6 ms·
A Man Who Thought Too Fast (2020)
- woleium 2y agoneeds (2020)
- woleium 2y agohttps://archive.is/zimBx https://archive.is/zimBx
- fallinditch 2y ago> Ramsey theory tells us, for instance, that among any six users of Facebook there will always be either a trio of mutual friends or a trio in which none are friends. AKA Ramsey's Theory of the Bleedin' Obvious.
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- Animats 2y agoAlso see Ramsey Pricing.[1] "Ramsey pricing (for a monopolist) says to mark up most the goods with the least elastic (that is, least price-sensitive) demand or supply." That's kind of obvious. [1] https://en.wikipedia.org/wiki/Ramsey_problem https://en.wikipedia.org/wiki/Ramsey_problem
- valicord 2y agoWhy is that obvious?
- lcnPylGDnU4H9OF 2y agoIf you try, you'll find that you can't think of a configuration for which neither is true. (Maybe it's not "obvious" but it's easy to arrive at that conclusion after some thought.) This comment from the other linked discussion describes what was actually proved; the narrow case of facebook friends is just a given example. https://news.ycombinator.com/item?id=23028299 https://news.ycombinator.com/item?id=23028299
- 01HNNWZ0MV43FF 2y agoCorrect my math if I'm wrong 6 people have 6 * (6 - 1) / 2 = 15 possible connections. If you toggle those off and on you have about 2 ^ 15 = 32,768 combinations. Some of those are surely redundant but I don't know stats well enough to de-dupe them. Either way I feel like it's higher than I can count in my head while also doing anything else mathematical
- bongodongobob 2y agoAll that is completely irrelevant, you're overcomplicating it. Each person has 1 of 2 states, connected or not. If 4 are connected, 2 aren't, if 4 aren't 2 are, etc.
- kevinwang 2y agoNo, each edge in the graph has 2 states. There are 15 edges between the 6 people.
- bongodongobob 2y agoBut the edge states don't matter, the node states do. They're either connected friends or they aren't. What's the third state?
- 01HNNWZ0MV43FF 2y ago2 groups of 3 mutual friends is a different graph than 3 pairs of friends
- lmm 2y agoNo, the connections are what matters. A may be friends with both B and C, but that doesn't mean that B is necessarily friends with C.
- bongodongobob 2y ago
- bongodongobob 2y agoThink about it and try to come up with a counterexample.
- valicord 2y agoJust because you can't come up with a counterexample easily doesn't make it obvious or even true. Of course in this case it is indeed true, but I'm asking what makes it obvious
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- bongodongobob 2y agoBecause it's saying "you have 6 blocks, they can be either green or blue. No matter what colors they are, you'll always have either 3 blue or 3 green."
- yongjik 2y agoYou misunderstood the problem. That's not what it's saying at all.
- bongodongobob 2y agoHow isn't it? There's no way to have 3 connected friends and 4 unconnected friends.
- Dylan16807 2y agoYou can have all 6 connected and still be in either category. For example, 6 friends in a loop can be divided into even and odd trios, where each trio has no direct friend connections within it. But six people in two separate triangles fall into the other category, with two trios of mutual friends. Notice how both of these cases give each person 2 friends. It's not nearly as simple as assigning everyone a color at the start. Another thing to think about: If you have a triangle of 3, and a line of 3, then you can find 3 mutual friends and three mutual non-friends. How would you use colors to differentiate this situation from the two-triangles situation?
- bee_rider 2y agoWell it’s written down right there, isn’t it? Who’d have written it down if it wasn’t true? Obviously nobody would do such a thing!
- omnicognate 2y agoThe proof of that particular case goes like this: Pick one of the six users. Split the other 5 users into friends and non-friends of the chosen user. There will either be at least 3 friends or at least 3 non-friends of the chosen user. If you can pick 3 friends of the chosen user then if any two of them are friends they plus the original user form a trio of mutual friends. If none of the 3 are friends then you have a trio in which none are friends. Similarly, if you can pick 3 non-friends of the original user then either two of them are non friends and you have a trio of non friends or all three are friends, forming a trio of mutual friends. It's easy to prove but I don't think I'd quite call it "bleedin' obvious". Ramsey's theorem [1], of course, is more general than that and isn't at all obvious (although it's still not very hard to prove). [1] https://en.m.wikipedia.org/wiki/Ramsey%27s_theorem https://en.m.wikipedia.org/wiki/Ramsey%27s_theorem
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- ChrisArchitect 2y ago(2020) Discussion then: https://news.ycombinator.com/item?id=23011233 https://news.ycombinator.com/item?id=23011233
- TMWNN 2y agoThe headline reminds me of why Isaac Asimov was bad at chess (start at "In 1994, Isaac Asimov's last autobiography"): <http://billwall.phpwebhosting.com/articles/Asimov_chess.htm http://billwall.phpwebhosting.com/articles/Asimov_chess.htm>
- earthboundkid 2y ago> The conclusion that I finally came to (right or wrong) was that I was unwilling to study the chessboard and weigh the consequences of each possible move I might make. Even people who couldn't see complex patterns might at least penetrate two or three moves ahead, but not I. I moved entirely on impulse, if not at random, and could not make myself do anything else. That meant I would almost certainly lose. > And again - why? To me, it seems obvious. I was spoiled by my ability to understand instantly, my ability to recall instantly. I expected to see things at once and I refused to accept a situation in which that was not possible. Interesting. I was also fairly bad at chess as a kid, and my self-diagnosis was similar. I found "checking" all the possible moves too tedious, so I just did the first one that looked promising, to terrible effect. I've tried to play a little as an adult, and now I have more patience, but that means I find chess too stressful! I don't want to go until I'm fairly sure I have a good move, but I still can't quite "check" all the possible moves, so I sweat the whole time and don't enjoy playing.
- mihaic 2y agoThanks for sharing, I have the same experience with chess, and have never liked it for the same reasons. There's no mediocre version of chess that you can do instinctively, and I don't really like to do tedious and repetitive mental work when playing games. I just suck at chess, and have given it up, as it's something that's not for me. I can play some decent poker though (not really great, just better than chess).
- tudorconstantin 2y agoI think “puzzle storms” on lichess can be considered a way of fiddling with chess instinctively. I found it also helps with not giving pieces for free when playing full games and also seeing the blunders your opponent makes way faster
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- IronWolve 2y agoQuestion, Do they have to measure a persons brain speed? I wonder what the bell curve is on society, there has to be some very fast outliers.
- danwills 2y agoSuch a multidimensional question, I think there's probably little point in being able to boil brain speed down to one number (or even small set of numbers). There's so many parts that all combine: The rate of propagation of activation-potentials along neural dendrites (well known that caffeine speeds this up, until your brain adapts to make it 'normal speed' again by extra myelination!), there's stuff like the frequency of brainwaves, which surely must have some effect.. and then the conscious-perception of the passage of time could kinda feed in too, not to mention memory-speed or speed at tasks that have been very highly practiced. Very interesting to think about though, thanks!
- BizarroLand 2y agoAlso, to follow up, the question is speed in what domain? Speed reading, speed computation, speed memorization, there are many domains of the mind in which the question of speed could be applied.
- noisem4ker 2y ago> caffeine speeds this up, until your brain adapts to make it 'normal speed' again by extra myelination Any more info or source on this? Thanks.