Re: [math-fun] Random slice of a cube
I googled for five minutes and looked at half a dozen articles; I saw various models of random polytopes, but none of them appeared to be of the form "choose a random plane that intersects a polytope nontrivially and then take that intersection". I also didn't find the resilt about random polygons that Warren mentioned (the one that says that a random polygon has 4 sides on average). So if Warren or anyone else can provide more leads, I'd appreciate it! Jim Propp On Sunday, October 15, 2017, Warren D Smith <warren.wds@gmail.com> wrote:
I believe there is a known published model of "random polytopes" in which the exact expected value is 4.
So, I'm not ready to sit down and try to do the calculus, but surely the probabilities for 3, 4, 5, and 6 are all the results of the sums of a few definite triple integrals. For 3 in particular, there's really just one triple integral, whose value has to be multiplied by 8 to get P(3). That integral gives the probability of a triangle being clipped off by any particular corner. The integral is a little bit hairy because it's a spherical one, and needs to have the "d theta d phi" term adjusted to give constant density over the sphere. This problem would have been fine on an 18.01 problem set, no? On Sun, Oct 15, 2017 at 9:53 PM, James Propp <jamespropp@gmail.com> wrote:
I googled for five minutes and looked at half a dozen articles; I saw various models of random polytopes, but none of them appeared to be of the form "choose a random plane that intersects a polytope nontrivially and then take that intersection".
I also didn't find the resilt about random polygons that Warren mentioned (the one that says that a random polygon has 4 sides on average). So if Warren or anyone else can provide more leads, I'd appreciate it!
Jim Propp
On Sunday, October 15, 2017, Warren D Smith <warren.wds@gmail.com> wrote:
I believe there is a known published model of "random polytopes" in which the exact expected value is 4.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I believe that if you take a random line in the plane that intersects a square, the probability that it intersects both diagonals of the square is 2 ln(2) / pi. Jim Propp On Sun, Oct 15, 2017 at 10:00 PM, Allan Wechsler <acwacw@gmail.com> wrote:
So, I'm not ready to sit down and try to do the calculus, but surely the probabilities for 3, 4, 5, and 6 are all the results of the sums of a few definite triple integrals. For 3 in particular, there's really just one triple integral, whose value has to be multiplied by 8 to get P(3). That integral gives the probability of a triangle being clipped off by any particular corner. The integral is a little bit hairy because it's a spherical one, and needs to have the "d theta d phi" term adjusted to give constant density over the sphere. This problem would have been fine on an 18.01 problem set, no?
On Sun, Oct 15, 2017 at 9:53 PM, James Propp <jamespropp@gmail.com> wrote:
I googled for five minutes and looked at half a dozen articles; I saw various models of random polytopes, but none of them appeared to be of the form "choose a random plane that intersects a polytope nontrivially and then take that intersection".
I also didn't find the resilt about random polygons that Warren mentioned (the one that says that a random polygon has 4 sides on average). So if Warren or anyone else can provide more leads, I'd appreciate it!
Jim Propp
On Sunday, October 15, 2017, Warren D Smith <warren.wds@gmail.com> wrote:
I believe there is a known published model of "random polytopes" in which the exact expected value is 4.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
That's essentially the same as it intersecting two opposite sides of the square, isn't it? If your point is that that value is discouragingly abstruse, and doesn't bode well for a simple analysis in 3-space, I guess I agree. On Sun, Oct 15, 2017 at 10:36 PM, James Propp <jamespropp@gmail.com> wrote:
I believe that if you take a random line in the plane that intersects a square, the probability that it intersects both diagonals of the square is 2 ln(2) / pi.
Jim Propp
On Sun, Oct 15, 2017 at 10:00 PM, Allan Wechsler <acwacw@gmail.com> wrote:
So, I'm not ready to sit down and try to do the calculus, but surely the probabilities for 3, 4, 5, and 6 are all the results of the sums of a few definite triple integrals. For 3 in particular, there's really just one triple integral, whose value has to be multiplied by 8 to get P(3). That integral gives the probability of a triangle being clipped off by any particular corner. The integral is a little bit hairy because it's a spherical one, and needs to have the "d theta d phi" term adjusted to give constant density over the sphere. This problem would have been fine on an 18.01 problem set, no?
On Sun, Oct 15, 2017 at 9:53 PM, James Propp <jamespropp@gmail.com> wrote:
I googled for five minutes and looked at half a dozen articles; I saw various models of random polytopes, but none of them appeared to be of the form "choose a random plane that intersects a polytope nontrivially and then take that intersection".
I also didn't find the resilt about random polygons that Warren mentioned (the one that says that a random polygon has 4 sides on average). So if Warren or anyone else can provide more leads, I'd appreciate it!
Jim Propp
On Sunday, October 15, 2017, Warren D Smith <warren.wds@gmail.com> wrote:
I believe there is a known published model of "random polytopes" in which the exact expected value is 4.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
More specifically, it doesn't bode well for a nice answer for the tetrahedron. Tetrahedra don't tile space, so Warren's type of argument does not seem to be available. The square with its four sides and two diagonals is a two-dimensional cartoon of a tetrahedron with its six edges. Jim On Sunday, October 15, 2017, Allan Wechsler <acwacw@gmail.com> wrote:
That's essentially the same as it intersecting two opposite sides of the square, isn't it? If your point is that that value is discouragingly abstruse, and doesn't bode well for a simple analysis in 3-space, I guess I agree.
On Sun, Oct 15, 2017 at 10:36 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
I believe that if you take a random line in the plane that intersects a square, the probability that it intersects both diagonals of the square is 2 ln(2) / pi.
Jim Propp
On Sun, Oct 15, 2017 at 10:00 PM, Allan Wechsler <acwacw@gmail.com <javascript:;>> wrote:
So, I'm not ready to sit down and try to do the calculus, but surely the probabilities for 3, 4, 5, and 6 are all the results of the sums of a few definite triple integrals. For 3 in particular, there's really just one triple integral, whose value has to be multiplied by 8 to get P(3). That integral gives the probability of a triangle being clipped off by any particular corner. The integral is a little bit hairy because it's a spherical one, and needs to have the "d theta d phi" term adjusted to give constant density over the sphere. This problem would have been fine on an 18.01 problem set, no?
On Sun, Oct 15, 2017 at 9:53 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
I googled for five minutes and looked at half a dozen articles; I saw various models of random polytopes, but none of them appeared to be of the form "choose a random plane that intersects a polytope nontrivially and then take that intersection".
I also didn't find the resilt about random polygons that Warren mentioned (the one that says that a random polygon has 4 sides on average). So if Warren or anyone else can provide more leads, I'd appreciate it!
Jim Propp
On Sunday, October 15, 2017, Warren D Smith <warren.wds@gmail.com <javascript:;>> wrote:
I believe there is a known published model of "random polytopes" in which the exact expected value is 4.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
-
Allan Wechsler -
James Propp