Multiplying Matrices with two for loops in C++ [duplicate] - c++

Question

I came up with this algorithm for matrix multiplication. I read somewhere that matrix multiplication has a time complexity of o(n^2).
But I think my this algorithm will give o(n^3).
I don't know how to calculate time complexity of nested loops. So please correct me.
for i=1 to n
for j=1 to n
c[i][j]=0
for k=1 to n
c[i][j] = c[i][j]+a[i][k]*b[k][j]

Answer 1

Using linear algebra, there exist algorithms that achieve better complexity than the naive O(n3). Solvay Strassen algorithm achieves a complexity of O(n2.807) by reducing the number of multiplications required for each 2x2 sub-matrix from 8 to 7.
The fastest known matrix multiplication algorithm is Coppersmith-Winograd algorithm with a complexity of O(n2.3737). Unless the matrix is huge, these algorithms do not result in a vast difference in computation time. In practice, it is easier and faster to use parallel algorithms for matrix multiplication.

Answer 2

The naive algorithm, which is what you've got once you correct it as noted in comments, is O(n^3).
There do exist algorithms that reduce this somewhat, but you're not likely to find an O(n^2) implementation. I believe the question of the most efficient implementation is still open.
See this wikipedia article on Matrix Multiplication for more information.

Answer 3

The standard way of multiplying an m-by-n matrix by an n-by-p matrix has complexity O(mnp). If all of those are "n" to you, it's O(n^3), not O(n^2). EDIT: it will not be O(n^2) in the general case. But there are faster algorithms for particular types of matrices -- if you know more you may be able to do better.

Answer 4

In matrix multiplication there are 3 for loop, we are using since execution of each for loop requires time complexity O(n). So for three loops it becomes O(n^3)

Answer 5

I recently had a matrix multiplication problem in my college assignment, this is how I solved it in O(n^2).
import java.util.Scanner;
public class q10 {
public static int[][] multiplyMatrices(int[][] A, int[][] B) {
int ra = A.length; // rows in A
int ca = A[0].length; // columns in A
int rb = B.length; // rows in B
int cb = B[0].length; // columns in B
// if columns of A is not equal to rows of B, then the two matrices,
// cannot be multiplied.
if (ca != rb) {
System.out.println("Incorrect order, multiplication cannot be performed");
return A;
} else {
// AB is the product of A and B, and it will have rows,
// equal to rown in A and columns equal to columns in B
int[][] AB = new int[ra][cb];
int k = 0; // column number of matrix B, while multiplying
int entry; // = Aij, value in ith row and at jth index
for (int i = 0; i < A.length; i++) {
entry = 0;
k = 0;
for (int j = 0; j < A[i].length; j++) {
// to evaluate a new Aij, clear the earlier entry
if (j == 0) {
entry = 0;
}
int currA = A[i][j]; // number selected in matrix A
int currB = B[j][k]; // number selected in matrix B
entry += currA * currB; // adding to the current entry
// if we are done with all the columns for this entry,
// reset the loop for next one.
if (j + 1 == ca) {
j = -1;
// put the evaluated value at its position
AB[i][k] = entry;
// increase the column number of matrix B as we are done with this one
k++;
}
// if this row is done break this loop,
// move to next row.
if (k == cb) {
j = A[i].length;
}
}
}
return AB;
}
}
#SuppressWarnings({ "resource" })
public static void main(String[] args) {
Scanner ip = new Scanner(System.in);
System.out.println("Input order of first matrix (r x c):");
int ra = ip.nextInt();
int ca = ip.nextInt();
System.out.println("Input order of second matrix (r x c):");
int rb = ip.nextInt();
int cb = ip.nextInt();
int[][] A = new int[ra][ca];
int[][] B = new int[rb][cb];
System.out.println("Enter values in first matrix:");
for (int i = 0; i < ra; i++) {
for (int j = 0; j < ca; j++) {
A[i][j] = ip.nextInt();
}
}
System.out.println("Enter values in second matrix:");
for (int i = 0; i < rb; i++) {
for (int j = 0; j < cb; j++) {
B[i][j] = ip.nextInt();
}
}
int[][] AB = multiplyMatrices(A, B);
System.out.println("The product of first and second matrix is:");
for (int i = 0; i < AB.length; i++) {
for (int j = 0; j < AB[i].length; j++) {
System.out.print(AB[i][j] + " ");
}
System.out.println();
}
}
}