Calculating # or Rows and Columns

Question

I have a # of images that I\'m stitching together into a sprite sheet, how can I calculate the number of rows and columns to fit equally in an even rectangle (no blank space

Answer 1

Since it's unlikely you'll have a large number to begin with there are a lot of ways you can proceed in factoring it.

See Best way to find all factors of a given number in C# for some of them.

The simplest is: - Loop from 1 to the square root of the number, call the index "i".

• if number mod i is 0, add i and number / i to the list of factors.

This will give you all the integers that divide your number N. The "other" number is, of course, obtained by dividing N by that integer.

Then, you need to pick the best pair according to some rule. You can chose the ones with the smallest difference: if a * b = N, choose the ones with the smallest absolute value of (a-b)

Answer 2

Here's a very fast and easy algorithm (where N is the number of images)

rows = floor(sqrt(N))
while(N % rows != 0)
     rows = rows - 1

And rows will be the number of rows needed. Columns can obviously be found with N / rows.

I hope this helps!

Answer 3

Well looking at it if the number is prime you want to have 1 row with x columns where x is the prime number. Else if the number is a perfect square the rows would be the square root of the number by the square root of the number (9 == 3x3) . Else factor the remainder.

Answer 4

What's your priority? Do you want it to have the difference between height, width to be minimum, or anything like that?

Given the number n of images. You should take every number i from 1 to sqrt(n). If n can be divided by i (n%i ==0), divide and increment an array power[i] each time it divides. If n cannot be divided any more by i (aka n%i != 0) increment i else divide again.

You should get all the divisors and their biggest power in a given number n.

Make combinations of these and you will get the dimensions of your square.

Answer 5

Take a look into Integer Factorization

Maybe that is what you need.

Answer 6

Seems like you have to find all factor pairs of the number, then pick the pair the gives you the most 'desirable' row:column ratio.

So for example:

bestRows = 1
bestRatio = ((double) 1) / N;
for (int i : 1 to N) {
  if ((N % i) == 0) {
    r = N % i
    c = N / i
    ratio = ((double) r) / N;
    if (firstIsBetter(ratio, bestRatio)) {
      bestRows = r;
      bestRatio = ratio;
    }
  }
}