
矩阵乘法
现在讨论矩阵乘法的过程。矩阵乘法只有满足一定的条件才能进行。假设两个矩阵是 P 和 Q,它们的维度是 P (a x b) 和 Q (z x y),当且仅当 b = x 时才能找到结果矩阵。那么所得矩阵 R 的阶将为 (m x q)。
算法
matrixMultiply(P, Q):
Assume dimension of P is (a x b), dimension of Q is (z x y)
Begin
if b is not same as z, then exit
otherwise define R matrix as (a x y)
for i in range 0 to a - 1, do
for j in range 0 to y – 1, do
for k in range 0 to z, do
R[i, j] = R[i, j] + (P[i, k] * Q[k, j])
done
done
done
End矩阵标准化
Suppose we have a 2x3 matrix: 4 5 6 1 2 3 The normalized matrix would be: 4/sqrt(pow(5,2) + pow(6,2)) 5/sqrt(pow(5,2) + pow(6,2)) 6/sqrt(pow(5,2) + pow(6,2)) 1/sqrt(pow(2,2) + pow(3,2)) 2/sqrt(pow(2,2) + pow(3,2)) 3/sqrt(pow(2,2) + pow(3,2))
示例
#include#include int main() { int row, col, row1, col1; int assignMatrix[50][50], rowAdd[100] = {0}; long long int a, square[50] = {0}; double norm[50][50], k; printf("Enter size of a matrix "); scanf("%d %d", &row, &col); printf("Enter matrix of size %dX%d
", row, col); for ( row1 = 0; row1 < row; row1++) { for (col1 = 0; col1 < col; col1++) { scanf("%d", &assignMatrix[row1][col1]); } } printf("
rows: %d cols: %d elements:
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",row,col); for( row1 = 0; row1 < row; row1++) { for( col1 = 0; col1 < col; col1++) { printf("%d ", assignMatrix[row1][col1]); } printf("
"); } for (row1 = 0; row1 < row; row1++) { for (col1 = 1; col1 < col; col1++) { a = assignMatrix[r][c]; square[row1] += a * a; } printf("Sum of squares of row %d: %lld
",row1,square[row1]); } for ( row1 = 0; row1 < row; row1++ ) { k = 1.0 / sqrt(square[row1]); for( col1 = 0; col1 < col; col1++ ) { norm[row1][col1] = k * assignMatrix[row1][col1] ; } } printf("
Normalized Matrix:
"); for( row1 = 0; row1 < row; row1++) { for( col1 = 0; col1 < col; col1++) { printf("%.3lf ", norm[row1][col1]); } printf("
"); } return 0; }
输出
Enter size of a matrix 2 3 Enter matrix of size 2X3 4 5 6 1 2 3 rows: 2 cols: 3 elements: 4 5 6 1 2 3 Sum of squares of row 0: 61 Sum of squares of row 1: 13 Normalized Matrix: 0.512 0.640 0.768 0.277 0.555 0.832 Process returned 0 (0x0) execution time : 12.446 s Press any key to continue.










