5 ms·
How to fit an elephant (2011)
- triclops200 9y agoNow can you add one more and make him wiggle his trunk?
- dEnigma 9y agoIn the actual paper they do exactly that https://www.google.at/url?sa=t&source=web&rct=j&url=https://publications.mpi-cbg.de/Mayer_2010_4314.pdf&ved=0ahUKEwiqzNeRmuvWAhWFb1AKHXVsBDYQFggtMAI&usg=AOvVaw1AmrjYniYTf8n9dYp405VW https://www.google.at/url?sa=t&source=web&rct=j&url=https://...
- deleted 9y ago[deleted]
- oddeyed 9y agoI can't help but feel like a complex number is two parameters (real&imag / mod&arg) - so really this is 8 parameters.
- Y_Y 9y agoCantor showed that these sets have the same cardinality. You can represent a complex (in the form of two reals) by interleaving digits or using a space-filling curve for example.
- Bromskloss 9y agoSo, it's all actually just one parameter?
- deleted 9y ago[deleted]
- wolfgke 9y ago> Cantor showed that these sets have the same cardinality. You can represent a complex (in the form of two reals) by interleaving digits or using a space-filling curve for example. But this (set) isomorphism between R and C is not continuous. Indeed one can show that there exists no continuous epimorphism f: R^n -> R^m, where m > n, since for every such continuous map f the image f(R^n) has a measure of 0 with respect to the Borel measure in R^m.
- waqf 9y ago> there exists no continuous epimorphism f: R^n -> R^m, where m > n Really? Then what is a https://en.wikipedia.org/wiki/Space-filling_curve https://en.wikipedia.org/wiki/Space-filling_curve?
- dahart 9y agoRight from that article you linked: "A non-self-intersecting continuous curve cannot fill the unit square because that will make the curve a homeomorphism from the unit interval onto the unit square (any continuous bijection from a compact space onto a Hausdorff space is a homeomorphism). But a unit square has no cut-point, and so cannot be homeomorphic to the unit interval, in which all points except the endpoints are cut-points."
- waqf 9y agoYeah, a non-self-intersecting map cannot, but OP didn't specify that, only "epimorphic" which certainly can. Moreover OP's argument specifically proves too much, because space-filling curves (as described in the article) have a range with positive Borel measure.
- Y_Y 9y agoSure, but did we need continuity? Also, if you want to be awkward, you can get around this by using the discrete topology, I don't think we needed the metric structure of R^n.
- 9y ago
- dahart 9y agoPairing functions only work on countable sets. This is funny because Cantor is the same person who proved real numbers are uncountable, and that there is no pairing function between 1 real number and naturals, let alone 2. https://en.m.wikipedia.org/wiki/Countable_set https://en.m.wikipedia.org/wiki/Countable_set
- Y_Y 9y agoI'm claiming a "pairing function" between single reals and pairs of reals. They have respective cardinalities 2^N0 and 2*2^N0=2^N0 where N0<2^N0 is the cardinality of the naturals.
- dahart 9y agoYou're right. I was wrong. I found an explanation of how to make interleaving method work. https://math.stackexchange.com/a/183383 https://math.stackexchange.com/a/183383
- Aardwolf 9y agoHe could have used an octonion, then it was 1 parameter.
- xelxebar 9y agoThis is a good point. If we're posing Occam's Razor-like arguments against models for having "too many free parameters", then we should probably really be comparing something more precise, like the models' Kolmogorov complexities. Otherwise, it's just too easy to hide a lot of complex machinery inside a "single parameter". In fact, from this perspective it's arguable that an arbitrary real number is actually a (countably) infinite set of parameters, since it takes that many bits to uniquely specify any real number.
- rkroondotnet 9y agoHow do you fit an elephant? One parameter at a time.
- pvg 9y agoThe only thing missing here is a little anecdote about how Von Neumann, when challenged on this, did it in his head and started rattling off the parameters. For arbitrary animals.
- scott00 9y agoSource please. Must learn more.
- pvg 9y agoVon Neumann was renowned for his great prowess at mental maths. A famous (if also not entirely serious) story: "When posed with a variant of this question involving a fly and two bicycles, John von Neumann is reputed to have immediately answered with the correct result. When subsequently asked if he had heard the short-cut solution, he answered no, that his immediate answer had been a result of explicitly summing the series (MacRae 1992, p. 10; Borwein and Bailey 2003, p. 42)." From http://mathworld.wolfram.com/TwoTrainsPuzzle.html http://mathworld.wolfram.com/TwoTrainsPuzzle.html
- mamon 9y agoThe article you linked to provides a trivial solution: "the trains take one hour to collide (their relative speed is 100 km/h and they are 100 km apart initially). Since the fly is traveling at 75 km/h and flies continuously until it is squashed (which it is to be supposed occurs a split second before the two oncoming trains squash one another), it must therefore travel 75 km in the hour's time." So if von Neumann was solving it by explicitly summing the series, as the anecdote claims, then he was doing it wrong :)
- pvg 9y agoI assume you're trying to beat the original responder in brazen literalism - now we have a citation request for a throwaway joke, an explanation (with citation) and a literal interpretation of the joke in the citation.
- 9y ago