< Earlier Kibitzing · PAGE 82 OF 82 ·
|Oct-04-19|| ||al wazir: <johnlspouge>: You may recall my saying that I thought I might have seen this problem before. Antipodes play an important role in your argument. In my first, abortive attempt to derive a formula I mumbled something about antipodes. That might have been an instance of what psychologists call recrudescence. If so, it was very incomplete.|
I followed your derivation of the formula f(N) = N^2 - 2N + 2. But when I got to the paragraph beginning "Third," I had some difficulty, because the terminology is confusing. You have introduced a plethora of poles. In ordinary parlance every sphere has but one equator and two poles at a time. (It is possible to carry out a coordinate transformation, but in the new coordinate system there will still be only one pair of poles.) I don't understand what is meant by a "successful choice." A choice of N points that lie within a given hemisphere? A choice of hemispheres that contain all of N given random points? I see that there is a one-to-one correspondence between the f(N) spherical areas and ... something.
I glanced at Wendel's paper, but it seemed to add a whole new set of complexities, so I quit reading after that glance. While I was pondering this, a glimmering -- I won't say a light, that would be immodest -- a faint adumbration of an idea struck me.
As I said, I have been unable to find anything wrong in my code since the last bug I fixed in the counting routine.
And I am too respectful of your mathematical acumen to suppose that you would post a mistaken demonstration of a theorem -- not to mention the countless other mathematicians and Putnam contestants who have solved the problem, and the examiners who checked their work.
That leaves one logical possibility: We are both right.
We are calculating two *different* probabilities!
|Oct-04-19|| ||al wazir: Gastropod is singular. Pseudopod is singular. Therefore the singular of antipodes must be antipod.|
|Oct-04-19|| ||johnlspouge: < <al wazir> wrote: Gastropod is singular. Pseudopod is singular. Therefore the singular of antipodes must be antipod. >|
In English, the singular is "antipode", derived by back-formation from "antipodes" in the early 17th century. You are have etymology on your side, but the first page of a Google search on "antipod" has no hits in English, and just one in a Serbo-Croation dictionary.
[ https://en.wiktionary.org/wiki/anti... ]
|Oct-04-19|| ||beatgiant: The singular of <kibitzers> is <pedant>.|
|Oct-04-19|| ||johnlspouge: < <beatgiant> wrote: The singular of <kibitzers> is <pedant>. >|
|Oct-04-19|| ||johnlspouge: < <al wazir> wrote: That leaves one logical possibility: We are both right.|
We are calculating two *different* probabilities! >
I would not be surprised. After all, I entered the thread by pointing everyone to Bertrand's Paradox.
[ https://en.wikipedia.org/wiki/Bertr... ]
My screed abstracted the problem. The probability distribution you are calculating lacks one of the following properties: (1) it chooses points independently from a continuous distribution; or (2) if it does, it does not choose antipodes with equal probability densities.
|Oct-04-19|| ||johnlspouge: The case N=4 is simpler than the general case and may aid in understanding. To start, choose any point on the sphere. Now, choose any three pairs of antipodes in general position (so no two pairs lie on the same great circle with the point). Now from each pair of antipodes, choose one with probability 1/2. The antipodes form 8 spherical triangles, all triangles have equal probability when one from each pair is chosen, so the probability that the initial point lies inside any one of the triangles is 1/8.|
The case N=4 is unusually simple, because the great circles through pairs of antipodes form spherical polygons that are triangles with the antipodes for vertices.
|Oct-04-19|| ||al wazir: <johnlspouge: The probability distribution you are calculating lacks one of the following properties: (1) it chooses points independently from a continuous distribution; or (2) if it does, if it does, it does not choose antipodes with equal probability densities.>|
The distribution is as "continuous" as a discrete numerical calculation performed by a digital computer using 32-bit arithmetic can be. The choice is as "independent" as an operation carried out deterministically (i.e., repeatably) can be.
My random number generator is ran2, copied from Numerical Recipes. The authors say this: <We think that, within the limits of its floating-point precision, ran2 provides perfect random numbers; a practical definition of “perfect” is that we will pay $1000 to the first reader who convinces us otherwise (by finding a statistical test that ran2 fails in a nontrivial way, excluding the ordinary limitations of a machine’s floating-point representation).> https://pdfs.semanticscholar.org/cc...
As for your second point, I tried inserting a branch (pseudo)randomly replacing the calculated point with its antipod in the loop where I calculate n (pseudo)random points inside the unit sphere. I didn't think it would change the results, aside from statistical error. It didn't.
Also, as <Tiggler> correctly noted, it doesn't make any difference whether or not I project the points onto the surface of the sphere. (I tried it both ways.) A plane drawn through the center of a sphere and two other points is independent of where those points are on their respective radii.
Let me try to pin down exactly what you are doing in your derivation.
Suppose the sphere is the Earth. I choose a set of n points on the surface and find their n antipodal antipoints. Together the points and antipoints comprise an augmented set of 2n distinct elements. Each point and its antipoint are associated with the great circle equidistant from the two, their equatorial great circle (EGC). All the EGCs are different.
Let's suppose that n=5, and five elements of the ten in the augmented set are at Chicago, Detroit, New York City, Pittsburgh, and Washington. I don't know which of these five are points and which are antipoints; I just know they are all in the augmented set.
It's clear that any hemisphere must contain five points of the augmented set. This applies to a hemisphere centered on one of those five cities (equivalently, a hemisphere for which one of those cities is a point or antipoint, or equally equivalently, a hemisphere bounded by the EGC of one of those cities). So what?
Since f(5) = 22, the five EGCs demarcate 22 "spherical areas" on the Earth's surface. But the five cities are so close together that all five lie in the *same* spherical area. I can travel between any two of them along a geodesic without crossing an EGC.
What is the one-to-one correspondence of which you wrote?
|Oct-04-19|| ||al wazir: <johnlspouge: In English, the singular is "antipode", derived by back-formation from "antipodes" in the early 17th century.>|
Bah! I suppose you think that "media" is a singular noun. Wrong. It is the plural of "medium."
But Media is a city in southeastern Pennsylvania.
|Oct-04-19|| ||johnlspouge: @<al wazir>: You have put the 10 = 2*5 antipodes into 2 clusters, but there are still 5-subsets of the 10 antipodes that lie in a single hemisphere and contain points from both clusters. Despite your clustering, the ECGs of antipodes still partition the sphere, but some spherical areas become very thin and short. In other words, the thinness of such an area corresponds to the precision required to position any hemisphere so it contains 5 fixed points from both of your clusters. If an arc joins pairs of points in such an area, _that_ arc is not bisected by a ECG.|
|Oct-04-19|| ||johnlspouge: < <al wazir> wrote: Bah! I suppose you think that "media" is a singular noun. >|
No. I do not agree with the majority usage and use "media" as a plural. At my biological work daily, I use "data" as a plural. Kindly do not lecture me on the plural of 2nd declension neuter Latin nouns. We pedants pride ourselves on the speed with which we can recite the declension.
If you are fond enough of "antipod" to use it, I can only suggest you learn Serbo-Croatian. Perhaps regrettably, "antipod" is not any standard of English.
|Oct-04-19|| ||johnlspouge: < <johnlspouge> wrote: We pedants pride ourselves on the speed with which we can recite the declension. >|
Still under 3 seconds on the first try.
Not bad :)
|Oct-04-19|| ||al wazir: <johnlspouge: You have put the 10 = 2*5 antipodes into 2 clusters, but there are still 5-subsets of the 10 antipodes that lie in a single hemisphere and contain points from both clusters.> I'm afraid that doesn't answer the question. In my example, where is the one-to-one correspondence? What corresponds to what?|
|Oct-04-19|| ||al wazir: <johnlspouge>: "Gastropod, "pseudopod," and "tripod" are English words. "Gastropode" and "pseudopode" are not. They are French. I don't know if "tripode" is a word in any language.|
"Antipod" may be Croatian, but it is certainly not Serbian. Антипод might be.
I'm a little surprised that someone who knows the Latin second declension -- and maybe the third, fourth, and fifth as well -- has fallen under the sway of a 17th-century solecism.
|Oct-04-19|| ||beatgiant: <al wazir> English got "gastropods", "pseudopods" and "tripods" via Latin, but "antipodes" directly from Greek, or so say a couple of dictionaries I checked. And the singular in Greek would be "antipus," like "platypus" (although the plural of "platypus" is not "platypodes").|
English is notoriously full of all kinds of non-uniformity.
|Oct-04-19|| ||al wazir: <beatgiant: English got "gastropods", "pseudopods" and "tripods" via Latin>. I doubt if the Romans had any use for the first two, though educated Romans spoke Greek. More likely medieval or Renaissance savants coined them. "Pseudo" is definitely Greek, not Latin.|
|Oct-05-19|| ||beatgiant: <al wazir>
The sources say "Latin", but not necessarily "ancient Latin". As you suggested, it's not unusual for much later scholars to coin new words based on Latin. And it's also not unusual for Latin to be based on Greek.
But my main point is, we don't need to speculate. There are good quality specialized works called etymological dictionaries, where a person like you has already done the job of finding out and publishing the answer.
Would you believe... etymonline.com?
|Oct-05-19|| ||al wazir: <johnlspouge> and <Tiggler>: I found a typo in my code -- this despite my having gone over and over it so many times previously.|
My numbers are now consistent, to ~1%, with the ones your formula gives:
All 4 pts in 1 half in 0.8810 of trials.
All 5 pts in 1 half in 0.6859 of trials.
All 6 pts in 1 half in 0.5042 of trials.
All 7 pts in 1 half in 0.3440 of trials.
All 8 pts in 1 half in 0.2308 of trials.
All 9 pts in 1 half in 0.1408 of trials.
All 10 pts in 1 half in 0.0924 of trials.
All 11 pts in 1 half in 0.0537 of trials.
All 12 pts in 1 half in 0.0318 of trials.
All 13 pts in 1 half in 0.0180 of trials.
All 14 pts in 1 half in 0.0116 of trials.
All 15 pts in 1 half in 0.0066 of trials.
All 16 pts in 1 half in 0.0030 of trials.
All 17 pts in 1 half in 0.0011 of trials.
All 18 pts in 1 half in 0.0012 of trials.
All 19 pts in 1 half in 0.0011 of trials.
All 20 pts in 1 half in 0.0004 of trials.
Pleas accept my crestfallen apology. All I can say in my defense is that I never swore that there could not possibly be any mistakes.
|Oct-05-19|| ||johnlspouge: Cool.
Apparently, all we have left to argue about is the singular of “antipodes”.
And where you are concerned, I am more than happy being right 1/2 the time :)
|Oct-05-19|| ||Tiggler: <al wazir> Apology accepted. And I'll stop calling you <Al>, but please note: though my name is not <tiggler> you can continue to call me that anyway.|
|Oct-07-19|| ||johnlspouge: < <Tiggler> wrote: And I'll stop calling you <Al> >|
Hmm. I don't know why, but this scene came to mind...
[ https://www.youtube.com/watch?v=rO1... ]
|Oct-09-19|| ||Tiggler: I was expecting this one: https://www.youtube.com/watch?v=uq-...|
|Oct-09-19|| ||Tiggler: <johnl> works near Chevy Chase, I believe.|
|Oct-10-19|| ||johnlspouge: < <Tiggler> wrote: I was expecting this one >|
Yours is more on the mark for the nickname; but less, for the message ;>)
|Oct-10-19|| ||johnlspouge: < <Tiggler> wrote: <johnl> works near Chevy Chase, I believe. >|
Nahh, but that other <johnl> is one hell of a guy ;>)
< Earlier Kibitzing · PAGE 82 OF 82 ·