如何在方形网格中查找具有单元格编号的坐标

Question

I'm struggling to figure this out.. my brain is fogging up.

I have a 1000x1000 grid with the 0 in the bottom left (0,0) and top right would be (1000,1000)

I have 40000 squares (cells) that measure 5x5 and would like to calculate the coordinates given a number from 1-40000...

Let's say square 13227 would be coordinates x and y (starting points)

Thanks:)

Answer 1

I would probably approach it in this way. I'd first start the calc based on the 5x5 grids, which really means that you've got 200x200 of those in your 1000x1000 grid.

To find the row , get the floor - ie Math.floor() - of your number divided by 200.
In your example: Math.floor((13227/200)) = 66 . So the corresponding 5x5 grid (for 13227) is going to be in row 66 (of the 5x5 grids)

Then to get the column , subtract the (row number x 200) from the original number.
In your example: 13227 - (66 * 200) = 27 . So it's going to be the 27th column (of the 5x5 grids)

Then I believe you should just be able to multiply those values by 5 to give you the bottom left-hand coordinate within your 1000x1000 grid

That is: x = 27 * 5 (135), y = 66 * 5 (330)

(Now that's just me roughing something out on paper and in my head, so you'll no doubt be able to tell me if I'm right or not, The only thing I'm not 100% sure on is whether my method is actually giving the bottom left-hand coordinate of the 5x5 square, but I'm pretty sure it is)

Answer 2

It is a classic problem in programming, converting 2D coordinates to 1D, it is usually done from 2D to 1D where indices start from 0 to length-1

First you have to figure out if you are in row-major or column-major:

Most common case is row major. In this case imagine you have a 2D array with coordinates (x,y) in [0, width-1][0, height-1] and you want to convert to indices i in [0, width*height-1] .

In row major, 2D->1D:

i = y * width + c

In row major, 1D->2D:

y = i / width  (euclidian division)
x = i % width  (modulo operation which compute the remainder of euclidian division)

Answer 3

Let's say you start numbering cells at (0,0) (botton left). That's cell number 0 .
Next cell is the contiguous at right: c1(0,1). And next is c2(0,2).

When this row is full last cell is c1000(0,999), located at bottom right.

So, next cell begins in a new row: c1000(1,0). Next is c1001(1,1), etc.

Where is located cell c12345? Because you have a grid with rows length=1000 you must find how many rows fit in this number:

numberOfRows= Math.floor(12345/1000)

which gives numberOfRows=12. There's a rest in the division, rest=345 . So cell 12345 is in the row number 13 (the other 12 are already filled). And is located at column=345

Until now we've been calculating cells, now we want coordinates measured in distances.

Each cell is 5 units (inches or whatever). So cell 1 goes from 0 to 5, cell 2 from 5 to 10, and cell n goes from (n-1)*5 to n*5 . same calculus for vertical distance.

So cell 12345 corners, expressed in units, are (345-1)*5, 13*5 . Notice I've used the "rest" (current column) for the left side and "13" (current row) for the upper side.

The opposite problem is that you have coordinates (x,y) (in units, not in cells) and you want to know which cell corresponds to.
Dividing by the size of each cell you get row and column:

row = Math.floor(x/5)
col = Math.floor(y/5)

And knowing you have 1000 cells in a row you can get the cell number by:

cellNum = (row - 1)*1000 + col