|Just another Perl shrine|
I feel like this should be an episode of Doctor Who with incongruent time lines.
At the time I responded to the clarification of what I meant by inversely proportional, I had already moved on from thinking it was an appropriate solution. I only clarified for the sake of completeness. What I should have said was something along the lines of:
Not that it matters since I now realize it does not solve my problem but there is a difference between having an inverse relationship and being inversly proportional. The idea that the more popular an animal is the fewer questions it should take to identify is an inverse relationship. When I said inversly proportional I meant that the product of popularity to questions should be a constant defining exactly how many questions should be asked. In the end, I was wrong.
As for what I am trying to achieve - I am attempting to build on top of Huffman coding. Let's say you have a file that you have done single byte frequency analysis on and generated a Huffman code tree. You notice that a few of the branches only have 1 leaf entry instead of 2. You decide you want to fill in those "holes" with with the highest frequency 2 byte pairs in the file. You fill in the first hole but before moving on to the next one, you realize a problem. The frequency analysis of the single bytes requires recalculating which means rebuilding the tree which means different holes.
I am not sure if that makes any more sense. I took the weekend off from thinking about it in hopes that I would have clarity today but it is still a jumbled pile of mud in my mind.
Cheers - L~R
In reply to Re^6: Challenge: Optimal Animals/Pangolins Strategy