Talk:Problem of Apollonius: Difference between revisions
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::: 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? --[[User:Rdm|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 <code>[(-4, 0), 3], [(4, 0), 3]</code>. --[[User:Ledrug|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. --[[User:Ledrug|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 <math>0 \div 0</math>. --[[User:Rdm|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. --[[User:Ledrug|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. --[[User:Rdm|Rdm]] 19:50, 16 September 2011 (UTC)
:::::::: <code>[(0, -3), 2], [(0, 0), 1], [(0, 3), 2]</code> doesn't have infinite number of answers. Nor does it have an answer involving an inf: there are 8 solutions, all are finite circles. --[[User:Ledrug|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 <code>;(_1^#:i.8) <@apollonius 0 _3 2, 0 0 1,: 0 3 2</code> 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.) --[[User:Rdm|Rdm]] 20:27, 16 September 2011 (UTC)
::::::::: If I turn the <code>while.</code> to <code>whilst.</code> in math/misc/amoeba, then: <lang j> ;(_1^#:i.8) <@apollonius 0 _3 2, 0 0 1,: 0 3 2
0 0 1</lang> I'll have to talk with Henry Rich to see if he agrees that this is a good change. Thanks! --[[User:Rdm|Rdm]] 20:31, 16 September 2011 (UTC)
==Turbines==
In general there are eight solutions to this problem. [http://en.wikipedia.org/wiki/File:Apollonius8ColorMultiplyV2.svg 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.
--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 11:46, 26 July 2013 (UTC)
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