optimized grid for rectangular items

Question

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)).

Answer 1

Good question. If your view has dimensions A x B (fixed) and your items have dimensions a x b (variable, to be maximized), then you need:

trunc(A / a) * trunc(B / b) >= N

I have no idea how to solve this though - the trunc is the tricky part, as it's non-linear.

Answer 2

Your solution can be easily improved to handle the generic case :

If we (temporary) forget the need to have an integer number of rows and columns, we have

rows * columns = N

x = aitem * y

aview = rows * x = rows * aitem * y

1 = columns * y = (N/rows) * ( aview / [aitem*rows]) = N * aview /(aitem * rows²)

hence rows=sqrt(N *aview/aitem) and columns = N/rows = sqrt(N * aitem / aview)

Then ceil(rows) and ceil(columns) is a solution while floor(rows) and floor(columns) are too small to be a solution together (if rows and columns are not integers). This leaves 3 possible solutions :

• floor(rows) ceil(columns)
• ceil(rows) floor(columns)
• ceil(rows) ceil (columns)

edited to correct the equations. The first result was wrong (see comments)

Answer 3

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)