How do you find the product of 2 matrices in C plus plus?

Answer 1

To multiply two matrices, the matrices must support the dot product format. That is, given the multiplication of matrices A * B = C, the dimensions of each array must be A[x][y], B[y][z] and C[x][z]. Note that the number of columns in A must be the same as the number of rows in B.

To calculate the results of elements in C, you are effectively summing the products of corresponding elements in a row of A and a column of B, hence there must be as many columns in A as there are rows in B.

This method of multiplication may seem odd at first, however let's begin with a simple example. Imagine you have a matrix containing the prices for a range of products:

Apple: 30p

Banana: 25p

Orange: 29p

This matrix would normally be represented as a one-dimensional array with three elements.

double A[3] {0.3, 0.25, 0.29};

But as a matrix, it is a two-dimensional array with one row:

double A[1][3] {{0.3, 0.25, 0.29}};

Now imagine you have another array that contains the daily sales quantities for each product:

Day: M T W T F S S

Apple: 5 6 4 3 8 6 7

Banana: 6 9 5 7 3 5 4

Orange: 7 9 4 7 3 5 6

This would be represented as a two dimensional array with three rows and seven columns:

size_t B[3][7] {{5, 6, 4, 3, 8, 6, 7},{6, 9, 5, 7, 3, 5, 4},{7, 9, 4, 7, 3, 5, 6}};

What you want is the total sales value of each day. Thus the total sales for Monday would be (0.3*5) + (0.25*6) + (0.29*7). Repeating this for the remaining days would reveal a one-dimensional array of 7 elements. But as a matrix, it is a two-dimensional array with just one row. Thus we have the following array declarations:

double A[1][3] {{0.3, 0.25, 0.29}};

size_t B[3][7] {{5, 6, 4, 3, 8, 6, 7},{6, 9, 5, 7, 3, 5, 4},{7, 9, 4, 7, 3, 5, 6}};

double C[1][7] {};

Where; x=1, y=3 and z=7.

To work out the totals, you need three nested loops:

for (size_t x=0; x!=1; ++x)

{

for (size_t z=0; z!=7; ++z)

{

for (size_t y=0; y!=3; ++y)

{

c[x][z] += a[x][y] * b[y][z];

}

}

}

Note the order of the loops: x, z and y (not x, y and z).

Now let's see a working example of this:

#include<iostream>

#include<random>

#include<time.h>

template<typename T>

void print_matrix (const T& a, const size_t rows, const size_t cols)

{

for (size_t r=0; r!=rows; ++r)

{

for (size_t c=0; c!=cols; ++c)

{

std::cout << a[r][c] << '\t';

}

std::cout << std::endl;

}

std::cout << std::endl;

}

int main()

{

double a[1][3] {{0.3, 0.25, 0.29}};

size_t b[3][7] {{5, 6, 4, 3, 8, 6, 7},{6, 9, 5, 7, 3, 5, 4},{7, 9, 4, 7, 3, 5, 6}};

double c[1][7] {};

for (size_t x=0; x!=1; ++x)

{

for (size_t z=0; z!=7; ++z)

{

for (size_t y=0; y!=3; ++y)

{

c[x][z] += a[x][y] * b[y][z];

}

}

}

std::cout << "\nProduct prices:\nApples\tOranges\tBananas\n"; print_matrix (a,1,3);

std::cout << "\nWeekly sales:\nMon\tTue\tWed\tThu\tFri\tSat\tSun\n"; print_matrix (b,3,7);

std::cout << "\nTotal sales:\nMon\tTue\tWed\tThu\tFri\tSat\tSun\n"; print_matrix (c,1,7);

}

Output:

Product prices:

Apples Oranges Bananas

0.3 0.25 0.29

Weekly sales:

Mon Tue Wed Thu Fri Sat Sun

5 6 4 3 8 6 7

6 9 5 7 3 5 4

7 9 4 7 3 5 6

Total sales:

Mon Tue Wed Thu Fri Sat Sun

5.03 6.66 3.61 4.68 4.02 4.5 4.84

Answer 2

The two input matrices must be of the same order and the return value must be another matrix of the same order. Each element of the return value is the sum of the corresponding elements of the input matrices.

#include<array>

// define the matrix type (an table of order 4)

const size_t order = 4;

using matrix = std::array<order, std::array<order, int>>;

// overload operator+ to cater for matrices: time complexity O(order*order)

matrix operator+ (const matrix& a, const matrix& b) {

matrix sum {}; // default initialised

for (size_t row=0; row<order; ++row)

for (size_t col=0; col<order; ++col)

sum[row][col] = a[row][col] + b[row][col];

return sum;

}

int main () {

matrix a {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}};

matrix b {{4, 3, 2, 1}, {8, 7, 6, 5}, {12, 11, 10, 9}, {16, 15, 14, 13}};

matrix c = a + b;

}

Answer 3

Use a valarray. A valarray is a specialised vector optimised for numeric computation.