tags:

views:

1252

answers:

4

I have two arrays for a chess variant I am coding in java...I have a console version so far which represents the board as a 1D array (size is 32) but I am working on making a GUI for it and I want it to appear as a 4x8 grid, so I have a 2-dimensional array of JPanels...

Question is, is there any formula that can convert the array[i][j] index into array[i] given the fact its a 4x8 array?

+3  A: 

i*8+j (assuming 8 is the horizontal width)

mr grumpy
+8  A: 

Given 4 columns by 8 rows then:

i = row * 4 + col

EDIT: My bad, nobody caught me on this mistake apparently. But it should actually be row * 4 + col. row * 8 + col would leave unnecessary gaps in the possible indexes.

Steve Wortham
I can't believe I didn't get that lol...I had (i*4)+(i*8) no wonder it was out of bounds, thankyouu!
Becky
You're very welcome. I had to do the exact same thing recently with a Tic Tac Toe game that learns. Storing past moves as a single index was a step towards improving efficiency.
Steve Wortham
+3  A: 

Every row in your 2D array is placed end to end in your 1D array. i gives which row you are in, and j gives the column (how far into that row). so if you are in the ith row, you need to place i complete rows end to end, then append j more onto that to get your single array index.

So it will be something like
singleDimIndex = array[0].length * i + j

Matt Greer
+1  A: 

Think of it this way:

You have one array that happens to be a 1 dimensional array, which really, is just a long concatenation of items of a two dimensional array.

So, say you have a two dimensional array of size 5 x 3 (5 rows, 3 columns). And we want to make a one dimensional array. You need to decide whether you want to concatenate by rows, or by columns, for this example we'll say the concatenation is by rows. Therefore, each row is 3 columns long, so you need to think of your one-dimensional array as being defined in "steps" of 3. So, the lengthy of your one dimensional array will be 5 x 3 = 15, and now you need to find the access points.

So, say you are accessing the 2nd row and the 2nd column of your two dimensional array, then that would wind up being 3 steps (first row) + the number of steps in the second row, or 3 + 2 = 5. Since we are zero-based indexing that is -1, so that would be at index 4.

Now for the specific formulation:

int oneDindex = (row * length_of_row) + column; // Indexes

So, as an example of the above you would wind up having

oneDindex = (1 * 3) + 1

And that should be it

Michael