Talk:Problem of Apollonius

From Rosetta Code

Suggest inline/block quoting the problem and algebraic solutions. (With citations, of course) --Michael Mol 17:34, 13 August 2010 (UTC)

Meta error propagation[edit]

Here's the danger of everyone copying everyone else's code: how many of the examples can handle the following three circles? x=0, y=2, r=1; x=0, y=-2, r=1; x=4, y=0, r=1 --Ledrug 23:38, 14 September 2011 (UTC)

OK great. What do we do to fix it? --Mwn3d 01:46, 15 September 2011 (UTC)
Problems which are simple translations are usually indications that the translator is unfamiliar with the algorithm and/or the target language. Perhaps rewrites (or audits) should be preferred. Perhaps translations should show up on the unimpl pages as ENAs? --Michael Mol 02:20, 15 September 2011 (UTC)
The problem is a division by zero, specifically when v11 is 0. AFAIK it's in all the examples. There may be other problems, for instance I wouldn't be surprised if all the solutions fail to handle some degenerate cases.
On a not quite related note, I do think line-by-line translations should be discouraged. Even if you have to pick up the algorithm from some existing examples, it's probably better to rewrite the code from semi-scratch, if only to make sure it's more idiomatic to your language. As to a solution -- I have no idea. --Ledrug 02:34, 15 September 2011 (UTC)

In searching of a proper fix, I scribbled many circles on a piece of paper, and reached the conclusion: I don't want any part of it. First off, the current problem of dividing by zero is just a minor annoyance. There are two fundamental problems:

  1. Circles can't always be represented by [(x, y), r]. Even if we ignore degenerate input circles, one still has to deal with degenerate solution circles, e.g. [(0, -2), 1], [(0, 0), 1], [(0, 2), 1] has two solutions that are straight lines.
  2. Touching each circle internally or externally isn't enough to identify a unique solution. [(0, -3), 2], [(0, 0), 1], [(0, 3), 2] has no solution that contains all circles inside, but has two solutions that's outside every circle. Essentially this involves whether solution is allowed to have negative radius: if it is, there's the risk of double counting; if not, there's the risk of missing solutions.

Besides the above, there are other special cases: [(0, 0), 1], [(0, 0), 2], [(0, 0), 3] has no solution; [(0, 1), 1], [(0, 2), 2], [(0, 3), 3] has infinitely many solutions, to name just two. Someone once said, "programming is all about special cases." Unfortunately, for Problem of Apollonius, there seem to be nothing but special cases. --Ledrug

For what it's worth: the J implementation treats all these degenerate cases as having no solutions. --Rdm 20:11, 15 September 2011 (UTC)
Wait, not even for [(0, -3), 2], [(0, 0), 1], [(0, 3), 2]? --20:29, 15 September 2011 (UTC)
Yes. For that case, it starts at the 0,-3 point (which is the center of the first circle), finds that the excess volume is negative infinity, and so stops there. But the error for 0,-3 is 112 which is larger than the acceptable error volume. So it reports no result. I suppose the question is: what criteria should be used to determine when simplex has completed, if it's not the presence of excess volume? Alternatively, what value should be reported for this case? And, why? --Rdm 20:54, 15 September 2011 (UTC)
I don't quite understand your question... assuming it's about solutions of the three circles, two of them are [(-4, 0), 3], [(4, 0), 3]. --Ledrug 21:04, 15 September 2011 (UTC)
Huh so I guess math/misc/amoeba is some kind of adaptive general equation solver? If so then your comment makes sence, but it also makes the J solution sound inadequate. --Ledrug 21:12, 15 September 2011 (UTC)
It's not clear to me why this should be any worse than languages which return "Not a Number" for . --Rdm 12:36, 16 September 2011 (UTC)
But we all seem to agree that other solutions are incorrect (I haven't looked at the recent D edit yet): being no more incorrect than other incorrect solutions doesn't say much, I think. --Ledrug 19:30, 16 September 2011 (UTC)
My comparison was not between "J solution" and "Incorrect solution implemented in language X". My comparison was between "J solution" and "languages such as Javascript which yield NaN for 0/0". If the J solution is incorrect for giving a non-answer when there are an infinity of answers, then the Javascript language must also be incorrect. --Rdm 19:50, 16 September 2011 (UTC)
[(0, -3), 2], [(0, 0), 1], [(0, 3), 2] doesn't have infinite number of answers. Nor does it have an answer involving an inf: there are 8 solutions, all are finite circles. --Ledrug 20:12, 16 September 2011 (UTC)
Oh, I misunderstood then. Ok, yes, looking at the J solution, it explicitly checks for the cases where all circles are interior tangent and logically, something like ;(_1^#:i.8) <@apollonius 0 _3 2, 0 0 1,: 0 3 2 should treat all the cases, but that's not working for me, and I am going to have to do some debugging to figure out why. (Note, by the way, that I did not write the J implementation here.) --Rdm 20:27, 16 September 2011 (UTC)
If I turn the while. to whilst. in math/misc/amoeba, then:
   ;(_1^#:i.8) <@apollonius 0  _3 2, 0 0 1,: 0 3 2
0 0 1
I'll have to talk with Henry Rich to see if he agrees that this is a good change. Thanks! --Rdm 20:31, 16 September 2011 (UTC)

Turbines[edit]

In general there are eight solutions to this problem. see for a picture showing the eight solutions for a configuration similar to the one depicted in the task description.

Circles are of course passé‎, in modern geometry they are replaced by an abstract object called a turbine. A turbine is made of modern points, which are like old fashioned points but have an added direction property. A turbine is the set of points which are equidistant from an origin point. If the points point towards the origin it looks like a turbine. If the points point at 90deg to the direction to the origin (tangential) it looks like a circle. If the origin is directed to a particular point then the structure is called a clock. If all the points are tangential in the same direction it is called a cycle.

If we say that two cycles touch only if their directions are the same at this point then if we replace the three circles with three cycles then the problem has a unique cycle as a solution. Of course there are eight ways to replace the three circles with three cycles each of the eight solutions (converted to a cycle) in the picture will solve one of these arrangements.

--Nigel Galloway (talk) 11:46, 26 July 2013 (UTC)