I have N rectangular items with an aspect ratio Aitem (X:Y).
I have a rectangular display area with an aspect ratio Aview
The items should be arranged in a table-like layout (i.e. r rows, c columns).
what is the ideal grid rows x columns, so that individual items are largest? (rows * colums >= N, of course - i.e. there may be "unused" grid places).
A simple algorithm could iterate over rows = 1..N, calculate the required number of columns, and keep the row/column pair with the largest items.
I wonder if there's a non-iterative algorithm, though (e.g. for Aitem = Aview = 1, rows / cols can be approximated by sqrt(N)).
Note: I couldn't quite understand Frédéric's answer, so I worked the problem out myself and came up with what appears to be the same solution. I figured I might as well explain what I did in case it is helpful.
First I normalized the aspect ratio of the view to that of the items. (I'm assuming you don't want to rotate the items.)
a = (view_width/view_height) / (item_width/item_height)
Now packing a rectangle of width/height ratio a with squares is equivalent to packing the view with items. The ideal case would be for our grid (of squares now) to fill the rectangle completely, which would give us
a = c/r
where r and c are the numbers of rows and columns:
N = r*c
Multiplying/dividing these two equations gives us
N*a = c^2 N/a = r^2
c = sqrt(N*a) r = sqrt(N/a)
If the grid is perfect, r and c will be integers, but if not, you have to try the three options Frédéric mentioned and keep the one where r*c is smallest but still more than N:
floor(r), ceil(c)ceil(r), floor(c)ceil(r), ceil(c)