Fully cover a rectangle with minimum amount of fixed radius circles

Question

I\'ve had this problem for a few years. It was on an informatics contest in my town a while back. I failed to solve it, and my teacher failed to solve it. I haven\'t met anyone

Answer 1

The hexagon is better than the diamond. Consider the percent area of the unit circle covered by each:

#!/usr/bin/env ruby

include Math

def diamond
  # The distance from the center to a corner is the radius.
  # On a unit circle, that is 1.
  radius = 1

  # The edge of the nested diamond is the hypotenuse of a
  # right triangle whose legs are both radii.
  edge = sqrt(radius ** 2 + radius ** 2)

  # The area of the diamond is the square of the edge
  edge ** 2
end

def hexagon
  # The hexagon is composed of 6 equilateral triangles.
  # Since the inner edges go from the center to a hexagon
  # corner, their length is the radius (1).
  radius = 1

  # The base and height of an equilateral triangle whose
  # edge is 'radius'.
  base = radius
  height = sin(PI / 3) * radius

  # The area of said triangle
  triangle_area = 0.5 * base * height

  # The area of the hexagon is 6 such triangles
  triangle_area * 6
end

def circle
  radius = 1
  PI * radius ** 2
end

puts "diamond == #{sprintf "%2.2f", (100 * diamond / circle)}%"
puts "hexagon == #{sprintf "%2.2f", (100 * hexagon / circle)}%"

And

$ ./geometrons.rb 
diamond == 63.66%
hexagon == 82.70%

Further, regular hexagons are highest-vertex polygon that form a regular tessellation of the plane.

Answer 2

According my calculations the right answer is:

D=2*R; X >= 2*D, Y >= 2*D,
N = ceil(X/D) + ceil(Y/D) + 2*ceil(X/D)*ceil(Y/D)

In particular case if the remainder for X/D and Y/D equal to 0, then

N = (X + Y + X*Y/R)/D

Case 1: R = 1, X = 2, Y = 2, then  N = 4

Case 2: R = 1, X = 4, Y = 6, then  N = 17

Case 3: R = 1, X = 5, Y = 7, then  N = 31

Hope it helps.

Answer 3

This site attacks the problem from a slightly different angle: Given n unit circles, what is the largest square they can cover?

As you can see, as the number of circles changes, so does the covering pattern.

For your problem, I believe this implies: different rectangle dimensions and circle sizes will dictate different optimal covering patterns.

Answer 4

When the circles are disposed as a clover with four leafs with a fifth circle in the middle, a circle will cover an area equal to R * 2 * R. In this arrangement, the question becomes: how many circles that cover an area of R * 2 * R will cover an area of W * H?, or N * R * 2 * R = W * H. So N = W * H / R * 2 * R.

Answer 5

For X and Y large compared to R, a hexagonal (honeycomb) pattern is near optimal. The distance between the centers of the circles in the X-direction is sqrt(3)*R. The distance between rows in the Y-direction is 3*R/2, so you need roughly X*Y/R^2 * 2*/(3*sqrt(3)) circles.

If you use a square pattern, the horizontal distance is larger (2*R), but the vertical distance is much smaller (R), so you'd need about X*Y/R^2 * 1/2 circles. Since 2/(3*sqrt(3) < 1/2, the hexagonal pattern is the better deal.

Note that this is only an approximation. It is usually possible to jiggle the regular pattern a bit to make something fit where the standard pattern wouldn't. This is especially true if X and Y are small compared to R.

In terms of your specific questions:

1. The hexagonal pattern is an optimal covering of the entire plane. With X and Y finite, I would think it is often possible to get a better result. The trivial example is when the height is less than the radius. In that case you can move the circles in the one row further apart until the distance between the intersecting points of every pair of circles equals Y.

2. Having a regular pattern imposes additional restrictions on the solution, and so the optimal solution under those restrictions may not be optimal with those restrictions removed. In general, somewhat irregular patterns may be better (see the page linked to by mbeckish).

3. The examples on that same page are all specific solutions. The solutions with more circles resemble the hexagonal pattern somewhat. Still, there does not appear to be a closed-form solution.