On Fri, 16 Sep 2011 22:16:59 +0200, Timon Gehr <timon.g...@gmx.ch> wrote:
I suggest:
* Introduce the algorithm "extremum" with a required predicate.
What would that do? The range has more than one extremum if it is
non-constant.
extremum!"a<b"([1,2,3,1,2,3]) would be equal to [1,1].
* Introduce extremumCount and extremumPos, both with required predicate.
And those will be minCount+maxCount and merge(minPos,maxPos) ?
Nope.
extremumCount!"a<b"([1,2,3,1,2,3]) == 2.
extremumPos!"a>b"([1,2,3,1,2,3]) == [3,1,2,3].
* Keep minCount and minPos without a predicate. They'd in all likelihood
use extremum.
How?
alias extremumCount!"a<b" minCount;
alias extremumPos!"a<b" minPos;
--
Simen
