Algorithm to get the maximum size of n squares that fit into a rectangle with a given width and height

Question

I am looking for an algorithm that can return the number of size of n squares that fit into a a rectangle of a given width and height, maximizing the use of space (thus, leaving the least amount of leftover space for squares that do not fit). Neither the rectangle nor the squares can be rotated.

For example, let's say I have a rectangle that is 5 inches by 7 inches, and I need it to fit 35 squares. The algorithm needs to tell me that the 35 squares fit if they are 1 inch wide/tall (as they could be laid out inside the rectangle in a 5 x 7 grid). Another example is if I need to divide a rectangle 35 inches by 1 inch into 35 squares. It should still tell me that the squares will fit if they are 1-inch wide/tall (as they could be laid out in the rectangle in a 35 x 1 grid).

The tricky part is that sometimes there may be leftover space, as the squares cannot be divided into partial squares. Let's say for either of the two examples above I need to lay out 34 squares and not 35 (in which case the answers would still be 1 inch), or maybe 33, or 7 squares. Or, perhaps the rectangle width and height aren't whole numbers. With the number of squares being a variable I need an algorithm that can tell me the size of the squares for a given rectangle width and height.

Thanks for your help!

Answer 1

Let's analyze a single case:

If we want to cover a $6x5$ rectangle with $7$ squares, we first note that each square at most has an area of $30/7$. We also know that for our filling to be optimal, we need to fill at least an axis.

So now we know that we will fill either the $6$ or the $5$ completely. We will first try with the 5. Since the maximal side of our squares is $\sqrt{30/7}$, we now need the smallest integer more than $5/\sqrt{30/7}$, that is equal to $\lceil5/\sqrt{30/7}\rceil=3$

So we would divide the $5$ in $3$ parts $p_x$, so each side will have $5/3$ . If so, I would cover $(5/3)^2*7=19\frac49$ of the thirty squares. If they don't fit vertically, we try with more parts $p_x$, until the squares fit in the other axis. $$\lfloor6/(5/p_x)\rfloor*\lfloor5/(5/p_x)\rfloor=\lfloor6/(5/p_x)\rfloor*p_x\ge7 \,\,\,\,\,\,\,\,\,\, (5/p_x=\text{the actual size of our squares)}$$ Also, since $p_x$ is integer $\lfloor p_x \rfloor =p_x$

If we do the same with the $6$, we get $\lceil6/\sqrt{30/7}\rceil$ again covering $19\frac49$ of the surface. We then do the same as with the $5$ and save it as $p_y$

Let's now recall the formulas we had.

We had an $x∗y$ rectangle and filled it with $N$ squares. We started with $p_x=⌈x/\sqrt{xy/N}⌉$=$⌈\sqrt{Nx/y}⌉$ or $p_y=⌈y/\sqrt{xy/N}⌉=⌈\sqrt{Ny/x}⌉$ and then we made them fit by shrinking them until they fit in the other axis. But we know we want the area maximised, so $Side=max(x/p_x,y/p_y)$.

A better method:

Once we have the start value $p_x=⌈x/\sqrt{xy/N}⌉=⌈\sqrt{Nx/y}⌉$,if it does not fit in the $y$ axis, we need to make it fit in the $y$ axis. For that, we use that $$a=x/p_x$$ where $a$ is the value of the actual side of our squares. Then we know that for our side to fit we need to reduce it: $$S_x=y/(\lceil y/a \rceil)$$ We do the same with $S_y$ and calculate $max(S_x,S_y)$ The advantage of this is that it needs a constant number of operations, the most expensive one being the square root.

Plain, unoptimized C Code

Some multiplications may be reused but for code readability and practical uses this is enough. Input: values of $x$,$y$, and $n$. Handcoded.

Output: The optimal side of our squares

#include <math.h>
#include <stdio.h>
#define MAX(x,y)    (x)>(y)?(x):(y) 
int main(){
    double x=5, y=6, n=7;//values here
    double px=ceil(sqrt(n*x/y));
    double sx,sy;
    if(floor(px*y/x)*px<n)  //does not fit, y/(x/px)=px*y/x
            sx=y/ceil(px*y/x);
    else
            sx= x/px;
    double py=ceil(sqrt(n*y/x));
    if(floor(py*x/y)*py<n)  //does not fit
            sy=x/ceil(x*py/y);
    else
            sy=y/py;
    printf("%f",MAX(sx,sy));
    return 0;
}

Answer 2

Credit for the algorithm is entirely from chubakueno's answer. I'm just posting a C# version of it here in case anyone else needs it.

/// Calculates the optimal side of squares to be fit into a rectangle
/// Inputs: x, y: width and height of the available area.
///         n: number of squares to fit
/// Returns: the optimal side of the fitted squares
double FitSquares(double x, double y, int n)
{
    double sx, sy;

    var px = Math.Ceiling(System.Math.Sqrt(n * x / y));
    if (Math.Floor(px * y / x) * px < n) {
        sx = y / Math.Ceiling(px * y / x);
    } else {
        sx = x / px;
    }

    var py = Math.Ceiling(System.Math.Sqrt(n * y / x));
    if (Math.Floor(py * x / y) * py < n) {
        sy = x / Math.Ceiling(x * py / y);
    } else {
        sy = y / py;
    }

    return Math.Max(sx, sy);
}

Answer 3

chubakueno's solution doesn't always work. Take for example $x=10$, $y=2$ and $n=8$. His algorithm returns 1 whereas the biggest square size is obtained by fitting one single line of eight squares, and is therefore $10/8=1.25$.

Below is an algorithm that works, written in Javascript. It also has the advantage of computing the minimum number of rows and columns, on top of the square size. Inputs are x, y and n.

// Compute number of rows and columns, and cell size
var ratio = x / y;
var ncols_float = Math.sqrt(n * ratio);
var nrows_float = n / ncols_float;

// Find best option filling the whole height
var nrows1 = Math.ceil(nrows_float);
var ncols1 = Math.ceil(n / nrows1);
while (nrows1 * ratio < ncols1) {
    nrows1++;
    ncols1 = Math.ceil(n / nrows1);
}
var cell_size1 = y / nrows1;

// Find best option filling the whole width
var ncols2 = Math.ceil(ncols_float);
var nrows2 = Math.ceil(n / ncols2);
while (ncols2 < nrows2 * ratio) {
    ncols2++;
    nrows2 = Math.ceil(n / ncols2);
}
var cell_size2 = x / ncols2;

// Find the best values
var nrows, ncols, cell_size;
if (cell_size1 < cell_size2) {
    nrows = nrows2;
    ncols = ncols2;
    cell_size = cell_size2;
} else {
    nrows = nrows1;
    ncols = ncols1;
    cell_size = cell_size1;
}