Talk:Entropy

What about a less dull example?

What about using "Rosetta code" as an example instead of the dull "1223334444"?--Grondilu 17:53, 22 February 2013 (UTC)

Or even funnier: write a program who computes its own entropy :) --Grondilu 12:40, 25 February 2013 (UTC)

Better yet, a bonus task of computing the entropy of each solution on the page. :) --TimToady 19:23, 25 February 2013 (UTC)

I like computing the entropy of “Rosetta Code” (it's about 3.08496, assuming my code is right); a more self-referential one is fine too, except it involves features that might block some languages from participating. (The draft child task is a better place for that.) –Donal Fellows 09:31, 26 February 2013 (UTC)

Alternate form

Not sure this is very useful, but I was wondering if one could not find a more concise way of writing this.

If we call $N$ the length of the string and $n_{c}$ the number of occurrences of the character c, we have:

$H=\sum _{c}-p_{c}\ln p_{c}=\sum _{c}-{\frac {n_{c}}{N}}\ln {\frac {n_{c}}{N}}=\ln N-{\frac {1}{N}}\sum _{c}n_{c}\ln n_{c}$

In perl6, this allows a slightly simpler formula, i.e. not using hyperoperators:

<lang Perl 6>sub entropy(@a) {

   log(@a) - @a R/ [+] map -> \n { n * log n }, @a.bag.values

}</lang>

For what it's worth.--Grondilu 18:14, 4 March 2013 (UTC)