Determining continuity of regions/curves from inequalities
in [Mathematica]
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From: Daniel Lichtblau on 15 Jun 2006 03:43 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. > One approach would be to see if any pair of boundary curves intersect, and, if so, whether intersection points are in both regions. This reduces (if you will) the problem to solving bivariate polynomials over the reals. This will only tell you if it is connected. Actual computation of paths would be a separate matter. Daniel Lichtblau Wolfram Research
From: Andrzej Kozlowski on 16 Jun 2006 00:29 On 15 Jun 2006, at 16:25, Carl K. Woll wrote: > 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. >> > > The function SemialgebraicComponents in the package > Algebra`AlgebraicInequalities may help. From the help browser: > > "The package provides a function for solving systems of strong > polynomial inequalities in one or more unknowns. To be precise, > SemialgebraicComponents[ineqs, vars] gives a finite set of > solutions of > the system of inequalities. That is, within the set of solutions, any > solution can be connected by a continuous path to a solution in the > finite set. The variable ineqs is a list of strong inequalities, where > both sides of each inequality are polynomials in variables vars with > rational coefficients. In other words, SemialgebraicComponents[ineqs, > vars] gives at least one point in each connected component of the open > semialgebraic set defined by inequalities ineqs. " > > Needs["Algebra`AlgebraicInequalities`"] > > First example: x>2 && y>0 > > SemialgebraicComponents accepts a list of inequalities: > > In[6]:= > SemialgebraicComponents[{x>2,y>0},{x,y}] > > Out[6]= > {{3,1}} > > Only one point is returned, so the region is connected. > > Second example: (x<2 && y<0) || (x>2 && y>0) > > This example is not a list of inequalities due to the Or. However, in > this case it is easy to construct an equivalent inequality that > encompasses the same region: > > In[5]:= > SemialgebraicComponents[(x-2)y>0,{x,y}] > > Out[5]= > {{0,-1},{3,1}} > > There are two components, so the region is not connected. > > Carl Woll > Wolfram Research > This is right, but the package relies on Cylindrical Algebraic Decomposition and will return at least one point in each component, but usually more than that. So just like my first answer, it only provides an upper bound. Of course when the upper bound is 1 we know that the region is connected. Andrzej Kozlowski
From: Adam Strzebonski on 20 Jun 2006 02:28
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" >>> <akoz(a)mimuw.edu.pl> >>> Sent: Tuesday, June 13, 2006 10:37 AM >>> Subject: Re: 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> >> >> > |