/// <summary>
/// 计算矩阵的秩
/// </summary>
/// <param name="matrix">矩阵</param>
/// <returns></returns>
private static int Rank(double[][] matrix)
{
//matrix为空则直接默认已经是最简形式
if (matrix == null || matrix.Length == 0) return 0;
//复制一个matrix到copy,之后因计算需要改动矩阵时并不改动matrix本身
double[][] copy = new double[matrix.Length][];
for (int i = 0; i < copy.Length; i++)
{
copy[i] = new double[matrix[i].Length];
}
for (int i = 0; i < matrix.Length; i++)
{
for (int j = 0; j < matrix[0].Length; j++)
{
copy[i][j] = matrix[i][j];
}
}
//先以最左侧非零项的位置进行行排序
Operation1(copy);
//循环化简矩阵
while (!isFinished(copy))
{
Operation2(copy);
Operation1(copy);
}
//过于趋近0的项,视作0,减小误差
Operation3(copy);
//行最简矩阵的秩即为所求
return Operation4(matrix);
}
/// <summary>
/// 判断矩阵是否变换到最简形式(非零行数达到最少)
/// </summary>
/// <param name="matrix"></param>
/// <returns>true:</returns>
private static bool isFinished(double[][] matrix)
{
//统计每行第一个非零元素的出现位置
int[] counter = new int[matrix.Length];
for (int i = 0; i < matrix.Length; i++)
{
for (int j = 0; j < matrix[i].Length; j++)
{
if (matrix[i][j] == 0)
{
counter[i]++;
}
else break;
}
}
//后面行的非零元素出现位置必须在前面行的后面,全零行除外
for (int i = 1; i < counter.Length; i++)
{
if (counter[i] <= counter[i - 1] && counter[i] != matrix[0].Length)
{
return false;
}
}
return true;
}
/// <summary>
/// 排序(按左侧最前非零位位置自上而下升序排列)
/// </summary>
/// <param name="matrix">矩阵</param>
private static void Operation1(double[][] matrix)
{
//统计每行第一个非零元素的出现位置
int[] counter = new int[matrix.Length];
for (int i = 0; i < matrix.Length; i++)
{
for (int j = 0; j < matrix[i].Length; j++)
{
if (matrix[i][j] == 0)
{
counter[i]++;
}
else break;
}
}
//按每行非零元素的出现位置升序排列
for (int i = 0; i < counter.Length; i++)
{
for (int j = i; j < counter.Length; j++)
{
if(counter[i]>counter[j])
{
double[] dTemp = matrix[i];
matrix[i] = matrix[j];
matrix[j] = dTemp;
}
}
}
}
/// <summary>
/// 行初等变换(左侧最前非零位位置最靠前的行,只保留一个)
/// </summary>
/// <param name="matrix">矩阵</param>
private static void Operation2(double[][] matrix)
{
//统计每行第一个非零元素的出现位置
int[] counter = new int[matrix.Length];
for (int i = 0; i < matrix.Length; i++)
{
for (int j = 0; j < matrix[i].Length; j++)
{
if (matrix[i][j] == 0)
{
counter[i]++;
}
else break;
}
}
for (int i = 1; i < counter.Length; i++)
{
if (counter[i] == counter[i - 1] && counter[i] != matrix[0].Length)
{
double a = matrix[i - 1][counter[i - 1]];
double b = matrix[i][counter[i]]; //counter[i]==counter[i-1]
matrix[i][counter[i]] = 0;
for (int j = counter[i] + 1; j < matrix[i].Length; j++)
{
double c = matrix[i - 1][j];
matrix[i][j] -= (c * b / a);
}
break;
}
}
}
/// <summary>
/// 将和0非常接近的数字视为0
/// </summary>
/// <param name="matrix"></param>
private static void Operation3(double[][] matrix)
{
for (int i = 0; i < matrix.Length; i++)
{
for (int j = 0; j < matrix[0].Length; j++)
{
if (Math.Abs(matrix[i][j]) <= 0.00001)
{
matrix[i][j] = 0;
}
}
}
}
/// <summary>
/// 计算行最简矩阵的秩
/// </summary>
/// <param name="matrix"></param>
/// <returns></returns>
private static int Operation4(double[][] matrix)
{
int rank = -1;
bool isAllZero = true;
for (int i = 0; i < matrix.Length; i++)
{
isAllZero = true;
//查看当前行有没有0
for (int j = 0; j < matrix[0].Length; j++)
{
if (matrix[i][j] != 0)
{
isAllZero = false;
break;
}
}
//若第i行全为0,则矩阵的秩为i
if (isAllZero)
{
rank = i;
break;
}
}
//满秩矩阵的情况
if (rank == -1)
{
rank = matrix.Length;
}
return rank;
}
static void Main(string[] args)
{
//示例矩阵1:秩为3
double[][] matrix1 = new double[][]
{
new double[] { 1, 1, 1 },
new double[] { 1, 1, 0 },
new double[] { 0, 1, 1 }
};
Console.WriteLine(Rank(matrix1));
//示例矩阵2:秩为3
double[][] matrix2 = new double[][]
{
new double[] { 3, 2, 0, 5, 0 },
new double[] { 3, -2, 3, 6, -1 },
new double[] { 2, 0, 1, 5, -3 },
new double[] { 1, 6, -4, -1, 4 }
};
Console.WriteLine(Rank(matrix2));
//示例矩阵3:秩为3
double[][] matrix3 = new double[][]
{
new double[] { 2, 3, 1, -3, -7 },
new double[] { 1, 2, 0, -2, -4 },
new double[] { 3, -2, 8, 3, 0 },
new double[] { 2, -3, 7, 4, 3 }
};
Console.WriteLine(Rank(matrix3));
Console.ReadLine();
}
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