From: rjh@cpax.org.uk   
      
   On 24/12/2022 4:53 pm, Mike Terry wrote:   
   > On 24/12/2022 14:21, Tim Rentsch wrote:   
   >> Ben Bacarisse  writes:   
   >>   
   >>> Tim Rentsch  writes:   
   >>>   
   >>>> ram@zedat.fu-berlin.de (Stefan Ram) writes:   
   >>>>   
   >>>>> ram@zedat.fu-berlin.de (Stefan Ram) writes:   
   >>>>>   
   >>>>>> Given n times of the 24-hour day, print their   
   >>>>>> average. For example, the average of "eight o'clock"   
   >>>>>> and "ten o'clock" (n=2) would be "nine o'clock".   
   >>>>>> (You can choose any representation, for example   
   >>>>>> "HH:MM" or "seconds since midnight".)   
   >>>>>   
   >>>>> Thanks for all replies!   
   >>>>>   
   >>>>> I waited a few days before answering to allow   
   >>>>> sufficient time to think about the problem.   
   >>>>>   
   >>>>> There were not enough tests written and run.  As a   
   >>>>> result, the puzzle has not yet been solved (unless I   
   >>>>> have overlooked a contribution or misworded   
   >>>>> expectations).   
   >>>>>   
   >>>>> So, here are two possible test cases.   
   >>>>>   
   >>>>> average( 23.5,  1.5 )==  0.5 average( 11.5, 13.5 )==   
   >>>>> 12.5   
   >>>>>   
   >>>>> (I use hours as units, so "0.5" means, "half past   
   >>>>> midnight".)   
   >>>>>   
   >>>>> I hope that these test cases encode sensible   
   >>>>> expectations for an average of two times on a 24-hour   
   >>>>> clock in the spirit of the example given in the OP,   
   >>>>> which was, "the average of eight o'clock and ten   
   >>>>> o'clock would be nine o'clock", since these test cases   
   >>>>> just have rotated that example by 3.5 and 15.5 hours.   
   >>>>>   
   >>>>> I believe that I have not seen an algorithm so far in   
   >>>>> this thread that would pass these tests.   
   >>>>   
   >>>> As before, the problem is underspecified.   
   >>>   
   >>> Some remarks not specifically in reply to you, Tim...   
   >>   
   >> I didn't really say very much.  Your comments are a lot   
   >> more interesting.   
   >>   
   >>> The input is a collection, t(n), of n > 1 numbers in [0,   
   >>> 24). The average should be a number, A, in [0, 24) that   
   >>> minimises   
   >>>   
   >>> Sum_{i=1,n} distance(A, t(i))   
   >>>   
   >>> (or Sum_{i=1,n} difference(A, t(i))^2 if you prefer to   
   >>> think in terms of variance).  So far, this is just what an   
   >>> average is.  The key point is what is the distance (or   
   >>> difference) whose sum (or sum of squares) we want to   
   >>> minimise?  For times, I would say it is the length of the   
   >>> shorter arc round an imaginary 24-hour clock face.   
   >>   
   >> Minimizing the square of the distance (aka length of the   
   >> shorter arc) is one metric, and I think a reasonable one.   
   >> (Minimizing the absolute value of the distance gives a   
   >> result more like the median than the mean, IIANM.)   
   >   
   > Yes - perhaps Ben was thinking of making   
   >   
   > Sum_{i=1,n} signed_distance(A, t(i))  =  0   
   >   
   > rather than minimising the (absolute) distance.   
      
   And so perhaps you are closer to what Ben was thinking of, but as   
   far as we know you are no closer to knowing what Stefan was   
   thinking of.   
      
   --   
   Richard Heathfield   
   Email: rjh at cpax dot org dot uk   
   "Usenet is a strange place" - dmr 29 July 1999   
   Sig line 4 vacant - apply within   
      
   --- SoupGate-Win32 v1.05   
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