CODEWITHSUNDEEP
java
Ryne Ignelzy
Ryne Ignelzy

Reputation: 137

Multiplication in Java with recursion

I am sorry, I am not sure if the title is correct, if it is not I will correct it once someone tells me what this is called. As you can understand I am new to programming...

I want to accomplish the following: I have a cycle:

for(int i=0; i < this.matrix.length; i++)

I will have a matrix for example like this:

1, 2, 2
2, 2, 3
0, 1, 2

I want to multiply the diagonal elements 1*2*2 I know how to get those elements each step of the cycle, but how can I used a temp variable, that every step will be multiplied by the new element? or is this not possible?

For example i make a variable: 

double temp;

and each cycle step I want the new value to multiply by the old, but keeping the value, not sure if I am explaining this well. But if we use this matrix i would want something like this:

temp = 1;

next step it

temp = 2;

next step

temp = 4;

I tried doing this myself but in the end would get the wrong results, I understand I am doing the multiplication wrong, because when i changed the 2 2 element of the matrix to 3 instead of 2 my end result would be 9 instead of 6. I am sorry if this is badly explained...

Upvotes: 1

Views: 458

Answers (6)

Frank
Frank

Reputation: 881

As your question title says , you wanna moltiply with recursion .

The possibility of a function to call itself is known as recursion.

public static void main(String[] args) {
    // TODO Auto-generated method stub

    int[][]  array= {
            {1, 2, 2},
            {2, 2, 3},
            {0, 1, 2}
            };

            if(array[0].length==array.length)   // check if numbers of  columns == rows 
            System.out.println("Result "+multiD(0, array));
            else
            System.out.println("No matrix NxN");
}

public static int multiD(int pos, int [][] m) { 
        if (pos == m.length ) {//get out after the last element (multiply for 1 )
            return 1;
        } else
            if ( m[pos][pos] == 0 ) {// get out if we found a 0 value , so we don't need to go forward
                return 0;
            } else
        return  m[pos][pos] * multiD(pos+1,m); //calculate the result 
    }

Assuming you have a matrix "m" 1000x1000 and the m[0][0] is 0 , you will iterate 1000 times when the result is known at the beginning . To prevent this you should write something likethis :

(Missing in the other answers)

int result = 1;
    for (int i=0; i<matrix.length; i++) {
        result *= matrix[i][i];
        if(result == 0) 
             break;
    }

Upvotes: 0

albert_nil
albert_nil

Reputation: 1658

  1. In your question you are only requesting for the main left-to-right diagonal output, so i'm assuming this is your only goal.
  2. Also, you are not specifying if the matrix will always be square or not; i will assume yes.
  3. Lastly, you are not specifying how this matrix is stored exactly in the variable. I'm assuming we are talking about a bidimensional array.

Here we go:

public static void main (String[] args) throws Exception {
    int[][] matrix = new int[3][];
    matrix[0] = new int[] {1, 2, 2};
    matrix[1] = new int[] {2, 2, 3};
    matrix[2] = new int[] {0, 1, 2};

    int result = 1;
    for (int i=0; i<matrix.length; i++) {
        result *= matrix[i][i];
    }
    System.out.println(result);
}

Edit: If you want also to include right-to-left:

public static void main (String[] args) throws Exception {
    int[][] matrix = new int[3][];
    matrix[0] = new int[] {1, 2, 2};
    matrix[1] = new int[] {2, 2, 3};
    matrix[2] = new int[] {0, 1, 2};

    int resultL2R = 1;
    int resultR2L = 1;
    for (int i=0; i<matrix.length; i++) {
        resultL2R *= matrix[i][i];
        resultR2L *= matrix[i][matrix.length-1-i];
    }
    System.out.println("left-to-right: " + resultL2R);
    System.out.println("right-to-left: " + resultR2L);
}

Upvotes: 2

Mayank Raj
Mayank Raj

Reputation: 32

Ok I think this code does what you want for the above matrix:

int temp=1;
for(int r=0; r<3; r++)//traversing through row
{
  for(int c=0; c<3; c++)//traversing through column
   {
     if(r==c)// condition for diagonal
       temp*=array[r][c];
   }// c close
  System.out.println("Multiplication value after row "+(r+1)+" = "+temp);
}// r close

Upvotes: 0

Kushagra Saxena
Kushagra Saxena

Reputation: 71

For Diagonal multiplication you can use below mentioned code

public static void main(String[] args) {
    //2D Array
    int a[][]={{1,2,3},{2,3,4},{3,4,5}};
    int multiplier=1;
    for(int i=0;i<a.length;i++){
        multiplier=multiplier*a[i][i];  
    }
    System.out.println(multiplier);
}

Upvotes: 1

Federico Paparoni
Federico Paparoni

Reputation: 722

You have to multiply in the loop for the [i][i] element

int[][]  array= {
   {1,2,2},
   {2,2,3},    
   {0,1,2}
};

int result=1;
for ( int i = 0; i < array.length ; i++) {
    result=result*(array[i][i]);
}
System.out.println("Result "+result);

Upvotes: 1

Alexey R.
Alexey R.

Reputation: 8676

I guess you would like to have the solution like this:

public static void main (String [] args)
{
    int[][] matrix = new int[][] {
            {1, 2, 2},
            {2, 2, 3},
            {0, 1, 2}
    };

    int result = 1;

    for(int i = 0; i < matrix.length; i++){
        result = result * matrix[i][i];
    }

    System.out.println("Result: " + result);

}

Since you're declaring the result variable before you get into for loop, it will preserve the value evaluated inside the loop.

Upvotes: 1

Related Questions