So, that's 3 more circles with 6 points each, or 18 more points.Īdding this result to the previous 9, we get 27 points. So, the first circle has given us 9 such points.Įach of the other circles is going to add another 6 such points (not 9, because the last set of 3 from the previous circle is the same as the first set of 3 from the new circle). Each of these would be exactly 25 from its nearest neighbor. Similarly, you could have another set of 3 points going through the center of the uppermost circle, and along the bottom of the uppermost circle. It might look something like this (You have to assume they're touching):Īlong the top of the uppermost circle, you could have 3 points (upper left corner of the region, 25 units to the right of that point, and then another 25 units to the right). 4 such circles could be placed one on top of each other in a 64-unit wide by 200-unit high area. Since any two points cannot be closer than 25 units, picture each point as the center of a circle with a radius of 25 units, which is the same as a circle with a diameter of 50 units. Since this is for a game, I'd expect the above known absolute upper limit to suffice, and the actual number of obstacles to vary. I am not aware of any methods of finding the actual absolute number of circles/disks/points, because there is an infinite number of positions for each point within the area. (Even the shape, or aspect ratio of the rectangle, matters.) In practice, it means that the true, absolute number is somewhat less. $N \le 40$.įrom circle packing, we know that not all "area" is used in this way that you cannot cover all of a surface with non-overlapping circles.
If the size of the rectangular region is $w = 64$ by $h = 200$, and $R = 25$ is the minimum distance between any pair of points, and points are allowed to reside on the perimeter of the region, then the number of points you can put within the rectangular region cannot exceed $N$,
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