On 02/06/2016 01:42 PM, tn wrote:
On Saturday, 6 February 2016 at 13:06:37 UTC, Andrei Alexandrescu wrote:
Could you please add two simple calculations? Assuming uniform random
distribution of data, compute the average number of swaps as a
weighted average of orderings. Also, show the number of lines (one
test or one swap per line) as a proxy for generated code size. That
should provide good insights.
This now prints the numbers of comparison and swap lines and computes
the average numbers of swaps. In addition, it also runs the same tests
for permutations of 5 elements some of which might be equal (an example
of such permutation would be [1, 2, 0, 2, 3], on the other hand [1, 2,
0, 2, 4] should not be included since it is identical). However, in this
case the average number of swaps is perhaps not so meaningful.
http://dpaste.dzfl.pl/2012caf872ec
Awesome, thanks. Will need to try at least a few of these out. At a
quick glance, partition5d seems to be the sweet spot - it's small and
does only 3.13/2.57 swaps on average. -- Andrei
