4 ms·

Oh,I misread this title and thought it was going to be about maths in the year 2048.

Al-Khwarizmi 9y ago
Me too. I kind of expected a dystopian article where people in 2048 are so badly educated that the most advanced math they can do is count U.S. states one by one! Kind of disappointed it's not that, to be honest :)
- tomahunt 9y ago
  I was expecting a maths where quantum computation had taken over analytic number theory and theorems could be proved using brute force enumeration.Or something
  - Filligree 9y ago
    Quantum computers can't do that. Not even in principle. :p
    - whatshisface 9y ago
      If you take just one piece of information from this blog: Quantum computers would not solve hard search problems instantaneously by simply trying all the possible solutions at once (The subtitle of Aaronson's blog.) https://www.scottaaronson.com/blog/ https://www.scottaaronson.com/blog/
shaunxcode 9y ago
Yeah sorry working on that one now. Stay tuned though!
- tomahunt 9y ago
  I don't think it needs to be changed. It's actually really interesting to wonder what Mathematics is going to be like in 2048.
raverbashing 9y ago
And that they were counting (US) states by exaustive enumeration. Well, yeah, how are you going to count them?
- jameshart 9y ago
  First, capture, tag, and release a number of US senators. 15 or so should work. Then, reset your traps and capture some more. Count how many of the second capture set were tagged in the first group. This tells you what proportion of the population of Senators you captured the first time, and by dividing the number of senators you tagged by this proportion you can estimate the total population of senators. Divide that number by two to obtain the number of states.
- jameshart 9y ago
  Knowing that a bijection exists between states and stars on the US flag, first obtain a US flag. Observe that the stars in the top left corner fall in a rectangular n x m grid with stars located at every node, and also in the rows and columns in between, forming a smaller (n-1) x (m-1) rectangle, so the number of stars is nm + (n-1)(m-1). Then you can just count n and m - 6 and 5 respectively - and substitute to obtain 30+20=50 stars.

Me too. I kind of expected a dystopian article where people in 2048 are so badly educated that the most advanced math they can do is count U.S. states one by one! Kind of disappointed it's not that, to be honest :)

tomahunt 9y ago
I was expecting a maths where quantum computation had taken over analytic number theory and theorems could be proved using brute force enumeration.Or something
- Filligree 9y ago
  Quantum computers can't do that. Not even in principle. :p
  - whatshisface 9y ago
    If you take just one piece of information from this blog: Quantum computers would not solve hard search problems instantaneously by simply trying all the possible solutions at once (The subtitle of Aaronson's blog.) https://www.scottaaronson.com/blog/ https://www.scottaaronson.com/blog/

shaunxcode 9y ago

Yeah sorry working on that one now. Stay tuned though!

tomahunt 9y ago
I don't think it needs to be changed. It's actually really interesting to wonder what Mathematics is going to be like in 2048.

raverbashing 9y ago

And that they were counting (US) states by exaustive enumeration. Well, yeah, how are you going to count them?

jameshart 9y ago
First, capture, tag, and release a number of US senators. 15 or so should work. Then, reset your traps and capture some more. Count how many of the second capture set were tagged in the first group. This tells you what proportion of the population of Senators you captured the first time, and by dividing the number of senators you tagged by this proportion you can estimate the total population of senators. Divide that number by two to obtain the number of states.
jameshart 9y ago
Knowing that a bijection exists between states and stars on the US flag, first obtain a US flag. Observe that the stars in the top left corner fall in a rectangular n x m grid with stars located at every node, and also in the rows and columns in between, forming a smaller (n-1) x (m-1) rectangle, so the number of stars is nm + (n-1)(m-1). Then you can just count n and m - 6 and 5 respectively - and substitute to obtain 30+20=50 stars.