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Comment by tzs | Hacker News Reader

I think blog.wolfram.com is probably the best company blog I've seen, from a marketing perspective. A large number of the entries are basically of the form "Here is an interesting problem, and here's how I solved it with Wolfram products". They generally let the problem have the spotlight rather than focus on the Wolfram products, so it doesn't feel like you are getting pitched.

Here's a related problem, but for e: using each digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 at most once each, and only the operators +, -, x, /, and ^ (exponentiation), and parenthesis for grouping, how close can you get to e? Digits may not be concatenated--for instance, you cannot get a 23 by simply placing the 2 next to the 3.

My best, after about 15 minutes of fiddling, were:

   (3x(4x7+1))/2^5 = 2.71875 ≈ e + 0.000468172

and

    2x(9x6-1)/(3x(8+5)) = 2.717948... ≈ e - 0.000333111

but it turns out you can do FAR better.

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((8^(5/4))^(((2/(0-(7+1)))^9)^(3-6))) = 2.718281848685499 ~ e +/- 2.0226453845140213e-08

That was fun. Although I'm having trouble confirming the answer... - Also I'm meant to be packing.

http://play.golang.org/p/G_Y5SblSuv for some brute force eval :)

Edit: This could be a nice demonstration for a genetic algorithm: A clearly defined fitness function yet an unknown (unknowable?) goal, and a distinct representation of the genes. I'm not sure how hereditary the fitness is though.

Further edit: Just 4 digits: (6-(8^(4/7))) 2.718658575969448 0.00037674751040306376 - I hope this is a) correct, b) interesting to someone else.

You beat me by about 5 orders of magnitude, but it is possible to do better, by another 200000 orders of magnitude or so! :-)

Best I've seen is given in this discussion: http://www.reddit.com/r/math/comments/zakqh/using_the_number...

Oh well, I'm sure it would have got there eventually...

Interesting problem nonetheless.

I have one of those, a symbolic regressor (not genetic) lying around in a broken state. The last working version gave:

  short: 2981^ (1/8)
  less short: (2574 + 4903 ^ (1/4))/950

I used: http://apod.nasa.gov/htmltest/gifcity/e.1mil

Perhaps I am misunderstanding your output but it is only correct up on till: 2.7182818

Digit concatenation was not allowed in the original challenge. If digit concatenation is allowed, then I believe that the best known is:

   (1+9^(-4^(7x6))^(3^(2^85))

which gives e to 18457734525360901453873570 decimal digits.

"Digits may not be concatenated--for instance, you cannot get a 23 by simply placing the 2 next to the 3."

I think that renders both of your solutions invalid, but cool nonetheless!

Oh sorry guys missed the digit concatenation bit. In that case I get:

  (9 - (5/7)^(1/2)) / 3

I was tired when I saw that post, it was late/early. rmccue, I also seemed to have missed that you gave an error bound =(

This has been Wolfram's strategy since the beginning. They have their own journal, books, user groups, all showing how to solve problems with mathematica.

I can do it with only 9.

FlipH is not a legal operator.

No idea what you're talking about. I'm merely improving the typography.

From the original spec of the problem: "only the operators +, -, x, /, and ^ (exponentiation), and parenthesis for grouping"

It has to be presentable. Typography matters.

Yeah agree, they know how to market their stuff. (Maybe Alpha is a marketing instrument after all? ;))

However I think for this problem, and many others, mastering a general purpose programming language is much more efficient. After all you can pipeline symbolic expressions to Mathematica - or Maple >:) - and you have the best of both worlds. Or you just use Ruby or another highly expressive language - the syntax of simple symbolic expressions is basically the same.

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