角数之和即可,此处的判断需要为一个方阵才可以进行求迹操作。
void tr(double a[][N + 1], int row1, int col1) //矩阵的迹
{
if (row1 != col1)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
double sum = 0;
for (int i = 1; i <= row1; i++)
sum += a[i][i];
cout << sum << endl;
cout << endl;
}
2.5 矩阵的对角元素之积
将该操作即为mul,则有
mul
\begin{pmatrix}
a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3}\\
a_{3,1}&a_{3,2}&a_{3,3}\\
\end{pmatrix}=
a_{1,1} * a_{2,2} * a_{3,3}
类似于求迹,将加法换成乘法即可。此处需要特别注意的是,不可以将存储答案的变量初始化为0,这里采用了将其初始化为第一个元素的方法。同样需要判断是否为方阵。
void mul(double a[][N + 1], int row1, int col1) //矩阵对角线积
{
if (row1 != col1)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
double sum = a[1][1];
for (int i = 2; i <= row1; i++)
sum *= a[i][i];
cout << sum << endl;
cout << endl;
}
2.6 矩阵的1-范数
ans = max(\sum\limits_{j = 1}^{rol} |a_{i,j}|) (i = 1, 2,3, ······,row)
使用自行构造的求最大值函数,以简化程序。
double max(double x, double y) { return x > y ? x : y; }
求1-范数采用两个循环,求出每一行的绝对值之和再取最大值即可。
void F1(double a[][N + 1], int row1, int col1) //矩阵的1-范数
{
double sum = 0, ans = 0;
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
sum += abs(a[i][j]);
ans = max(ans, sum);
sum = 0;
}
cout << ans << endl;
cout << endl;
}
2.7 矩阵的无穷-范数
ans = max(\sum\limits_{j = 1}^{row} |a_{j,i}|) (i = 1, 2,3, ······,rol)
和1-范数求法类似,将i,j简单对调即可。
void F2(double a[][N + 1], int row1, int col1)//矩阵的无穷-范数
{
double sum = 0, ans = 0;
for (int i = 1; i <= col1; i++)
{
for (int j = 1; j <= row1; j++)
sum += abs(a[j][i]);
ans = max(ans, sum);
sum = 0;
}
cout << ans << endl;
cout << endl;
}
2.8 矩阵的Frobenius范数
ans=\sqrt{\sum\limits_{i=1}^{row}\sum\limits_{j=1}^{col}|a_{i,j}|^2}
直接求出每个元素的平方和再使用cmath中的sqrt函数求开方即可。
void frobenius(double a[][N + 1], int row1, int col1)//矩阵的Frobenius 范数
{
double sum = 0;
for (int i = 1; i <= row1; i++)
for (int j = 1; j <= col1; j++)
sum += a[i][j] * a[i][j];
cout << sqrt(sum) << endl;
cout << endl;
}
下面给出完整程序,并附上样例输入输出。
#include <stdio.h>
#include <iostream>
#include <cmath>
#include <string.h>
using namespace std;
#define N 100
double max(double x, double y) { return x > y ? x : y; }
double a[N + 1][N + 1], b[N + 1][N + 1];
int max1, max2;
void dot(double a[][N + 1], double b[][N + 1], int row1, int col1, int row2, int col2)//矩阵乘法
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
if (row2 != col1)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col2; j++)
for (int k = 1; k <= col1; k++)
c[i][j] += a[i][k] * b[k][j]; //答案存储矩阵
}
for (int i = 1; i <= row1; i++)//输出
{
for (int j = 1; j <= col2; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
void p(double a[][N + 1], double b[][N + 1], int row1, int col1, int row2, int col2) //矩阵加法
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
if (row1 != row2 || col1 != col2)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
c[i][j] = a[i][j] + b[i][j]; //答案存储矩阵
}
for (int i = 1; i <= row1; i++)//输出
{
for (int j = 1; j <= col2; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
void d(double a[][N + 1], double b[][N + 1], int row1, int col1, int row2, int col2) //矩阵减法
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
if (row1 != row2 || col1 != col2)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
c[i][j] = a[i][j] - b[i][j]; //答案存储矩阵
}
for (int i = 1; i <= row1; i++)//输出
{
for (int j = 1; j <= col2; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
void T(double a[][N + 1], int row1, int col1) //矩阵转置
{
double c[N + 1][N + 1];
memset(c, 0, sizeof(c));
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
c[j][i] = a[i][j];
}
for (int i = 1; i <= col1; i++)//输出
{
for (int j = 1; j <= row1; j++)
cout << c[i][j] << ' ';
cout << endl;
}
cout << endl;
}
void tr(double a[][N + 1], int row1, int col1) //矩阵的迹
{
if (row1 != col1)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
double sum = 0;
for (int i = 1; i <= row1; i++)
sum += a[i][i];
cout << sum << endl;
cout << endl;
}
void mul(double a[][N + 1], int row1, int col1) //矩阵对角线积
{
if (row1 != col1)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
double sum = a[1][1];
for (int i = 2; i <= row1; i++)
sum *= a[i][i];
cout << sum << endl;
cout << endl;
}
void F1(double a[][N + 1], int row1, int col1) //矩阵的1-范数
{
double sum = 0, ans = 0;
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col1; j++)
sum += abs(a[i][j]);
ans = max(ans, sum);
sum = 0;
}
cout << ans << endl;
cout << endl;
}
void F2(double a[][N + 1], int row1, int col1)//矩阵的无穷-范数
{
double sum = 0, ans = 0;
for (int i = 1; i <= col1; i++)
{
for (int j = 1; j <= row1; j++)
sum += abs(a[j][i]);
ans = max(ans, sum);
sum = 0;
}
cout << ans << endl;
cout << endl;
}
void frobenius(double a[][N + 1], int row1, int col1)//矩阵的Frobenius 范数
{
double sum = 0;
for (int i = 1; i <= row1; i++)
for (int j = 1; j <= col1; j++)
sum += a[i][j] * a[i][j];
cout << sqrt(sum) << endl;
cout << endl;
}
int main()
{
int row1, col1, row2, col2;
cin >> row1 >> col1;
for (int i = 1; i <= row1; i++)
for (int j = 1; j <= col1; j++)
cin >> a[i][j];
cin >> row2 >> col2;
for (int i = 1; i <= row2; i++)
for (int j = 1; j <= col2; j++)
cin >> b[i][j];
printf("矩阵乘法\n");
dot(a, b, row1, col1, row2, col2);
printf("矩阵加法\n");
p(a, b, row1, col1, row2, col2);
printf("矩阵减法\n");
d(a, b, row1, col1, row2, col2);
printf("矩阵转置\n");
T(a, row1, col1);
printf("矩阵的迹\n");
tr(a, row1, col1);
printf("矩阵对角线的积\n");
mul(a, row1, col1);
printf("矩阵的1-范数\n");
F1(a, row1, col1);
printf("矩阵的无穷-范数\n");
F2(a, row1, col1);
printf("矩阵的Frobenius 范数\n");
frobenius(a, row1, col1);
}
1.in
2 3
1 2 3
3 2 1
3 2
4 5
6 8
7 8
1.out
矩阵乘法
37 45
31 39
矩阵加法
Can't do this!
矩阵减法
Can't do this!
矩阵转置
1 3
2 2
3 1
矩阵的迹
Can't do this!
矩阵对角线的积
Can't do this!
矩阵的1-范数
6
矩阵的无穷-范数
4
矩阵的Frobenius 范数
5.2915
2.in
3 3
1 2 3
4 5 6
7 8 9
3 3
1 2 3
3 2 1
2 1 3
2.out
矩阵乘法
13 9 14
31 24 35
49 39 56
矩阵加法
2 4 6
7 7 7
9 9 12
矩阵减法
0 0 0
1 3 5
5 7 6
矩阵转置
1 4 7
2 5 8
3 6 9
矩阵的迹
15
矩阵对角线的积
45
矩阵的1-范数
24
矩阵的无穷-范数
18
矩阵的Frobenius 范数
16.8819
三、空间优化
对于任意的一个二维数组,我们可以采用一个一维数组进行定义,那么这个一维数组的大小即为row*col,我们可以在mian()函数中将其定义为。
double a[(row + 1) * (col + 1)];
此时,如果对于一个矩阵A我们要访问它的元素A_{i,j}可以通过数组a[i * (j - 1) + j] 来访问对应元素。
同时,在进行值传递的时候,可以略去空间声明,节省空间。下面给出其在矩阵乘法中的实现。
void dot(double a[], double b[], int row1, int col1, int row2, int col2)//矩阵乘法
{
double c[(row1 + 1)*(col2 + 1)];
memset(c, 0, sizeof(c));
if (row2 != col1)
{
cout << "Can't do this!" << endl;
cout << endl;
return;
}
for (int i = 1; i <= row1; i++)
{
for (int j = 1; j <= col2; j++)
for (int k = 1; k <= col1; k++)
c[i * (j - 1) + j] += a[i * (k - 1) + k] * b[k * (j - 1) + j]; //答案存储矩阵
}
for (int i = 1; i <= row1; i++)//输出
{
for (int j = 1; j <= col2; j++)
cout << c[i * (j - 1) + j] << ' ';
cout << endl;
}
cout << endl;
}
对于其他部分的函数,转化方式类似,在此不进行赘述。
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