1
$\begingroup$

I wrote a program to find all the circles with at least three points located at the integers grid. I started with smallest boxes (width >= height) calculating the radius of a circle with points located at:

  • (0, 0)
  • (x, width) where 0 < x <= height
  • (height, y) where 0 < y <= width

Each box gives (w+1)*(h+1) - 1 real circles, several with repeated radius (circles with 4 or more points in the lattice) and avoid the one with radius infinite. I checked all circles with width <= W, avoid repetitions and sort them. Then I find the smallest radius for boxes of width W+1 and reject from the sorted list all the findings greater than W+1 smallest. This way I assure my reduced list has no missing circles.

enter image description here

First circle's radii (squared) I got are:

1/2, 1/1, 5/4, 25/18, 25/16, 2/1, 5/2, 25/9...

Each circle and radius is identified by four integers (width, height, x, y), width >= height, x >= width, y >= height.

My javascript program and HTML/SVG results can be seen here: jsfiddle

  1. Has this sequence already a name or another use?
  2. Is my procedure correct (no missing circles)?
$\endgroup$
3
  • $\begingroup$ "Then I find the smallest radius for boxes of width W+1 and reject from the sorted list all the findings greater than W+1 smallest." I'm not sure I understand what this part is saying. $\endgroup$
    – Mark S.
    Mar 15, 2015 at 20:18
  • $\begingroup$ Also, it's not quite the same thing, but you may be interested in the rational parametrization of the unit circle and the consideration of those points as a group. $\endgroup$
    – Mark S.
    Mar 15, 2015 at 20:21
  • $\begingroup$ I made a list of radii of all the rectangles with width <= 6, height <= width. The list excluded repeated radii but it has very large ones, to say one r=15.51 (width=6, height=1, x=5, y=1). This particular value(s) clearly don't belong to a consecutive list. So I assumed if a go to rectangles family with width=7 and found the smallest radius of this group (r=3.50), this values helps me to reduce the list (removing the greaters) and leave it as a complete (at the moment, width <= 6). This explanation is "seen" at the program and tables presented in the jsfiddle link. $\endgroup$
    – Jolu Mij
    Mar 17, 2015 at 1:06

1 Answer 1

0
$\begingroup$

See "The On-line Encyclopedia of Integer Sequences":

http://oeis.org/A192494 and http://oeis.org/A192493

$\endgroup$
1
  • $\begingroup$ Thanks! I noticed is enough to enter few terms in oeis site to get both series though I didn't realize I'll find them separated as integers denominators and numerators. The series site has a reference to the document Circles Passing through 3 Distinct Points of the Square Lattice. I'll check if my algorithm works the same for bigger circles. $\endgroup$
    – Jolu Mij
    Apr 2, 2015 at 20:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .