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I've posted in this thread,
http://www.codeguru.com/forum/showth...ht=acute+angle
but it was a while ago and the OP never came back and it's more of an algorithm question so I continue in a new thread here.
Say you generate the three corners of a triangle in a plane at random. What's the chance the triangle is acute?
I've presented a solution already but I think I have a better one this time.
Some consider the problem bad-posed.
A random triangle is generated by the way of three random points and one cannot really generate random points in an infinite plane, or can one? I think so. Maybe not in practice but conceptually.
Say you have an urn filled with N black and white balls. You start drawing balls from the urn. You notice the color and put it back. After a while you detect that one in four is white. According to the frequentist definition of probability this means the chance of a white ball is 1/4. Now someone tells you there aren't N balls in the urn but an infinite number. Does this change anything? Not in my view. The chance of drawing a white ball still is 1/4.
In my view it's not a problem that there are an infinite number of triangles in the plane. From a frequency probabilistic viewpoint the important thing is the number of acute triangles in relation to the total number of triangles. This relation must have a limit but it doesn't matter whether the number of triangles are finite or infinite, just as the urn example shows.
So in my view the problem is not bad-posed. You can conceptually draw triangles at random from an infinite plane and the frequentist probability measure still will exist as the number of acute triangles in relation to total number of triangles.
My first approach to a solution was to limit the infinite plane to a region. I tried two versions, a unit square and a unit circle. This allowed me to perform computer simulations. I generated triangels at random within the region and then counted the acute ones in relation to the total number.
The basis for this approach is the assumption that the acute triangle fraction is the same wherever you look in the plane. This would mean that the triangles were "fractal" in nature. Regardless of where you looked and how large a region you looked at the acute triangle fraction would be constant. Unfortunately this is not true. As soon as a region is introduced certain triangles are exclude, namely those that don't fit within the region. Specifically those that have one or two corner points inside but the rest outside.
Now to my new approach.
Instead of limiting the plane to a region, the set of all triangles in the infinite plane are transformed while keeping the acute triangle fraction intact. The transformations are those which don't change the inner angles of a triangle, namely rotation, translation and scaling. In other words those transformations don't change whether a triangle is acute or not.
The important thing here is that the acute triangle fraction doesn't change during the transformation. If it does the approach fails. After the transformations the acute triangle fraction must be the same it were before transformations started
The first transformation is that of scale. We have the set of all triangles in the infinite plane. Now all triangles are scaled so the longest side becomes 1.
The second transformation is that of rotation and translation. Now all triangles are rotated and translated so that their length 1 sides becomes aligned, and with the opposite corner in the same direction.
After this a coordinate system is introduced. The left corner of the length 1 side is put in [0, 0] (origo) and the right corner in [1,0].
Remember it was the longest side of all triangles that become length 1 and that means the other two sides are shorter. This allow us to plot the area within which the opposite corner (from the length 1 side) of all triangles will be. The best way to picture this area is to think of an American football. Cut it in half where it's at the narrowest and look at it from the side.
You get this by drawing two circles with radius 1. One circle has its centre in [0,0] and the other in [1,0]. The upper half of their intersection area looks like the half American football I pictured.
To recapitulate.
We have transformed the set of all triangles in the infinite plane to the set of all alligned 1 length sided triangles with the opposite corner in the same direction. And most importantly we have done that while keeping the acute triangle fraction intact.
We've fixed the transformed triangle set in a coordinate system so all the triangles have their 1 length sides at [0,0] and [1,0] and their opposite corner within a "half American football" limiting area.
To simulate a solution to the problem we now only have to generate opposite corner points within the limiting area and count how many of these triangles are acute in relation to the total number. That's the probability of getting an acute triangle.
I'll write that program but not today.
My specific question is whether you think my reasoning holds.
Nice. This should include all the possible types of triangles. Here's something interesting: if you draw a circle with the radius of 0.5 and with the center at (0, 0.5) - when you pick any point on this line and connect it with the ends of [0, 1] segment, you'll get a right triangle. All possible right triangles have the third point on that line. All the triangles with the third point under, or inside this circle are obtuse.
I realized this while I used Paint to visualize what you said, but I just saw that the idea is not new - in fact, there's a text on this topic on the page linked in the thread where all this started (Wolfram Math World: http://mathworld.wolfram.com/ObtuseTriangle.html).
On that page there is given the chance of obtaining an obtuse triangle, and it's proportional to wholeSurface/obtuseAreaSurface. Since the chance of obtaining a right triangle is proportional to the surface of the line defining the circle centered at (0,0.5), which is 0, the chance for a acute triangle would be 1 - obtuseChance.
1 - 0.63938... = 0.36062... (according to Wolfram Math World)
But, what does this mean. That it's impossible to have true right triangles? That there's an inherent (but generally insignificant) flaw in our number system? (Remember how irrational numbers have no finite decimal representation?) Those triangles do exist, I believe - but the chance of obtaining one is close to 0, but not 0. Or I'm wrong?
However, some might argue that this view of the original problem somewhat alters the nature of the problem at hand: this is more like - what percentage of all existing triangles is acute? The original problem was: what is the chance of getting an acute triangle if you randomly pick 3 points in a plane.
Now, this might or may not be one and the same question depending of what the reader considers to be the core of the problem - the triangles themselves, or the process of picking the points?
In the original thread, when you replied to superbonzo you said: "Math is nothing but a tool you know. Maybe you're applying the wrong yardstick. A problem doesn't become nonsensical just because your favorite yardstick cannot measure it. I wouldn't be too categorical I were you."
Naturally, you have a point, but some problems are not solvable in nature - i.e. the specific question (although it may not seem like that) is absurd because it is in conflict with the very nature of the problem. It cannot be answered.
There are examples in math and science where someone proved that something cannot be proved.
It's like Heisenberg's uncertainty principle in physics - stating that pairs of physical properties, like position and momentum, cannot both be known to arbitrary precision. This has nothing to do with the imperfection of the tool used for the measurement - it's how things are, it's in the nature of a quantum particle.
Now, I'm not familiar with the works related to the problem of picking the random points for the triangle, and I think I can assume that you are not either, so it is our right to have doubts about this conclusion. Wolfram Math World page says that "it is impossible to pick random variables which are uniformly distributed in the plane", and provides a reference to the paper where this is discussed: Eisenberg, B. and Sullivan, R. "Random Triangles n Dimensions." Amer. Math. Monthly 103, 308-318, 1996.
If one perceives the core of the problem to be the operation of picking the points, one could argue that the result might depend on how the points are picked.
It's really all about how you see it.
Another thing - computer simulations can help, but note that there are limitations other than not being able to represent an infinite plane; since the precision of the variables is limited, within the finite surface you would use for the simulation, there exist triangles that the computer is unable to generate. This may or may not affect the final result, but it's certainly something to consider.
This time I've tried to cut beyond that.Quote:
Originally Posted by TheGreatCthulhu
I've argued that this problem is well-posed. It is sensical to talk about "the chance of an acute triangle" although it may be non-sensical to talk about "random points in an infinite plane".
1. From what I can see the basic axioms of probability don't exclude infinite sets.
2. From what I can see it's possible to transform the set of all triangles, to the set I suggested, with the acute triangle fraction intact.
As long as no one disputes the above I'm home. It's how math works.
Thank you for the references and your effort. I'll consider everything in detail.
The problem of bad posedness has nothing to do with the use of infinite sets or infinite dimensional geometries of any cardinality; two examples: the interval [0,1] is an infinite non numerable set with a natural uniform probability distribution; the set of right continuous left limited paths in R->R^n is an inifinite dimensional topological space that (equipped with a proper filtration) has a natural uniform probability dstribution (the Wienier process aka Brownian motion) and has a cardinality even greater then the set of "triangles in an infinite plane" as you have defined it ( that, BTW, has the same cardinality of an interval ) !
Here we are not speaking of introducing a probability in the set of all triangles. Of course, this is possible. The problem comes when you try to define a natural notion of "probability uniformity" with infinite sets.
Actually, the same issue happens with finite sets; but finite sets come equipped with a natural measure (the counting measure) that allows you speaking about "chance". This measure is the invariant measure coming from its permutation group.
Conversely, when the set is infinite every point that is contained in an infinite strictly decreasing chain of measureable sets ( like, say, any point in the interval [0,1] ) will have measure 0. This means that if this happens for every point of the set then there can be no probability measure invariant with respect to its permutation group. There's no counting measure for such sets. There's no natural "uniformity" nor "chance".
Of course, you can arbitrarely introduce symmetries and/or probability measure on such sets. But there are inifinitely possible inequivalent choices.
Unless you add them to their geometry structure a priori: examples are compact Lie group's (the circle, the set of rotations of any dimenions,...) which always have a left invariant probability measure (that comes from the set of its diffeomorphisms ) (but note that to each Lie group structure on the same set corresponds a different prob.measure); non-compact Lie groups (the set of translation, the Lorentz group, the Heinsenberg group ,...) that admit "uniform" prob.measure represented in their Banach duals ( for example, plane waves in ordinary quantum mechanics ) and so on.
So, if the problem is "find a family of probability measures over the set of triangles whose limiting measure is translation invariant and calculate the limit of the measure of the subsets of acute triangles" then the problem is well posed and solvable (but there are many solutions dependeing on the way you choose that family); conversely, if the problem is "find the chance of a triangle being acute" then the problem is bad posed (unless you have a very naive notion of a probability theory that reduce the problem to the formulation above...)
I haven't defined anything really.Quote:
Originally Posted by superbonzo
I've assumed that the traditional axioms of probability are valid also with infinite sets. This would mean that the notion of probability in the frequentists sense also exists for infinite sets. And this would mean that it's indeed possible to determine the fraction of acute angles in the infinite set of all triangles.
I haven't chosen any "family of probability measures". I've used the standard frequentist notion of probability.Quote:
So, if the problem is "find a family of probability measures over the set of triangles whose limiting measure is translation invariant and calculate the limit of the measure of the subsets of acute triangles" then the problem is well posed and solvable (but there are many solutions dependeing on the way you choose that family);
Consider the urn example I gave. It doesn't matter whether the urn has a finite or an infinite number of balls. By drawing balls you can determine what fraction is white and that's the same as the probability of getting a white ball. The same with the triangles. By drawing from a transformed set which is equivalent to the infinite set of all triangles and counting the acute ones you can determine the probability of an acute triangle.
Thank you for your reply but I would appreciate if you were a little more specific in your criticisms.
I depend heavily on the urn example. Do you agree that it doesn't matter whether the urn contains a finite or an infinite number of balls? Do you agree that the notion of a fraction of white balls also exists in the infinite case?
I've assumed it's possible to transform the infinite set of all triangles to another set thereby preserving the original acute angle fraction. Do you agree?
>> Thank you for your reply but I would appreciate if you were a little more specific in your criticisms.
I'd appreciate it too :), but there are so many foundational issues involved here, ranging from the foundation of mathematics to physics, probability and plain natural philosophy... it would take pages only for introducing the general problem rigorously...
>> I've assumed it's possible to transform the infinite set of all triangles to another set thereby preserving the original acute angle fraction. Do you agree?
So, are you conjecturing that any tranformation f:R^2*3->R^2*3 that preserves angles and whose image is bounded give the same result ?
for example, given a triangle translate and scale it in such a way that the circumscribed circle is centered in the origin and has radius one. Then rotate it in such a way one of its side is vertical. Obviously our assumptions are satisfied; also, the image is representable as a simple pair of angles, thus it have the same "dimensionality" of the image of your transformation. Let's take such pair of angles uniformly on the circle. Is the result the same ?
a quick calculation (non tested, but I suppose is correct) with wolfram mathematica gives me 0.360+-0.001 for your transformation and 0.248+-0.001 for the transformation above. Both angle preseving, different "chances".
>>I depend heavily on the urn example. Do you agree that it doesn't matter whether the urn contains a finite or an infinite number of balls? Do you agree that the notion of a fraction of white balls also exists in the infinite case?
in your urn example the cardinality of the set of balls is inessential a priori; that is you assume from the beginning that the empirical ratio of white to black balls is fixed. The number of balls does not play any role in that specific probabilistic model.
BTW, you could think that there exist a "physically" intuitive notion of "uniformly random"; you'd be wrong; I'll try to give you an example ( that does not nearly exhaust the many foundational issues involved here ):
take, say, a linear array of N leds; under each led put a small circuit that blinks its led randomly at a fixed rate R (say, a random number generator with Poisson statistics ); suppose also that R is small compared to N in such a way that the probability of two leds blinking simultaneously is low. Under these assumptions you'll see a uniformly distributed flash over the array length.
Now, put yourself in the center of the array and start spinning at angular speed w (I mean you, not the array); given that the blinking rate depends only on the local time of the led+circuit system you'll observe the rates of leds at a distance r to be decreased to ~ R * sqrt( 1 - (r*w/c)^2) ; thus the distribution won't be uniform anymore even if the phenomena generating the random events (the led array) have not been touched. Thus, uniformity is an observer dependent property (and things get even worse in general relativity and/or quantum field theory ). Of course, you can define locally a uniform point-distribution. But you can do that only because the "local" representation of space ( eg a subset of an euclidean/Minkowsky/... space ) has a priori a definition of uniformity (actually, a local symmetry group strictly related to its geometrical structure ).
Actually, it's the other way round. It doesn't matter how many balls there are in the urn, as long as 1/4th of them are white and you draw balls uniform randomly (meaning each ball has the same chance of being draw each time) then the probability that you draw a white ball is independent of the number of balls in the urn. That's why it is reasonable to state that it also works with an infinite number of balls.Quote:
Originally Posted by nuzzle
The problem would be very different if 1/4th of the top 100 balls are white and everything else black. If you then draw uniform randomly from the top 100 balls, the probability of drawing a white ball is 1/4. But that doesn't mean that 1/4th of all the balls are white.
That's the whole point. These two problems look the same, but they are very different. I believe your reasoning is flawed, because you are generalizing a consequence of one problem into a general rule.
D_Drmmr:
As for the urn example, I think that it is implied that nuzzle was not talking about an urn with any sort of special arrangement of the balls ("top 100 balls are white"). Also, the urn is just a more "reader-friendly" term for a set, so you shouldn't think of it in terms of it's physical properties, like only being able to pick the topmost balls - imagine that, somehow, you are able to pick any of the balls.
But, the urn-balls form a countable set, while the set of points is uncountable; perhaps this is not the right parallel to draw?
superbonzo:
Would it make any difference if the points could somehow... "pop-up" on their own? If there wasn't any sentient-being-introduced picking involved? Or this is just the same?
All:
What we need to do here is to correctly define the actual problem that nuzzle is trying to solve, and to make a distinction between it and all other discussion that is actually philosophical in nature.
So, essentially: what is the percentage of acute triangles in the set of all possible triangles? Is there an inherent flaw in such a problem proposition, too?
Disk Triangle Picking: http://mathworld.wolfram.com/DiskTrianglePicking.html
It says:
"The probability P_2 that three random points in a disk form an acute triangle is
P_2=4/(pi^2)-1/8=0.280284... "
But it also uses those nasty words "Pick three points [...] distributed independently and uniformly"...
Yes there are two ways of seeing it depending on whether you know the white ball fraction or whether you want to determine it by drawing balls.Quote:
Originally Posted by D_Drmmr
You claim the triangles are not drawn with equal probability. How is that?
I may be mathematically naive but if you can define the infinite plane as a set of points then you can also pick points from this set at random. Why not? Each point has an equal probability of being picked and the probability of a specific point being picked is zero.
And if you pick three points independently at random in the infinite plane you've picked a triangle at random in the infinite plane. If you pick triangles like that doesn't each triangle have an equal chance of being picked? If not, exactly what triangles are favoured?
I mentioned this in the orginal thread post #9.Quote:
Originally Posted by TheGreatCthulhu
http://www.codeguru.com/forum/showth...ht=acute+angle
I perform simulations there on a unit square and got the same results (and I also used a unit circle).
But as I mentioned in #3 in this thread, as soon as you limit the infinite plane to a region you skew the fraction of acute triangles so that's a fawlty approach I wanted to avoid this time.
I'll try being more specific. This is what nuzzle did: he defined an angle preserving transformation f from the set of triangles T to the set of triangles; he parametrized the image of that transformation as a subset of the plane (what he describes as an "American football"); then he uniformly picked points from the football taking the probability that such points represent acute triangles. Recapitulating, he argues that : (*) being the transformation angle preserving and being triangle acuteness an angle dependent property then such probability is "the chance that a triangle is acute":
now, if A is the set of acute triangles, F is the "American football" set in R^2, you have
A -j-> T -f-> T -p-> F
( where j is the natural inclusion and p a function sending each transformed triangle to the corresponding point in the american football set )
the probability he computed is equal to L{ p(f(j(A))) } / L{ F } , where L{} is the Lebesgue measure in the plane ( aka the "area" ).
now, the statement (*) tantamounts saying that:
if f and f' preserve acuteness and F and F' are bounded then L{ p(f(j(A))) } / L{ F } = L{ p'(f'(j(A))) } / L{ F' }
this statement is false ( I provided a counter example in post #10, but it should be evident from the very definition of a measure ).
what do you mean ?Quote:
Originally Posted by TheGreatCthulhu
That would remove the seemingly problematic issue of whether one can pick triangles at random in the plane.Quote:
Originally Posted by TheGreatCthulhu
It would give the chance of an acute triangle an a-priory probability just like there's an a-priori probability attached to each side of a coin (1/2) and each side of a dice (1/6) for example.
I know and that's what makes if fun. -:)Quote:
Originally Posted by superbonzo
I have argued that limiting the plane to a region skews the acute triangle fraction because not all triangles will fit within the region. Transforming all triangles to a region will also skew the triangle fraction in a different way. In this case the region won't be evenly populated so the acute triangle fraction will vary within the region.Quote:
for example, given a triangle translate and scale it in such a way that the circumscribed circle is centered in the origin and has radius one. Then rotate it in such a way one of its side is vertical. Obviously our assumptions are satisfied; also, the image is representable as a simple pair of angles, thus it have the same "dimensionality" of the image of your transformation. Let's take such pair of angles uniformly on the circle. Is the result the same ?
a quick calculation (non tested, but I suppose is correct) with wolfram mathematica gives me 0.360+-0.001 for your transformation and 0.248+-0.001 for the transformation above. Both angle preseving, different "chances".
My transformation is not to a region. It preserves both the internal angles and the percentage of acute triangles in relation to all triangles. Fixing the triangles to a region takes place after the transformation.
My transformation makes the longest side of every triangle equal to 1 and all triangles are aligned along this side with the opposite corner in the same direction. This means all opposite corners will fall within a half American football shaped region. If the opposite corners are not evenly distributed my solution falls apart of course.
There doesn't even exist a physically intutive notion of the infinite plane. All physical planes are regions. But I argue that if one accepts the notion of an infinite plane as an abstraction, nothing stands in the way of accepting the notion of random points in it.Quote:
BTW, you could think that there exist a "physically" intuitive notion of "uniformly random"; you'd be wrong; I'll try to give you an example ( that does not nearly exhaust the many foundational issues involved here ):
Quote:
My transformation is not to a region. It preserves both the internal angles and the percentage of acute triangles in relation to all triangles. Fixing the triangles to a region takes place after the transformation.
so your making the assumption that your "transformation preserves ... the percentage of acute triangles in relation to all triangles" to prove that you have found "the percentage of acute triangles in relation to all triangles" without ever having defined what "the percentage of acute triangles in relation to all triangles" is ?Quote:
...If the opposite corners are not evenly distributed my solution falls apart of course.
in order to compute a "percentage" you need to count two sets A and B and take the ratio between them, right ? well, how do you count sets ?
there are essentially two ways:
1) if you want your "counting" to produce a unique object representing the "number of its elements" you'll came up with the concept of cardinality: two sets have the same cardinality if there exist a bijection between them. For example, every finite set is in bijective corrispondence with a set {1,2,..,n} for some natural number n, so we simply say its cardinality is 'n'. The next cardinal greater then every finite set is the cardinality of the set of natural numbers. Note that the sets of integers, pair of integers, tuple of intergers, rational numbers, odd numbers, prime numbers, algebraic numbers,... have all the cardinality of the set of natural number. So things like the ratio of (the cardinality of) odd numbers to the ratio of (the cardinality of) even numbers is not defined.
2) the other way is to literally count them, that is attach labels 1,2,3,.. to elements of the two sets. This tantamounts defining a sequence of strictly increasing finite subsets of A x B. Each such subset has finite cardinality, thus, you can take their ratios. Finally, you take the limit of that ratio. For example, you could count odd to even numbers as 1,2,3,4,5,6,... and concluding that their "inifinte" ratio is 1/2. The problem is that if you count them in a different way you'll get different limits (including non converging ones): for example, 1,3,2,5,7,4,.. is a well defined such sequence but the ratio converges to 1/3.
Now you could add an ad hoc rule to impose that, say, the "right" counting sequences are those that preserve ordering, in this way you could demonstrate that for such sequences the ratio always converge to 1/2. Fine. But what happens if a set has no canonical definition of order (the set of triangles is such a set) ?
Of course, you can always add ad hoc rules more and more restricting the way of counting elements; but at the end you'll end up with tautological results simply restating your starting assumptions.
Anyway, can you describe what do you mean exactly by "Fixing the triangles to a region takes place after the transformation." ?
that is, try writing down what properties should a transformation satisfy in order to be considered "acute triangle percentage preserving".
There's no need to "accept" the idea of an infinite plane which is a perfectly defined concept in every set theory that includes the natural number system and the axiom of choice ( all this is related but totally indipendent from any tenet coming from other philosophical fields, including existentiality conditions of actual infinities ).Quote:
There doesn't even exist a physically intutive notion of the infinite plane. All physical planes are regions. But I argue that if one accepts the notion of an infinite plane as an abstraction, nothing stands in the way of accepting the notion of random points in it.
Regarding the concept of "physical plane" you're again taking things too naively. In physics and mathematics the geomertical structure of the plane (and more often its higher dimensional analogues) happens in many different contexts; in some of them they indeed have an intuitive physical interpretation: for example, if you assume the realism of special relativity then spacetime is a Minkowsky space whose elements have a physically well defined interpretation. if you assume the realism of general relativity then the Einstein field is a 2-2 tensor field defined over tangent spaces that in turn have a well defined physical interpretation.
BTW, you could think that objects described in physical theories can never corresponds exactly to the "real" things they represents, they are always approximated. Sometimes this is the case ( for example, a rigid body in classical mechanics is never assumed really rigid, or a electric AC current is never assumed really sinusoidal ), sometimes not. Notable examples are the solutions of Einstein equations, that are not equations in its common sense, because its solutions are geometrical structures. So, if you assume the realism of general relativity then, say, the Kerr black holes are in a strict sense isomorphic to their geometrical abstract representations ( that is, if you take a region of space time surrounding it and whose density satisfy certain conditions, then the Einstein field has exactly that form, no matter how energy is microscopically distributed there ).
I mean, since the problem is related to and since the result depends on the process of picking (or the procedure used to pick) the points, would it make a difference, would the problem still exist if no one would actually do the picking, but if the points that would form the triangle emerged on their own, in some natural, random way. Like... I don’t know... raindrops hitting the ground, thus defining random points on a plane, or random impacts on the surface of the Moon.Quote:
Originally Posted by superbonzo
Maybe my understanding of the notion of randomness lacks some important aspect of it.
I’ll quote the Wolfram Math World article again:
"Unfortunately, the solution of the problem depends on the procedure used to pick the "random" points (Portnoy 1994)." (http://mathworld.wolfram.com/ObtuseTriangle.html; Portnoy 1994 paper PDF: http://projecteuclid.org/DPubS?servi....ss/1177010497).
Note that the word 'random' is between quotation marks, indicating that there’s no true randomness involved once the pick procedure is defined.
But, back to the idea of the points emerging on their own. As I’ve asked before - is this jut the same as picking the points? If there’s no person, or rather, if there’s no method defined that is used to pick them, but if the points emerged in some natural (truly random?) process - would anything change, would the question still be ill-posed? Or there would be some underlying rules/laws that would govern this emergence, and thus these rules could be considered equivalent to the pick method? Are these just two ways of expressing the same thing, or there’s a fundamental difference.
In nature, processes that would be considered random are actually the result of existing starting conditions and forces at work, they are the result of the laws of physics. Even a simple act of throwing a coin is the result various physical interactions governed by these laws. So, one could argue that everything that happens is just a result of a long cause & effect chain, and there’s nothing truly random. But I refuse to believe that. There must be some truly random process in nature, at the very fundamental level of it (maybe at quantum level?). I won’t go further into it, since with this we enter the realm of philosophy.
As you’ve said: “you could think that there exist a "physically" intuitive notion of "uniformly random"; you'd be wrong”.
I guess I had it coming but this kind of sophisticated math reasoning is above my head. Still your argumentation is beginning to sink in.Quote:
Originally Posted by superbonzo
You've showed that two kinds of angle preserving transformations from infinite plane to finite plane leads to different results. One was to a unit circle and one was to a unit line segment (as I suggested).
There are two ways to continue here for me.
1. You say there's no difference between the two but I claim there is.
2. I have to admit I cannot come up with why one should be considered more natural or logical than the other.
How much I would like it to be 1 it seems I will have to go for 2. I'll have to regroup and see if I can come up with something. If not, at least now I have a much better understanding of the fundamental issues at play here.
To use my urn analogy. An urn can have an infinite number of balls and the fraction of white balls exists and can be measured. The problem is that there are many urns and the white ball fraction varies between them.
I definately recognize my "half American football" in that link. And when I simulate it I get the analytically expected result.Quote:
Originally Posted by TheGreatCthulhu
So although "my" solution came 100 years too late there is at least some consolation in that I arrived at it independently. :)
I'll just have to accept that when one goes from infinite to finite, strange things happens to probability.
Here are the simulations I promised.
They're modifications of the code I used in the other thread.
There are two. The first is the situation where all triangles in the infinite plane are transformed to the unit circle. In the second all triangles are transformed with their longest side on the unit line segment. The latter is "my" solution in this thread.
Code:double rnd() { // random double between 0.0 and 1.0 (inclusive)
return double(rand()) / double(RAND_MAX);
}
void test1() {
srand(unsigned(time(0)));
int N=1000000; // number of tries
double PI = 3.1415926535897932384626;
int n = 0; // hit counter
// unit circle
for (int i=0; i<N; i++) {
double fi = 2.0*PI*rnd();
double ax = sin(fi); // 3 random points on unit circle
double ay = cos(fi);
fi = 2.0*PI*rnd();
double bx = sin(fi);
double by = cos(fi);
fi = 2.0*PI*rnd();
double cx = sin(fi);
double cy = cos(fi);
double a2 = (ax-bx)*(ax-bx) + (ay-by)*(ay-by); // side lengths squared
double b2 = (bx-cx)*(bx-cx) + (by-cy)*(by-cy);
double c2 = (cx-ax)*(cx-ax) + (cy-ay)*(cy-ay);
if (a2+b2>c2 && b2+c2>a2 && c2+a2>b2) n++; // count acute triangles
}
std::cout << double(n) / double(N) << "\n"; // print frequency (probability)
n = 0; // hit counter
// unit line segment
for (int i=0; i<N; i++) {
double ax = 0.0;
double ay = 0.0;
double bx = 1.0;
double by = 0.0;
double cx;
double cy;
do { // repeat until point is within half American football
cx = rnd();
cy = rnd();
} while ((cx*cx + cy*cy > 1.0) || ((cx-1.0)*(cx-1.0) + cy*cy > 1.0));
double a2 = (ax-bx)*(ax-bx) + (ay-by)*(ay-by); // side lengths squared
double b2 = (bx-cx)*(bx-cx) + (by-cy)*(by-cy);
double c2 = (cx-ax)*(cx-ax) + (cy-ay)*(cy-ay);
if (a2+b2>c2 && b2+c2>a2 && c2+a2>b2) n++; // count acute triangles
}
std::cout << double(n) / double(N) << "\n"; // print frequency (probability)
}
So, regarding the "percentage of all acute triangles" issue: if I understood this well - because both the set of all triangles and the set of acute triangles have the same cardinality, since there exits a corresponding bijection, it doesn't make sense to compare this sets in terms of percentage/ratio? (Same goes for the set of obtuse triangles, or any subset of the set of all triangles on which a similar bijection may be defined.)
Let us step back from the triangle problem a bit, and take a look at this simple example.
Consider the segment [0,1], and the following question: What is the percentage of all elements x e [0,1] that satisfy x <= 0.5, in relation to all the elements of [0, 1]?
Following the above logic, asking this question would also make no sense. Clearly, these elements are all the elements of [0, 0.5] (there are infinitely many), and one could intuitively say that the answer is 50%, but, since there's a bijection from [0,1] to [0, 0.5], the cardinality is the same, so the question is nonsensical.
Am I right?
If so, then this provided me with a new insight in the properties of infinite sets, and I'm rather glad about it.
"The more you understand what is wrong with a figure, the more valuable that figure becomes."
— Lord Kelvin
But, is there some property of these sets that would enable as to quantify the relation between a set of all triangles and it's subset of acute triangles? Does mathematics define such a property? Something that would enable as to speak of these sets in a fashion similar to "percentage"? A numerical value that can tell us in more detail how one set relates to the other, enabling us to know more than just the fact that one is a subset of the other?
If nothing like that is defined, is there any reason that such a property doesn't exist? Maybe our (human) knowledge of sets can be further expanded.
"When you can measure what you are speaking about, and express it in numbers, you know something about it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts advanced to the stage of science."
— Lord Kelvin (again)
Not exactly. As I sead in post #19, the concept of "numbers that count things" is modeled as cardinal numbers. Cardinal numbers are "usual" integers only for finite sets, thus the usual notion of "ratio" can be defined only for them.Quote:
... because both the set of all triangles and the set of acute triangles have the same cardinality, since there exits a corresponding bijection, it doesn't make sense to compare this sets in terms of percentage/ratio? (Same goes for the set of obtuse triangles, or any subset of the set of all triangles on which a similar bijection may be defined.)
Now, you can do two things:
1) try defining what is a ratio of general cardinal numbers; you can do that excatly in the same way you construct the set of rational numbers from the set of natural numbers ( with some added tecnicalities, it all depends on the axiomatic system you choose... ).
But as you noted, cardinality is a very coarse grained way of looking at sets: the interval [0,1] is the same as the interval [0,0.5], but it's also the same as any subset of R^n, like the interior of a square, a cube or a 100-dimensional sphere...
2) conclude that probability has to do with ratios and countings only when finite sets are involved. With infinite sets you need different concepts. The most common way of representing a probabilistic model is through a "probability space" ( that in turn is a special case of a measure space ) and of "random variables" ( that are simply measurable functions between probability spaces ).
In this way you can see the interval [0,1] as a probability space and you'll be able to conclude that the measure (ie the probability) of the interval [0,0.5] is 0.5.
yes, you can define a probability structure over the set of triangles ( as an exotic example, you can send bijctively each triangle to a point in the interval [0,1] and use its natural probability measure directly ).Quote:
But, is there some property of these sets that would enable as to quantify the relation between a set of all triangles and it's subset of acute triangles?
What you cannot do is defining a translation invariant probability measure over that set, because such a thing does not exist ( eg. there's no such probability space in the class of probability spaces ).
I was pretty convinced there existed a percentage of acute triangles among all triangles in the infinite plane, but arguments in this thread has lead me to believe that may not be the case. At least to be able to calculate the percentage you need to transform all triangles to the finite plane and by doing that you influence the percentage. I though I had found a transform which would keep the percentage intact but appearantly that's not even possible.Quote:
Originally Posted by TheGreatCthulhu
This is disturbing and my belief is that if you "instead of taking triangles to finity took probability to infinity" you could find the one true percentage. Maybe some matematician can figure that out but I can't.
:):):)Quote:
Originally Posted by nuzzle
Funny how, when the foundations of our own beliefs are shaken, or when we need to readjust our way of seeing the world, we find it disturbing...
My intention was not to obfuscate the example with practical issues, but to show the importance of implicit assumptions that were not mentioned. In the case of the urn drawing uniformly random is the implicit assumption. In case of the original problem, there are a lot of implicit assumptions along the way. I think superbonzo has done a great job to bring them to light.Quote:
Originally Posted by TheGreatCthulhu
If I can change your example a bit to make it precise. The probability that a uniformly random point on the interval [0, 1) lies in the interval [0, 0.5) is 0.5. I agree that you cannot count the number of points in either of these intervals, but you can show that for any point in the interval [0, 0.5) there is exactly one uniquely corresponding point in the interval [0.5, 1). The function f(x) = 0.5 + x defines this relation. This property, combined with the definition of a uniform random probability measure (that every possible outcome has the same probability) yields the probability of 0.5.Quote:
Originally Posted by TheGreatCthulhu
So, even though we can't count the number of points in either of the intervals above, we can deduce properties for these points that lead to the conclusion that the probability is 0.5. That's the cool thing about mathematics. :cool:
Yes I find it disturbing. I want an infinite-sided dice and I'm sure I will have one some day. :)Quote:
Originally Posted by TheGreatCthulhu
But for the time being I have to accept that such a dice would be sphere shaped and wouldn't have a side to stop on when thrown. It would keep rooling forever. In short, a round dice has no outcome. This is at the heart of this problem isn't it.
I've noticed that there's a more probabilistically sound approach to this problem called Gaussian triangles. The Gaussian distribution is defined in the plane so it's possible to generate triangles according to this distribution. When I have some time to spare I'll check out so called heavy-tail distributions which I believe also are defined in the plane.
I’ve done some reading and I think I’ve got it.
But, first a little intro to some concepts, since it might of help to someone who is interested, but nor all too familiar with the concepts that were discussed here.
Others may skip next few passages (unless you want to check if I didn’t misinterpret something).
Probability space is a mathematical tool (or one of available tools) used to create a model of a real-world probabilistic process, or experiment (or a mind experiment – why not?). If we shift our focus form all the mathematical details and definitions, the concept is really quite simple.
Probability space, also called probability triple (O, E, p) is simply a following construction:
O: A set of all possible outcomes - called the sample space.The concept of outcomes is straightforward; anything that can happen and is relevant to the problem is an outcome - if you throw a die, an outcome would be getting a specific number. You can model this with {1, 2, 3, 4, 5, 6}. An outcome is the result of a single execution of the model.
E: A set of events, where each event is a set of zero or more outcomes (a set of other sets containing elements o, such that o e O). It is with these that we work with - I’ll expand on this in a moment.
p: Is the probability measure. A moment to explain this to anyone who might be reading this, and is not familiar with the topic; this is a special kind of a measure, which is simply a function that assigns numbers to sets, with the limitations that these numbers are >= 0, that p(emptySet) = 0, and that p(unionOfDisjointSets) = p(set1) + p(set2) + ... The numbers that a probability measure assigns are in the range [0, 1], or 0% - 100% if you prefer, and represent the probabilities of events.
But, since individual outcomes are rarely the point of interest, the events are used to model a wider range of possibilities. For example, what if, instead of being interested in the probability of getting a specific number when you roll a die, you wanted to know the possibility of getting an even number. An event {2, 4, 6} e E models this – and it is considered that it has happened if any of its outcomes occurred. Only those events that are interesting or relevant in relation to the problem are considered. There are related rules, but I’ll left them out.
The probability measure p, being a function, returns the probabilities of events.
p: E --> [0, 1]
Now we enter the core of our discussion. The probabilities of the events, returned by the probability measure, are related to the probability mass function (in a discrete case - like with dice) or to a probability distribution function (in a continuum case - like with points of a plane). These ascribe probabilities to the outcomes.
First, some of us have (or had?) this notion that there is some natural probability related to each outcome (that a result of a die roll, or of a point selection has some inherent probability), but this was wrong. The probabilities are given (ascribed!) so that they fit a specific problem. You might say - but a chance for a specific outcome of a die roll is 1/6 and that’s it, this cannot be changed! But, these chances are "ascribed" by the physical properties of a die. Let us abstract this object. It becomes {1, 2, 3, 4, 5, 6}. With this, you can do whatever you want. You can use the same object to model a rigged die, say a one that is weighted so that a specific number has a better chance of appearing, by ascribing different probabilities.
Now, this becomes even more apparent in the point-picking problem. Why would any outcome have any inherent chance in this case?
The second important aspect of the problem which most of us failed to see is that, in order to solve the triangle problem, we approached it by transforming our infinite plane into something more manageable. Different approaches yielded different results, all seemingly valid. But, as superbonzo said, there's no way to define this transformation in such a way that our probability measure p still gives the same results for the same events (such an event specifically being "an acute triangle is formed").
Someone somewhere proved it. Period.
By applying the transformation, we change the output of p.
If I'm right, this is what it's all about.
There might be some technicalities here that I overlooked, but I'm sure others will point them out if it's necessary.
A side thought:
About nuzzle’s infinite sided die - yes, it would be a sphere, but why would it roll forever? Introduce friction, and it would eventually stop. The topmost point is the result of the roll. But, the surface of a sphere is a continuum, so what would be the right way to model such a die? Would the sample space be the set of, say, non-negative reals [0, +inf. ), in respect to its physical surface, or the set of natural numbers {1, 2, 3, ...}, in relation to the concept of "infinite-sided" object?
Some replies:
Of course, I agree. A bit of irony: I assumed that you are trying to obfuscate it with practical issues, which lead to this conversation - another proof that there's a good reason to be explicit :)Quote:
Originally Posted by D_Drmmr
So, if I tied (1), I’d get some ratio-like expressions that wouldn’t really tell me much more that what I had without them?Quote:
Originally Posted by superbonzo
If I tried (2), someone might argue that this is only one of possible results, since there's really no "correct" probability distribution function.
Yes. But, I wasn't talking about cardinality in general, I was referring to the cardinal number of the specific sets used in my example. (Being the same as the cardinal number of the set of reals, unless I'm mistaken.)Quote:
Originally Posted by superbonzo
Right, I suppose you known how to construct the set of rational numbers from the set of natural numbers ? you can do something similar, but with some technical difficulties. Anyway, as you said, the resulting "ratios" would be useless in modeling the concept of probability.Quote:
Originally Posted by TheGreatCthulhu
>> If I tried (2), someone might argue that this is only one of possible results, since there's really no "correct" probability distribution function.
Yes that's totally right, but...
>> ...this notion that there is some natural probability related to each outcome (that a result of a die roll, or of a point selection has some inherent probability), but this was wrong.
the key is what do you mean by "natural". Actually, modern mathematics has a well defined and sophisticated meaning of that word ( that comes from category theory ) capable of modeling a very general concept of structure.
Anyway, intuitively, given an object and its structure you can sometimes specify a set of natural properties or naturally related objects. These might include a probability structure.
For example, every scientist will understand you if you say "consider a uniform probability density over the interval [0,1]" or "over a sphere" or "over a finite set" or "over a thorus" or over "a projective space", "a flag manifold", "a Moebius strip" or even over the set of "right continuous left bounded paths"... but every scientist will NOT understand you if you say "consider a uniform probability density over the set of reals", or over "the set of triangles", "the set of continuous paths", "the set of chords of a circle", "the set of natural numbers"...
what does the former geometrical objects have in common that the latter have not ?
it's not cardinality: excluding finite sets, common cardinalities appear in both situations ...
it's not an intuitive notion of "boundedness": like, bounded interval vs the whole real line; but the set of chords of a circle is bounded ( or better, compact would be the right term ) ...
Most geometrical objects in both lists ( and every object in the first ) specify (maybe implicitly, sometimes it depends on the context in which they appear) a group of symmetries ("compatible" with their structure, this might imply continuity, measurability, differentiability, or whatever): a finite set has its permutation group, a sphere its group of rotations, the real line its group of translations, a bounded interval its semigroup of local translations, a 2-dimensional projective space its group of Moebius transformation and so on...
Now, each symmetry carries a notion of "uniformity", in the sense that "uniform things" will be somehow compatible with and invariant with respect to that symmetry.
Finally, using nuzzle's words, this is the heart of the problem: Does every geometrical object admit a probability structure that is compatible with and invariant with respect to its symmetry group ?
The answer is NO.
And this is why we cannot get an unanimous/unique answer to the question "what's the probability that of a triangle being acute?" : because we consider rotated\scaled\translated triangles naturally equivalent AND because the set of triangles does not admit a probability measure invariant with respect to its group of "natural" transformations.
I think the round dice quite well illustrates the problem here. It has no outcome just like drawing uniformly distributed points at random in the infinite plane has no outcome.Quote:
Originally Posted by TheGreatCthulhu
To get an outcome you need to introduce different measures, such as friction, to get the dice to stop. It will be the stopping measures that determines the outcome because the dice is still round no matter what you do. You get different results depending on how you make the dice stop. It's the same with the triangles.
I'll have to respectfully disagree - because I feel that, if things are left as they are, your example can be rather misleading.Quote:
Originally Posted by nuzzle
You mean - if the friction is stronger, the roll will end sooner? You think that with cube-shaped dice is different, that no "external" factors influence the result? Actually, there's no real difference. What if you threw a standard 6-sided die underwater? This would certainly shorten the time it takes for the roll to end. And if you could roll it in space, in vacuum? It would spin forever, at least in an idealized case. Just like your sphere if there are no "stopping measures". So, a die stops and gives a specific result not because it's cube-shaped, but because there's gravity, and because it has a certain effect on that specific shape. This is clearly a "stopping measure", as you called it.Quote:
Originally Posted by nuzzle
If you agree with what I said above, then you'll also have to agree that if a sphere can stop, there is an outcome.Quote:
Originally Posted by nuzzle
The problem here has little to do with the plane being infinite. It has to do with the fact that we need to transform the set of triangles into something that we can work with (like your half American football). The transforms used are translation/rotation/scale; this is because these specific transformations preserve something that we deemed important - and that is the shape of the triangle - they give images that are similar to the original triangles.
But this is where our problem comes into play. There is no way to define such a transformation, apply it, and still get the same probability of acute triangles. This transformation changes our probability measure, and each specific transformation method changes it in a different way (so, in the end, we get different results).
I'll try to create an analogy here. (But, I'll admit on the start that it could be better.)
Imagine if you made a cube out of paper, something like those that kids might make at school. After you've written some dots/numbers, you can use it as a 6-sided die. Now, a chance of getting an even number is 1/2. So, you have a set of outcomes O = {1, 2, 3, 4, 5, 6}, and an event that you're interested in, where an even number is the result E_even = {2, 4, 6}. This is a subset of O. With our triangles, you have a set of outcomes (sample space) where each represents a formation of a triangle after 3 points have been picked, and you have an event "an acute triangle is formed", which is a subset of sample space containing all the elements that form such a triangle. An event has occurred if any of its outcomes is picked.
Now we apply the transformation. Imagine you take your paper cube and you unwrap or open it a bit. Imagine also that this is some strange or super-quality paper that keeps it shape. (See the attached image for reference.) Although this transformation doesn't have a real purpose, it's still a valid transformation - some translation, and some rotation. It's analogous to the one we apply in the other case; only, with our set of triangles, we apply the transformation for a reason - in order to solve the problem, and in such a way that the shape of the triangle doesn't change. With our paper cube, we keep the shape of the sides (square). However, what is the chance of getting an even number now? We altered the shape of our paper die, and it's not a cube anymore! We changed the probability measure. With triangles, we shape the set of all triangles (which doesn't really have a shape - the set, I mean) into a half American football, or into a circular area, or a rectangular area, or something else.
Again, this may not be the best analogy, but I guess it illustrates the point.
Also, there's another thing I'd like to point out. Consider the half American football approach. Here, the set of all triangles is mapped onto the half American football shape, and then one point is picked from there. What does this has to do with the act of picking points in a plane? There might be some correlation between the two, but you'll note that the plane doesn't figure anywhere in this approach.
Finally, I'll try to compile everything that superbonzo said in a few lines of quotes.
You'll see that he tries to tell, from the start, that the resulting probability is not invariant to the transformation that is applied.
You'll also see that he argues there's no "correct" probability assignment over the sample space, although there might be some "natural" (or natural) probability distribution arising from the symmetries of the structure in question (thus being more appealing to the human mind).
Quote:Quote:
"But again "picking randomly" from any unbounded set of an euclidean space in a translation invariant way is pure-non-sense." (src >)
Quote:
"The problem of bad posedness has nothing to do with the use of infinite sets or infinite dimensional geometries of any cardinality [...]" (src >)
Quote:
"Here we are not speaking of introducing a probability in the set of all triangles. Of course, this is possible. The problem comes when you try to define a natural notion of "probability uniformity" with infinite sets.
Actually, the same issue happens with finite sets; but finite sets come equipped with a natural measure (the counting measure) that allows you speaking about "chance"." (src >)
Quote:
"Of course, you can arbitrarely introduce symmetries and/or probability measure on such [infinite] sets. But there are inifinitely possible inequivalent choices." (src >)
Quote:
"a quick calculation (non tested, but I suppose is correct) with wolfram mathematica gives me 0.360+-0.001 for your transformation and 0.248+-0.001 for the transformation above. Both angle preseving, different "chances"." (src >)
Quote:
"BTW, you could think that there exist a "physically" intuitive notion of "uniformly random"; you'd be wrong" (src >)
Quote:
"if f and f' preserve acuteness and F and F' are bounded then L{ p(f(j(A))) } / L{ F } = L{ p'(f'(j(A))) } / L{ F' }
this statement is false ( I provided a counter example in post #10, but it should be evident from the very definition of a measure )." (src >)
Quote:
"What you cannot do is defining a translation invariant probability measure over that set [the set of triangles], because such a thing does not exist ( eg. there's no such probability space in the class of probability spaces )." (src >)
Quote:
"[qouting me] >> ...this notion that there is some natural probability related to each outcome (that a result of a die roll, or of a point selection has some inherent probability), but this was wrong.
the key is what do you mean by "natural". Actually, modern mathematics has a well defined and sophisticated meaning of that word ( that comes from category theory ) capable of modeling a very general concept of structure.
Anyway, intuitively, given an object and its structure you can sometimes specify a set of natural properties or naturally related objects. These might include a probability structure.
[...]
each symmetry carries a notion of "uniformity", in the sense that "uniform things" will be somehow compatible with and invariant with respect to that symmetry.
Finally, using nuzzle's words, this is the heart of the problem: Does every geometrical object admit a probability structure that is compatible with and invariant with respect to its symmetry group ?
The answer is NO.
And this is why we cannot get an unanimous/unique answer to the question "what's the probability that of a triangle being acute?" : because we consider rotated\scaled\translated triangles naturally equivalent AND because the set of triangles does not admit a probability measure invariant with respect to its group of "natural" transformations." (src >)
Well, in my opinion the fundamental problem is the infinite plane. The infinite plane simply has no notion of a uniform random distribution of points so picking a triangle at random using this distribution just doesn't have an outcome, just like a round dice doesn't.Quote:
Originally Posted by TheGreatCthulhu
There is nothing wrong with the transformations. The problem is that we apply them to an acute angle fraction that doesn't exist. We're using the transformations to "fix" a fraction in a region of the plane from a nonexisting fraction in the infinite plane. We're using the transformations as ways of making a round dice stop by giving it sides. It will be the transformations that determine the acute angle fraction.
The mathematically correct way to deal with this is to not assuming uniformly randomly distributed points in the plane. Instead one must switch to a probability distribution that is defined for the infinite plane such as the Gaussian.
I don't want to be to insistent, and I'm not claiming I'm absolutely right.
But, I said "little to do with", not "nothing to do with". So I agree, but IMO, it's only a part of the story.
The way I see it, a similar problem might arise even with finite sets, if you had to apply a transformation for some reason (the word "transformation" having a rather broad sense here), and if the probability measure was not invariant to this particular transformation.
So, yes, there's nothing wrong with the transformation - there's something wrong with the probability measure - it's not invariant to it.
Anyway, whatever probability distribution you used, can you avoid translation/rotation/scaling in order to solve the problem?
I think one has to accept that the problem lacks a solution. The reason is that the uniform probability distribution doesn't exist in the infinite plane. Picking triangles from this distribution simply has no outcome (again compare with the round dice). The transformations are just different ways to obscure this fact (in the round dice analogy they would artificially add sides to enforce an outcome).Quote:
Originally Posted by TheGreatCthulhu
The only way this problem has a solution is to use a distribution which does exist in the infinite plane, such as the Gaussian. Then there's no need for transformations. I've seen in the litterature that for such so called Gaussian triangles the acute triangle fraction is 1/4.