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Re: Determining continuity of regions/curves from inequalities

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67366] Re: [mg67216] Determining continuity of regions/curves from inequalities
  • From: Adam Strzebonski <adams at wolfram.com>
  • Date: Tue, 20 Jun 2006 02:14:55 -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> <90D474EF-9989-4847-9060-306CC851F3F2@mimuw.edu.pl>
  • Reply-to: adams at wolfram.com
  • Sender: owner-wri-mathgroup at wolfram.com

The problem has been well researched (for instance search for
'adjacency algorithm for CAD' on Google), however at present
there is no implementation in Mathematica. If the number of
variables is <= 3, InequalityPlot(3D) can be used to find
the connected components, but of course plots are subject to
numeric computation error.

Best Regards,

Adam Strzebonski
Wolfram Research

Andrzej Kozlowski wrote:
> 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
>>> <akoz at mimuw.edu.pl>
>>> Sent: Tuesday, June 13, 2006 10:37 AM
>>> Subject: [mg67366] Re: [mg67216] Determining continuity of regions/curves  from 
>>> inequalities
>>>
>>>
>>>> (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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