
I wrote the program and i want some help pleas this program check between if they are related or not . i have no problem of reading file and inserted in 2d array but i have wrote program to sum the ones each rows and compared to the second matrix and so that about coulmns , note : two matrix are not equal ! ,, so i will compared the small one with the bigger one so that i use start point and end point , i need only help to write the two method which sum rows and coulmns,
4
**0 0 0 0
1 1 1 1
1 1 0 0
1 1 1 1**
6
0 1 1 1 1 0
0 0 0 0 0 0
1 1 1 1 1 0
1 1 1 0 0 0
1 1 1 1 1 1
1 1 1 1 1 1
my work is
public static void main(String[] args)
{
try
{
Scanner sc = new Scanner(new File("zoo"));
int caseOne = sc.nextInt();
int count=0 ;
while(caseOne>=count)
{
//read the first matrix size.
int length1=sc.nextInt();
int sum=0;
System.out.println(length1);
int[][] m1 = new int[length1][length1];
// print array in rectangular form
for (int r=0; r<m1.length; r++)
{
for (int c=0; c<m1[r].length; c++)
{
m1[r][c]=sc.nextInt();
System.out.print(" " + m1[r][c]);
}
System.out.println("");
}
// case two
int length2=sc.nextInt();
System.out.println(length2);
int[][] m2 = new int[length2][length2];
// print array in rectangular form
for (int r=0; r<m2.length; r++)
{
for (int c=0; c<m2[r].length; c++)
{
m2[r][c]=sc.nextInt();
{
System.out.print(" " + m2[r][c]);
}
}
for (int a=0; a<m2.length; a++)
{
for (int c=0; c<m2[a].length; c++)
{
int startPoint= m2[r][c]-m1[r][c]/2;
System.out.print(" " + startPoint);
int EndPoint= m2[r][c]-startPoint;
System.out.print(" " + EndPoint);
}
System.out.println("");
}
count++;
}
}
}
catch (Exception e)
{
System.out.println(e);
}
}
public static int sumRow( int[][] m1,int startPoint,int ei, int A[])
{
return A[i];
}
public static int sumcol(int m2[][],int si,int ei, int B[])
{
int i=0 ;
return B[i];
}
}

sorry for my weak english
If you are facing any programming issue, such as compilation errors or not able to find the code you are looking for.
Ask your questions, our development team will try to give answers to your questions.