Arithmetic operation so that 0, 1, & 2 return 0 | 3, 4, & 5 return 1, etc.

Question

Hello SO community!

I'm trying to take 9x9, 12x12, 15x15, etc. arrays and have the program interpret them as multiple 3x3 squares.

For example:

0 0 1 0 0 0 0 0 0
0 0 0 0 0 2 0 0 0
0 0 0 0 0 0 0 3 0
0 0 0 0 0 0 6 0 0
0 0 4 0 0 0 0 0 0
0 0 0 0 0 5 0 0 0
0 0 0 0 0 0 0 0 0
0 7 0 0 0 0 0 0 0
0 0 0 0 8 0 0 0 9

Will be understood as:

0 0 1 | 0 0 0 | 0 0 0
0 0 0 | 0 0 2 | 0 0 0
0 0 0 | 0 0 0 | 0 3 0
------+-------+------
0 0 0 | 0 0 0 | 6 0 0
0 0 4 | 0 0 0 | 0 0 0
0 0 0 | 0 0 5 | 0 0 0
------+-------+------
0 0 0 | 0 0 0 | 0 0 0
0 7 0 | 0 0 0 | 0 0 0
0 0 0 | 0 8 0 | 0 0 9

Where:

"1" @ [0][2] is in box "[0][0]"
"2" @ [1][5] is in box "[0][1]"
...
"6" @ [3][6] is in box "[1][2]"
...
"9" @ [8][8] is in box "[2][2]"

.

I can use row % 3 and column % 3 to determine the row and column values within the box, but how can I determine which box a given value in the array is stored in?

This formula could be used in a method such as the one below.

public int[] determineCoordinatesOfBox(int rowInArray, int columnColumnInArray) {
    // determine row value
    // determine column value

    // return new int[2] with coordinates
}

It seems possible and I've been beating my head over this. Perhaps I'm making a simple problem too difficult?

Many thanks for the help!

• Justian

Answer 1

You're looking for the / operator:

box[0] = rowInArray / 3;
box[1] = columnInArray / 3;

Answer 2

If I understand correctly, it's just simple integer division.

Since you're coding Java (it would be the same in at least C, C++ and C#), it's simply / operator:

int rowInArray = 3;
int columnInArray = 7;

int boxY = rowInArray / 3;    // will evaluate to 1
int boxX = columnInArray / 3; // will evaluate to 2

int rowInBox = rowInArray % 3;       // will evaluate to 0
int columnInBox = columnInArray % 3; // will evaluate to 1

Just keep both the arguments of division integer - 7 / 3 is 2, but 7 / 3.0 or 7.0 / 3 will be 2.5.