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An easy way to understand the unit n-dimensional hypercube is to consider the region of space in n dimensions that you can get if every coordinate component lies in [0, 1]. So for one dimension it's the line segment from 0 to 1, for two dimensions it's the square with corners (0, 0) and (1, 1), etc.

Write a program or function that given n returns the average Euclidean distance of two points uniformly random selected from the unit n-dimension hypercube. Your answer must be within 10-6 of the actual value. It's ok if your answer overflows your language's native floating point type for big n.

Randomly selecting a 'big' number of points and calculating the average does not guarantee such accuracy.

Examples:

1 → 0.3333333333...
2 → 0.5214054331...
3 → 0.6617071822...
4 → 0.7776656535...
5 → 0.8785309152...
6 → 0.9689420830...
7 → 1.0515838734...
8 → 1.1281653402...

Data acquired from MathWorld.

This is , lowest byte-count wins.

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  • \$\begingroup\$ Reiterated winning criterion in an edit. \$\endgroup\$
    – Magic Octopus Urn
    Feb 1, 2017 at 20:54
  • \$\begingroup\$ Just to be clear: distance refers to the Euclidean distance, yes? \$\endgroup\$
    – Dennis
    Feb 1, 2017 at 20:56
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    \$\begingroup\$ @carusocomputing What's the point of the challenge if you want me to solve it for you? \$\endgroup\$
    – orlp
    Feb 1, 2017 at 21:39
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    \$\begingroup\$ @orlp registering my objection to a challenge that is a math puzzle until someone figures out the math, then it becomes a programming puzzle when everyone copies the formula in different languages. I need to ask a meta question about this. \$\endgroup\$
    – Sparr
    Feb 1, 2017 at 22:22
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    \$\begingroup\$ When you say 5 digits of accuracy, do you mean to within 1e-5, or would an estimate of 1.500000000000001 be wrong when the output should be 1.499999999999999? \$\endgroup\$
    – xnor
    Feb 1, 2017 at 22:49

1 Answer 1

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Mathematica, 68 bytes

NIntegrate[(1-((E^-u^2+u*Erf@u√π-1)/u^2)^#)/u^2,{u,0,∞}]/√π&

Implementation of the formula using NIntegrate to approximate its value.

Image

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  • \$\begingroup\$ Beat me by 58 seconds :( \$\endgroup\$
    – JungHwan Min
    Feb 1, 2017 at 23:30
  • \$\begingroup\$ Where did you find the formula? \$\endgroup\$
    – feersum
    Feb 2, 2017 at 22:58
  • \$\begingroup\$ @feersum It's formula (8) from Hypercube Line Picking. (10) is a typo and shouldn't have the double integral or du. \$\endgroup\$
    – miles
    Feb 2, 2017 at 23:03
  • \$\begingroup\$ If you actually input this as text it parses u√π as one token, so you need a space between u and √. \$\endgroup\$
    – feersum
    Feb 2, 2017 at 23:41

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