[HN Gopher] A Man Who Thought Too Fast (2020)
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A Man Who Thought Too Fast (2020)
Author : Anon84
Score : 48 points
Date : 2024-07-22 18:38 UTC (4 hours ago)
(HTM) web link (www.newyorker.com)
(TXT) w3m dump (www.newyorker.com)
| woleium wrote:
| needs (2020)
| woleium wrote:
| https://archive.is/zimBx
| fallinditch wrote:
| > 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.
| Animats wrote:
| Also 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
| valicord wrote:
| Why is that obvious?
| lcnPylGDnU4H9OF wrote:
| If 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
| 01HNNWZ0MV43FF wrote:
| Correct 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 wrote:
| All 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.
| bongodongobob wrote:
| Think about it and try to come up with a counterexample.
| valicord wrote:
| Just 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
| bongodongobob wrote:
| Because 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 wrote:
| You misunderstood the problem. That's not what it's
| saying at all.
| omnicognate wrote:
| The 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
| ChrisArchitect wrote:
| (2020)
|
| Discussion then: https://news.ycombinator.com/item?id=23011233
| TMWNN wrote:
| The 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>
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