Multiformat Listening Test @ 64 kbps  FINISHED 
Multiformat Listening Test @ 64 kbps  FINISHED 
Aug 16 2007, 00:00
Post
#1


Group: Members Posts: 3642 Joined: 14May 03 From: Bad Herrenalb Member No.: 6613 
The much awaited results of the Public, Multiformat Listening Test @ 64 kbps are ready  partially. So far, I only uploaded an overall plot along with a zoomed version. The details will be available tomorrow. You can also download the encryption key on the results page that is located here:
http://www.listeningtests.info/mf641/results.htm http://www.listeningtests.info/mf641/resultsz.png Nero and WMA Professional 10 are tied and WMA Professional 10 is tied to Vorbis. Vorbis however performed worse than Nero. Of course, High Anchor is best and Low Anchor loses. This one goes to the experts: How would you rank codecs in such a situation, where A=B and B=C, but C<A?  http://listeningtests.hydrogenaud.io/sebastian/



Aug 19 2007, 15:44
Post
#2


Group: Members Posts: 216 Joined: 20July 03 Member No.: 7896 
This one goes to the experts: How would you rank codecs in such a situation, where A=B and B=C, but C<A? That is actually not a contradiction as such (though further expert opinion on the actual statistical metric used is needed). You think that is a contradiction, because such situation doesn't happen in "normal" number systems, like integers, reals, etc. What you noticing is the property of total order breaking. However not all valuations have that property. Take integers, if you take 2 integers at random there is the way to count from one to the other, precisely because there is a total order and you know what is less/greater than what, what equals what and what follows what. On the other hand take Complex numbers, this is the first number system students usually exposed to in school that doesn't have total order on its elements (though school teachers don't usually mention that). Given 2 random Complex numbers there isn't "the" way to count from one to the other, in fact there are infinitely many ways, all correct in some sense. So while I don't know if the underlying statistical measure produces set of values that has total order, your example (if not subject to some freaky error) shows that it doesn't, and should be read as raking: 1) HE = WMA 2) Ogg My immediate intuition would be to use equivalence classes to solve this problem. 1) Make individual comparison between every possible pair 2) Look at the ones with strict inequalities 3) Pick the largest of them all (strictly greater, not >=) 4) Rank that first 5) Add all who are directly equal to it to its equivalence class (not ones that are equal by some chain of equalities) 6) Removing them from further consideration 7) From remaining, rank the next largest as number 2 8) Go to 5 and repeat for the rest of the ranking. This post has been edited by lexor: Aug 19 2007, 16:00  The Plan Within Plans



LoFi Version  Time is now: 3rd June 2015  12:43 