The matrix G I4 A where is G 1 0 0 0 0 1 0 0 0 0 1 0 0 0

The matrix G = [I_4 | A], where is G = [1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1| 0 1 1 1 0 1 1 1 0 1 1 1] is a generator matrix in standard form for a [7, 4] binary code that we denote by H_3. By Theorem 1.2.1 a parity check matrix for H_3, is H = [A^T |I_3] = [0 1 1 1 1 0 1 1 1 1 0 1| 1 0 0 0 1 0 0 0 1]. This code is called the [7, 4] Hamming code. Find at least four information sets in the [7, 4] code H_3 from Example 1.2.3. Find at least one set of four coordinates that do not form an information set.

Solution

import java.util.Scanner; class MatrixMultiplication { public static void main(String args[]) { int m, n, p, q, sum = 0, c, d, k; Scanner in = new Scanner(System.in); System.out.println(\"Enter the number of rows and columns of first matrix\"); m = in.nextInt(); n = in.nextInt(); int first[][] = new int[m][n]; System.out.println(\"Enter the elements of first matrix\"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) first[c][d] = in.nextInt(); System.out.println(\"Enter the number of rows and columns of second matrix\"); p = in.nextInt(); q = in.nextInt(); if ( n != p ) System.out.println(\"Matrices with entered orders can\'t be multiplied with each other.\"); else { int second[][] = new int[p][q]; int multiply[][] = new int[m][q]; System.out.println(\"Enter the elements of second matrix\"); for ( c = 0 ; c < p ; c++ ) for ( d = 0 ; d < q ; d++ ) second[c][d] = in.nextInt(); for ( c = 0 ; c < m ; c++ ) { for ( d = 0 ; d < q ; d++ ) { for ( k = 0 ; k < p ; k++ ) { sum = sum + first[c][k]*second[k][d]; } multiply[c][d] = sum; sum = 0; } } System.out.println(\"Product of entered matrices:-\"); for ( c = 0 ; c < m ; c++ ) { for ( d = 0 ; d < q ; d++ ) System.out.print(multiply[c][d]+\"\\t\"); System.out.print(\"\ \"); } } } }