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Cardinality of concave polygons' set.

  1. Dec 29, 2006 #1
    i need to find the cardinality of the set of all concave polygons.
    i know that each n-polygon is characterized by its n sides, and n angles, but i didn't find its cardinality, for example we can divide this set to disjoint sets of: triangles,quandrangulars, etc.
    we can characterize the triangles set by each triangle has (x,y,z) which are its sides, and (a,b,c) are its angles, the angles are between 0 and pi, but besides this i dont know how to procceed, from here.
    can someone advise?
     
  2. jcsd
  3. Dec 29, 2006 #2

    matt grime

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    I presume you mean as subsets of the plane. But up to what equivalence? I doubt you'll find any concave (though I'll admit I'm guessing what a concave polygon is - not a convex one, and convex means for any two points in the interior the line segment between them is contained in the interior) triangles, or quadrilaterals.

    Once you've fixed the sides of a triangle, its angles are fixed, by the way. It won't help here, but it's true.
     
  4. Dec 30, 2006 #3
    my mistake it's convex polygons/
     
  5. Dec 30, 2006 #4

    matt grime

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    So you have alread stated you know it is a subset of

    [tex] \coprod_{n \geq 3} R^n \times [0,\pi]^n[/tex]

    (n lengths, and n angles) and that has cardinality c. All you need to do is show it has cardinality at least c, which is trivial.
     
  6. Dec 30, 2006 #5
    you mean i need to find a 1-1 function from a set which has cardinality c to the set of convex polygons.
    well, let's take the function f:R^n-> the set of convex polygons and we are mapping an (x1,x2,...,xn) to ((x1,x2,...,xn),(a1,a2,...,an))
    where the xs are the lengths and the a's are the angles, is this a well defined function?
     
  7. Dec 30, 2006 #6

    matt grime

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    No. Just writing down a set of numbers that are supposed to be lengths and angles doesn't mean there is a polygon with that data. Actually - I take that back - that does work if we restrict to positiv x_i. However it is far more complicated than I meant. Why not fix n=3? You can show that the set of triangles has cardinality at least c trivially.
     
  8. Dec 30, 2006 #7
    yes, i can see that.
    btw, why couldn't i do the same for the whole set of convex polygons, i could take R^n+ (where the coordinates are positive), this is the only modification is it not.
    i could also mapp the [0,pi]^n to the set of polygons, pretty much the same way i did in previous post.
     
  9. Dec 30, 2006 #8

    matt grime

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    Why would you want to? Why make life more complicated that you need to? Why not just go for the simplest and most elegant solution, especially when a more complicated one adds nothing but complications. Focus on the important stuff, not the fancy unimportant stuff.

    When people do cardinality questions, for some reason, they tend to get far too unnecessarily focused on unimportant details. For example, they tend to attempt to show every countable set is in bijection with N by constructing a bijection. Don't, if you don't need to.
     
  10. Dec 30, 2006 #9
    what i wanted to know is why it doesnt work?
     
  11. Dec 30, 2006 #10

    matt grime

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    Sorry. I don't understand what you want to do (nor why). Write it out in full, preferably latexed.

    The possible issues I have are

    1. What is n? I used it as an index. What are you using it for.

    2. If you're going to write down an explicit funcion that maps, say, a 4-tuple, (w,x,y,z) in R^4 to a polygon, then which one are you going to map it to? There are infinitely many poylgons with the same w,x,y,z.
     
    Last edited: Dec 30, 2006
  12. Dec 30, 2006 #11
    ok, so we want to find a function from R+^3x[0,pi)^3 to the triangles set (R+ is the postive real numbers), i think the simple function here is a one that maps an element from R+^3x[0,pi)^3 to itself, cause every element in the triangle can actually be characterized by (x,y,z) lengths and (a,b,c) angles.
    but i feel something is missing cause there's obviously a connection between the angles and the lengths, so not every (x,y,z) we can choose any (a,b,c) we want, obviously there's a restriction.
    and that follows when there's the triangle inequality, i.e x+y>z z+y>x and z+x>y, but then how to write theses condinitons here as a function?
     
  13. Dec 30, 2006 #12

    matt grime

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    Why are youi creating a map from that domain? You might wish to create such a function, but I don't wish to, so I'm not sure 'we' is the best choice of words. If you really want to, then the cosine rule will help. But it is completely pointless. You just get a subset of the set you started with as the domain. If you do choose to go down this OTT route make sure you demonstrate the resulting set has cardinality c, won't you?

    Anyway, carry on. But I'll make do with the function: t --> (t,1,1) from R+ to the set of isoscoles triangles, thus demonstrating (with the original observation that the set of convex polys was a subset of a set with cardinality c) that the cardinality of the set of all convex polygons is c.
     
    Last edited: Dec 30, 2006
  14. Dec 30, 2006 #13
    you could make it even easier than that, you can make a function t->(t,t,t) from R+ to the equal sides triangles.
    well i must say, i didnt see this simplicity, and you did say it was trivial. (-:
     
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