Let's discuss Mathematics

[monikernemo]

I do not really get it, but they have figured out a bit more of Anaxagoras of Clazomenae's problem about squaring the circle?

Around 450 BCE, Anaxagoras of Clazomenae had some time to think. The Greek mathematician was in prison for claiming the sun was not a god, but rather an incandescent rock as big as the Peloponnese peninsula. A philosopher who believed that “reason rules the world,” he used his incarceration to grapple with a now-famous math problem known as squaring the circle: Using a compass and a straightedge, can you produce a square of equal area to a given circle?

The exact question posed by Anaxagoras was answered in 1882, when the German mathematician Ferdinand von Lindemann proved that squaring the circle is impossible with classical tools. He showed that pi — the area of a circle with a radius of 1 — is a special kind of number classified as transcendental (a category that also includes Euler’s number, e). Because a previous result had demonstrated that it’s impossible to use a compass and a straightedge to construct a length equal to a transcendental number, it’s also impossible to square a circle that way.

Tarski's Circle Squaring Problem from 1925 asks whether it is possible to partition a disk in the plane into finitely many pieces and reassemble them via isometries to yield a partition of a square of the same area. It was finally resolved by Laczkovich in 1990 in the affirmative. Recently, several new proofs have emerged which achieve circle squaring with better structured pieces: namely, pieces which are Lebesgue measurable and have the property of Baire (Grabowski-Máthé-Pikhurko) or even are Borel (Marks-Unger).

In this paper, we show that circle squaring is possible with Borel pieces of positive Lebesgue measure whose boundaries have upper Minkowski dimension less than 2 (in particular, each piece is Jordan measurable). We also improve the Borel complexity of the pieces: namely, we show that each piece can be taken to be a Boolean combination of Fσ sets. This is a consequence of our more general result that applies to any two bounded subsets of Rk, k≥1, of equal positive measure whose boundaries have upper Minkowski dimension smaller than k.​

Writeup Paper

They are answering a different question here. The classical question of squaring the circle (which means to construct a square with same area of circle via compass and straight-edge construction) is resolved and is deemed impossible. The question stated in the paper asks if you can cut a disk into finitely many pieces and reassemble them by rotation and translation to get a square of the same area. But cutting up a disk into pieces in mathematics can mean absolutely ridiculous things and not just a naive slicing and dicing. This does not contradict the classical question of squaring a circle since they are fundementally asking about different things.

 

[Samson]

Can someone explain this question from the ChatGPT does an MBA paper?

QUESTION 5

The Pennsylvania Department of State is implementing a new electronic voting system. Voters will now use a very simple self-service computer kiosk for casting their ballots. If that kiosk is busy, voters will patiently queue up and wait until it is there turn.

It is expected that voters will spend on average 5 minutes at the kiosk. This time will vary across voters with a standard deviation of 5 minutes. Voters are expected to arrive at a demand rate of 10 voters per hour. These arrivals will be randomly spread out over the hour (you can assume that the number of voters arriving in any time period follows a Poisson distribution).

What is the average amount of time that a voter will have to wait before casting their vote?

Answer

To find the right answer, one must look at a standard equation from queuing theory. The equation for the average waiting time states that:

Average Waiting Time = Average Processing Time x Utilization / (1-Utilization). Plugging in an average processing time of 5 minutes and an average utilization of 5/6, we get:

Average Waiting Time = 5 x (5/6) / (1 - 5/6) = 25 minutes.

So, the correct answer is 25 minutes waiting in line. If we add the 5 minutes at the kiosk, we obtain a total of 30 minutes.

Sniff Test

If you can process 12 an hour and expect 10 an hour I would intuitively expect the mean wait time will be closer to zero than five people in front of you in the queue.

The wait for the first person will always be zero. At a rate of 10/hour surely the time the poll is open will be significant to the overall mean waiting time?

However this is an apparently reputable paper that is ited all over in the discussion about AI and stuff, so one would expect me to be wrong rather than them.

 

[thetrooper]

Not sure where to ask; could be here or in science/math.

Is there an outlier test in Excel?

Asked in another part of the forum, copied here.

 

[Kyriakos]

Asked in another part of the forum, copied here.

This seems to give all the needed info, from a first glance:

How (and Why) to Use the Outliers Function in Excel

An outlier is a value that is significantly higher or lower than most of the values in your data.

www.howtogeek.com

It says you manually use prompts to define the set (eg in the example, from b2 to b14) and have the program divide the values into quartiles. Following that, it establishes outliers according to a specific ratio to the interquartile range 1,3.
In the end of the article, it also gives a function which is built-in to give the mean average while allowing you to define the percentage you want to identify as outlier.

 

[thetrooper]

Nifty, Thanks!

 

[Samson]

They have invented a new shape



Called ‘the hat’, the 13-sided shape can be arranged in a tile formation such that it never forms a repeating grid

One of mathematics’ most intriguing visual mysteries has finally been solved – thanks to a hobbyist in England.

The conundrum: is there a shape that can be arranged in a tile formation, interlocking with itself ad infinitum, without the resulting pattern repeating over and over again?

In nature and on our bathroom walls, we typically see tile patterns that repeat in “a very predictable, regular way”, says Dr Craig Kaplan, an associate professor of computer science at the University of Waterloo in Ontario. What mathematicians were interested in were shapes that “guaranteed non-periodicity” – in other words, there was no way to tile them so that the overall pattern created a repeating grid.

Such a shape would be known as an aperiodic monotile, or “einstein” shape, meaning, in roughly translated German, “one shape” (and conveniently echoing the name of a certain theoretical physicist).

Now, mathematicians appear to have found what they were looking for: a 13-sided shape they call “the hat”. The discovery was largely the work of David Smith of the East Riding of Yorkshire, who had a longstanding interest in the question and investigated the problem using an online geometry platform. Once he’d found an intriguing shape, he told the New York Times, he would cut it out of cardstock and see how he could fit the first 32 pieces together.

Spoiler A video, but I do not understand it :

 

[Kyriakos]

(massive edit)
Hm, I'd like to read Penrose's reaction to this!
Up to now I (falsely) thought he had actually proven that you need at least two tiles for aperiodic tiling.

 

[Samson]

I have to admit I do not know what is different between this story and the one above. The tile is different though.

This infinite tiling pattern could end a 60-year mathematical quest

After 60 years of searching, mathematicians might have finally found a true single ‘aperiodic’ tile — a shape that can cover an infinite plane, but never make a repeating pattern.

Periodic tilings have translational symmetry: a honeycomb pattern, for example, can be repeated forever and looks identical after being shifted in any of six directions by any number of cells. But in aperiodic tilings any such shift is impossible.

In March, a team announced an important breakthrough in the search for an aperiodic tile. David Smith, a hobbyist mathematician based in Bridlington, UK, discovered a shape that he suspected could be an aperiodic tile and, together with three professional mathematicians, Smith wrote up a proof that his tile — together with its mirror, or flipped, image — could be used to build infinite aperiodic tilings of the plane. (The proof has not yet been peer reviewed, although mathematicians have reportedly said that it seems to be rigorous.)

Smith’s shape was not a single aperiodic tile, because it and its mirror image are effectively two separate tiles — and both versions were required for tiling the entire plane. But now the same group of mathematicians has reported a modified version of their original tile that can build aperiodic tilings without being flipped2.This proof was posted on the preprint server arXiv and has not yet been peer reviewed.

The first aperiodic tilings were discovered in the 1960s, and they involved 20,426 tile types. After various improvements, Roger Penrose, a mathematician at the University of Oxford, UK — who won a Nobel Prize in Physics in 2020 for his foundational work on the theory of black holes — discovered the first aperiodic tiling made of only two tile types that were not merely mirror images of each other. Penrose’s tilings now adorn the patio of Oxford’s mathematics department.

Spoiler Images :

Writeup Single tile paper Mirror image tile paper

 

[Harv]

My FB feed gave part of a video featuring a prisoner math problem - and this forum appears to be full of mathematicians!

100 Prisoners, numbers 1 to 100.
Each prisoner searches 50 (of 100) cells to find his number.
If all 100 prisoners find their number, they win.
If one fails to find his number, they lose.

I am told there is a strategy to improve the probability of success from negligible to 31%. I stopped the video (I think it is from Veritasium) to avoid seeing the solution, but the FB algorithm will make the video disappear. I will take a few days and give this one a go.

EDIT 1: I will add information as I think of it...
I believe Prisoners 1 through 100 operate sequentially as opposed to independently. This might not matter because we can run a strategy based on the assumption that the previous prisoner succeeded. So if Prisoner 1 searched Cells 1 through 50 and found his number, then that provides additional information for Prisoner 2.

ADDED 2:
100 factorial is the number of permutations. That might become relevant later.
We can reduce the probability of success to 0. We can have all prisoners choose cells 1-50 and this guarantees that half of them do not find their number.

ADDED 3:
I would do better to reduce the value of N from 100 to a more manageable number such as 2, then 4, then 6, then 10. Whatever the solution is, it will extend to larger numbers.
Also I would do well to watch the first part of the video again to determine if I missed a key piece of information.

 

[Samson]

Is communication allowed? I guess not or it should be trivial.

 

[Harv]

Is communication allowed? I guess not or it should be trivial.

I grabbed a wording of the problem from the Widipedia article.

The director of a prison offers 100 death row prisoners, who are numbered from 1 to 100, a last chance. A room contains a cupboard with 100 drawers. The director randomly puts one prisoner's number in each closed drawer. The prisoners enter the room, one after another. Each prisoner may open and look into 50 drawers in any order. The drawers are closed again afterwards. If, during this search, every prisoner finds their number in one of the drawers, all prisoners are pardoned. If even one prisoner does not find their number, all prisoners die. Before the first prisoner enters the room, the prisoners may discuss strategy — but may not communicate once the first prisoner enters to look in the drawers. What is the prisoners' best strategy?

100 prisoners problem - Wikipedia

en.wikipedia.org

So far, I reduced N=4 and only got a 1/6 probability of success - so I am still missing something.

 

[Kyriakos]

Should be directly related to the difference between random (=non tied) chances of x participants, and tied ones. For the simplest example, while if you have only two boxes, two players, and one attempt to get the correct box, in a random/untied attempt you have 1/2 for each player and consequently 1/4 for both winning (or both losing), if you tie the strategy (say they agreed to choose the box that corresponds to who goes first, who goes second) you now have a 1/2 chance of both winning or both losing, since effectively only one attempt is made (no room for one winning and one losing).

I think the result for (untied/no strategy) attempts of number x, in a set of options of number n, should be (n+n-1+...+n-x)/[(n)(n-1)...(n-x)], which gets simplified to (2n-x)(1/2)/(n!/x!). Of course this is when the position of "winning number" is random itself. So in the problem you posted you can incorporate that it isn't random at all, but specific for each player, leading to cumulative states and thus strategies to exploit them.

 

[Kyriakos]

Scaling triangles is a very popular part of new proofs (of established geometrical theorems, such as Ptolemy's), because it is both fun and easy to do.
I applied it to an (otherwise trivial) issue inspired by a recent basic video.

The point in the video (which is way too forced, since you can immediately calculate the needed area by simpler methods; eg since in similar triangles the ratio of corresponding sides is stable, height to smaller side is still 9/6=3/2=>(3/2x)^2+x^2=36 etc ), is to use the square of the ratio of corresponding sides of non-linked similar triangles as the ratio of their area. But this can be proven more elegantly (imo), by scaling:
a) you scale one (not already linked) similar triangle so as to have a common side with the other similar triangle. This means their common side is scaled by sideA/sideB (A is of the other triangle, B of the one scaled).
b) since they are similar triangles, when they link up (as in the image) they will inevitably form a right angle (this allows for circular calculations too, since it is known -Thales theorem etc- that if two linked up triangles have a right angle formed, the over-triangle has its base be the diameter of a circle, and all its vertices be points on the periphery).
c)In such a linked similar triangles formation, obviously the ratio of their respective areas is proportional to just the ratio of their bases. But due to (a) this gets multiplied by the same ratio, so ends up being the square of the ratio.