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A Math Genius Blooms Late and Conquers His Field
- deleted 9y ago[deleted]
- failrate 9y agoIt is always worthwhile to introduce the techniques from one field of study into another.
- bmh100 9y agoI would just like to express my gratitude to [Kevin Hartnett](https://www.wired.com/author/kevin-hartnett/ https://www.wired.com/author/kevin-hartnett/) for making an enjoyable article that I could almost follow as a quantitatively minded programmer / non-mathematician. It makes sense saying that graphs are somehow a form of matroid. Even without knowing what a matroid is, I get a sense of the importance of spatial relationships.
- __mbm__ 9y agoKevin wrote a blog post on this topic here: https://www.quantamagazine.org/the-tricky-translation-of-mathematical-ideas-20170628/ https://www.quantamagazine.org/the-tricky-translation-of-mat...
- codepie 9y agoAs another user pointed out, why should be the chromatic polynomial of rectangle with deleted edge be: q^4 - 3q^3 + 2q^2 and not q * (q - 1)^3. A counter example: when q=2, we have two ways to color the rectangle with a deleted edge. Am I missing something? I think fixating q as the number of possible ways to color the end points of the deleted edge leads to the wrong result.
- omaranto 9y agoThis article was "reprinted" from the original at Quanta Magazine. The original article has the correct answer of q(q-1)^3. (It was probably corrected after Wired "reprinted" it.) https://www.quantamagazine.org/a-path-less-taken-to-the-peak-of-the-math-world-20170627/ https://www.quantamagazine.org/a-path-less-taken-to-the-peak...
- _asummers 9y agoIndeed. At the bottom: > Correction June 27, 2017: The original version of this article included an error in the calculation of the chromatic polynomial of a rectangle.
- trentmb 9y agoPrevious discussion: https://news.ycombinator.com/item?id=14646280 https://news.ycombinator.com/item?id=14646280
- davidcamel 9y agoI majored in math in undergrad, and I always daydreamed about solving difficult mathematical problems despite a lack of formal training. I even had a teacher that I had to "pretend to understand". Seeing a real-world example of this fantasy come true is fascinating. The article was also surprisingly well-written; most mention of higher mathematics in the media is oversimplified to death, but this was an honest and yet approachable presentation of the Rota conjecture (now theorem). By the way, here's another result on chromatic polynomials (proved first by I don't know, but re-discovered by my combinatorics class): Define a "gluing" operation by taking two graphs and connecting them along a common vertex. The chromatic polynomial, h(x), of the new graph, is the product of the chromatic polynomials of the subgraphs over x: h(x) = f(x)*g(x) / x.
- vlasev 9y agoFor more such properties, check out this paper [1], section 1.5. [1]: https://www.cs.elte.hu/blobs/diplomamunkak/mat/2009/hubai_tamas.pdf https://www.cs.elte.hu/blobs/diplomamunkak/mat/2009/hubai_ta...
- JadeNB 9y agoThat's neat! If you redundantly add an extra vertex (in its own component) whenever you glue, then you actually get h(x) = f(x)g(x). I wonder if there's some natural way of defining a multiplication of connected graphs so that you get equality on the nose?
- KKKKkkkk1 9y agoDoes getting an undergrad degree in astronomy and physics, and then a math PhD from UIUC, really count as "blooms late"? Sounds like a fairly standard career path to me.
- ianai 9y agoYes, calling him a late bloomer when he's 34 is pretty awful.
- vnchr 9y agoThere's a traditional view in math that most great accomplishments are had before 30, based on past mathematicians' successes.
- __mbm__ 9y agoYou're correct: Wired has hyped the original title of the article in Quanta Magazine: "A Path Less Taken to the Peak of the Math World". Any time you see the words "Math Genius" you should assume hyperbole. If you like this kind of story, I highly recommend putting Quanta (https://www.quantamagazine.org/ https://www.quantamagazine.org/) on your radar and avoiding most of the tabloid-science articles in Wired.
- foobaw 9y agoMaybe he's a late bloomer in the sense that he was neither a math prodigy or attended IMO. I personally know several IMO winners who started studying Math when they were 3~4 years old and it's been part of their life.
- vinhboy 9y ago> his father taught statistics and his mother became one of the first professors of Russian literature in South Korea I notice that really talented people, always have talented parents. Rarely do I read stories about poor blue collar parents producing science wiz. It leads me to believe that genetics play a much bigger role in our intelligence than nurture.
- screye 9y agoDoesn't it support nurture in a way though ? Children with genius parents usually expose their children to high level content very early into their childhood. They also pass on a way of thinking and intuition of their subjects that a non-expert in the field won't have.
- bradjohnson 9y agoWhy does it lead you to believe that? If they have talented parents, wouldn't the parents raise them in a way to encourage their talents to blossom? Would be interesting to see if the children of talented parents that are put up for adoption and raised by average parents are as successful. Or vice versa, talented parents raising children of average parents.
- vinhboy 9y agoI guess you're right. I am obviously not that smart... Haha.. But to your point, Steve Jobs is an example of that. He had blue collar adoptive parents, but his birth parents were PhD level people. I am not gonna put my foot in my mouth again and say this is proof of anything, but it is interesting to me.
- jacquesm 9y agoIt's the nature vs nurture thing. It mostly boils down to 'a bit of both' and if you are really lucky in either department then you can still very well manage to succeed.
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- samfisher83 9y agoUIUC is a really good school. Getting in there is not easy.
- vlasev 9y ago> "Every one of these graphs has a unique chromatic polynomial" This is incorrect. Two different graphs may have the same chromatic polynomial. For example, all trees of N vertices have the same chromatic polynomial: x(x-1)^(N-1)
- soVeryTired 9y agoI think they're trying to say that the graph uniquely determines the polynomial, rather than that the polynomial determines the graph. Or at least, that's how I read it. But I agree it's a bit ambiguous.
- JadeNB 9y ago> > "Every one of these graphs has a unique chromatic polynomial" > This is incorrect. Two different graphs may have the same chromatic polynomial. For example, all trees of N vertices have the same chromatic polynomial: x(x-1)^(N-1) As soverytired (https://news.ycombinator.com/item?id=14697626 https://news.ycombinator.com/item?id=14697626) points out, you're refuting the claim that the graphs have distinct chromatic polynomials. To say that a graph has a unique chromatic polynomial means that it has only one, not that no other graph has the same one. (For example, (almost?) everyone has a unique biological mother.)