Question

根据下面提到的pdf链接，我需要计算给定矩阵'A'和
的矩阵'x' 使用后向替换的上三角矩阵的矩阵'b'。

链接：http://www.mathcs.emory.edu/~haber/math315/chap3.pdf

实际上我只需要使用1维数组。我也为它开发了逻辑并尝试了编译它，但实际上它没有“表达式”和“初始化”的错误 | 1 2 3 | | X0 | | B0 | | 0 4 5 | x | x1 | = | b1 | | 0 0 6 | | X2 | | B2 | 公式：
1）6 * x2 = b2 2）4 * x1 + 5 * x2 = b1 3）1 * x0 + 2 * x1 + 3 * x3 = b0

This is my code:
//here: 's' is size of matrix eg: for 3x3 it is 3
//x[] i need to get the result
//b[] it is to be multiplied with cm to get 
//cm[] is the matrix which will be multiplied with b[] to get x[]

    for(int p=s-1; p>0; p--)
        { 
        if(p==s-1)
            {
            x[p] = b[p]/cm[s*s];                        // computing x2
            }
        else
        {
                for(int j=0; int k=s-p; j<s-i; k>0; j++; k--)
                    {
                        c[p]+ = cm[s*p - j]*x[p+k];
                }
            }

        x[p] = (b[p] - c[p])/cm[s*p-(s-i)];
        }
Errors: 
1) variable 'x' may not be initialized
2) variable 'c' may not be initialized
3) expression for(int j=10; intk=s-p;j<s-i;k>0;j++;k--) has no effect
4) expected a ")" in point 3.
5) expected an expression in line  c[p]+ = cm[s*p - j]*x[p+k];
6) variable 'x' was set but never used

Please help me how to solve these errors?
Also let me know is my logic correct?

Answer 1

我不知道你是否已声明你的变量，如果没有那么请你这样做，你应该写下你的第二个for循环，就像我在下面写的那样......

for(int j=0,int k=s-p; j<s-i, k>0; j++, k--)

因为多个初始化，条件和增量必须用逗号分隔。

Answer 2

我希望仍然需要答案。代码的粗略概念是正确的，但是您有一些未定义的变量（如 i ）以及有关使用列主矩阵的操作的一些错误：行

x[p] = b[p]/cm[s*s];

实际上必须是：

x[p] = b[p]/cm[s*s- 1];

只要我们在C谈话。无论如何，我只是发布有效的代码：

void backward_substitution (double *A, double *b, double *x, int rows)
{
//here: 'rows' is size of matrix eg: for 3x3 it is 3
//x[] where we put the result of backward substitution
//b[] it is the vector to be used.
//A[] is the square triangular matrix.

register int i,j;

// x(m,1) = b(m)/A(m,m); %% this is for the last row.
x[rows-1] = b[rows-1]/A[(rows-1)*rows + (rows-1)];

// for i = m-1:-1:1     % i starts at 2, end at 4, w/ steps of 1
//    x(i,1) = ( b(i)- A(i,i+1:m)*x(i+1:m))  /  A(i,i);
// end

 for (i=(rows-2); i>=0; i--)
    {
      x[i] = b[i];        
        for (j=i+1; j<rows; j++)
        {
             x[i] -= A[(j)*rows + i]*x[j];
        };             
      x[i] = x[i] / A[(i)*rows + i];
    }; //for (i=1; i<rows; i++)
}; //void forward_substitution(double *A, double b*, int n)

为方便起见，这些注释包含MATLAB中的实际实现。享受。

在给定矩阵'a'和矩阵'b'的情况下，使用向后替换上三角矩阵计算向量x

2 个答案: