Re: [math-fun] Random nonempty intersection of cube and plane, revisited
James Propp <jamespropp@gmail.com> wrote:
Does your cube have side-length 1 or side-length 2?
Sigh. Answered in the text of mine which you inexplicably quoted in full.
Since the volume of the cube of side-length 2 is 8, and since the mean width is (a+b+c)/2 = (2+2+2)/2 = 3, the average cross-sectional area should be 8/3 ~ 2.6667. Keith?s 2.8819 seems awfully far off the mark (especially in comparison with his suggestively accurate estimate 4.0004 from two years ago).
I don't follow your reasoning as to why it should be 8/3. I have fairly high confidence in my program, and I would give ten-to-one odds that the correct value of average area is between 2.881 and 2.883. Allan Wechsler <acwacw@gmail.com> wrote:
My suspicion immediately falls on the definition of "random plane intersecting a cube". Here are two possibilities:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
But a subtler difficulty concerns the choice of integration weight: ie. what probability distribution to assign to the phase space (here, planes in 3-space).
James Propp <jamespropp@gmail.com> wrote:
Geometric measure theory looks for measures that are invariant with respect to symmetries of the underlying space. For instance, we seek a continuous measure on the set of planes in 3-space that is invariant under translation and rotation. And it turns out to be unique (up to scaling)! That?s the measure I have in mind.
Likewise. You beat me to it. And the same measure applies to planes through any other solid, including one with no symmetries. Invariant under reflections, too. Allan Wechsler <acwacw@gmail.com> wrote:
Rotating a plane around a line in that plane by a tiny angle ought to be a tiny move, and invariance means that the choice of rotation axis should not matter. But then, translating a plane perpendicular to itself without changing its orientation can be shown to be a smaller move than any rotation, and is therefore 0, which violates positive definiteness.
I was envisioning leaving the planes alone and rotating or translating the *cube*. I will describe how I generate random planes, and then let you guys shoot the planes down. :-) I use the formula ax + by + cz + d = 0 for the plane. (The cube is defined as the volume between -1 and +1 for all three coordinates.) I (pseudo-)randomly, uniformly, and independently choose each of a, b, and c to be real numbers between -1 and +1. That gives me a random point in a cube. I then set r to sqrt(a^2 + b^2 + c^2), and abort if r > 1, so as to carve the cube into a sphere. I then divide each of a, b, and c by r, so as to get a random point on the surface of the sphere. I then set d equal to a random number from 0 to d_max, where d_max is some number that exceeds the square root of 3, that being the maximum "radius" of the cube. (I've been using 1.75 for the first run, 2.0 for the second.) If the resulting plane does not intersect the cube, I don't count it. I keep going until I have the desired number of planes that intersect the cube. Most recently, one billion for each of the two runs. By this time tomorrow, if nothing goes wrong, I should have the results for two runs with ten billion planes each.
http://mathworld.wolfram.com/SpherePointPicking.html See Eq. 16.
On Aug 7, 2019, at 8:13 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com> wrote:
Does your cube have side-length 1 or side-length 2?
Sigh. Answered in the text of mine which you inexplicably quoted in full.
Since the volume of the cube of side-length 2 is 8, and since the mean width is (a+b+c)/2 = (2+2+2)/2 = 3, the average cross-sectional area should be 8/3 ~ 2.6667. Keith?s 2.8819 seems awfully far off the mark (especially in comparison with his suggestively accurate estimate 4.0004 from two years ago).
I don't follow your reasoning as to why it should be 8/3. I have fairly high confidence in my program, and I would give ten-to-one odds that the correct value of average area is between 2.881 and 2.883.
Allan Wechsler <acwacw@gmail.com> wrote:
My suspicion immediately falls on the definition of "random plane intersecting a cube". Here are two possibilities:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
But a subtler difficulty concerns the choice of integration weight: ie. what probability distribution to assign to the phase space (here, planes in 3-space).
James Propp <jamespropp@gmail.com> wrote:
Geometric measure theory looks for measures that are invariant with respect to symmetries of the underlying space. For instance, we seek a continuous measure on the set of planes in 3-space that is invariant under translation and rotation. And it turns out to be unique (up to scaling)! That?s the measure I have in mind.
Likewise. You beat me to it. And the same measure applies to planes through any other solid, including one with no symmetries. Invariant under reflections, too.
Allan Wechsler <acwacw@gmail.com> wrote:
Rotating a plane around a line in that plane by a tiny angle ought to be a tiny move, and invariance means that the choice of rotation axis should not matter. But then, translating a plane perpendicular to itself without changing its orientation can be shown to be a smaller move than any rotation, and is therefore 0, which violates positive definiteness.
I was envisioning leaving the planes alone and rotating or translating the *cube*.
I will describe how I generate random planes, and then let you guys shoot the planes down. :-)
I use the formula ax + by + cz + d = 0 for the plane. (The cube is defined as the volume between -1 and +1 for all three coordinates.) I (pseudo-)randomly, uniformly, and independently choose each of a, b, and c to be real numbers between -1 and +1. That gives me a random point in a cube. I then set r to sqrt(a^2 + b^2 + c^2), and abort if r > 1, so as to carve the cube into a sphere. I then divide each of a, b, and c by r, so as to get a random point on the surface of the sphere. I then set d equal to a random number from 0 to d_max, where d_max is some number that exceeds the square root of 3, that being the maximum "radius" of the cube. (I've been using 1.75 for the first run, 2.0 for the second.) If the resulting plane does not intersect the cube, I don't count it.
I keep going until I have the desired number of planes that intersect the cube. Most recently, one billion for each of the two runs. By this time tomorrow, if nothing goes wrong, I should have the results for two runs with ten billion planes each.
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Jim Propp << Geometric measure theory looks for measures that are invariant with respect to symmetries of the underlying space. For instance, we seek a continuous measure on the set of planes in 3-space that is invariant under translation and rotation. And it turns out to be unique (up to scaling)! That’s the measure I have in mind. >> I'd accept that looks likely to be true, at any rate provided the statistic (arithmetic mean here) and distributions (uniform here) are specified; though I shouldn't like to have to write down a proof. And it's worth keeping in mind that the integrand and the weight are related via a Jacobian, so that the chinese wall we like to erect when thinking about these matters is actually porous. See the discussion later in https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) which emphasises that the situation remains far from cut-and-dried! The strategy of combining uniform distance from cube centre to plane (meeting cube) with uniform plane normal (point on sphere centred on cube) looks plausibly to satisfy Euclidean invariance. However Keith's algorithm for "picking" points on a sphere does _not_ deliver uniformity --- compare Brad Klee's useful citation http://mathworld.wolfram.com/SpherePointPicking.html particularly the elegant equn. 12--15 . Fred Lunnon On 8/8/19, bradklee@gmail.com <bradklee@gmail.com> wrote:
http://mathworld.wolfram.com/SpherePointPicking.html
See Eq. 16.
On Aug 7, 2019, at 8:13 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com> wrote:
Does your cube have side-length 1 or side-length 2?
Sigh. Answered in the text of mine which you inexplicably quoted in full.
Since the volume of the cube of side-length 2 is 8, and since the mean width is (a+b+c)/2 = (2+2+2)/2 = 3, the average cross-sectional area should be 8/3 ~ 2.6667. Keith?s 2.8819 seems awfully far off the mark (especially in comparison with his suggestively accurate estimate 4.0004 from two years ago).
I don't follow your reasoning as to why it should be 8/3. I have fairly high confidence in my program, and I would give ten-to-one odds that the correct value of average area is between 2.881 and 2.883.
Allan Wechsler <acwacw@gmail.com> wrote:
My suspicion immediately falls on the definition of "random plane intersecting a cube". Here are two possibilities:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
But a subtler difficulty concerns the choice of integration weight: ie. what probability distribution to assign to the phase space (here, planes in 3-space).
James Propp <jamespropp@gmail.com> wrote:
Geometric measure theory looks for measures that are invariant with respect to symmetries of the underlying space. For instance, we seek a continuous measure on the set of planes in 3-space that is invariant under translation and rotation. And it turns out to be unique (up to scaling)! That?s the measure I have in mind.
Likewise. You beat me to it. And the same measure applies to planes through any other solid, including one with no symmetries. Invariant under reflections, too.
Allan Wechsler <acwacw@gmail.com> wrote:
Rotating a plane around a line in that plane by a tiny angle ought to be a tiny move, and invariance means that the choice of rotation axis should not matter. But then, translating a plane perpendicular to itself without changing its orientation can be shown to be a smaller move than any rotation, and is therefore 0, which violates positive definiteness.
I was envisioning leaving the planes alone and rotating or translating the *cube*.
I will describe how I generate random planes, and then let you guys shoot the planes down. :-)
I use the formula ax + by + cz + d = 0 for the plane. (The cube is defined as the volume between -1 and +1 for all three coordinates.) I (pseudo-)randomly, uniformly, and independently choose each of a, b, and c to be real numbers between -1 and +1. That gives me a random point in a cube. I then set r to sqrt(a^2 + b^2 + c^2), and abort if r > 1, so as to carve the cube into a sphere. I then divide each of a, b, and c by r, so as to get a random point on the surface of the sphere. I then set d equal to a random number from 0 to d_max, where d_max is some number that exceeds the square root of 3, that being the maximum "radius" of the cube. (I've been using 1.75 for the first run, 2.0 for the second.) If the resulting plane does not intersect the cube, I don't count it.
I keep going until I have the desired number of planes that intersect the cube. Most recently, one billion for each of the two runs. By this time tomorrow, if nothing goes wrong, I should have the results for two runs with ten billion planes each.
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On Wed, Aug 7, 2019 at 11:39 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
However Keith's algorithm for "picking" points on a sphere does _not_ deliver uniformity -
Yes it does. Keith's algorithm is to choose a random point uniformly inside the unit sphere, and then choose a point on the sphere by drawing the line from the origin to this point and extending it until it intersects the sphere. Keith's algorithm for choosing a point uniformly inside the unit sphere is to choose uniformly random points inside the cube of side 2, discarding the point and trying again if the point is not inside the unit sphere. Keith's algorithm for choosing a point uniformly inside the cube of side 2 is to choose 3 independent uniform numbers between -1 and 1. Which of these three steps do you think is faulty? They all seem pretty obviously correct to me. It's possible that you didn't read Keith's description of his algorithm carefully, and thought he was just choosing a point in the unit cube, and then drawing a line from the origin to that point and seeing where it intersected the sphere. That would indeed give a distribution that is not rotationally symmetric, but it's not what Keith's doing.
compare Brad Klee's useful citation http://mathworld.wolfram.com/SpherePointPicking.html particularly the elegant equn. 12--15 .
There's more than one algorithm that produces an element of the same distribution. The algorithm given in mathworld has the advantage that it doesn't involve a "discard and try again" step, so it has bounded execution time. But it does involve calculation of trig functions, so it's far from clear whether it's faster on average than Keith's algorithm. Andy
Fred Lunnon
On 8/8/19, bradklee@gmail.com <bradklee@gmail.com> wrote:
http://mathworld.wolfram.com/SpherePointPicking.html
See Eq. 16.
On Aug 7, 2019, at 8:13 PM, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com> wrote:
Does your cube have side-length 1 or side-length 2?
Sigh. Answered in the text of mine which you inexplicably quoted in full.
Since the volume of the cube of side-length 2 is 8, and since the mean width is (a+b+c)/2 = (2+2+2)/2 = 3, the average cross-sectional area should be 8/3 ~ 2.6667. Keith?s 2.8819 seems awfully far off the mark (especially in comparison with his suggestively accurate estimate 4.0004 from two years ago).
I don't follow your reasoning as to why it should be 8/3. I have fairly high confidence in my program, and I would give ten-to-one odds that the correct value of average area is between 2.881 and 2.883.
Allan Wechsler <acwacw@gmail.com> wrote:
My suspicion immediately falls on the definition of "random plane intersecting a cube". Here are two possibilities:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
But a subtler difficulty concerns the choice of integration weight: ie. what probability distribution to assign to the phase space (here, planes in 3-space).
James Propp <jamespropp@gmail.com> wrote:
Geometric measure theory looks for measures that are invariant with respect to symmetries of the underlying space. For instance, we seek a continuous measure on the set of planes in 3-space that is invariant under translation and rotation. And it turns out to be unique (up to scaling)! That?s the measure I have in mind.
Likewise. You beat me to it. And the same measure applies to planes through any other solid, including one with no symmetries. Invariant under reflections, too.
Allan Wechsler <acwacw@gmail.com> wrote:
Rotating a plane around a line in that plane by a tiny angle ought to be a tiny move, and invariance means that the choice of rotation axis should not matter. But then, translating a plane perpendicular to itself without changing its orientation can be shown to be a smaller move than any rotation, and is therefore 0, which violates positive definiteness.
I was envisioning leaving the planes alone and rotating or translating the *cube*.
I will describe how I generate random planes, and then let you guys shoot the planes down. :-)
I use the formula ax + by + cz + d = 0 for the plane. (The cube is defined as the volume between -1 and +1 for all three coordinates.) I (pseudo-)randomly, uniformly, and independently choose each of a, b, and c to be real numbers between -1 and +1. That gives me a random point in a cube. I then set r to sqrt(a^2 + b^2 + c^2), and abort if r > 1, so as to carve the cube into a sphere. I then divide each of a, b, and c by r, so as to get a random point on the surface of the sphere. I then set d equal to a random number from 0 to d_max, where d_max is some number that exceeds the square root of 3, that being the maximum "radius" of the cube. (I've been using 1.75 for the first run, 2.0 for the second.) If the resulting plane does not intersect the cube, I don't count it.
I keep going until I have the desired number of planes that intersect the cube. Most recently, one billion for each of the two runs. By this time tomorrow, if nothing goes wrong, I should have the results for two runs with ten billion planes each.
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participants (4)
-
Andy Latto -
bradklee＠gmail.com -
Fred Lunnon -
Keith F. Lynch