All right. I think I'm starting to understand the reasoning that went into the spec as written. Thank you for giving me the necessary pointers.
Now, that leads me inevitably to the obvious followup question: Is there a way to run "my" version of this operator? Regards, Elias On 21 March 2014 21:45, Jay Foad <jay.f...@gmail.com> wrote: > On 21 March 2014 13:39, Elias Mårtenson <loke...@gmail.com> wrote: > > On 21 March 2014 19:55, Jay Foad <jay.f...@gmail.com> wrote: > >> Associativity matters because the last item of the result is 3>1>2, > >> which means 3>(1>2). But your explanation would lead to (3>1)>2 as the > >> last item. > > > > > > You are right, but that can be addressed by defining my solution to > simply > > evaluate the pairs from the right. > > The items of the result of f\A B C D ... are: > > A > A f B > A f (B f C) > A f (B f (C f D)) > ... > > If f is not associative, then knowing the Nth item of the result is of > no use when calculating the (N+1)th item. > > Jay. >
