nested loop??

Hi,
Im new to programming and ive just set up a sudoku GUI in Java.
Basically my goal is to be able to solve a sudoku puzzle, but i like the graphics supported by java. I have a 9x9 array holding instances of JCombobox.

I have a set up JButton, that when pressed loops through the array and if any items have been selected in an instance of JCombobox it sets the Enable() methos to false.

Then at the moment my "solve" button loops through each row and column to see if any JComboBox has been deactivate, if so it gets the selected item from that instance of the combobox and removes the selected item from the ArrayList of every other JComboBox in that row/column.

i.e.
for(int i=0; i<9; i++)
for(int j=0; j<9; j++)
{
if(!cb_Array[i][j].isEnabled())
for(int k=0; k<9; k++)
cb_Array[i][k].avail_list((String)cb_Array[i][j].getSelectedItem());
}

searches through the rows and removes selected items form the available number list( (Integer)ArrayList ) of every other JComboBox istance in that particular row.

What Im having trouble with is doing something similar for the 3x3 squares.

If the [i][j] coordinates for the first (top left corner) 3x3 are as follows:

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

how do i loop throught it?

Welcome to the 'ranch. This forum is for fun little puzzles. This kind of question ought to go in Java In General( beginner ).

Now, I assume you want to loop through all 9 3x3's- you already know how to loop through the example you gave.

The most self-explanatory way would just be

for( int x = 0; x < 3; x++ ){ for( int y = 0; y < 3; y++ ){ //these loop through the all the 3x3s // all code past this point is per 3x3 box. for( int xx = 0; xx < 3; xx++ ){ for( int yy = 0; yy < 3; yy++ ){ // these loop through a 3x3 box. cb_Array[ x * 3 + xx ][ y * 3 + yy ].doFunStuff(); } } } }


I think this is what you're asking.


(See Jim? I'm giving back to the community for your answers)
(Me an' Jim are on a first-name basis)
[ January 17, 2006: Message edited by: Nick George ]

Who is this "Nick" person, anyway?

This question already appears here in JiG(b). We should continue the discussion there if necessary, but I think the OP seems to have figured it out.

Layne