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Re: Determining continuity of regions/curves from inequalities

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67254] Re: [mg67216] Determining continuity of regions/curves from inequalities
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 14 Jun 2006 06:29:20 -0400 (EDT)
  • References: <200606130506.BAA23751@smc.vnet.net> <E473D604-0718-42F2-B6F4-2444A2712032@mimuw.edu.pl> <001f01c68f2c$1520ec50$6501a8c0@bblaptop> <DFFCE756-9C3D-4848-B4F6-E6DC38750D4B@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

I think I now have an idea of how it might be done with Mathematica  
in a special case, when the region described by the inequalities is  
bounded. Perhaps the method can be extended to a unbounded region,  
but so far I have not thought  about this. I have thought about it on  
the train on my way to my university and I have to give a lecture in  
less than an hour, so I can't work out the details not to mention  
trying to implement it. But the idea is as follows:

Problem. Given a set of algebraic inequalities with a bounded,  
possibly disconnected  solution region and two points in the region,  
determine if the points are in the same connected component.

The idea is this. Find the cylindrical decomposition of the region  
and the cylinders that contain the two points. Now, triangulate these  
cylinders in such a way that the points become vertices of the  
triangulation and the whole region becomes a planar graph, with these  
two points as vertices. Construct the Adjacency matrix of the  
triangulation. Now you can use the function ConnectedComponents of  
the Combinatorica package to answer your question.

In principle I can see (I think) how to implement this, although it  
would take some work and after all might not be computationally  
viable except in simple cases. There may also be well known and more  
efficient algorithms that do this: in particular there should be an  
algorithm for triangulating any semi-algebraic set (actually such  
algorithms are well known - the issue is if there are efficient  
implementations-which is something I do no know). However, I am sure  
Adam Strzebonski knows ;-)

Andrzej Kozlowski


On 14 Jun 2006, at 06:40, Andrzej Kozlowski wrote:

> I think I gave a misleading impression. As it name suggests,  
> CylindricalDecomposition shows the number of cylinders, not  
> components. The number of cylinders in cylindrical decomposition is  
> never smaller than the number of components, but it will often be  
> larger, as in your example.
> I do not think it will be easy to write a function, or at least an  
> efficient functions,  that will counts the number of components  in  
> Mathematica. If you need to do this for complicated cases you will  
> need to use software that computes the homology  (the zeroth  
> homology group is generated by the components) of semi-algebraic  
> sets. However, I do not know, just off hand, of any program that  
> can do it.
>
> Andrzej Kozlowski
>
>
>
> On 14 Jun 2006, at 05:58, Bonny Banerjee wrote:
>
>> Hi Andrzej,
>>
>> Thanks for your response.
>>
>> I am not sure that I made my problem clear enough. I was looking  
>> for a way to figure out whether there exists any path to travel  
>> from one point to another lying entirely within a given region. If  
>> such a path exists, I would call that region continuous, otherwise  
>> not.
>>
>> The procedure that you specified -- to count the number of Or's in  
>> the CylindricalDecomposition of a logical expression -- does not  
>> serve that purpose. An example is provided in the attached file  
>> where the region is continuous but CylindricalDecomposition gives  
>> 8 components. I would have liked the output to be one component.
>>
>> A possible way of doing this would be to grow regions from each of  
>> the two points with different colors (say blue and green) in all  
>> directions. If, at any point, the two colors meet, then the region  
>> is continuous. If the entire boundary is filled but the colors  
>> haven't met, then the region is discontinuous with respect to  
>> those two points, i.e. those two points cannot be joined by a  
>> path. However, I do not know whether Mathematica has any functions  
>> suitable for such computations.
>>
>> --Bonny.
>>
>>
>> ----- Original Message ----- From: "Andrzej Kozlowski"  
To: mathgroup at smc.vnet.net
>> Sent: Tuesday, June 13, 2006 10:37 AM
>> Subject: [mg67254] Re: [mg67216] Determining continuity of regions/curves  
>> from inequalities
>>
>>
>>> *This message was transferred with a trial version of CommuniGate 
>>> (tm) Pro*
>>>
>>> On 13 Jun 2006, at 14:06, Bonny Banerjee wrote:
>>>
>>>> Is there an easy way in Mathematica to determine whether the  
>>>> region  or curve
>>>> formed by a system of inequalities is continuous or not?
>>>>
>>>> For example, the output of some function (e.g. Reduce) might be  
>>>> as follows:
>>>>
>>>> x>2 && y>0
>>>>
>>>> which forms a continuous region. Again, the following output
>>>>
>>>> (x<2 && y<0) || (x>2 && y>0)
>>>>
>>>> is not continuous. Similarly, for curves.
>>>>
>>>> Given such a system of inequalities, how to determine whether the
>>>> region/curve it forms is continuous or not? Or in other words,  
>>>> if I  pick any
>>>> two random points, say P1 and P2, lying on the output curve/ 
>>>> region,  does
>>>> there exist a continuous path lying entirely within the output  
>>>> curve/region
>>>> from P1 to P2?
>>>>
>>>> Any help will be appreciated.
>>>>
>>>> Thanks,
>>>> Bonny.
>>>>
>>>
>>> In general the answer is complicated, but in cases like the above  
>>> you  can tell how many components there are buy looking at how  
>>> many Or's  appear in the CylindricalDecomposition of your set of  
>>> inequalities.  For example consider
>>>
>>> ss = CylindricalDecomposition[x^3 + y^2 + z^4 < x^5,
>>>    {x, y, z}]
>>>
>>>
>>> (-1 < x < 0 && -Sqrt[x^5 - x^3] < y < Sqrt[x^5 - x^3] &&
>>>    -(x^5 - x^3 - y^2)^(1/4) < z < (x^5 - x^3 - y^2)^
>>>      (1/4)) || (x > 1 && -Sqrt[x^5 - x^3] < y <
>>>     Sqrt[x^5 - x^3] && -(x^5 - x^3 - y^2)^(1/4) < z <
>>>     (x^5 - x^3 - y^2)^(1/4))
>>>
>>> so there are two components. Take the point (2,0,0}. We have
>>>
>>>
>>> ss[[1]]/.Thread[{x,y,z}->{2,0,0}]
>>>
>>> Out[11]=
>>> False
>>>
>>> In[12]:=
>>> ss[[2]]/.Thread[{x,y,z}->{2,0,0}]
>>>
>>> Out[12]=
>>> True
>>>
>>> So (2,0,0) lies in the first component. Now consider (-1/2,0,0)
>>>
>>>
>>> ss[[1]] /. Thread[{x, y, z} -> {-2^(-1), 0, 0}]
>>>
>>>
>>> True
>>>
>>>
>>> ss[[2]] /. Thread[{x, y, z} -> {-2^(-1), 0, 0}]
>>>
>>> False
>>>
>>> so (-1/2,0,1} lies in the second component. The two cannot be   
>>> connected by a curve lying within the region satisfying the   
>>> inequality. You can see the two components as follows:
>>>
>>>
>>> <<Graphics`InequalityGraphics`
>>>
>>>
>>> InequalityPlot3D[x^3+y^2+z^4<x^5,{x,-3,3},{y},{z}]
>>>
>>> Andrzej Kozlowski
>>>
>>>
>>> <Trial.nb>
>


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