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MathGroup Archive 2006

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Re: Re: Searching for a function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70877] Re: [mg70854] Re: Searching for a function
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 30 Oct 2006 05:33:10 -0500 (EST)
  • References: <ehv8vp$g6f$1@smc.vnet.net> <200610290340.XAA08628@smc.vnet.net> <0352689D-3226-4F4C-BE3C-8ED8F651C7D5@mimuw.edu.pl> <077d01c6fb9a$4337e5f0$1e0110ac@bblaptop> <96EFF006-FE33-4D5C-BB08-BD57CA39E8DB@mimuw.edu.pl> <086e01c6fbba$baee8340$1e0110ac@bblaptop>

I still find your message confusing or even contradictory.  You write:

But how do I prove that there does not exist any function for Path1  
and Path2 to be homotopic?

Well, of course the argument I presented shows that there is no  
homotopy between the two paths you gave in your example, and that  
means not just a linear homotopy but any homotopy whatever.

  Perhaps what you are really asking is something quite different?  
Maybe have two paths which are homotopic, but not by a linear  
homotopy, and you would like to find some other homotopy between  
them? It is easy to create such examples, but I won't bother with  
doing it here. Well, unfortunately, I am pretty sure, there is fully  
algorithmic method to solve this problem. Even though you can prove  
that some homotopy must exist, you still have to construct candidates  
"by hand", and you can only use Mathematica to decide if your  
candidate works or not.
In most simple numerical applications of homotopy methods linear  
homotopies, like in your examples are used, and it is not difficult  
to check if they work or not. You can also prove, as I pointed out  
earlier, that no homotopy exists, in the cases when it does not  
exist. But the troublesome cases are when a homotopy is known to  
exist  but a simple linear one does not work. I do not believe  
Mathematica or any other computer program can deal with such cases  
without human direction.

Andrzej Kozlowski


On 30 Oct 2006, at 09:31, Bonny Banerjee wrote:

> Thanks Andrzej. I am aware of the solution you mentioned. What you  
> presented is an "algorithm" for computing whether two paths are  
> homotopic or not. What I want is a function with the properties I  
> mentioned. I want a function because I am going to analyze the  
> properties of that function for further reasoning. An algorithm  
> will not serve that purpose.
>
> An example of F is:
>
> F(u,t) = u*Path2(t) + (1-u)*Path1(t), where 0<=t<=1 and 0<=u<=1
>
> There exists some value of u and t for which F(u,t) belongs to the  
> set of obstacles P. This proves that this particular function F is  
> not a homotopy function between Path1 and Path2. Or in other words,  
> Path1 and Path2 are not homotopic under the function F.
>
> But how do I prove that there does not exist any function for Path1  
> and Path2 to be homotopic? That is where I am stuck.
>
> Any help will be appreciated.
>
> Thanks,
> Bonny.
>
>
>
>
> ----- Original Message ----- From: "Andrzej Kozlowski"  
To: mathgroup at smc.vnet.net
> Subject: [mg70877] Re: [mg70854] Re: Searching for a function
>
>
>> This is a rather different prolem form the one you originally  
>> posted  (or appeared to).  Actually, it amounts to the problem  
>> that has been frequently discussed on this list over the years:  
>> that of deciding if  a point lies inside a polygon or not. Your  
>> two piecewise linear paths  will define a polygon, and the paths  
>> will be homotopic in R^2 with  your set of points removed if and  
>> only if none of the "obstacles"  lies inside this polygon. There  
>> are many ways to decide if a point  lies inside a polygon or not  
>> but I think the most efficient uses  winding numbers. Such a code  
>> has been also posted several times; so I  will just quote  
>> something that was posted in 2000 by William Self:
>>
>> windingNumber[point_, vList_] := Module[
>> {g, ang},
>> g[t_] := Mod[t + Pi, 2Pi] - Pi;
>> ang[p_, a1_, a2_] := g[ArcTan @@ (a2 - p) - ArcTan @@ (a1 - p)];
>> (Plus @@ MapThread [ ang[point, #1, #2] & , {vList, RotateLeft 
>> [vList]}
>> ])/2/Pi]
>>
>> So, making use of that,  here is a function which will decide if  
>> two paths, described just as you have done, are homotopic in the  
>> plane  with the obstacles removed:
>>
>> homotopicQ[path1_, path2_, obstacles_] := With[{poly =
>>      poly = Join[path1, Most[Rest[
>>   Reverse[path2]]]]}, And @@ (Map[windingNumber[#,
>>             poly] &, obstacles] /. {0 -> True, _Integer -> False})]
>>
>> For your example:
>>
>>
>> path1={{0,0},{10,0},{10,10}};path2={{0,0},{0,10},{2,12},{5,20}, 
>> {10,10}};
>> obstacles={{5,5},{7,5},{7,10},{5,10},{12,12},{15,12},{15,20}, 
>> {12,20}};
>>
>>
>> homotopicQ[path1,path2,obstacles]
>>
>> False
>>
>>
>> We can check that this is indeed correct by looking at a picture:
>>
>> Show[Graphics[{Red, Line[path1], Green, Line[path2], Blue, Map[
>>       Point[#] &, obstacles]}]]
>>
>>
>> Andrzej Kozlowski
>> Tokyo, Japan
>>
>>
>>
>>
>>
>> On 30 Oct 2006, at 05:38, Bonny Banerjee wrote:
>>
>>> Thanks Andrzej. I was trying to simplify the problem and  
>>> eventually  made it trivial.
>>>
>>> I was looking at the problem of finding whether two paths are   
>>> homotopic or not. A path in 2D space (R^2) is a continuous  
>>> mapping Path:[0,1]->R^2, where Path(0) and Path(1) are the two  
>>> terminal  points. Two paths, Path1 and Path2, are said to be  
>>> homotopic with  respect to a set of obstacles P if they share the  
>>> same starting and  ending points and one can be continuously  
>>> deformed into the other  without crossing any obstacle. Formally,  
>>> Path1 and Path2 are  homotopic to each other if there exists a  
>>> continuous function F: [0,1]*[0,1]->R^2 such that
>>>
>>> 1. F(0,t) = Path1(t) and F(1,t) = Path2(t), 0<=t<=1
>>> 2. F(u,0) = Path1(0) = Path2(0) and F(u,1) = Path1(1) = Path2(1),  
>>> 0<=u<=1
>>> 3. F(u,t) does not belong to P, 0<=u<=1 and 0<=t<=1
>>>
>>> The problem: I have two paths, Path1 and Path2, which are finite   
>>> curves that start at some point and end at another point. The   
>>> domain is piecewise linear i.e. a path is a sequence of linear   
>>> segments. I also have a set of obstacle points P. How do I  
>>> compute,  using Mathematica, whether these two paths are  
>>> homotopic or not?
>>>
>>> Here is an example:
>>>
>>> Path1 is the sequence of points {(0,0), (10,0), (10,10)}. Thus,
>>> Path1(t) = (20t, 0)   for 0<=t<=1/2
>>> Path1(t) = (10, 20t-10)   for 1/2<=t<=1
>>>
>>> Path2 is the sequence of points {(0,0), (0,10), (2,12), (5,20),  
>>> (10,10)}. Thus,
>>> Path2(t) = (0, 40t)   for 0<=t<=1/4
>>> Path2(t) = (8t-2, 8t+8)   for 1/4<=t<=1/2
>>> Path2(t) = (12t-4, 32t-4)   for 1/2<=t<=3/4
>>> Path2(t) = (20t-10, 50-40t)   for 3/4<=t<=1
>>>
>>> The obstacles are the set of points P = {(5,5), (7,5), (7,10),   
>>> (5,10), (12,12), (15,12), (15,20), (12,20)}.
>>>
>>> In this example, Path1 and Path2 are not homotopic because the   
>>> obstacles {(5,5), (7,5), (7,10), (5,10)} have to be crossed in   
>>> order for one to be deformed into the other. Thus, there does  
>>> not  exist any function F that satisfies all the conditions for  
>>> this  example.
>>>
>>> Given two paths and a set of obstacles, how can I use  
>>> Mathematica  to compute whether there exists any function F or not?
>>>
>>> Thanks,
>>> Bonny.
>>>
>>>
>>>
>>>
>>>
>>> ----- Original Message -----
>>> From: "Andrzej Kozlowski" <akoz at mimuw.edu.pl>
To: mathgroup at smc.vnet.net
>>> Subject: [mg70877] Re: [mg70854] Re: Searching for a function
>>>
>>>
>>> (tm) Pro*
>>> > This adds nothing to the information or the lack of it, in your
>>> previous
>>> > post. If your sets are finite and they have the discrete
>>> topology then
>>> > every function is continuous, so you actually have n^m
>>> functions, where n
>>> > is the cardinality of the soure and m the  cardinality of the
>>> target. One
>>> > can also imagine that what you meant  is a continuous function
>>> from an
>>> > interval to another interval, which  takes a given finite set of
>>> points to
>>> > certain specified images. In  this case you will have infinitely
>>> many
>>> > functions. In fact, the same  is true if you restrict yourself to
>>> > polynomials of arbitrary degrees.  On the other hand if you  
>>> restrict
>>> > yourself to polynomial of a given  degree you can make sure that
>>> the answe
>>> > r is finite. For example, in  your case:
>>> >
>>> > A=Range[2,10,2];B=Range[1,10,2]; ls = Transpose[{A, B}];
>>> >
>>> >
>>> > InterpolatingPolynomial[ls,x]
>>> >
>>> > x-1
>>> >
>>> > In general a polynomial of degree n is determined by its values  
>>> at n
>>> > points so this answer is  unique for polynomials of degree <=5.  
>>> For
>>> > example, here is how we can get another polynomial which
>>> satisfies  all
>>> > your conditions:
>>> >
>>> >
>>> > AppendTo[ls, {3, 9/2}];
>>> >
>>> >
>>> > f[x_] = Expand[InterpolatingPolynomial[ls, x]]
>>> >
>>> > x^5/42 - (5*x^4)/7 + (170*x^3)/21 - (300*x^2)/7 +
>>> >   (2213*x)/21 - 647/7
>>> >
>>> > You can check that this polynomial satisfies all your  
>>> conditions, in
>>> > fact:
>>> >
>>> >
>>> > f /@ A == B
>>> >
>>> >
>>> > True
>>> >
>>> > Andrzej Kozlowski
>>> > Tokyo, Japan
>>> >
>>> > On 29 Oct 2006, at 12:40, Bonny Banerjee wrote:
>>> >
>>> >> I am looking for continuous functions only from set A to set B.
>>> Sorry
>>> >> for
>>> >> not making it clear.
>>> >>
>>> >> --Bonny.
>>> >>
>>> >>
>>> >>
>>> >>
>>> >> "Bonny Banerjee" <banerjee.28 at osu.edu> wrote in message
>>> >> news:ehv8vp$g6f$1 at smc.vnet.net...
>>> >>> Is it possible for Mathematica to solve this problem:
>>> >>>
>>> >>> Given sets A and B, does there exist a function from A to B?
>>> If  yes,
>>> >>> what
>>> >>> is
>>> >>> the function?
>>> >>>
>>> >>>
>>> >>> Here is an example:
>>> >>>
>>> >>> Let, A = {x such that 0<x<11 and Mod[x,2]==0}
>>> >>>
>>> >>> B = {y such that 0<y<11 and Mod[y+1,2]==0}
>>> >>>
>>> >>> Then, there exists a function from A to B
>>> >>>
>>> >>> y = x - 1
>>> >>>
>>> >>>
>>> >>> Thus, is there a way to specify arbitrary sets A, B, and use
>>> >>> Mathematica
>>> >>> to
>>> >>> figure out whether there exits a function from A to B or not?
>>> >>>
>>> >>> Thanks,
>>> >>> Bonny.
>>>
>


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